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12, GENERAL, CHEMISTRY 2, QUARTER 2, , LEARNING ACTIVITY SHEET
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Sample Table of Contents, Code, , Page, number, , STEM_GC11CTIVa-b-140, , 1 – 16, , Explain the second law of, thermodynamics and its significance, , STEM_GC11CTIVa-b-142, , 17 – 27, , Use Gibbs’ free energy to determine the, direction of a reaction, , STEM_GC11CTIVa-b-143, , 28 – 59, , Explain chemical equilibrium in terms of, the reaction rates, of the forward and the reverse reaction, , STEM_GC11CEIVb-e-145, , 60 – 88, , Compentency, Predict the spontaneity of a process, based on entropy, , Calculate equilibrium constant and the, pressure or, concentration of reactants or products in, an equilibrium mixture, , STEM_GC11CEIVb-e-148, , State the Le Chatelier’s principle and, apply it qualitatively, to describe the effect of changes in, pressure,, concentration and temperature on a, system at equilibrium, , STEM_GC11CEIVb-e-149, , 97 – 106, , Define Bronsted acids and bases, , STEM_GC11ABIVf-g-153, , 107 – 111, , 89 – 96, , Discuss the acid-base property of water, , 112 - 125, , STEM_GC11ABIVf-g-154, Calculate ph from the concentration of, hydrogen ion or, hydroxide ions in aqueous solutions, , STEM_GC11ABIVf-g-156, , 126 – 140, , Describe how a buffer solution maintains, its ph, , STEM_GC11ABIVf-g-160, , 141 – 153, , Calculate the ph of a buffer solution, using the Henderson, Hasselbalch equation, , STEM_GC11ABIVf-g-161, , 154 – 172, , Define oxidation and reduction reactions, , STEM_GC11ABIVf-g-169, , 173 – 180
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Balance redox reactions using the, change in oxidation, number method, , STEM_GC11ABIVf-g-170, , 181 – 189, , Identify the reaction occurring in the, different parts of the cell, , STEM_GC11ABIVf-g-172, , 190 – 204, , Define reduction potential, oxidation, potential, and cell potential, , STEM_GC11ABIVf-g-176, , 205 – 217, , Calculate the standard cell potential, , STEM_GC11ABIVf-g-178, , 218 – 233, , Relate the value of the cell potential to, the feasibility of, using the cell to generate an electric, current, , STEM_GC11ABIVf-g-179, , 234 – 243, , Describe the electrochemistry involved in, some common, batteries:, a. Leclanche dry cell, b. Button batteries, c. Fuel cells, d. Lead storage battery, , STEM_GC11ABIVf-g-180, , 244 – 256, , Apply electrochemical principles to, explain corrosion, , STEM_GC11ABIVf-g-181, , 257 – 267, , Explain the electrode reactions during, electrolysis, , STEM_GC11ABIVf-g-182, , 268 – 277, , Describe the reactions in some, commercial electrolytic processes, , STEM_GC11ABIVf-g-183, , 278 - 287
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Republic of the Philippines, , Department of Education, COPYRIGHT PAGE, Learning Activity Sheet in EARTH SCIENCE, (Grade 12), Copyright © 2020, DEPARTMENT OF EDUCATION, Regional Office No. 02 (Cagayan Valley), Regional Government Center, Carig Sur, Tuguegarao City, 3500, “No copy of this material shall subsist in any work of the Government of the Philippines. However,, prior approval of the government agency or office wherein the work is created shall be necessary, for exploitation of such work for profit.”, This material has been developed for the implementation of K to 12 Curriculum through the, Curriculum and Learning Management Division (CLMD). It can be reproduced for educational, purposes and the source must be acknowledged. Derivatives of the work including creating an, edited version, an enhancement of supplementary work are permitted provided all original works, are acknowledged and the copyright is attributed. No work may be derived from this material for, commercial purposes and profit., Consultants:, Regional Director, : ESTELA L. CARIÑO, EdD., CESO IV, Assistant Regional Director, : RHODA T. RAZON, EdD., CESO V, Schools Division Superintendent, : ORLANDO E. MANUEL, PhD, CESO V, Asst. Schools Division Superintendent(s): WILMA C. BUMAGAT, PhD., CESE, CHELO C. TANGAN, PhD., CESE, Chief Education Supervisor, CLMD : OCTAVIO V. CABASAG, PhD, Chief Education Supervisor, CID, : ROGELIO H. PASINOS, PhD., Development Team, Writers, , Content Editor, , Language Editor, Illustrators, Layout Artists, Focal Persons, , : MIRAFLOR GARMA, GRACE ANN CALIBOSO, DOLORES LIBAN,, WILLIAM ERRO, ROSEMARIE C. FERNANDEZ, JOVELYN, BANGANYAN CLETO ABBIDO, IVON ADDATU, SHAROLYN GALURA,, KIMBERLY PAGDANGANAN, : CHRISTOPHER S. MASIRAG- SDO CAGAYAN, LEAH DELA CRUZSDO SANTIAGO, ROSELLE MENDOZA-SDO NUEVA VIZCAYA, MARK, KENNETH SUMBILLO- TUGUEGARAO CITY, : MARIBEL S. ARELLANO- SDO CAGAYAN, : Name, School, SDO, : Name, School, SDO, : GERRY C. GOZE, PhD., Division Learning Area Supervisor, NICKOYE V. BUMANGALAG, PhD. Division LR Supervisor, ESTER T. GRAMAJE, Regional Learning Area Supervisor, RIZALINO CARONAN, PhD. Regional LR Supervisor, Printed by: DepEd Regional Office No. 02, Regional Center, Carig Sur, Tuguegarao City, , Address: Regional Government Center, Carig Sur, Tuguegarao City, 3500, Telephone Nos.: (078) 304-3855; (078) 396-9728, Email Address: region2@deped.gov.ph
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GENERAL CHEMISTRY 2, Name: ____________________________, Grade Level: _______________________, , Date: _____________, Score: ____________, , LEARNING ACTIVITY SHEET, Spontaneous Process and Entropy, Background Information for the Learners (BIL), Chemical Thermodynamics is the study of the interrelation of heat and, work with chemical reactions or with physical changes of state within the, confines of the laws of thermodynamics., Thermodynamics is a scientific discipline that deals with the, interconversion of heat and other forms of energy. It has traditionally recognized, three fundamental laws: First Law - Energy of the universe is constant. “Energy, can be converted from one form to another, but it can never be created nor, destroyed”; Second Law - Entropy of universe increases. “The entropy of the, universe increases in a spontaneous process and remains unchanged in the, equilibrium process”; and Third Law - At absolute zero, the entropy of a perfect, crystal is 0. “The entropy of the perfect crystalline substance is zero at the, absolute zero of temperature (T = 0, K = -273.150C).”, Spontaneous process as stated in the second law is a physical or, chemical change that occurs by itself. A process that takes place without energy, from an external source. It is the time-evolution of a system which releases free, energy and it moves to a lower, more thermodynamically stable energy state., If heat flows into surroundings (exothermic) the random motion of the, molecules in the surroundings increases. Thus, the entropy of the surroundings, increases. Entropy is a thermodynamic quantity that is a measure of, randomness and disorder. It measures how spread out or dispersed the energy, of a system is among the different possible ways that system can contain, energy. It tells whether a process or chemical reaction can occur. The, connection between entropy and the spontaneity of a reaction is expressed by, the second law of thermodynamics., The change in entropy for a given amount of heat absorbed also, depends on temperature. If the temperature of the surroundings is high, the, molecules are already quite energetic. Therefore, the absorption of heat from, an exothermic process in the system will have relatively little impact on the, motion of the molecules and the resulting increase in entropy of the, surroundings will be small. However, if the temperature of the surroundings is, , 1, Practice Personal Hygiene protocols at all times.
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low, than the addition of the same amount of heat will cause a more drastic, increase in molecular motion and hence a larger increase in entropy., Consider the phase changes illustrated bellow. Raising the temperature, of a substance will result in more extensive vibrations of the particles in solids, and more rapid translations of the particles in liquids and gases. At higher, temperatures, the distribution of kinetic energies among the atoms or molecules, of the substance is also broader (more dispersed) than at lower temperatures., Thus, the entropy for any substance increases with temperature., , Source: https://openstax.org/resources/f3f96b7f897d7cd062c326b3e451634f6c3d5a20, , The entropy of a substance increases (ΔS > 0) as it transforms from a, relatively ordered solid, to a less-ordered liquid, and then to a still less-ordered, gas. The entropy decreases (ΔS < 0) as the substance transforms from a gas, to a liquid and then to a solid., Consider the illustration bellow., , Source: https://encryptedtbn0.gstatic.com/images?q=tbn%3AANd9GcSU3fNtyjImEWHO5JmAuogCgNibjoAR2urP5w&, usqp=CAU, , What did you observe? Water is always flows downward on its own., But never goes upward automatically., , Learning Competency:, Predict the spontaneity of a process based on entropy., (STEM_CG11CT-IVa-b-140), , 2, Practice Personal Hygiene protocols at all times.
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2. How do you compare the pictures? __________________________________, _____________________________________________________________________, , 3. Which one is more spontaneous? Why? _____________________________, _____________________________________________________________________, , Activity 2. I’M EVERYWHERE!, A spontaneous process is one that takes place without energy from an, external source. For a chemical reaction to be spontaneous, it should proceed, as written (from left to right), without an input of energy., An endothermic process absorbs heat from the surroundings and has a, positive value, whereas an exothermic process release heat to its surroundings, and has a negative value., Examples of reactions, 1. Combustion of methane, CH4, , +, , 2O2, , →, , 2. Acid-base neutralization, H+(aq) + OH-(aq), , 6CO2 +, →, , H2O(l), , 2H2O ∆H0 = -890.4 kJ/mol, ∆H0 = -56.2 kJ/mol, , *Both of these reactions are very exothermic and are not reversible., 3. Solid to liquid phase transition of water, H2O(s), →, H2O(l), ∆H0 = 6.01 kJ/mol, 4. Dissolution of ammonium nitrate in water, NH4NO3(s) →, NH4+(aq), +, NO3-(aq) ∆H0 = 6.01 kJ/mol, *Ice melting above 00C and ammonium nitrate dissolving in water are both, spontaneous process yet endothermic., Exercise:, Directions: Classify the given situations below whether the process is, spontaneous or non-spontaneous., ______ 1. Rusting of iron in moist air, , ______ 6. Drying of leaves, , ______ 2. Decaying of radioisotopes, , ______ 7. Dissolving of salt, , ______ 3. Oxidation of gold, , ______ 8. Radioactive atom splits up, , ______ 4. Spoilage of food, water, , ______ 9. Dissolution of sand in, , ______ 5. Burning of chlorine, , ______ 10. Fireworks, , 4, Practice Personal Hygiene protocols at all times.
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Activity 3. I’M A PART OF YOU!, Entropy, S, is the thermodynamic quantity that is a measure of how, spread out or dispersed the energy of a system is among the different possible, ways that system can contain energy. It is a quantity that is generally used to, describe the course of a process, that is, whether it is a spontaneous process, and has a probability of occuring in a definite direction, or a non-spontaneous, process and will not proceed in the defined direction, but in the reverse, direction., Most processes are accompanied by entropy change. The following are, processes that lead to an increase in entropy of the system, Process, , Order, , →, , Disorder, , Melting, , Solid, , →, , Liquid, , Vaporization, , Liquid, , →, , Vapor, , Dissolving, , Solute, , →, , Solution, , Heating, , System at T1, , →, , System at T2 (T2 > T1), , Entropy change examples:, 1. Gas in balloon spreads out into room and deflates but never a balloon, spontaneously filled with air., ►The molecules of gas at a high pressure always spread to lower, pressure regions., 2. Hot coffee in a room gets cooler and the heat spreads out into the room,, but never a cold cup of coffee being spontaneously warmed up., ►Heats always goes from high temperature into cooler regions., The spreading out of more concentrated molecules and the spreading, out of more concentrated energy are changes from more order to more random., , Exercise:, FACT OR BLUFF!, Directions: Write Fact on the blank if the condition illustrates entropy and, write Bluff if does not illustrates entropy., _____________1. Oxidation of nitrogen, _____________2. Sublimation of mothballs, _____________3. Reduction of silicon, _____________4. Lighting of candles, _____________5. Flow of water up hill, , 5, Practice Personal Hygiene protocols at all times.
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_____________6. Digestion of food, _____________7. Boiling water for tea, _____________8. Flow of heat from a cold body to a hot body, _____________9. Diffusion of LPG, _____________10. Making popcorn, , Activity 4. WORD SEARCH, Directions: Search and encircle the important terms being described in the, sentences below. Words can be forward, backward, vertical, horizontal, or, diagonal., 1. The scientific discipline that deals with the interconversion of heat and, other forms of energy., 2. A process of a physical or chemical change that occurs by itself., 3. The measure of randomness and disorder., 4. Process that gives off heat to the surroundings., 5. Process that absorbs heat from the surroundings., 6. The value of the product during endothermic process., 7. The value of the product during exothermic process., 8. The change of phase from solid to liquid., 9. The change of phase from liquid to gas., 10. The change of phase from solid to gas., , 6, Practice Personal Hygiene protocols at all times.
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Activity 5. CORRECT ME IF I’M WRONG!, Directions: Write TRUE if the statement is correct but if it’s false, change the, underlined word or group of words to make the whole statement true., ____________1. If heat flows into the surroundings, the random motion of the, molecules in the surroundings decreases., ____________2. In a chemical reaction, the heat change is positive if the heat, product is lower than the heat reactant., ____________3. The heat change is negative if the heat product is greater than, the heat reactant., ____________4. Spontaneous process is reversible reaction., ____________5. Entropy changes occur when gas molecules inside the LPG, tank escape and spread out into room., , 7, Practice Personal Hygiene protocols at all times.
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____________6. Heat flows from hotter objects to a colder one is a, spontaneous process., ____________7. Burning of fuel is an example of endothermic reaction., ____________8. Coffee granules dissolve faster in hot water than in cold water., ____________9. Melting of ice cream left on top of a table is an example of, exothermic reaction., ____________10. For a chemical reaction to be spontaneous, it should, proceed without an input of energy., , SUM UP!, , 1. What characterize a spontaneous process? ____________________, ________________________________________________________, ________________________________________________________, ________________________________________________________, ______, 2. How does spontaneity apply to a chemical reaction? ______________, _______________________________________________________, ________________________________________________________, _______________________________________________________, 3. How do entropy changes occur? _____________________________, ________________________________________________________, ________________________________________________________, _______________________________________________________, 4. How is hot object in an open area gets cooler? ___________________, _______________________________________________________, ________________________________________________________, ________________________________________________________, , 8, Practice Personal Hygiene protocols at all times.
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References, Commission on Higher Education. (2016). General Chemistry 2 (Teaching, Guide for Senior High School). Quezon City: Commission on Higher, Education., https://www.teacherph.com/general-chemistry-2-teaching-guide/, Jessie A. Key. Introductory Chemistry-1st Canadian Edition. Chapter 18., Chemical Thermodynamics., https://opentextbc.ca/introductorychemistry/chapter/entropy-and-the-secondlaw-of-thermodynamics-2/, , 10, Practice Personal Hygiene protocols at all times.
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ANSWER KEY, Activity 1:, 1. Pictures at the left is rock rolling and a ski sliding down a hill while at the, right is a rock pushing and a ski sliding up a hill., 2. The rock and ski at the left rolling and skiing down a hill occurs without, any intervention while rock and ski at the right need to apply force to, make the rock roll and ski up the hill., 3. Pictures at the left is more spontaneous because it moves without any, outside intervension. The speed at which is occurs governed by kinetics., Activity 2:, 1., 2., 3., 4., 5., , Spontaneous, Spontaneous, N0n-spontaneous, Spontaneous, N0n-spontaneous, , 6. Spontaneous, 7. Spontaneous, 8. Spontaneous, 9. Non-spontaneous, 10. Spontaneous, , Activity 3:, 1., 2., 3., 4., 5., , Bluff, Fact, Bluff, Fact, Bluff, , 6. Fact, 7. Fact, 8. Bluff, 9. Fact, 10. Fact, , Activity 4:, 1., 2., 3., 4., 5., , Thermodynamics, Spontaneous, Entropy, Exothermic, Endothermic, , 6. Positive, 7. Negative, 8. Melting, 9. Vaporization, 10. Sublimation, , Activity 5:, 1., 2., 3., 4., 5., , Increase, Negative, Lower, Irreversible, TRUE, , 6. TRUE, 7. Exothermic, 8. TRUE, 9. Endothermic, 10. TRUE, , 11, Practice Personal Hygiene protocols at all times.
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Sum Up:, 1. Spontaneous processes are characterized by a decrease in the system’s, free energy, they do not need to be driven by an outside source of, energy. It is characterized by an increase in entropy., 2. In a chemical reaction, heat change of reaction is equal to heat of, product minus the heat of reactant. If it is exothermic, then heat change, of reaction is equal to negative. The heat product must be lower than, the heat reactant., 3. Increasing the temperature will increase the entropy. Change in volume, will lead to change in entropy. The larger the volume, the higher the, entropy., , (1) Prepared by:, MIRAFLOR T. GARMA, Cordova National High School, , 12, Practice Personal Hygiene protocols at all times.
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GENERAL CHEMISTRY 2, Name: ________________________________, , Date: ________, , Grade Level: ___________________________, , Score: _______, , LEARNING ACTIVITY SHEET, THE SECOND LAW OF THERMODYNAMICS, Background Information for the Learners (BIL), Chemical Thermodynamics is the study of the interrelation of heat and, work with chemical reactions or with physical changes of state within the, confines of the laws of thermodynamics., Thermodynamics is a scientific discipline that deals with the, interconversion of heat and other forms of energy. It has traditionally recognized, three fundamental laws: First Law - Energy of the universe is constant. “Energy, can be converted from one form to another, but it can never be created nor, destroyed”; Second Law - Entropy of universe increases. “The entropy of the, universe increases in a spontaneous process and remains unchanged in the, equilibrium process”; and Third Law - At absolute zero, the entropy of a perfect, crystal is 0. “The entropy of the perfect crystalline substance is zero at the, absolute zero of temperature (T = 0,, K = -273.150C).”, The Second Law of Thermodynamics deals with entropy, the quantity, that measures how spread out or dispersed the energy of a system is among, the different possible ways that system can contain energy. It tells whether a, process or chemical reaction can occur. The connection between entropy and, the spontaneity of a reaction is expressed by the second law of, thermodynamics. This law says that when energy changes from one form to, another form, or matter moves freely, entropy (disorder) in a close system, increases. Differences in temperature, pressure and density tend to even out, horizontally after a while., The change in entropy for a given amount of heat absorbed also, depends on temperature. If the temperature of the surroundings is high, the, molecules are already quite energetic. Therefore, the absorption of heat from, an exothermic process in the system will have relatively little impact on the, motion of the molecules and the resulting increase in entropy of the, surroundings will be small. However, if the temperature of the surroundings is, low, then the addition of the same amount of heat will cause a more drastic, increase in molecular motion and hence a larger increase in entropy., The significance of this law is that, it tells us about the direction of heat, transfer and what process are impossible even if they satisfy the first law., 13, Practice Personal Hygiene protocols at all times.
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Examples are: engine can’t have an efficiency of 100%, a fridge can’t work, without a power supply. Another example is a human body. We eat food (high, temperature reservoir). The coffee eventually cools down showing that the heat, only flows from high temperature to low temperature without the aid of any, external agent. A cold object in contact with a hot one never gets colder,, transferring heat to the hot object and making it hotter furthermore. Mechanical, energy, such as kinetic energy, can be completely converted to thermal energy, by friction, but the reverse is impossible., Because the universe is made up of the system and the surroundings,, the entropy change in the universe (∆Suniv) for any process is the sum of the, entropy changes in the system (∆Ssys) and in the surroundings (∆Ssur)., ∆Suniv, , =, , ∆Ssys, , +, , ∆Ssur, , >, , 0, , Process is spontaneous, , ∆Suniv, , =, , ∆Ssys, , +, , ∆Ssur, , =, , 0, , Process tends not to occur,, equilibrium is attained, , ∆Suniv, , =, , ∆Ssys, , ∆Ssur, , +, , <, , 0, , Reverse process occurs, spontaneously, , Learning Competency:, Explain the second law of thermodynamics and its significance., (STEM_GC11CT-IVa-b-142), , Activity 1. IT’S GETTING HOTTER IN HERE!, , Directions: Calculating Entropy Changes in the system: Standard Entropy of, Reaction, ∆S0rxn . Data needed in calculating the entropy change:, 1. Suppose that the system is represented by the following reactions:, aA, , +, , bB, , →, , cC, , +, , dD, , 2. The standard entropy of reactions ∆S0rxn is given by the difference in, standard entropies between the products and the reactants., ∆S0, , =, , ∑nS0 (products), , -, , ∑nS0 (reactants), , 3. • Where m and n are the stoichiometric coefficients in the reaction., , ∆S0rxn =, , [cS0(C), , +, , dS0(D)] - [aS0(A), , +, , bS0(B)], , 14, Practice Personal Hygiene protocols at all times.
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Note: Standard entropy values: J/K mol; 1atm; 250C, Selected Thermodynamic Values (at 250C)*, ∆Hf°, S°, ∆Gf°, (kJ/mol), (J/K·mol), (kJ/mol), C(s), 0, 5.7, 0, CO2(g), -393.509, 213.6, -394.359, H2(g), 0, 130.6, 0, PbO(s,), -217.32, 69.54, -187.89, NO(g), 90.25, 210.6, 86.55, NO2(g), 33.18, 240.5, 51.31, SO2(s), -296.83, 248.5, -300.194, SO3(g), -395.72, 256.2, -371.06, H2SO4(l), -813.989, 156.904, -690.003, H2O(l), -285.83, 69.91, -237.129, Pb(s), 0, 64.89, 0, I(s), 0, 116.7, 0, *Taken from "The NBS Tables of Chemical Thermodynamic Properties" (1982), and "CRC Handbook of Chemistry and Physics", 1st Student Edition (1988), Species, , From the standard entropy values in the Thermodynamic Data table, calculate, ∆S0 for the following reaction., , Study Me!, H2(g), , +, , I2(s), , →, , 2HI(g), , →, , 2HI(g), , Step 1. Write the standard entropy below each formula, H2(g), From the table, S0(J/K.mol), , +, , 130.6, , I2(s), 116.7, , 206.3, , Step 2. Using the equation for the standard entropy of reaction, ∆S0, , =, , ∑nS0 (products), , =, , [(2) S0 HI] - [(1) S0 H2 + (1) S0 I2], , -, , ∑nS0 (reactants), , Step 3. Substitute the entropy values., =, , ∆S0, , [(2) (206.3)] - [(1) (130.6) + (1) (116.7)], , =, , [412.6] - [247.3], , =, , +165.3 J/K-mol, , 15, Practice Personal Hygiene protocols at all times.
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SOLVE ME!, 1. Determine S for the reaction:, SO3(g), Given: S0(J/K.mol):, , 256.2, , 2. Calculate S for the reaction, SO2(s), Given: S0(J/K.mol), , +, , H2O(l) → H2SO4(l), 69.9, , +, , 248.5, , NO2, 240.5, , 156.9, , →, , SO3(g), 256.2, , +, , NO(g), 210.6, , 3. Calculate S at 250C for the reduction of a given these absolute, entropies:, 2PbO(s) +, C(s) →, 2Pb(s) + CO2(g), 0, Given: S (J/K.mol) 69.54, 5.7, 64.89, 213.6, , 16, Practice Personal Hygiene protocols at all times.
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Activity 2. PREDICT ME!, , General rules in predicting entropy change of the system:, 1. If the reaction produces more gas molecules than it consumes, ∆S0 is, positive., 2. If the total number of gas molecules diminishes, ∆S0 is negative., 3. If there is no net change in the total number of gas molecules, ∆S0 may, be positive or negative, but will be relatively small numerically., POSITIVE, Increase in number of gas molecules, Increase in entropy, Solid converted to liquid, Solid converted to gas, Liquid converted to gas, , NEGATIVE, Decrease in number of gas molecules, Decrease in entropy, Gas converted to solid, Gas converted to liquid, Liquid converted to solid, , Directions: Predict whether the entropy change of the system in each of the, following is positive or negative., Chemical Reaction, , 1.O2(g), , →, , Positive/, Negative, , Reason, , 2O(g), , 2. N2(g, 10atm) →, , N2(g, 1atm), , 3.6CO2(g) + 6H2O(g) →, C6H12O6(g) + 6O2(g), 4.2H2(g) + O2(g) → 2H2O(l), 5.NH4Cl(s) →, NH3(g) + HCl(g), , 17, Practice Personal Hygiene protocols at all times.
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Activity 3. CROSSWORD PUZZLE, Directions: Solve the crossword using the clues., , 18, Practice Personal Hygiene protocols at all times.
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Activity 4. FACT OR BLUFF, Directions: Write Fact on the blank if the statement is correct and Bluff if the, statement is incorrect., ____________1. The second law of thermodynamics deals with entropy., ____________2. The entropy of the universe decreases in a spontaneous, process., ____________3. When the entropy change in the universe is greater than, zero, the process is spontaneous., ____________4. When the entropy change in the universe is less than zero,, the process is spontaneous., ____________5. When the entropy change in the universe is equal to zero,, the process is equilibrium., ____________6. The second law of thermodynamics tells us that heat always, flows from a body at a higher temperature to a body at the lower temperature, ____________7. If the reaction produces more gas molecules than it, consumes, the entropy change is positive., ____________8. If there is no net change in the total number of gas, molecules, the entropy change is negative., ____________9. If the total number of gas molecules diminishes, the entropy, change is either negative or positive., ____________10. Digestion of food is a spontaneous process., , Activity 5: RESEARCH TIME!, Directions: Explain how second law of thermodynamics applies in the, following., 1. Eating food, ________________________________________________________, ________________________________________________________, ________________________________________________________, 2. Hot cup coffee in a table, ________________________________________________________, ________________________________________________________, ________________________________________________________, ________________________________________________________, , 19, Practice Personal Hygiene protocols at all times.
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References:, Commission on Higher Education. (2016). General Chemistry 2 (Teaching, Guide for Senior High School). Quezon City: Commission on Higher, Education., https://www.teacherph.com/general-chemistry-2-teaching-guide/, "The NBS Tables of Chemical Thermodynamic Properties" (1982) and "CRC, Handbook of Chemistry and Physics", 1st Student Edition (1988), https://www2.chem.wisc.edu/deptfiles/genchem/netorial/modules/thermodyna, mics/table.htm, Jessie A. Key. Introductory Chemistry-1st Canadian Edition. Chapter 18., Chemical Thermodynamics., https://opentextbc.ca/introductorychemistry/chapter/entropy-and-the-secondlaw-of-thermodynamics-2/, , 21, Practice Personal Hygiene protocols at all times.
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ANSWER KEY, Activity 1:, 1. -169.2 J/K·mol, 2. -22.2 J/K·mol, 3. +198.8 J/K·mol, Activity 2:, Chemical Reaction, 1.O2(g), , →, , 2O(g), , Positive/Negative, , Reason, , Positive, , Increase in number of gas, molecules., Decrease in pressure of the, system, will, increase, entropy., Decrease in number of gas, particles., Net decrease in number of, molecules and gases are, converted to solids., A solid is converted to two, gaseous products., , 2. N2(g, 10atm) → N2(g,, 1atm), , Positive, , 3.6CO2(g) + 6H2O(g) →, C6H12O6(g) + 6O2(g), 4.2H2(g) + O2(g) →, 2H2O(l), , Negative, , 5.NH4Cl(s) →, NH3(g) + HCl(g), , Positive, , Negative, , Activity 3:, 1. Product, 2. Exothermic, 3. Endothermic, 4. Pressure, 5. Positive, , 6. Reactant, 7. Temperature, 8. Negative, 9. Thermodynamics, 10. Entropy, , Activity 4:, 1. Fact, 2. Bluff, 3. Fact, 4. Bluff, 5. Fact, , 6. Fact, 7. Fact, 8. Bluff, 9. Bluff, 10. Fact, , Activity 5:, 1. We eat food (high temperature reservoir). The chemical energy of food, is utilized by body to maintain its temperature and to do work. When we, do some work our body warms up and rejects heat into the environment, (low temperature reservoir). Even if we are not doing work our body still, rejects a lot of heat to environment due to metabolism., 22, Practice Personal Hygiene protocols at all times.
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2. When a hot cup of coffee left on a table the coffee will eventually cool, down showing that heat only flows from high temperature to low, temperature without the aid of any external agent., Sum Up:, 1. The Second Law of Thermodynamics is the Entropy of universe, increases, “states that the entropy of the universe increases in a, spontaneous process and remains unchanged in an equilibrium process., 2. When the entropy change in the universe is greater than zero, the, process is spontaneous; if it is less than zero, reverse process occurs, spontaneously; and when it is equal to zero, process tends not to occur,, equilibrium is attained., 3. If the reaction produces more gas molecules than it consumes, ∆S0 is, positive; If the total no of gas molecules diminishes, ∆S0 is negative; and, If there is no net change in the total no of gas molecules, ∆S 0 may be, positive or negative, but will be relatively small numerically., 4. The second law of thermodynamics plays the most important role in, making our life easier, i.e. heat transfer from one medium to another., , Prepared by:, (2) MIRAFLOR T. GARMA, Cordova National High School, , 23, Practice Personal Hygiene protocols at all times.
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GENERAL CHEMISTRY 2, Name: __________________________ Grade Level: ____________, Date: ___________________________ Score:__________________, , LEARNING ACTIVITY SHEET, GIBBS’ FREE ENERGY, Background Information for the Learners (BIL), According to the second law of thermodynamics, a spontaneous, reaction increases the entropy of the universe: that is ∆𝑆𝑢𝑛𝑖𝑣 > 0. In, determining the direction of the ∆𝑆𝑢𝑛𝑖𝑣 , calculation of ∆𝑆𝑠𝑦𝑠 and ∆𝑆𝑠𝑢𝑟𝑟 are both, necessary. The mathematical expression is given by, ∆𝑆𝑢𝑛𝑖𝑣 = [∆𝑆𝑠𝑦𝑠 + ∆𝑆𝑠𝑢𝑟𝑟 ] > 0, Since ∆𝑆𝑠𝑢𝑟𝑟 =, , −∆𝐻𝑠𝑦𝑠, 𝑇, , eq. 1, , , we have, , ∆𝑆𝑢𝑛𝑖𝑣 = [∆𝑆𝑠𝑦𝑠 +, , −∆𝐻𝑠𝑦𝑠, 𝑇, , ] >0, , eq. 2, , Multiplying both sides of the equation by T gives, 𝑇∆𝑆𝑢𝑛𝑖𝑣 = [−∆𝐻𝑠𝑦𝑠 + 𝑇∆𝑆𝑠𝑦𝑠 ] > 0 eq. 3, The equation expressing only the properties ∆𝐻𝑠𝑦𝑠 𝑎𝑛𝑑 𝑇∆𝑆𝑠𝑦𝑠 can now be used, as a criterion in determining if a spontaneous reaction occurs., For convenience, multiply both sides of the equation in equation 3 by -1 and, change the > sign with <:, −𝑇∆𝑆𝑢𝑛𝑖𝑣 = [∆𝐻𝑠𝑦𝑠 − 𝑇∆𝑆𝑠𝑦𝑠 ] < 0 eq. 4, Equation 4 says that as a reaction proceeds at a constant pressure and, temperature, the reactants forms products and if the changes in H and S of the, system is less than zero, the process is spontaneous., The relationship between H and S was introduced by the American, Physicist J. Willard Gibbs. He introduced another thermodynamic quantity, known as the Gibb’s free energy (G) or simply free energy. The mathematical, relationship is given by, 𝐺 = 𝐻 − 𝑇𝑆, , eq. 5, , 24, Practice Personal Hygiene protocols at all times.
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where m and n are stoichiometric coefficients. 𝑮𝒐𝒇 is the standard free-energy, of formation of a compound. It is the free-energy change that occurs when 1, mole of the compound is synthesized from its elements in their standard states., , EXAMPLES, 3. Calculate the standard free-energy changes for the following, reactions at 250C., a. CH4(g) + O2(g) → CO2(g) + H2O(g), b. MgO(s) → Mg(s) + O2(g), , a. Solution, •, , Write the balanced chemical equation and placed below each, formula the values of ∆𝐺. Use Table 1 to locate for the needed, values., CH4(g, , ∆𝐺, , +, , 2O2(g), , 𝑘𝐽, , →, , 𝑘𝐽, , -50.8𝑚𝑜𝑙, , CO2(g), , + 2H2O(l), , 𝑘𝐽, , 0 𝑚𝑜𝑙, , -, , -394.4𝑚𝑜𝑙, , 𝑘𝐽, , 237.2𝑚𝑜𝑙, •, , Use equation 10 or 11 to compute for the free energy, 𝑘𝐽, 𝑘𝐽, ) + (−394.4, )], 𝑚𝑜𝑙, 𝑚𝑜𝑙, 𝑘𝐽, 𝑘𝐽, 𝑘𝐽, ) + (0, )] = − 818.0, − ∑ [(−50.8, 𝑚𝑜𝑙, 𝑚𝑜𝑙, 𝑚𝑜𝑙, , 𝑜, ∆𝐺𝑟𝑥𝑛, = ∑ [(2 ∗ −237.2, , •, , Determine the spontaneity of the reaction, 𝑘𝐽, , ∆𝐺 = −818.0 𝑚𝑜𝑙 which is lesser than 0. This means that the, reaction is spontaneous in the forward direction., b. Solution, •, , Write the balanced chemical equation and placed below each, formula the values of ∆𝐺. Use Table 1 to locate for the needed, values., 2MgO(s), , →, , 2Mg(s), , +, , O2(g), , 28, Practice Personal Hygiene protocols at all times.
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b. Solution, •, , Write the balanced chemical equation and placed below each, formula the values of ∆𝐺. Use Table 1 to locate for the needed, values., 2C2H6(g), , +, , 7O2(g), , 𝑘𝐽, , ∆𝐺, , 𝑘𝐽, , -32.89𝑚𝑜𝑙, •, , →, , 4CO2(g), , + 6H2O(l), , 𝑘𝐽, , 0 𝑚𝑜𝑙, , 𝑘𝐽, , -394.4𝑚𝑜𝑙, , -237.2𝑚𝑜𝑙, , Use equation 10 or 11 to compute for the free energy, 𝑘𝐽, 𝑘𝐽, ) + (4 ∗ −394.4, )], 𝑚𝑜𝑙, 𝑚𝑜𝑙, 𝑘𝐽, 𝑘𝐽, ) + (7 ∗ 0, )] =, − ∑ [(2 ∗ −32.89, 𝑚𝑜𝑙, 𝑚𝑜𝑙, 𝑘𝐽, − 2935.02, 𝑚𝑜𝑙, , 𝑜, ∆𝐺𝑟𝑥𝑛, = ∑ [(6 ∗ −237.2, , •, , Determine the spontaneity of the reaction, 𝑘𝐽, , ∆𝐺 = −2935.02 𝑚𝑜𝑙 which is lesser than 0. This means that, the reaction is spontaneous in the forward direction., , FACTORS AFFECTING THE SIGN OF G, ∆𝐻, , 𝑇∆𝑆, , ∆𝐺, , Significance, , +, , +, , X, , At high temperature, reactions proceeds, spontaneously while at low temperature,, the reaction is spontaneous in the reverse, direction, , +, , -, , +, , ∆𝐺 is always positive. Reaction proceeds, spontaneously in the reverse direction at, all temperatures., , -, , +, , -, , ∆𝐺 is always positive. Reaction proceeds, spontaneously direction at all temperatures, , -, , -, , X, , At high temperatures, reaction proceeds, spontaneously in the reverse direction and, at low temperatures, the reaction proceeds, spontaneously, , 30, Practice Personal Hygiene protocols at all times.
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References, , Chang, R. (2010). Chemistry (10th ed., pp. 814-835). McGraw-Hill, Inc., USA., Ebbing, D.D & Gammon, S.D. (2017). General Chemistry (9th ed., 745-769)., Houghton Mifflin Company, USA., , 39, Practice Personal Hygiene protocols at all times.
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ACTIVITY 4: TEMPERATURE AND CHEMICAL REACTIONS, Direction: For the chemical reactions that are non-spontaneous in Problem, Set 3, determine at what temperature will the reaction occur spontaneously?, Item 1. The reaction will occur spontaneously at a temperature higher than, 770C., Item 2. The reaction will occur spontaneously at a temperature equal to 15.930C or lower., Item 4. The reaction will occur spontaneously at a temperature lower than, −161.570 𝐶, Item 6. The reaction will occur spontaneously at a temperature lower than, 341.530 𝐶, Item 7. The reaction will occur spontaneously at a temperature higher than, 707.950 𝐶, Item 8. The reaction will occur spontaneously at a temperature equal to, 42.280C or higher., Item 9. The reaction will occur spontaneously at a temperature equal to, 55.620C or higher., , 50, Practice Personal Hygiene protocols at all times.
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Table 1. THERMODYNAMIC DATA AT 1 ATM AND 250C, , 51, Practice Personal Hygiene protocols at all times.
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52, Practice Personal Hygiene protocols at all times.
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53, Practice Personal Hygiene protocols at all times.
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Prepared by:, GRACE ANN M. CALIBOSO - AGCAOILI, David M. Puzon Memorial National High School, , 55, Practice Personal Hygiene protocols at all times.
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GENERAL CHEMISTRY 2, Name: _____________________________________ Grade Level: _______, Date: _______________________________________ Score:____________, , LEARNING ACTIVITY SHEET, CHEMICAL EQUILIBRIUM AND REACTION RATE, , BACKGROUND INFORMATION FOR THE LEARNERS (BIL), Few chemical reactions proceed in only one direction. Most do not, proceed in only one direction and proceed essentially to completion. This type, of reaction is called reversible reaction. At first, the reaction proceeds in the, formation of products. As soon as some of the products are formed, the reverse, process occurs and reactant molecules are formed from the product molecules., Reversible reactions use a double headed arrow indicating that forward and, backward reactions occur concurrently. The general form of a reversible, reaction is, 𝑎𝐴 + 𝑏𝐵, 𝑐𝐶 + 𝑑𝐷, where the lower cases letters represent stoichiometric coefficients of the, reactants and products., To be in equilibrium is to be in a state of balance. In a reversible reaction,, physical equilibrium and chemical equilibrium occurs., Equilibrium between two phases of the same substance is called, physical equilibrium because the changes that occur are physical processes., Consider the scenario during a hot summer day when you buy some, solidified water (ice) then you put it in a glass jar. After some time, the ice will, turn to liquid because of the very hot temperature. The vaporization of water in, a closed container at a given temperature is an example., H2O(l), , H2O(g), , Chemical equilibrium occurs when the rates of the forward and reverse, reactions are equal and the concentrations of the reactants and products, remain constant. Chemical equilibrium is a dynamic process. It is dynamic, because 1) when the reactants start to form the products, the product would, then start to reform the reactants. The two opposing processes happen at, 56, Practice Personal Hygiene protocols at all times.
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different rates but a certain point in the reaction will be reached where the rates, of the forward and backward reactions are the same (Figure 1) and 2) the, concentrations of the reactants and products remain becomes constant (Figure, 2), , Figure 1. Changes in the rate of the, forward and backward reactions in a, reversible reaction, , Chang, R. (2010). Chemistry (10th ed., pp. 617., McGraw-Hill, Inc., USA., , Figure 2. Changes in the amount of, reactants and products in a reversible, reaction, , Chang, R. (2010). Chemistry (10th ed., pp. 617., McGraw-Hill, Inc., USA., , Consider the dissociation of N2O4, a colourless gas use in rocket engines, that dissociates to form brown NO2. The frozen N2O4 is left to vaporize as it is, warmed above its boiling point (2120C) in a sealed container, the gas turns, darker as the colourless N2O4 gas dissociates into brown NO2. The gas inside, will cease to become darker because the system reaches equilibrium. The, reaction is represented by, , 57, Practice Personal Hygiene protocols at all times.
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The important things to keep in mind in a reversible reaction are:, •, , At equilibrium, the concentrations of reactants and products no, longer change with time., , •, , For equilibrium to occur, neither reactants nor products can escape, from the system., , •, , At equilibrium, a particular ratio of concentration terms equals a, constant., , EQUILIBRIUM CONSTANT, Kc, In 1864, Guldberg and Waage postulated the Law of Mass Action, which expresses for any reaction, the relationship between the concentrations, of the reactants and products present at equilibrium. The general equilibrium, equation is given by, 𝑎𝐴 + 𝑏𝐵, , 𝑐𝐶 + 𝑑𝐷, , where A, B, C and D are the chemical species. According to the law of mass, action, the equilibrium condition is given by, 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠, 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠, , [𝐶]𝑐 [𝐷]𝑑, , 𝐾𝑐 = [𝐴]𝑎[𝐵]𝑏, , This relationship is the equilibrium-constant expression for the reaction. The, K c is the equilibrium constant, which is the numerical value obtained when [ ], is substituted by the molar concentrations of the chemical species. K c is also, referred to as K eq . The equilibrium-constant expression depends only on the, stoichiometry of the reaction, not on its mechanism., Reactions with reacting species that are in the same phase are known, to be homogenous equilibrium. Consider the following reactions., N2O4 (g), , 2NO2 (g), , Both are in the gas phase, , [𝑁𝑂2 ]2, 𝐾𝑐 =, [𝑁2 𝑂4 ], , 59, Practice Personal Hygiene protocols at all times.
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EXAMPLES, 1. Gaseous hydrogen iodide is placed in a closed container at 4250C, where it, partially decomposes to hydrogen and iodine. At equilibrium, it is found that, [HI] = 3.53 * 10-3 M, [H2] = 4.79 * 10-4 M and [I2] = 4.79 * 10-4 M. What is the, value of Kc at this temperature? At what side will the equilibrium lie? The, chemical reaction is given by:, 2HI, , H2 (g) + I2 (g), , Solution, 𝐾𝑐 =, , [𝐻2 ] [𝐼2 ], [𝐻𝐼]2, , =, , [4.79∗10−4 ] [4.79∗10 −4 ], [3.53∗10−3 ]2, , = 0.0184, , Since the computed Kc is lesser than 1, the equilibrium will lie to, the left and it favors the reactant., 2. Carbonyl chloride (phosgene) was used in WWI as a poisonous gas. At, 740C, [CO] = 1.2 * 10-2 M, [Cl2] = 0.054 M, and [COCl2] = 0. 14 M. Calculate, Kc and determine at what side will the equilibrium lie. The chemical reaction, is given by:, CO (g) + Cl2 (g), , COCl2, , Solution, [𝐶𝑂𝐶𝑙 ], , [0.14], , 𝐾𝑐 = [𝐶𝑂][𝐻2 ] = [1.2 ∗ 10−3] [0.054] = 2165.49, 2, , Since the computed Kc is greater than 1, the equilibrium will lie to, the right and it favours the product., , SUMMARY OF GUIDELINES FOR WRITING EQUILIBRIUM CONSTANT, EXPRESSION, KC, 1. The concentrations of the reacting species in the condensed phase, are expressed in, be expressed in, , 𝑚𝑜𝑙, 𝐿, , 𝑚𝑜𝑙, 𝐿, , ; in the gaseous phase, the concentrations can, , ., , 62, Practice Personal Hygiene protocols at all times.
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2. The concentrations of pure solids, pure liquids (in heterogeneous, equilibria), and solvents (in homogenous equilibria) do not appear in, the equilibrium constant expressions., 3. The equilibrium constant Kc is a dimensionless quantity., 4. In quoting a value for the equilibrium constant, always apply the, balanced equation and the temperature., , REACTION QUOTIENT AND THE DIRECTION OF A REACTION, The direction of a reaction is determined using the reaction quotient, Qc. A reaction quotient is a number obtained by substituting reactant and, product concentrations at any point during a reaction into an equilibriumconstant expression. The general reaction is given by, 𝑎𝐴 + 𝑏𝐵, , 𝑐𝐶 + 𝑑𝐷, , where A, B, C and D are the chemical species and lower case letters are the, stoichiometric coefficients of the reactants and the products. The reaction, quotient in terms of molar concentrations is given by, [𝐶]𝑐 [𝐷]𝑑, , 𝑄𝑐 = [𝐴]𝑎[𝐵]𝑏, , 𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑠, 𝑟𝑒𝑎𝑐𝑡𝑎𝑛𝑡𝑠, , To determine the direction in which the net reaction will proceed to achieve, equilibrium, Kc and Qc are used. The three possible cases (Figure 6) are as, follows:, •, , Qc < Kc, , The ratio of initial concentrations of products to reactants is, too small. To reach equilibrium, reactants must be converted, to products. The system proceeds from left to right, (consuming reactants, forming products) to reach equilibrium., , •, , Qc = Kc, , The initial concentrations are equilibrium concentrations. The, system is at equilibrium., , •, , Qc > Kc, , The ratio of initial concentrations of products to reactants is, too large. To reach equilibrium, products must be converted, to reactants. The system proceeds from right to left, (consuming products, forming reactants) to reach equilibrium., 63, , Practice Personal Hygiene protocols at all times.
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Figure 6. The direction of reversible reaction to reach equilibrium depends on the relative, magnitudes of Qc and Kc., Chang, R. (2010). Chemistry (10th ed., pp. 633. McGraw-Hill, Inc., USA., , EXAMPLES:, 1. At the start of a reaction, there are 0.249 mol N2, 3.21 * 10-2 mol H2 and 6.42, * 10-4 mol NH3 in a 3.50-L reaction vessel. If the equilibrium constant (Kc) for, the reaction is 1.2 at this temperature, decide whether the system is at, equilibrium. If it is not, predict which way the net reaction will proceed. The, reaction is given by, N2 (g) + 3H2 (g), , 2NH3 (g), , Solution, ▪, , Compute for the initial concentration of the reacting species, [N2] =, [N2] =, , 0.249 𝑚𝑜𝑙, 3.50 𝐿, , 3.21 ∗ 10−2 𝑚𝑜𝑙, 3.50 𝐿, , [NH3] =, ▪, , = 9.17 ∗ 10−3 M, , 6.42 ∗ 10−4 𝑚𝑜𝑙, 3.50 𝐿, , = 1.83 ∗ 10−4 M, , Compute for Qc, [𝑁𝐻3 ]2, 3, 2 ][𝐻2 ], , Qc = [𝑁, ▪, , = 0.0711 𝑀, , [1.83∗10−4 ]2, , = [0.0711][[9.17∗10−4]3 = 0.56, , Decision, Qc = 0.56 > 1.2, the system is not in equilibrium. The net, result will be an increase in the concentration of NH3 and, a decrease in the concentrations of N2 and H2. The net, reaction will proceed from left to right until equilibrium is, reached., 64, , Practice Personal Hygiene protocols at all times.
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2. The Kc for the formation of nitrosyl chloride, an orange-yellow compound,, from nitric oxide and molecular chlorine is 6.5 * 104. In a certain experiment,, 2.0 * 10-2 mole of NO, 2.0 * 10-3 mole of Cl2, and 6.8 moles of NOCl are, mixed in 2.0-L flask. In which direction will the system proceed to reach, equilibrium? The reaction is given by, 2NO (g) + Cl2 (g), , 2NOCl (g), , Solution, ▪, , Comput for the initial concentration of the reacting species, [NO] =, [Cl2] =, , 2.0 ∗ 10−2 𝑚𝑜𝑙, 2.0 𝐿, 8.3 ∗ 10−3 𝑚𝑜𝑙, , [NOCl] =, ▪, , 2.0 𝐿, 6.8 𝑚𝑜𝑙, 2.0 𝐿, , = 0.01 𝑀, = 0.00415 M, , = 3.4 M, , Compute for Qc, [𝑁𝑂𝐶𝑙]2, , [3.4]2, , Qc = [𝑁𝑂]2[𝐶𝑙 ] = [0.01] 0.00415]2 = 6.71 * 107, 2, , ▪, , Decision, , ▪, , Qc = 6.71 * 107 > 6.5 * 104, the system is not in equilibrium., The net result will be a decrease in the concentration of NOCl, and an increase in the concentrations of NO and Cl2. The net, reaction will proceed from right to left until equilibrium is, reached., , Learning Competency:, Explain chemical equilibrium in terms of the reaction rates of the forward and, reverse reaction (STEM_GC11CT-IVa-b-145), , 65, Practice Personal Hygiene protocols at all times.
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Activity 1: PROBLEM SETS, , A. Directions: Read carefully each of the following items. Choose the letter, that correspond to the best answer and write it on the answer sheet., 1. Which of the following is TRUE for a chemical reaction at equilibrium?, a. only the forward reaction stops, b. only the reverse reaction stops, c. both the forward and reverse reactions stops, d. the rate constant for the forward and backward reactions are equal, e. the rates of the forward and backward reactions are equal, 2. Which of the following is TRUE regarding the concentration of products,, for a chemical reaction that is already at equilibrium assuming no, disruptions to the equilibrium?, a. The concentrations of products will not change because there are, no more reactants., b. The concentrations of products will not change because the, limiting reagent is gone., c. The concentrations of products will not change because the, forward and reverse rates are equal, d. The concentrations of products will change continually because, of reversibility., 3. Which of the following are equal for a chemical system at equilibrium? If, all are equal, answer e., a. The concentrations of reactants and products are equal, b. The rate constant for the forward and reverse reactions are equal, c. The time that a particular atom or molecule spends as a reactant, and product are equal, d. The rate of the forward and reverse reaction, e. All of the above are equal, , 66, Practice Personal Hygiene protocols at all times.
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4. A chemical equilibrium may be established by starting a reaction with, ____________________., a. reactants only, b. products only, c. equal quantities of reactants and products, d. any quantities of reactants and products, e. all of the above, 5. An equilibrium that strongly favors products has ______________., a. a value of K≪ 1, , d. a value of Q ≪ 1, , b. a value of K ≫ 1, , e. K = Q, , c. a value of Q ≫ 1, 6. The equilibrium constant for the acid ionization of mercaptoethanol is, 1.91 * 10-10. The reaction is given by, HSCH2CH2OH (aq), , H+(aq) + SCH2CH2OH- (aq), , Which of the following statements is true regarding this equilibrium?, I., , The reaction is product favored., , II., , The reaction is reactant favored, , III., , Equilibrium lies far to the right, , IV., , Equilibrium lies far to the left, , a. I and III, , d. II and IV, , b. I and IV, , e. None are true, , c. II and III, 7. The equilibrium constant for the formation of hydrogen iodide from iodine, is 45 at a certain temperature., H2 (g) + I2 (g), , 2HI, , Which of the following statements is true regarding this equilibrium?, I., , The reaction is product favored., , II., , The reaction is reactant favored, , III., , Equilibrium lies far to the right, , IV., , Equilibrium lies far to the left, , a. I and III, , d. II and IV, , b. I and IV, , e. None are true, , c. II and III, , 67, Practice Personal Hygiene protocols at all times.
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8. If the reaction quotient Qc has a smaller value than the related, equilibrium constant Kc, ______________________., a. the reaction is at equilibrium, b. the reaction is not at equilibrium, and will make more products at, the expense of the reactants, c. the reaction is not at equilibrium, and will make more reactants at, the expense of the products, d. the value of Kc will decrease until it is equal to Q, e. the reaction favors the products, 9. If the reaction quotient Qc has a smaller value than the related, equilibrium constant Kc, ______________________., a. the reaction is at equilibrium, b. the reaction is not at equilibrium, and will make more products at, the expense of the reactants, c. the reaction is not at equilibrium, and will make more reactants at, the expense of the products, d. the value of Kc will decrease until it is equal to Q, e. the reaction favors the products, 10. If the equilibrium is established by initially adding 0.10 mol each of A, and B to a 1L container, then which of the following must be true once, the mixture achieves equilibrium?, A + 2B, , 2C, , Kc = 320, , a. [A] = [B], , d. [A] > [B], , b. [A] = [B] = [C], , e. [A] < [B], , c. [B] = 2[C], 11. Which of the following is the correct equilibrium expression for the, following reaction:, man + woman, [𝑚𝑎𝑛][𝑤𝑜𝑚𝑎𝑛], , a. Kc=, , [𝑐𝑜𝑢𝑝𝑙𝑒], , [𝑚𝑎𝑛]+ [𝑤𝑜𝑚𝑎𝑛], , b. Kc=, , [𝑐𝑜𝑢𝑝𝑙𝑒], , couple, [𝑐𝑜𝑢𝑝𝑙𝑒], , d. [𝑚𝑎𝑛][𝑤𝑜𝑚𝑎𝑛], e. Kc = [couple] – [man] –, , [woman], c. Kc = [man] [woman] [couple], , 68, Practice Personal Hygiene protocols at all times.
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12. Which of the following is the correct equilibrium expression for the, following reaction assuming homogeneity:, fool (money)10, , fool + 10money, , a. Kc= [fool(money)10] [fool] [money], [𝑓𝑜𝑜𝑙][𝑚𝑜𝑛𝑒𝑦]10, , b. Kc = [𝑓𝑜𝑜𝑙(𝑚𝑜𝑛𝑒𝑦), , 10 ], , c. Kc = [fool(money)10] [fool] [money], d. Kc =, , [𝑓𝑜𝑜𝑙(𝑚𝑜𝑛𝑒𝑦)10 ]10, [𝑓𝑜𝑜𝑙][𝑚𝑜𝑛𝑒𝑦], , e. Kc = [fool] + 10[money] – [fool(money)10], 13. For the following hypothetical equilibrium, what is the value of the, equilibrium constant if the concentrations at equilibrium are shown as, A(g) + 2B(g), , 2C(g), , when A = 4.5 * 10-5M; B = 2.2 * 10-2M; and C = 2.2 * 10-3M, d. 2.3 * 108, , a. 0. 22, , e. 9.5 * 103, , b. 9.9, c. 4.3 * 105, , 14. For the following hypothetical equilibrium, what is the value of the, equilibrium constant if the concentrations at equilibrium are shown as, A(aq) + 2B(aq), , 2C(aq) + D(aq), , when A = 4.5 * 10-5M; B = 2.2 * 10-2M; C=2.2 * 10-3M; and, D = 1.2 * 10-2M, a. 52, , d. 65, , b. 32, , e. 49, , c. 67, 15. All of the following are NOT seen in a equilibrium-constant expression, EXCEPT _______ of the reacting species, a. amount, b. molar concentration, , d. state, e. molal concentration, , c. phase, , 69, Practice Personal Hygiene protocols at all times.
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when 1.05 moles of Br2 are put in a 0.980-L flask, 1.20 percent of the, Br2 undergoes dissociation. Calculate the equilibrium constant for the, reaction. At what side will the equilibrium lie?, 4. Methanol is manufactured industrially by the reaction, CO (g) + 2H2 (g), , CH3OH (g), , A gaseous mixture at 500K is 0.020 M CH3OH; 0.10 M CO; and 0.10, M H2. Calculate the equilibrium constant for the reaction if the, temperature is 300K. At what side will the equilibrium lie?, 5. Pure phosgene gas was placed in a 435-mL container with a, temperature of 767K. The phosgene gas dissociates to CO and Cl 2., It was found out that the concentrations of the reacting species at that, temperature are [COCl2] = 1.6 M; [CO] = 1.3 M; [Cl2] = 0.06 M. What, is the equilibrium constant for the reaction? At what side will the, equilibrium lie? The reaction is given by, COCl2 (g), , CO (g) + Cl2 (g), , 6. In a 1767.3-mL sealed container was a mixture of 0.13 mole Na2CO3;, 1.21 moles SO2; 0.53 mole O2; 1.93 moles Na2SO4; and 0.07 mole, CO2 at 670C. What is the equilibrium constant for the reaction at, 450K? At what side will the equilibrium lie? The reaction is given by, 1, , Na2CO3 (s) + SO2 (g) + 2O2 (g), , Na2SO4 (s) + CO2 (g), , 7. Ammonium carbanate decomposes as follows, NH4CO2NH2 (s), , 2NH3 (g) + CO2 (g), , If the mixture is placed in a sealed steel 7.69-dm3 vessel and the, amount of the reacting species are [NH4CO2NH2] = 0.45 mole; [NH3], = 0.21 mole; [CO2] = 1.32 moles, what is the equilibrium constant for, the reaction? At what side will the equilibrium lie?, 8. Consider the reaction of carbon and hydrogen gas to form methane., Experiment shows that at 3630C, there are 4.0 moles C; 0.4 mole H2;, and 1.3 moles CH4 in a 10 dm3. What is the equilibrium constant for, the reaction? At what side will the equilibrium lie? The reaction is, given by, C (s) + 2H2 (g), , CH4 (g), , 72, Practice Personal Hygiene protocols at all times.
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9. At 657K, the following reaction occurs in a 1.25-L sealed container., CH4 (g) + 2H2S (g), , CS2 (g) + 4H2 (g), , It was found out that the mixture contains 1.25 moles CH4; 0.89 mole, H2S; 2.41 moles CS2; and 0.2 mole H2. What is the equilibrium, constant for the reaction at that temperature? At what side will the, equilibrium lie?, 10. At 5650C, the following reaction happens, 2HgO (s), , 2Hg (l) + O2 (g), , If the reaction mixture is placed in a 2354.21-mL sealed steel tank, container and it contains [HgO] = 1.2 M; [Hg] = 2.3 M; and [O2] = 0.21, M. What is equilibrium constant for the reaction at that temperature?, At what side will the equilibrium lie?, , Activity 4: CALCULATING REACTION QUOTIENT, QC, Directions: Read and solve the following problems. Show your complete, solution. Final answer must be rounded-off in two decimal places., 1. A 50-L reaction vessel contains 1.0 moles N2; 3.0 moles H2 and 0.5 mole, NH3. Will more NH3 be formed or it will dissociate when the mixture goes, to equilibrium at 4000C. the equation is given by, N2 (g) + 3H2 (g), , 2NH3 (g), , Calculate for the reaction quotient of the reaction. At what side will the, equilibrium shift if the equilibrium constant of the reaction is 0.50 at, 4000C?, 2. A 10-L vessel contains 0.0015 mole CO2; 1.0 mole C; and 0.10 mole CO., If a small amount of carbon is added to this vessel and the temperature, is reached to 10000C, will more carbon monoxide form? What is the, reaction quotient of the reaction. At what side will the equilibrium shift if, the equilibrium constant of the reaction is 1.17 at that temperature? The, reaction is given by, C (s) + CO2 (g), , 2CO (g), , 3. The following equilibrium process has been studied at 4580C., 1, , Na2CO3 (s) + SO2 (g) + 2O2 (g), , Na2SO4 (s) + CO2 (g), , 73, Practice Personal Hygiene protocols at all times.
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In one experiment, the amount of the reacting species at equilibrium are, found to be [Na2CO3] = 0.645 mole; [SO2] = 0.65 mole; [O2] = 0.67 mole;, [Na2SO4] = 1.3 moles; and [CO2] = 0.015 mole. Assuming that the, volume of the gas in the container is 988 mL at 4980C, what is the, reaction quotient of the reaction? At what side will the equilibrium shift if, the equilibrium constant of the reaction is 1.3 at that temperature?, 4. The equilibrium concentration for the reaction between hydrochloric acid, and oxygen gas to form liquid water and chlorine gas is given by, 4HCl (g) + O2 (g), , 2Cl2 (g) + 2H2O (l), , At 7600C, the concentration of the reacting species are [HCl] = 1.79 M;, [O2] = 0.14 M; [H2O] = 0.45 M; and [Cl2] = 3.45 M. Calculate the reaction, quotient of the reaction if it is contained in a 675-mL container and Kc is, 5.22. At what side will the equilibrium shift?, 5. At the start of the reaction, there are 0.167 mole C2H4; 0.394 mole H2O;, 1.17 mole C2H6 and 0.23 mole O2 in a 7963- mL tank at 9800C. If the, equilibrium constant for this reaction at 5000C is 2.67, what is the, reaction quotient? Predict at which direction side the net reaction will, proceed. The reaction is given by, 2C2H4 (g) + 2H2O (g), , 2C2H6 (g) + O2 (g), , 6. A mixture of 0.16 mole hydrogen gas and 1.3 moles sulfur gas to form, 0.46 mole dihydrogen sulfide is placed in a 54.36-mL steel container., Assuming that the equilibrium constant of the reaction at 706.45K is, 0.76, in what direction will the system proceed? The reaction is, 2H2 (g) + S2 (g), , 2H2S (g), , 7. The equilibrium constant for the formation of calcium sulphate from the, reaction of calcium oxide, sulphur dioxide and oxygen gas is 2.67 at, 790C. In a certain experiment, 0.10 mole CaO; 1.2 moles SO2; 0.12 mole, O2; and 1.8 moles CaSO4 are mixed in a 3.45-L container. In which, direction will the system proceed? The reaction is, 2CaO (s) + 2SO2 (g) + O2 (g), , 2CaSO4 (s), , 8. At 12800C, the Kc for the reaction, 2ZnS (s) + 3O2 (g), , 2ZnO (s) + 2SO2 (g), , is 1.68. if the reacting species is placed in a 3.21-L sealed container and, the concentrations are as follows: [ZnS] = 1.2 M; [O 2] = 0.17 M; [ZnO] =, 74, Practice Personal Hygiene protocols at all times.
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1.2 M; and [SO2] = 3.1 M, what will be the reaction quotient of the, reaction? To which direction will the system proceed?, 9. The reaction, 2NOBr (g), , 2NO (g) + Br2 (g), , has a Kc of 0.018 at 670C. The reacting species are placed in a 1700cm3 sealed container with amounts [NOBr] = 3.94 moles; [NO] = 1.34, moles; and [Br2] = 1.67 moles. What will be the value of Qc? To which, direction will the system proceed?, 10. A mixture of 1.12 moles NO2; 0.2 moles H2; 0.16 mole NH3 and 0.04, mole H2O is placed in a 473.0-cm3 steel container. Calculate Qc, assuming that Kc of the reaction is 1.29 at 980K. To which direction will, the system proceed? The reaction is given by, 2NO2 (g) + 3H2 (g), , 2NH3 (g) + H2O (l), , Reflection, , 1. I learned that ___________________________________________, ___________________________________________________________, ___________________________________________, , 2. I enjoyed most on _________________________________________, ____________________________________________________________, ____________________________________________________________, , 3. I want to learn more on __________________________________________, ____________________________________________________________, ____________________________________________________________, ________________________________________________________, , 75, Practice Personal Hygiene protocols at all times.
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References, Brown, T.L. et al. Chemistry: The Central Science (12th ed., pp. 610-649)., Pearson Prentice Hall. USA, Chang, R. (2010). Chemistry (10th ed., pp. 614-657). McGraw-Hill, Inc.,, USA., Ebbing, D.D & Gammon, S.D. (2017). General Chemistry (9th ed., 580-622)., Houghton Mifflin Company, USA., , 76, Practice Personal Hygiene protocols at all times.
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ANSWER KEY, , ACTIVITY 1, , A. Multiple Choice, 1. e, 2. c, 3. d, 4. e, 5. b, 6. d, 7. a, 8. b, 9. c, 10. d, 11. d, 12. b, 13. c, 14. e, 15. b, B. Homogenous or Heterogeneous, 1. HM, 2. HM, 3. HT, 4. HM, 5. HT, 6. HM, 7. HT, 8. HT, 9. HT, 10. HT, , 77, Practice Personal Hygiene protocols at all times.
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ACTIVITY 3, 1. H2 = 2.5 moles = 0.21 M, S2 = 1.35 * 10-5 moles = 1.125 * 10-6 M, H2S = 8.70 moles = 0. 73 M, 𝐾𝑐, , [0.73]2, = 1.07 ∗ 107, [0.21]2 [1.125 ∗ 10−6 ], , Kc ≫ 1, the equilibrium will lie to the right and favours the product, 2. [NH3] = 0.25 M, [N2] = 0. 11 M, [H2] = 1.91 M, 𝐾𝑐 =, , [0.25], 1, 3, [0.11]2 [1.91]2, , = 0.29, , Kc ≪ 1, the equilibrium will lie to the left and favours the reactant, 3. Br2 = 1.05 moles = 1.07 M, Br = 1.05 moles * 1.20% = 0.0126 mole = 0.01 M, 𝐾𝑐 =, , [0.01]2, = 9.35 ∗ 10−5, [1.07], , Kc ≪ 1, the equilibrium will lie to the left and favours the reactant, 4. CH3OH = 0.020 M, CO = 0.10 M, H2 = 0.10, 𝐾𝑐 =, , [0.020], = 20, [0.10][0.10]2, , Kc ≫ 1, the equilibrium will lie to the right and favours the product, 5. [COCl2] = 1.6 M, [CO] = 1.3 M, [Cl2] = 0.06 M, 𝐾𝑐 =, , [1.3][0.06], = 0.05, [1.6], , Kc ≪ 1, the equilibrium will lie to the left and favours the reactant, 6. Na2CO3 = 0.13 mole = 0.07 M, SO2 = 1.21 moles = 0.68 M, O2 = 0.53 mole = 0.30, Na2SO4 = 1.93 moles = 1.09, 80, Practice Personal Hygiene protocols at all times.
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CO2 = 0.07 mole = 0.04, 𝐾𝑐 =, , [0.04], 1, [0.68][0.30]2, , = 0.11, , Kc ≪ 1, the equilibrium will lie to the left and favours the reactant, 7. [NH4CO2NH2] = 0.45 mole = 0.06 M, [NH3] = 0.21 mole = 0.03 M, [CO2] = 1.32 moles = 0.17 M, 𝐾𝑐 = [0.03]2 [0.17] = 1.53 ∗ 10−4, Kc ≪ 1, the equilibrium will lie to the left and favours the reactant, 8. C = 4.0 moles = 0.4 M, H2 = 0.4 mole = 0.04 M, CH4= 1.3 moles = 0.13 M, 𝐾𝑐 =, , [0.13], = 81.25, [0.04]2, , Kc ≫ 1, the equilibrium will lie to the right and favours the product, 9. CH4 = 1.25 moles = 1 M, H2S = 0.89 mole = 0. 71 M, CS2 = 2.41 moles = 1.93 M, H2 = 0.2 mole = 0. 16 M, [1.93][0.16]4, 𝐾𝑐 =, = 2.51 ∗ 10−3, [1][0.71]2, Kc ≪ 1, the equilibrium will lie to the left and favours the reactant, 10. [HgO] = 1.2 M, [Hg] = 2.3 M, [O2] = 0.21 M, 𝐾𝑐 = [0.21] = 0.21, Kc ≪ 1, the equilibrium will lie to the left and favours the reactant, , 81, Practice Personal Hygiene protocols at all times.
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ACTIVITY 4, 1. N2 = 1.0 moles = 0.02 M, H2= 3.0 moles = 0.06, NH3 = 0.5 mole = 0.01, Kc = 0.50, [0.01]2, 𝑄𝑐 =, = 23.15, [0.02][0.06]3, Qc > Kc, the system is not in equilibrium. To attain equilibrium,, products must be converted to reactants., 2. CO2 = 0.0015 mole = 0.00015 M, C = 1.0 mole = 0.1 M, CO = 0.10 mole = 0.01, Kc = 1.17, 𝑄𝑐 =, , [0.01]2, = 0.67, [0.00015], , Qc < Kc, the system is not in equilibrium. To attain equilibrium,, reactants must be converted to products., 3. [Na2CO3] = 0.645 mole = 0.65 M, [SO2] = 0.65 mole = 0.66 M, [O2] = 0.67 mole = 0.68 M, [Na2SO4] = 1.3 moles = 1.32 M, [CO2] = 0.015 mole = 0.02, Kc = 1.3, 𝑄𝑐 =, , [0.02], 1, , = 0.04, , [0.66][0.68]2, Qc < Kc, the system is not in equilibrium. To attain equilibrium,, reactants must be converted to products., 4. [HCl] = 1.79 M, [O2] = 0.14 M, [H2O] = 0.45 M, [Cl2] = 3.45 M, Kc = 5.22, 𝑄𝑐 =, , [3.45]2, = 8.28, [1.79]4 [0.14], 82, , Practice Personal Hygiene protocols at all times.
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Qc > Kc, the system is not in equilibrium. To attain equilibrium,, products must be converted to reactants., 5. C2H4 = 0.167 mole = 0.21 M, H2O = 0.394 mole = 0.49 M, C2H6 = 1.17 mole = 1.47, O2 = 0.23 mole = 0.29 M, Kc = 2.67, [1.47]2 [0.29], 𝑄𝑐 =, = 59.18, [0.21]2 [0.49]2, Qc > Kc, the system is not in equilibrium. To attain equilibrium,, products must be converted to reactants., 6. S2 = 1.3 moles = 23.91 M, H2 = 0.16 mole = 2.94 M, H2S = 0.46 mole = 8.46 M, Kc = 0.76, [8.46]2, 𝑄𝑐 =, = 0.35, [2.94]2 [23.91], Qc < Kc, the system is not in equilibrium. To attain equilibrium,, reactants must be converted to products., 7. CaO = 0.10 mole = 0.03 M, SO2 = 1.2 moles = 0.35 M, O2 = 0.12 mole = 0.03 M, CaSO4 = 1.8 moles = 0.52 M, KC = 2.67, 𝑄𝑐 =, , 1, = 272.11, [0.35]2 [0.03], , Qc > Kc, the system is not in equilibrium. To attain equilibrium,, products must be converted to reactants., 8. [ZnS] = 1.2 M, [O2] = 0.17 M, [ZnO] = 1.2 M, [SO2] = 3.1 M, KC = 1.68, , 83, Practice Personal Hygiene protocols at all times.
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[3.1]2, 𝑄𝑐 =, = 332.53, [0.17]2, Qc > Kc, the system is not in equilibrium. To attain equilibrium,, products must be converted to reactants., 9. [NOBr] = 3.94 moles = 2.32 M, [NO] = 1.34 moles = 0.79 M, [Br2] = 1.67 moles = 0.98 M, KC = 0.018, 𝑄𝑐 =, , [0.79]2 [0.98], = 0.11, [2.32]2, , Qc > Kc, the system is not in equilibrium. To attain equilibrium,, products must be converted to reactants., 10. NH3 = 0.16 mole = 0.34 M, H2O = 0.04 mole = 0.08 M, NO2 = 1.12 moles = 2.37 M, H2 = 0.2 moles = 0.42 M, Kc = 1.29, 𝑄𝑐 =, , [0.34]2, = 0.28, [2.37]2 [0.42]3, , Qc < Kc, the system is not in equilibrium. To attain equilibrium,, reactants must be converted to products., , Prepared by:, GRACE ANN M. CALIBOSO - AGCAOILI, David M. Puzon Memorial National High School, , 84, Practice Personal Hygiene protocols at all times.
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, CHEMICAL EQUILIBRIUM, Background Information for the Learners (BIL), In a chemical reaction, chemical equilibrium is the state in which both, reactants and products are present in concentrations which have no further, tendency to change with time, so that there is no observable change in the, properties of the system. Usually, this state results when the forward reaction, proceeds at the same rate as the reverse reaction. The reaction rates of the, forward and backward reactions are generally not zero, but equal. Thus, there, are no net changes in the concentrations of the reactant(s) and product(s)., Such a state is known as dynamic equilibrium., Historical Introduction, The concept of chemical equilibrium was developed and found that, some chemical reactions are reversible. For any reaction mixture to exist at, equilibrium, the rates of the forward and backward (reverse) reactions are, equal. In the following chemical equation with arrows pointing both ways to, indicate equilibrium,[5] A and B are reactant chemical species, S and T are, product species, and α, β, σ, and τ are the stoichiometric coefficients of the, respective reactants and products:, αA+βB⇌σS+τT, The equilibrium concentration position of a reaction is said to lie "far to the, right" if, at equilibrium, nearly all the reactants are consumed. Conversely the, equilibrium position is said to be "far to the left" if hardly any product is formed, from the reactants., Guldberg and Waage (1865), building on Berthollet's ideas, proposed the law, of mass action:, Forward reaction rate =k+ Aα Bβ, backward reaction rate = K_Sϭ Tϯ, where A, B, S and T are active masses and k+ and k− are rate constants., Since at equilibrium forward and backward rates are equal:, K+ {A}α {B}β = K_{S}ϭ {T}t, , 85, Practice Personal Hygiene protocols at all times.
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and the ratio of the rate constants is also a constant, now known as, an equilibrium constant., Kc= Kt = {S}ϭ {T}t, k, {A}ϭ {B}β, By convention the products form the numerator. However, the law of, mass action is valid only for concerted one-step reactions that proceed through, a single transition state and is not valid in general because rate equations do, not, in general, follow the stoichiometry of the reaction as Guldberg and Waage, had proposed (see, for example, nucleophilic aliphatic substitution by SN1 or, reaction of hydrogen and bromine to form hydrogen bromide). Equality of, forward and backward reaction rates, however, is a necessary condition for, chemical equilibrium, though it is not sufficient to explain why equilibrium, occurs., Despite the failure of this derivation, the equilibrium constant for a, reaction is indeed a constant, independent of the activities of the various, species involved, though it does depend on temperature as observed by, the van 't Hoff equation. Adding a catalyst will affect both the forward reaction, and the reverse reaction in the same way and will not have an effect on the, equilibrium constant. The catalyst will speed up both reactions thereby, increasing the speed at which equilibrium is reached., Although the macroscopic equilibrium concentrations are constant in, time, reactions do occur at the molecular level. For example, in the case, of acetic acid dissolved in water and forming acetate and hydronium ions,, CH3CO2H + H2O ⇌ CH3CO−2 + H3O+, a proton may hop from one molecule of acetic acid on to a water molecule and, then on to an acetate anion to form another molecule of acetic acid and leaving, the number of acetic acid molecules unchanged. This is an example of dynamic, equilibrium., Equilibria, like the rest of thermodynamics, are statistical phenomena,, averages of microscopic behavior., A French chemist, Henri Le Chatelier studied the changes made in the, system that affect its equilibrium. He found out that “ if a system at equilibrium, is subjected in to a stress, the stress is displaced in a direction so as to relieve, the stress. This is called Le Chatelier’s principle., Studies revealed that different chemical reactions behave differently. In, the same manner that equilibrium is reached at different points for different, chemical reactions. Equilibrium constant will determine when a given chemical, reaction reaches its equilibrium point. It is expressed in terms of the, 86, Practice Personal Hygiene protocols at all times.
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concentrations of the products divided by the concentrations of the reactants., Each concentration is raised to an exponent equal to the coefficient of the, substance in the balanced chemical equation. Mathematically, for a chemical, reaction,, aA(g) +, , bB(g) ==, , cC(g) +, , dD(g), , Keq = ([C]c) ([D]d), ([A]a) ([B]b), Where K eq is the equilibrium constant, [C] & [D] are the concentrations of the products, [A] & [B] are the concentrations of the reactants, a b,c,d are the coefficients of the respective products in the balanced, chemical, equation., , Illustrative example:, , Initial, At equilibrium, , HC2H3O2, =, 0.1 molar, 0.10-0.01(0.1), 0.099, , H+, , +, 0, , 0.01(0.01), 0.001, 0.001, , C2H3O2, 0, 0.01(0.01), 0.001, 0.001, , The concentration of the substance and the ions at equilibrium are, obtained as shown because only 1.0% of the solution was ionized and the rest, was unionized., Using the dissociation process, express the Keq of the system., Keq = [H+] [Ac-], [HAc], = (0.01) (0.1) (0.01) (0.1), (0.099), = 1.01 x 10-5, , Learning Competency:, Calculate the Equilibrium constant and the pressure and concentration of, reactants or products in an equilibrium mixture (STEM_11CE-IVb-e-148), , 87, Practice Personal Hygiene protocols at all times.
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Activity I. LET’S TEST YOUR UNDERSTADING!, Directions: Give the equilibrium constant expression of the following reactions:, 1., 2., 3., 4., 5., , 2HCl(g), 4Fe(s), 2NaCl(s), 2KClO3, H2(g), , =, +, =, =, +, , H2(g), 3O2(g), 2Na(s), 2KCl(s), Cl2(g), , +, =, =, =, , Cl(g), 2Fe2O3(s), +, Cl2(g), 3O2(g), 2HCl(g), , SOLUBILITY PRODUCT CONSTANT, Assuming that Lead (II) Chloride, PbCl2 is placed in a beaker with water., The crystals will star to dissolve and as the amount of lead ions and chloride, ions increases in the solution, the possibility of ions returning to the solid state, will also increase. The rate at which the crystals dissolve to form ions and the, rate at which the ions return to solid state is known as solubility equilibrium., If we try to determine the solubility product of lead (II) chloride given its, dissociation process, we have:, PbCl2(aq), , =, , Pb+(aq), , +, , 2Cl-(aq), , Note that [PbCl2] is not part of the solubility product expression since the, undissociated PbCl2, is not actually in the solution. The solubility equilibrium is, attained when the rate at which ions leave the solid state is equal to the rate at, which ions return to the solid state. The rate of formation of ions is dependent, on the area of a solids. The greater the surface area of the solid used, the, greater is the rate of formation of ions., , Illustrative example:, 1. Write the solubility product expression for the dissociation of sodium, chloride and silver nitrate., a. NaCl(s), =, Na+(aq), +, Cl-(aq), Ksp, =, [Na+]{Cl-], b. AgNO3(s), =, Ag+(aq), +, NO3-(aq), +, Ksp, =, [Ag ] [NO3 ], , Activity 2: SOLUBILITY PRODUCT EXPRESSION ., DIRECTION: Give the solubility product expression of the following reactions:, 1. MgSO4(s), =, Mg+(s) +, SO4-(aq), 2. AgCl(s), =, Ag+(aq), +, Cl-(aq), +, 3. CaCl2(s), =, Ca (aq), +, 2Cl(aq), 4. MgI2(s), =, Mg+(aq), +, 2I(aq), , 88, Practice Personal Hygiene protocols at all times.
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References:, Frank Brescia, John Arents, Herbert Meislich, amos Turk, Fundamentals of, Chemistry 4th Edition, Ma. Christina Padolina, PhD, Laboratory Manual and Workbook in Chemistry, Marasinghe, B.Dr. 2010. Upper Secondary Chemistry. A textbook of chemistry, for Grades 11 &12., Bettelheim, Brown, Campbell, Farrell, Introduction to General, Organic and, Biochemistry, 8th Edition, Sackheim, George I, Lehman, Dennis D., Chemistry for Health Sciences, 8 th, Edition, , 90, Practice Personal Hygiene protocols at all times.
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ANSWER KEY, ACTIVITY 1, 1. [H2] [Cl], [HCl], 2., , [Fe2O3]2, [Fe]4 [O2]3, , 3. [Na]2 [Cl2], [NaCl]2, 4. [KCl]2 [O2]3, [KClO3]2, 5., , [HCl]2, [H2] [Cl2], , ACTIVITY 2, 1. [Mg2] [SO4-2], 2. [Ag+] [Cl-], 3. [Ca+2] [Cl2-], 4. [Mg2] [I2-], , ACTIVITY 3, Problem #1, Answer: We calculate the reaction quotient and then compare it with K:, Q = P2SO3/ P2SO2PO2, = 0.572/0.412 x 0.16, = 12.1, Since Q is greater than K (12.1>3.40), the partial pressure of the product, (SO3), is too large relative to the partial pressures of the reactants (SO2 and O2). To, reach equilibrium, the reaction must go from right to left., 2SO2(g), +, O2(g) ←, 2SO3(g), , 91, Practice Personal Hygiene protocols at all times.
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Problem #2, Answer: We first calculate the reaction quotient for the given reaction and the, given mixture:, Q= P2I/PI2 = (1.0 x 10-5)2/ 1.0 = 1.0 x 10-10, We see that Q is less than K (1.0 x 10-10 < 2.3 x 10-10). Therefore the partial, pressure of I is too small and the partial pressure of I2 is too large. The reaction, can go from left to right., I2, →, 2I, Problem#3, Answer: At equilibrium, K = P2NO3/PN2O4, We substitute the given equilibrium pressure:, K=(0.41)2/ 2.3 x 10-7 = 7.3 x 105, This large equilibrium constant indicates that the equilibrium favors the product, NO2 in this case. Most N2O4 molecules are spilt up into NO2 molecules at 1000K, , Prepared by:, 5 DOLORES ARAGON-LIBAN, Magalalag National High School, , 92, Practice Personal Hygiene protocols at all times.
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GENERAL CHEMISTRY 2, Name: _______________________, Date: ________________________, , Grade Level: __________________, Score:, __________________, , LEARNING ACTIVITY SHEET, FACTORS THAT AFFECT CHEMICAL EQUILIBRIUM, Background Information for the Learners (BIL), Le Châtelier’s Principle states that if an external stress is applied to a, system at equilibrium, the system adjusts in such a way that the stress is, partially offset as the system reaches a new equilibrium position., Stress - means a change in concentration, pressure, volume, or temperature, that removes the system from the equilibrium state. We will use Le Châtelier’s, principle to assess the effects of such changes., , The Effect of Changes in Concentration, , Adding or removing a product or reactant disturbs equilibrium., a. Stress of an added reactant or product is relieved by reaction in the, , direction that consumes the added substance. Adding reactant, the, reaction shifts to the right (toward the product). Adding product, the, reaction shifts to the left (toward reactant), b. Stress of removing a reactant or product is relieved by reaction in the, direction that replenishes the removed substance. Removing reactant, will shift the reaction towards the reactant. Removing a product will shift, the reaction towards the product., Changes in the value of Q (reaction quotient) disturbs an equilibrium., a. Keq > Q – reaction shifts to the product, b. Keq < Q – reaction shifts to the reactant, , To show you how to predict the changes in concentration on a system at, equilibrium., Let us consider the synthesis of ammonia., , 𝑁2(𝑔) + 3𝐻2(𝑔) ⇌ 2𝑁𝐻3(𝑔), 93, Practice Personal Hygiene protocols at all times.
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Given the concentration of each of the molecule we have as follows;, [N2] =0.399 M [H2] = 1.197 M [NH3] = 0.202 M, , What if we are going to add 1.0 mol/L of N2, we can answer it by calculating the, Q., For the system which is no longer at equilibrium, 𝑄=, , [𝑁𝐻3 ]0 2, [𝑁2 ]0 [𝐻2 ]0 3, , =, , (0.202)2, = 1.70 × 10−2, (1.399)(1.197)3, , Using the first set of equilibrium concentrations, we are to calculate for the value, of K., , [ NH 3 ] 2, ( 0 . 202 ) 2, K =, =, = 5 . 96 x 10, [ N 2 ][ H 2 ] 3 ( 0 . 399 )( 1 . 197 ) 3, , −2, , Therefore, we can conclude that since the Q is less than the value of K because, of the added N2 the system will shift to the right to restore equilibrium., , According to Le Châtelier’s Principle the system will shift to a direction that, consumes Nitrogen., , The Effect of a Change in Pressure, , There are three ways to change the pressure of a reaction system involving, gaseous components:, 1. Add or remove a gaseous reactant or product., 2. Add an inert gas (one not involved in the reaction)., 3. Change the volume of the container., , Changes in pressure and volume affects the number of moles in the, gaseous reactants and gaseous products. Increase in pressure due to, decrease in volume results in a reaction in the direction of fewer number of, moles. Decrease in pressure due to increase in volume results in a reaction in, the direction of greater number of moles., 94, Practice Personal Hygiene protocols at all times.
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Example 1: in the production of ammonia, 𝑁2(𝑔) + 3𝐻2(𝑔) ⇌ 2𝑁𝐻3(𝑔) ., Increasing the pressure will shift the reaction towards the product because, there is fewer number of moles (2) in the product than in the reactant, (1+3=4moles), , Note:, - When an inert gas is added, there is no effect on the equilibrium position., -The addition of an inert gas increases the total pressure but has no effect on, the concentrations or partial pressures of the reactants or products., , The Effect of a Change in Temperature, , A change in concentration, pressure, or volume may alter the equilibrium, position, that is, the relative amounts of reactants and products, but it does not, change the value of the equilibrium constant., - Only a change in temperature can alter the equilibrium constant., To see why, let us consider the reaction, , N 2O4 ( g ) 2 NO2 ( g ), The forward reaction is endothermic (absorbs heat, ΔH° > 0):, , heat + N 2O4 ( g ) 2 NO2 ( g ) ΔH˚ = 58.0 kJ/mol, So the reverse reaction is exothermic (releases heat, ΔH° < 0):, , 2 NO2 ( g ) → N 2 O4 ( g ) + heat ΔH˚ = - 58.0 kJ/mol, - In summary, a temperature increase favors an endothermic reaction, and a, temperature decrease favors an exothermic reaction., , Using Le Châtelier’s Principle, , For each of the following reactions, predict how the value of K changes as the, temperature is increased., , 95, Practice Personal Hygiene protocols at all times.
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a.N 2 ( g ) + O2 ( g ) 2 NO( g ) H = 181 k J, b.2 SO2 ( g ) + O2 ( g ) 2 SO3( g ) H = −198 k J, SOLUTION, a. This is an endothermic reaction, as indicated by the positive value for ΔH˚., Energy, can be viewed as a reactant, and K increases (the equilibrium shifts to the, right) as the temperature is increased., b. This is an exothermic reaction (energy can be regarded as a product). As, the temperature is increased, the value of K decreases (the equilibrium shifts, to the left)., , The Effect of a Catalyst, , A catalyst enhances the rate of a reaction by lowering the reaction’s activation, energy., A catalyst lowers the activation energy of the forward reaction and the reverse, reaction to the same extent. However, they do not affect the amount reactants, and products in equilibrium., , Learning Competency:, State the Le Châtelier’s Principle and apply it qualitatively to describe the effect, of changes in pressure, concentration and temperature on a system at, Equilibrium. (STEM_GC11CEIVb-e-149), , Activity 1. PREDICT ME, Directions: Predict the direction of the following chemical equations given in, each items., 1) For the reaction below, which change would cause the equilibrium to shift, to the right?, CH4(g) + 2H2S(g) ↔ CS2(g) + 4H2(g), (a) Decrease the concentration of dihydrogen sulfide., (b) Increase the pressure on the system., (c) Increase the temperature of the system., (d) Increase the concentration of carbon disulfide., (e) Decrease the concentration of methane., , 96, Practice Personal Hygiene protocols at all times.
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2) What would happen to the position of the equilibrium when the following, changes are made to the equilibrium system below? 2SO3(g) ↔ 2SO2(g) + O2(g), (a) Sulfur dioxide is added to the system., (b) Sulfur trioxide is removed from the system., (c) Oxygen is added to the system., 3) What would happen to the position of the equilibrium when the following, changes are made to the reaction below? 2HgO(s) ↔ Hg(l) + O2(g), (a) HgO is added to the system., (b) The pressure on the system increases., 4) When the volume of container the following mixture of gases is increased,, what will be the effect on the equilibrium position? 4HCl(g) + O2(g) ↔ 2H2O(g) +, 2Cl2(g), 5) Predict the effect of decreasing the volume of the container for each, equilibrium., (a) 2H2O(g) + N2(g) ↔ 2H2(g) + 2NO(g), (b) SiO2(s) + 4HF(g) ↔ SiF4(g) + 2H2O(g), (c) CO(g) + H2(g) ↔ C(s) + H2O(g), , Activity 2: EQUILIBRIUM SHIFT, Directions: Fill in the blank, Give the direction of the chemical equation below., 1. Given the equilibrium equation: PCl5(g) ⇌ PCl3(g) + Cl2(g), a) If the [PCl5] is increased, the equilibrium will shift to the ________, b) If the [PCl5] is decreased, the equilibrium will shift to the ________, c) If the [PCl3] is increased, the equilibrium will shift to the ________, d) If the [PCl3] is decreased, the equilibrium will shift to the ________, e) If the [Cl2] is increased, the equilibrium will shift to the ________, f) If the [Cl2] is decreased, the equilibrium will shift to the ________, 2. Given the reaction at equilibrium: A(g) ⇌ B(g) ∆H = + 32.5 kJ C(g), a) If the temperature was increased, which way would this equilibrium, shift: _____________, b) If the temperature was decreased, which way would this equilibrium, shift: ______________, 3. Given the reaction: X(g) + Y(g) → W(g) + Z(g) ∆H = -75 kJ, a) Rewrite this as a thermochemical reaction (exothermic-heat term on right), ____________, b) If the temperature was increased, which way would this equilibrium shift:, ____________, c) If the temperature was decreased, which way would this equilibrium shift:, ______, , 97, Practice Personal Hygiene protocols at all times.
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4. Given the equilibrium equation: N2O4(g) ⇌ 2O2(g) + N2(g), a) If the total pressure of this system is increased, the equilibrium will shift, _____, b) If the total pressure of this system is decreased, the equilibrium will shift, _____, c) If the total volume of this system is increased, it means the same as, decreasing the pressure so the equilibrium will shift to the ______, d) If the total volume of this system is decreased, it means the same as, increasing the pressure so the equilibrium will shift to the______, 5) Given the equilibrium equation: X + Y + heat Z ⇌ XY, a) Increasing the temperature will cause the equilibrium to shift _______, Activity 3: APPLY ME, Directions: Apply Le Châtelier’s Principle to predict the following problems., 1. Consider the following equilibrium system involving SO2, Cl2, and SO2Cl2, (sulfuryl dichloride):, SO2(g) + Cl2(g) ↔ SO2Cl2(g), Predict how the equilibrium position would change if, (a) Cl2 gas were added to the system;, (b) SO2Cl2 were removed from the system;, (c) SO2 were removed from the system. The temperature remains constant., 2. Consider the following equilibrium systems:, (a) A ↔2B, ΔH° = 20.0 kJ/mol, (b) A + B ↔ C, ΔH° = - 5.4 kJ/mol, (c) A ↔ B, ΔH° = 0.0 kJ/mol, Predict the change in the equilibrium constant Kc that would occur in each, case if the temperature of the reacting system were raised., 3. What effect does an increase in pressure have on each of the following, systems at equilibrium? The temperature is kept constant and, in each case,, the reactants are in a cylinder fitted with a movable piston., (a) A(s) ↔ 2B(s), (b) 2A(l) ↔ B(l), (c) A(s) ↔ B(g), (d) A(g) ↔ B(g), (e) A(g) ↔ 2B(g), , 98, Practice Personal Hygiene protocols at all times.
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ANSWER KEY, ACTIVITY 1, 1) c, increase the temperature of the system because a decrease in, temperature favors the exothermic reaction., 2) a) Shifts left to counteract the increased concentration of SO2(g)., b) Shifts left to counteract the decrease in concentration of SO3(g)., c) Shifts left to counteract the increase in concentration of O2(g)., 3) a) No shift because pure liquids and solids have no effect on the equilibrium, position., b) Heterogeneous equilibrium, therefore increased pressure has no effect, on the system., 4) Shifts left to increase the number of gas molecules. The increased in volume, results to decreased in pressure. Therefore the reaction shifts to the left in the, direction of greater number of moles of gases., 5) a) Shifts left to produce fewer number of gas molecules., b) Shifts right to produce fewer number of moles of gas., c) Shifts right to produce fewer number of moles of gas., , ACTIVITY 2, 1. a. Right, b. Left, c. Left, d. Right, e. Left, f. Right, 2. a. Right (toward product), b. Left (towards reactant), 3. a. W(g) + Z(g) → X(g) + Y(g), b. Left (reactant), c. Right (product), 4. a. Left, b. Right, c. Right, d. Left, 5. a. Right, , ΔH˚ = +75 kJ, , 101, Practice Personal Hygiene protocols at all times.
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ACTIVITY 3, 1. According to Le Chateliers principle a system at equilibrium will react to an, external stress in such a way that the stress is involve. Hence,, a. The system will consume the added reactant Cl2 so it will shift to the right, b. The system will replace the removed SO2Cl2 so it will shift to the right, c. The system will replaces the moved reactant SO2 so it will shift to the left, 2. a.ΔH˚ is positive, so the direction will shift to the right as the temperature, increases., b.ΔH˚ is negative so the direction favors the reactant side and heat is in the, product side., c. ΔH˚ is equal to zero it has no effect on the equilibrium., 3. a. since no gas on the equation, there will be no effect on the equilibrium., b. no gas involve in the reactant (heterogeneous equilibrium) that is why, there’s no effect on the equilibrium., c. since no gas in the reaction, increasing the pressure will shift, d. since there is equal number of moles of reactants and products, increased, in pressure has no effect on the system., e. increased in pressure will shift the reaction equilibrium to the left., , Prepared by:, WILLIAM E. ERRO, Itawes National Agricultural and Technical School - Main Campus, , 102, Practice Personal Hygiene protocols at all times.
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GENERAL CHEMISTRY 2, Name: ______________________ Grade Level: ____________________, Date: ________________________ Score:, , ____________________, , LEARNING ACTIVITY SHEET, FACTORS THAT AFFECT CHEMICAL EQUILIBRIUM, Background Information for the Learners (BIL), The Bronsted - Lowry Theory is an acid- base theory proposed by Johannes, Nicolaus Bronsted and Thomas Martyn Lowry in 1923. this theory states that, any compound that can transfer a proton to any other compound is an acid and, the compound that accept proton is a base. A proton is a nuclear particle with, a unit positive electrical charge.This substance can function an acid with the, presence of a base. Likewise, the base can function with presence of an acid., In addition, an acidic substance loses a proton it forms base and this is called, conjugate base of an acid and when a substance gains proton it is called, conjugate acid of the base, , Bronsted - Lowry Definition of Terms:, Acids - a substance that is capable of donating Hydrogen ion H+ (proton)., Bases - it is a substance that is capable of accepting Hydrogen ion H+, Conjugate acid-base pair - is defined as an acid and its conjugate base or a, base and its conjugate acid., Amphoteric - species that can act as acids or bases (Ex. water)., Conjugate Base - species that remains after an acid donates a Hydrogen ion, (H+)., Conjugate Acid - an species that forms after a base accepts the Hydrogen ion, (H+)., , Acid ionization or Dissociation Equation:, , HA( a q) + H 2 O( l ) → A − ( a q ) + H 3O + ( a q), acid, , base, , conjugate base, , conjugate acid, , 103, Practice Personal Hygiene protocols at all times.
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Base ionization or Dissociation Equation, , B( a q) + H 2 O( l ) → BH, base, , acid, , +, , ( a q), , conjugate acid, , + OH, , −, , ( a q), , conjugate base, , Example 1:, , HCl + H 2O → H 3O + + Cl −, - The chloride ion (Cl-) is conjugate base formed from the acid HCl, and the, Hydronium ion (H3O+) is the conjugate acid of the base water (H2)., , Learning Competency:, Define Bronsted Acids and Bases. (STEM_GC11ABIVf-g-153 ), , Activity 1: CLASSIFY ME, 1. Classify the following as Brønsted acids, bases or both., a. H2O, b. OHc. NH3, d. NH4+, e. NH2f. CO32-, , Activity 2: TRY ME, 2. What is the conjugate base of the following:, a. HClO4, b. NH4+, c. H2O, d. HCO3-, , 104, Practice Personal Hygiene protocols at all times.
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, ACID-BASE PROPERTY OF WATER, Background Information for the Learners (BIL), Water has so many unique properties which makes it an exceptional, solvent and makes it essential for life on Earth. One of its unique properties is, its ability to act as either an acid or as a base in a solution. Water can be a, proton donor (an acid) or a proton acceptor (a base). Therefore, water, molecules exhibit amphiprotic property even in trace amounts., H3O+, , Water accepts a, proton, acts as a, base, , H2O, , Water donates a, proton, acts as, an acid, , OH-, , What does amphiprotic mean?, Think of a hammer. A hammer can be, used to drive nails into a wood, also it, can be used to remove nails from the, wood. It serves a dual purpose just like, amphoteric substances., https://www.distrelec.biz/Web/WebShopImages/l, andscape_large/2-/01/30110722-01.jpg, , Water, just like hammer, can, accept and donate a proton., , Note: The term amphoteric is a general term for substances that can react both, as an acid and a base. On the other hand, amphiprotic (protic refers to, hydrogen ion) is a more specific term used to describe a substance which can, both donate and accept hydrogen ions (protons). All amphiprotic substances, are amphoteric but not all amphoteric substances are amphiprotic. For, example, ZnO acts as a Lewis acid, which can accept an electron pair from OHbut cannot donate a proton., , 108, Practice Personal Hygiene protocols at all times.
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How does this happen? For example, the reaction of a strong acid like, hydrochloric acid (HCl) and water leads to the formation of chlorine and, hydronium (H3 O+ ) ions., 𝐻𝐶𝑙 + 𝐻2 𝑂 → 𝐻3 𝑂+ + 𝐶𝑙, In this case, water acts as a base because it accepts a proton which leads to, the formation of hydronium ion (H3 O+ ). On the other hand, the reaction between, a base like ammonia (NH3 ) and water leads to the formation of ammonium, (NH4+ ) and hydroxide (OH − ) ions., 𝑁𝐻3 + 𝐻2 𝑂 → 𝑁𝐻4+ + 𝑂𝐻 −, Water acts as an acid because it donates a proton which leads to the formation, of hydroxide ion (OH − )., , Auto-ionization of Water, Some substances undergo reactions without the addition of another, substance. This is just one of the unique properties of water. Under standard, conditions, experiments on the electrical conductivity of water have shown that, water ionizes on a very small extent. The auto-ionization (self-ionization) of, water refers to the reaction in which a water molecule donates one of its protons, (H atom) to a neighbouring water molecule, either in pure water or in aqueous, solution. The result is the formation of a hydroxide ion (OH-) and a hydronium, ion (H3O+). The auto-ionization of water molecules is shown as (also called, autoprotolysis of water):, , In chemical reaction equation,, , equation (1), , We often use the simplified form of the reaction:, equation (2), , 109, Practice Personal Hygiene protocols at all times.
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Among the above equations, it does not make any difference which, equation is used to explain the ionization of water. Even though, equation (1) is, more appropriate, the equation (2) offers a more simple view of the ionization, which then will be used on the next discussion., Note that this process is readily reversible. Because water is a weak acid, and a weak base, the hydronium and hydroxide ions exist in very, very small, concentrations relative to that of non-ionized water. Just how small are these, concentrations? Let's find out by examining the equilibrium constant for this, reaction (also called the auto-ionization constant), which has the special, symbol, Kw, , The Ion Product of Water, In the study of acid-base reactions, the hydrogen ion concentration is the, key because its value indicates the acidity or basicity of the solution. Because, only a very small fraction of water molecules are ionized, the concentration of, water, [H2O], remains virtually unchanged. Therefore, the equilibrium constant, (Kc) for the auto-ionization of water, using equation (1), is, 𝐾𝑐 = [𝐻3 𝑂 + ][𝑂𝐻 − ], where:, , equation (3), Kc, , = equilibrium constant, , [𝐻3 𝑂+ ] = concentration of hydronium ion, [𝑂𝐻 − ] = concentration of hydroxide ion, , Since we can use H+ and H3O+ interchangeably to represent the hydrated, proton, the equilibrium constant can also be expressed as, 𝐾𝑐 = [𝐻 + ][𝑂𝐻 − ], , equation (4), , To indicate that the equilibrium constant refers to the auto-ionization of water,, we replace Kc by Kw, 𝐾𝑤 = [𝐻3 𝑂+ ][𝑂𝐻 − ] = [𝐻 + ][𝑂𝐻 − ], , equation (5), , where Kw is called the ion-product constant of water, which is the product of, the molar concentrations of 𝐻 + and 𝑂𝐻 − ions at a particular temperature., , 110, Practice Personal Hygiene protocols at all times.
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In pure water at 250 C, the concentrations of 𝐻 + and 𝑂𝐻 − are equal., Pure water or any other aqueous solution in which this ratio holds is said to be, neutral., [𝐻 + ]= 1.0 x 10-7 M, [𝑂𝐻 − ] = 1.0 x 10-7 M, 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑒, , 𝑤ℎ𝑒𝑟𝑒: 𝑀 = 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛, 𝑚𝑜𝑙𝑎𝑟𝑖𝑡𝑦 (𝑙𝑖𝑡𝑒𝑟𝑠 𝑜𝑓 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛), Substituting these in equation (4), we have, 𝐾𝑤 = [𝐻 + ][𝑂𝐻 − ] = (1.0 x 10-7) (1.0 x10-7) = 1.0 x 10-14, Whether we have pure water or an aqueous solution of dissolved species, the, following relation always holds at 25°C., 𝐾𝑤 = [𝐻 + ][𝑂𝐻 − ] = 1.0𝑥 10−14, , equation (6), , This equation is very useful in calculating either H+ ion concentration knowing, OH- concentration or OH- ion concentration knowing H+ concentration through, the following equations:, , [𝐻+ ] =, [𝑂𝐻− ] =, , 𝐾𝑤, , =, [𝑂𝐻 − ], 𝐾𝑤, , =, [𝐻 + ], , 1.0 𝑥 10−14, [𝑂𝐻 − ], 1.0 𝑥 10−14, [𝐻 + ], , equation (7), equation (8), , A neutral solution is an aqueous solution in which the concentrations, of H3O+ ions and OH- ions are equal just like pure water at 25°C., An acidic solution is a solution in which the concentration of hydrogen, ions is greater than the concentration of hydroxide ions. For example,, hydrogen chloride ionizes to produce H+ and Cl− ions upon dissolving in water., , A basic solution is a solution in which the concentration of hydroxide, ions is greater than the concentration of hydrogen ions. Solid potassium, hydroxide dissociates in water to yield potassium ions and hydroxide ions., , 111, Practice Personal Hygiene protocols at all times.
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The relationship between [H+] and [OH-] is inversely proportional. The increase, in concentration of the OH− ions causes a decrease in the concentration of the, H+ ions and the ion-product of [H+][OH−] remains constant., Solution, , General Condition, , At 25°C, , acidic, , [𝐻3 𝑂+ ] > [𝑂𝐻 − ], , [𝐻3 𝑂 + ] > 1.0 𝑥 10−7, , [𝑂𝐻 − ] < 1.0 𝑥 10−7, , neutral, , [𝐻3 𝑂 + ] = [𝑂𝐻 − ], , [𝐻3 𝑂 + ] = 1.0 𝑥 10−7, , [𝑂𝐻 − ] = 1.0 𝑥 10−7, , basic, , [𝐻3 𝑂+ ] < [𝑂𝐻 − ], , [𝐻3 𝑂+ ] < 1.0 𝑥 10−7, , [𝑂𝐻 − ] > 1.0 𝑥 10−7, , Table showing the relationship between [H+] and [OH-] ions., , Sample Problem 1: Using Kw for an Aqueous Solution, The concentration of OH- ions in a certain household ammonia cleaning, solution is 0.0025 M. Calculate the concentration of H+ ions., , Solution:, In this problem, we are given the concentration of the OH- ions which is 0.0025, M and are asked to calculate the [H+]. The relationship between [H+] and [OH] in an aqueous solution is given by the ion-product equation,, 𝐾𝑤 = [𝐻 + ][𝑂𝐻 − ] = 1.0 𝑥 10−14 , rearranging this equation, [𝐻 + ] =, , 𝐾𝑤, 1.0 𝑥 10−14, 1.0 𝑥 10−14, =, =, = 4.0 𝑥 10−12 𝑀, [𝑂𝐻 − ], [𝑂𝐻 − ], 0.0025, , Because [H+] < [OH-], the solution is basic., Sample Problem 2. Calculating the [OH-] from a given [H+], Sufficient acidic solute is added to a quantity of water to produce a solution, with [H+] = 4.0 x 10-3 M. What is the [OH-] in this solution?, , Solution:, In this problem, we are given the concentration of the H+ ions which is 4.0 x, 10-3 M and are asked to calculate the [OH-]. The relationship between [H+] and, [OH-] in an aqueous solution is given by the ion-product equation,, , 112, Practice Personal Hygiene protocols at all times.
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Activity 2: DRILL YOUR BRAIN!, A. Directions. Draw a graph showing the relationship of [OH-] and [H+] on the, following situation. Explain the relationship., [𝐻3 𝑂 + ], , Neutral, Solution, , [𝑂𝐻 − ], , [𝐻3 𝑂 + ], , 10-14, 10-13, , 10-14, 10-13, , 10-12, 10-11, 10-10, , Acidic, Solution, , Basic, Solution, , [𝑂𝐻 − ], , [𝐻3 𝑂 + ], , 10-14, 10-13, , 10-14, 10-13, , 10-14, 10-13, , 10-14, 10-13, , 10-12, 10-11, 10-10, , 10-12, 10-11, 10-10, , 10-12, 10-11, 10-10, , 10-12, 10-11, 10-10, , 10-12, 10-11, 10-10, , 10-9, 10-8, 10-7, , 10-9, 10-8, 10-7, , 10-9, 10-8, 10-7, , 10-9, 10-8, 10-7, , 10-9, 10-8, 10-7, , 10-9, 10-8, 10-7, , 10-6, 10-5, 10-4, , 10-6, 10-5, 10-4, , 10-6, 10-5, 10-4, , 10-6, 10-5, 10-4, , 10-6, 10-5, 10-4, , 10-6, 10-5, 10-4, , 10-3, 10-2, 10-1, , 10-3, 10-2, 10-1, , 10-3, 10-2, 10-1, , 10-3, 10-2, 10-1, , 10-3, 10-2, 10-1, , 10-3, 10-2, 10-1, , 100, , 100, , 100, , 100, , 100, , 100, , [𝑂𝐻 − ], , ___________________________________________________________, ___________________________________________________________, ___________________________________________________________, B. Calculate the molar H+ ion concentration of a solution if the OH- ion, concentration is, a. vinegar, 1.0 x 10-11 M, b. ammonia, 5.6 x 10-3 M, c. NaOH solution, 2.5 x 10-2 M, C. Calculate the molar OH- ion concentration of a solution if the H+ ion, concentration is, a. coffee, 1.0 x 10-5 M, b. lemon juice, 2.5 x 10-2 M, c. cleanser, 5.0 x 10-10 M, , 114, Practice Personal Hygiene protocols at all times.
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Activity 3: HOW WELL DO YOU KNOW?, A. Answer the following briefly., 1. Write the chemical equation showing the ionization of water., ________________________________________________________, 2. Write the equilibrium constant expression for this equation., ________________________________________________________, 3. What is the relationship between [H3O+] and [OH-] in pure water at 25°C?, ________________________________________________________, 4. How can the relationship between [H3O+] and [OH-] be used to define or, identify if the solution is basic or acidic?, ________________________________________________________, ________________________________________________________, 5. If a base is added to pure water, why does the [H3O+] decrease?, _____________________________________________________________________, _______________________________________________________, , B. Solve for the following problems. Tell whether the solution is basic or, acidic., 1. Sufficient acidic solute is added to a quantity of water to produce a, solution with [H+] = 5.7 x 10-6 M. What is the [OH-] in this solution?, 2. The concentration of OH- ions in a particular household ammonia, cleaning solution is 0.075M. What is the H+ ion concentration?, , C. Selected information about five solutions, each at 25°C, is given in the, following table. Complete the table by filling in the missing information., [H+], , [OH-], , Acidic or Basic, , 2.2 x 10-2 M, 3.3 x 10-3 M, 7.7 x 10-2 M, 6.3 x 10-8 M, 4.2 x 10-6 M, , Practice Personal Hygiene protocols at all times., 115
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Activity 4: OPERATION: Crossword Puzzle, Directions. Complete the crossword by filling in the boxes to form a word that, fits each clue., , Practice Personal Hygiene protocols at all times., 116
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References:, Books, Stoker, H. Stephen.2010. Exploring General, Organic and Biological Chemistry., CENGAGE Learning. Philippine Edition., Teaching Guide for Senior High School. General Chemistry 2. 2016. Commission, on Higher Education., Timberlake, Karen C. Chemistry: An Introduction to General, Organic, and, Biological Chemistry.12th ed., Whitten, Kenneth W et.al. 2004. General Chemistry. Brooks/Cole. 7th ed., Online Sources, Acid-Base Properties of Water (n.d.). Retrieved June 15, 2020 from, https://courses.lumenlearning.com/introchem/chapter/acid-base-properties-ofwater/, Acids and Bases. (n.d). Retrieved June 15, 2020 from, https://www.khanacademy.org/science/chemistry/acids-and-bases-topic/acidsand-bases/a/water-autoionization-and-kw, Ionization of Acids and Bases (n.d.). Retrieved June 15, 2020 from, https://www.ck12.org/book/cbse-chemistry-book-class-11/section/8.12/, The Acid-Base Properties of Water and Ion Product (n.d.). Retrieved June 15,, 2020 from, https://faculty.ncc.edu/LinkClick.aspx?fileticket=xZ3v05k1gPc%3D&tabid=1896, , Practice Personal Hygiene protocols at all times., 118
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ANSWER KEY, Activity 1. Who Am I?, A., , 1. Water is said to be amphiprotic, that is, H2O molecules can both, , donate and accept protons. Yes. Water also exhibits amphoterism because it, can either act as an acid or a base in a reaction., 2. Answer varies. Some of which are:, a. Pencil. It is used to write and can also be used to erase., b. Screwdriver. It is used to drive/tighten or remove screws., B. In the following reactions, identify if water is playing the role of an acid, a, base, or neither., 1., , 𝑆 2− + 𝐻2 𝑂 → 𝐻𝑆 − + 𝑂𝐻 −, , ___acid_____, , 2., , 𝐻𝐶𝑙𝑂2 + 𝐻2 𝑂 → 𝐶𝑙𝑂2− + 𝐻3 𝑂+, , ___base____, , 3., , 𝐶𝑢2+ + 6 𝐻2 𝑂 → 𝐶𝑢(𝐻2 𝑂)2+, 6, , ___neither__, , 4., , 𝑂𝐶𝑙 − + 𝐻2 𝑂 → 𝐻𝑂𝐶𝑙 + 𝑂𝐻 −, , ___acid____, , 5., , 𝐻𝑆𝑂4− + 𝐻2 𝑂 → 𝑆𝑂42− + 𝐻3 𝑂+, , ___base__ _, , Activity 2: Drill it!, A. Draw a graph showing the relationship of [OH-] and [H+] on the following, situation. Explain the relationship., + Acidic, [𝑂𝐻 − ] [𝐻3 𝑂 ], Solution, 10-14 10-14, , [𝑂𝐻 − ], , [𝐻3 𝑂 + ], , 10-14, , 10-14, , 10-14, , 10-13, , 10-13, , 10-13, , 10-13, , 10-13, , 10-13, , 10-12, , 10-12, , 10-12, , 10-12, , 10-12, , 10-12, , 10-11, , 10-11, , 10-11, , 10-11, , 10-11, , 10-11, , 10-10, , 10-10, , 10-10, , 10-10, , 10-10, , 10-10, , 10-9, , 10-9, , 10-9, , 10-9, , 10-9, , 10-9, , 10-8, , 10-8, , 10-8, , 10-8, , 10-8, , 10-8, , 10-7, , 10-7, , 10-7, , 10-7, , 10-7, , 10-7, , 10-6, , 10-6, , 10-6, , 10-6, , 10-6, , 10-6, , 10-5, , 10-5, , 10-5, , 10-5, , 10-5, , 10-5, , 10-4, , 10-4, , 10-4, , 10-4, , 10-4, , 10-4, , 10-3, , 10-3, , 10-3, , 10-3, , 10-3, , 10-3, , 10-2, , 10-2, , 10-2, , 10-2, , 10-2, , 10-2, , 10-1, , 10-1, , 10-1, , 10-1, , 10-1, , 10-1, , 100, , 100, , 100, , 100, , 100, , 100, , [𝐻3 𝑂 + ], 10-14, , Neutral, Solution, , Basic, Solution, , [𝑂𝐻 − ], , Practice Personal Hygiene protocols at all times., 119
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The relationship between [𝐻3 𝑂+ ] and [𝑂𝐻 − ] in aqueous solution is an inverse, proportion; when [𝐻3 𝑂+ ] is increased, [𝑂𝐻 − ] decreases, and vice versa., B. Calculate the molar H+ ion concentration of a solution if the OH- ion, concentration is, a. vinegar, 1.0 x 10-11 M, , [H+] =1.0 x 10-3 M, , b. ammonia, 5.6 x 10-3 M, , [H+] =1.8 x 10-12 M, , c. NaOH solution, 2.5 x 10-2 M, , [H+] =4.0 x 10-13 M, , C. Calculate the molar OH- ion concentration of a solution if the H+ ion, concentration is, a. coffee, 1.0 x 10-5 M, , [OH-] = 1.0 x10-9 M, , b. lemon juice, 2.5 x 10-2 M, , [OH-] = 4.0 x10-13 M, , c. cleanser, 5.0 x 10-10 M, , [OH-] = 2.0 x10-5 M, , Activity 3: How well do you know?, A. Answer the following briefly., 1. Write the chemical equation showing the ionization of water., 𝐻2 𝑂 + 𝐻2 𝑂 ↔ 𝐻3 𝑂+ + 𝑂𝐻 −, 2. Write the equilibrium constant expression for this equation., 𝐾𝑤 = [𝑂𝐻 − ][𝐻3 𝑂+ ] = 1.0 𝑥 10−14, 3. What is the relationship between [H3O+] and [OH-] in pure water at 25°C?, The formation of H3O+ ion by ionization of pure water is always, accompanied by the formation of an OH- ion, thus [H3O+] is always equal to, [OH-] which is equal to 1.0 x 10-7 M., [𝑂𝐻 − ] = [𝐻3 𝑂+ ] = 1.0 𝑥 10−7, 4. How can the relationship between [H3O+] and [OH-] be used to define or, identify if the solution is basic or acidic?, If [𝐻3 𝑂+ ] > [𝑂𝐻 − ], then the solution is acidic, If [𝐻3 𝑂+ ] < [𝑂𝐻 − ], then the solution is basic, , Practice Personal Hygiene protocols at all times., 120
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5. If a base is added to pure water, why does the [H3O+] decrease?, The relationship between [H+] and [OH-] is inversely proportional. The, increase in concentration of the OH− ions causes a decrease in the, concentration of the H+ ions and the ion-product of [H+][OH−] remains constant., B. Solve for the following problems. Tell whether the solution is basic or acidic., 1. 1.8 𝑥 10−9 𝑀, acidic, 2. 1.3 𝑥 10−13 𝑀, basic, , C. Selected information about four solutions, each at 25°C, is given in the, following table. Complete the table by filling in the missing information., [H+], , [OH-], , Acidic or Basic, , 2.2 x 10-2 M, , 4.5 x 10-13 M, , Acidic, , 3.0 x 10-12 M, , 3.3 x 10-3 M, , Basic, , 1.3 x 10-13 M, , 7.7 x 10-2 M, , Basic, , 6.3 x 10-8 M, , 1.6 x 10-7 M, , Basic, , 4.2 x 10-6 M, , 2.4 x10-9 M, , Acidic, , Activity 4: Operation: Crossword Puzzle, 1.Hydroxide ion, 2.Autoionization, 3.Amphoteric, 4.Neutral, 5.Acid, 6.Molarity, 7.Basic, 8.Amphiprotic, 9.Base, 10. Acidic, , Prepared by:, Rosemarie C. Fernandez, Itawes National High School, , Practice Personal Hygiene protocols at all times., 121
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, pH: A MEASURE OF ACIDITY, , Background Information for the Learners (BIL), Personnel working in food processing,, medicine, agriculture, spa and pool, maintenance, soap manufacturing, and, wine making measure the [H3O+] and, [OH-] of solutions. The proper level of, acidity is necessary to evaluate the, functioning of the lungs and kidneys, to, , https://images.wisegeek.com/chemist-with-vials.jpg, , control bacterial growth in foods, and to prevent the growth of pests in food, crops., In aqueous solution, an acid is defined as any species that increases, the concentration of H+ ions, while a base increases the concentration of OHions. Typical concentrations of these ions in solution can be very small, and, they also span a wide range. Let’s take a look on the concentration of H+ ions, of pure water at 25°C and to the acid present in our stomach. We have learned, from our previous lesson that pure water has a value of 1.0 x 10-7 M of H+ and, the acid in our stomach has 1.0 x10-1 M. This means that the [H+] of acid in our, stomach is 6 orders of magnitude larger than in pure water., We have also learned how to quantify the concentration of H+ ions from, the given concentration of OH- ions or vice-versa using the ion-product, constant of water, Kw. Now we are going to learn the relationship between, these concentrations and the so-called, pH., , Practice Personal Hygiene protocols at all times., 122
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The pH Concept, Because the concentrations of H+ and OH- ions in aqueous solutions, are frequently very small numbers and therefore inconvenient to work with,, Soren Sorensen† in 1909 proposed a more practical measure called pH. The, pH scale is a scale of small numbers that is used to specify molar hydronium, ion concentration in an aqueous solution., , https://www.sciencenewsforstudents.org/wp-content/uploads/2019/11/860_SS_pH.png, , Figure 1: pH of some common substances. On the pH scale, values below 7.0 are acidic, a, value of 7.0 is neutral, and values above 7.0 are basic., , Timberlake, Karen C. Chemistry: An Introduction to General, Organic, and Biological Chemistry.12th ed., , https://www.sciencenewsforstudents.org/wp-content/uploads/2019/11/860_SS_pH.png, , Figure 2: The pH of a solution can be determined using (a) pH meter, (b) pH paper, (c), indicators that changes in color in response to pH values., , Practice Personal Hygiene protocols at all times., 123
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Relating pH and [H3O+], We can calculate the pH of a solution knowing the concentrations of H+, or OH- ions in that solution. The calculation of pH scale values involves the use, of logarithms. The pH is the negative logarithm of an aqueous solution’s molar, hydronium ion concentration. Expressed mathematically, the definition of pH, is, , 𝑝𝐻 = −𝑙𝑜𝑔[𝐻3 𝑂+ ] or 𝑝𝐻 = −𝑙𝑜𝑔[𝐻 + ], , equation (1), , The negative logarithm gives us a positive number for pH, which, otherwise would be negative due to the small value of [H+]. Like the equilibrium, constant, the pH of a solution is a dimensionless quantity., Because pH is simply a way to express hydrogen ion concentration,, acidic and basic solutions at 25°C can be distinguished by their pH values, as, follows:, , Notice that pH increases as [H+] decreases., Sometimes we may be given the pH value of a solution and asked to, calculate the H+ ion concentration. In that case, we need to take the antilog of, equation (1) as follows:, , [𝐻3 𝑂+ ] = 10−𝑝𝐻, , or, , [𝐻 + ] = 10−𝑝𝐻, , equation (2), , Integral pH Values, For any hydronium ion concentration expressed in exponential notation, in which the coefficient is 1.0, the pH is given directly by the negative of the, exponent value of the power of 10:, , [𝐻3 𝑂+ ] = 1.0 𝑥 10−𝑥, , equation (3), , 𝑝𝐻 = 𝑥, , equation (4), , Practice Personal Hygiene protocols at all times., 124
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Thus, if the hydronium ion concentration is 1.0 x 10-9, then the pH will be, 9.00. This simple relationship between pH and hydronium ion concentration is, valid only when the coefficient in the exponential notation expression for the, hydronium ion concentration is 1.0., Sample problem 1. Calculating pH from [H3O+] and [OH-], Calculate the pH for each of the following solutions., a. [H3O+] = 1.0 x 10-6 M, b. [OH-] = 1.0 x10-6 M, Solution:, a. Because the coefficient in the exponential expression for the molar, hydronium ion concentration is 1.0, the pH can be obtained from the, relationships (equations 3 and 4), , [𝐻3 𝑂 + ] = 1.0 𝑥 10−𝑥 = 1.0 𝑥 10−6, 𝑝𝐻 = 𝑥 = 6, The power of 10 is -6 in this case, so the pH will be 6.00., b. The given value is the concentration of OH- ions. Before we could, compute for the pH, we have to find first the concentration of H3O+ ions using, the ion product of water,, Kw = [H3O+] [OH-] = 1.0 x 10-14, rearranging we have, , [𝐻3 𝑂 + ] =, , 1.0 𝑥 10−14, 1.0 𝑥 10−6, , = 1.0 x 10-8, applying now equations 3 and 4, , [𝐻3 𝑂 + ] = 1.0 𝑥 10−𝑥 = 1.0 𝑥 10−8, 𝑝𝐻 = 𝑥 = 8, The power of 10 is -8, in this case, the pH will be 8.00., , Non-integral pH Values, If the coefficient in the exponential expression for the molar hydronium, ion concentration is not 1.0, then the pH will have a non-integral value; that is,, it will not be a whole number. For example, consider the following non-integral, pH values., Practice Personal Hygiene protocols at all times., 125
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Note: The easiest way to obtain non-integral pH values such as these involves, using an electronic calculator that allows for the input of exponential numbers, and that has a base-10 logarithm key (LOG). In using such an electronic, calculator, you can obtain logarithm values simply by pressing the LOG key, after having entered the number whose log is desired. For pH, you must, remember that after obtaining the log value, you must change signs because, of the negative sign in the defining equation for pH., , Sample Problem 2. Calculating pH from [H3O+], Aspirin, which is acetylsalicylic acid, was the first non-steroidal antiinflammatory drug used to alleviate pain and fever. If a solution of aspirin has, a [H3O+] = 1.7 x 10-3 M, what is the pH of the solution?, Solution: What is known is the [H3O+] = 1.7 x 10-3 M and we are asked to, compute for the pH of the solution of aspirin. We know that to get the value of, pH, we use equation (1),, , 𝑝𝐻 = −𝑙𝑜𝑔[𝐻3 𝑂+ ], substituting from the given, 𝑝𝐻 = −𝑙𝑜𝑔[1.7 𝑥 10−3 ] = 2.77, , Sample Problem 3. Calculating [H3O+] from pH, Determine the [H3O+] for solutions having a pH=8.25., Solution: We are given the value of pH=8.25 and we are asked to compute for, the concentration of H3O+. For this, we are going to use equation (2), , [𝐻3 𝑂 + ] = 10−𝑝𝐻 , substituting, [𝐻3 𝑂 + ] = 10−8.25 = 5.6 𝑥 10−9 𝑀, Relating pOH and [OH-], A pOH scale analogous to the pH scale can be devised using the, negative logarithm of the hydroxide ion concentration of a solution. Thus, we, define pOH as, , 𝑝𝑂𝐻 = −𝑙𝑜𝑔 [𝑂𝐻 − ], , equation (5), , Practice Personal Hygiene protocols at all times., 126
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If we are given the pOH value of a solution and asked to calculate the, OH- ion concentration, we can take the antilog of equation (5), , [𝑂𝐻 − ] = 10−𝑝𝑂𝐻, , equation (6), , Relating pH and pOH, Now consider again the ion-product constant for water at 25°C:, Kw = [H+] [OH-] = 1.0 x 10-14, Taking the negative logarithm of both sides and from the definitions of pH and, pOH, we obtain, -log [H+] + -log [OH-] = -log 1.0 x 10-14, pH +, , pOH = 14.00, , equation (7), , This provides us with another way to express the relationship between the, H+ ion concentration and the OH- ion concentration., Sample Problem 6. Calculating [OH-] from pOH, In a NaOH solution [OH-] is 2.9 x 10-4 M. Calculate the pH of the solution., Solution: Solving this problem takes two steps. First, we need to calculate pOH, using equation (5). Next, we use equation (7) to calculate the pH of the solution., First, we use equation (5),, 𝑝𝑂𝐻 = −𝑙𝑜𝑔 [𝑂𝐻 − ] = − log(2.9 𝑥 10−4 ) = 3.54, Then we use equation (7),, pH + pOH = 14, pH = 14 - pOH = 14 - 3.54 = 10.46, The answer shows that the solution is basic, Note: We can also use the ion-product constant of water, Kw= [H+][OH-] to, calculate [H+], and then we can calculate the pH from the [H+]., , Practice Personal Hygiene protocols at all times., 127
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Learning Competency:, Calculate pH from the concentration of hydrogen ions or hydroxide ions in, aqueous solutions. (STEM_GC11AB-IVf-g-156), , Activity 1. BRAINS OUT!, Directions: Read and analyze each problem carefully. Identify the known, variables and solve for the unknown following the steps in solving a problem., 1. The OH- ion concentration of a blood sample is 2.5 x 10-7 M. What is, the pH of the blood?, 2. The pH of an unknown solution is 5.68. What is the [OH-] and pOH of, the solution?, 3. Upon investigating the pH of different ponds, you found out that a, certain pond measures 4.2 which is below the recommended pH of 6.5., a. What are the [H3O+] and [OH-] of the pond?, b. What are the [H3O+] and [OH-] of the pond that has a pH of 6.5?, 4. Calculate pH from the following hydrogen ion concentration (M)., Identify each as an acidic pH or a basic pH., Hydrogen Ion Concentration, 1. 0.0015, 2. 5.0 x 10-9, 3. 1.0, 4. 1.0 x 10-12, 5. 0.0001, 6. 1.0 x 10-4, 7. 1.0 x 10-8, 8. 5.63 x 10-9, 9. 3.67 x 10-6, 10. 3.25 x 10-4, , pH, , Basic/Acidic, , 5. You decided to test the pH of your brand-new swimming pool. The, instruction manual advises to keep it between 7.2-7.6. Shockingly, you, found out that the pH of your pool is 8.3! What kind of chemical should you, add?, , Practice Personal Hygiene protocols at all times., 128
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Q4. How can you use the pH scale in differentiating acidic, neutral and, basic solutions?, ________________________________________________________, Activity 3: pH matters! 1 PIC, 4 SENTENCES, Directions: The pictures below illustrate the different effects of pH in our, environment. Using four (4) sentences, describe how pH affects our daily life., 1. Human Survival, , Commons.wikimedia.org, , 2. Animal Survival, ______________________________, ______________________________, ______________________________, ______________________________, , https://rain-acid.weebly.com/consequences.html, , 3. Plant Survival, , ____________________________, ____________________________, ____________________________, ____________________________, https://www.usgs.gov/special-topic/water-scienceschool/science/acid-rain-and-water, , Practice Personal Hygiene protocols at all times., 130
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4. Soil Acidity, ____________________________, ____________________________, ____________________________, ____________________________, , https://www.usgs.gov/special-topic/water-scienceschool/science/acid-rain-and-water, , Activity 4: SHOW ME WHAT YOU’VE GOT!, Directions: Investigate and analyze the given situation. Provide an explanation, for your answer., Hair stylists recommend slightly acidic and near neutral shampoo for, smoother hair. You find 5 brands that you like. The first has a pH of 3.6, the, second is 13, the third is 8.2, the fourth is 6.8 and the fifth is 9.7. Which one, should you buy? Explain why it is better., ______________________________________________________________, ______________________________________________________________, __________________________________________________________., , Reflection:, 1. I learned that ____________________________________, _______________________________________________________, _______________________________________________________, , 2. I enjoyed most on _________________________________, _______________________________________________________, _______________________________________________________, , 3. I want to learn more on _____________________________, _______________________________________________________, _______________________________________________________, , Practice Personal Hygiene protocols at all times., 131
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References:, Books, Stoker, H. Stephen. 2010. Exploring General, Organic and Biological, Chemistry. CENGAGE Learning. Philippine Edition., Teaching Guide for Senior High School. General Chemistry 2. 2016., Commission on Higher Education, Timberlake, Karen C. Chemistry: An Introduction to General, Organic, and, Biological Chemistry.12th ed., Whitten, Kenneth W et.al. 2004. General Chemistry. Brooks/Cole. 7th ed., Online Sources, Acid-Base Properties of Water (n.d.). Retrieved June 15, 2020 from, https://courses.lumenlearning.com/introchem/chapter/acid-base-properties-ofwater/, Ionization of Acids and Bases (n.d.). Retrieved June 15, 2020 from, https://www.ck12.org/book/cbse-chemistry-book-class-11/section/8.12/, , Practice Personal Hygiene protocols at all times., 132
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4. Calculate pH from the following hydrogen ion concentration (M). Identify, each as an acidic pH or a basic pH., Hydrogen Ion Concentration, 1. 0.0015, 2. 5.0 x 10-9, 3. 1.0, 4. 1.0 x 10-12, 5. 0.0001, 6. 1.0 x 10-4, 7. 1.0 x 10-8, 8. 5.63 x 10-9, 9. 3.67 x 10-6, 10. 3.25 x 10-4, , pH, 2.82, 8.30, 0, 12, 3, 4, 8, 8.25, 5.44, 3.49, , Basic/Acidic, acidic, Basic, Acidic, Basic, Acidic, Acidic, Basic, Basic, Acidic, acidic, , 5.. You decide to test the pH of your brand new swimming pool. The instruction, manual advises to keep it between 7.2-7.6. Shockingly, you found out that the, pH of your pool is 8.3! What kind of chemical should you add?, Add an acidic solution to the pool., Activity 2: FILL ME UP!, Directions: Read and analyze each problem carefully. Identify the known, variables and solve for the unknown following the steps in solving a problem., [𝐻3 𝑂 + ], , 𝑝𝐻, , [𝑂𝐻 − ], , 𝑝𝑂𝐻, , 1 𝑥 100, , 0, , 1 𝑥 10−14, , 14, , 1 𝑥 10−1, , 1, , 1 𝑥 10−12, , 13, , 1 𝑥 10−2, , 2, , 1 𝑥 10−11, , 12, , 1 𝑥 10−3, , 3, , 1 𝑥 10−10, , 11, , 1 𝑥 10−4, , 4, , 1 𝑥 10−9, , 10, , 1 𝑥 10−5, , 5, , 1 𝑥 10−9, , 9, , 1 𝑥 10−6, , 6, , 1 𝑥 10, , −8, , 8, , 1 𝑥 10−7, , 7, , 1 𝑥 10−7, , 7, , −8, , 8, , 1 𝑥 10, , −6, , 6, , 1 𝑥 10−9, , 9, , 1 𝑥 10−5, , 5, , 1 𝑥 10−10, , 10, , 1 𝑥 10−4, , 4, , 1 𝑥 10−11, , 11, , 1 𝑥 10−3, , 3, , 1 𝑥 10−12, , 12, , 1 𝑥 10−2, , 2, , −13, , 13, , 1 𝑥 10, , −1, , 1, , 1 𝑥 10−14, , 14, , 1 𝑥 100, , 1 𝑥 10, , 1 𝑥 10, , 0, , Practice Personal Hygiene protocols at all times., 134
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Q1. From the completed table, what relationship of pH and hydronium ion, concentration can be established?, As the concentration of hydronium ion increases, the pH value, decreases., Q2. From the completed table, what relationship of pH and hydroxide ion, concentration can be established?, As the concentration of hydroxide ion increases, the pH value increases., Q3. What is the relationship of hydroxide and hydronium ion, concentrations? How about the pH and pOH?, Hydronium and hydroxide ion concentrations are inversely proportional,, that is, as hydronium ion concentration increases, the hydroxide ion, concentration decreases. Same applies with pH and pOH., Q4. How can you use the pH scale in differentiating acidic, neutral and, basic solutions?, , Activity 3: pH matters! 1 PIC, 4 SENTENCES, Directions: The pictures below illustrate the different effects of pH in our, environment. Using four (4) sentences, describe how pH affects our daily life., , ANSWERS MAY VARY., 1. Human Survival, Human body works within the pH range of 7.0 to 7.8. Living organisms, can survive only in a narrow range of pH. Large amount of acids are being, released in the stomach every time you eat. It helps in the digestion of food, without harming the stomach. Tooth decay starts when the pH of the mouth is, lower than 5.5. Tooth enamel, made up of calcium phosphate, is corroded when, the pH in the mouth is below 5.5., , Practice Personal Hygiene protocols at all times., 135
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2. Animal Survival, When pH of rain water is less than 5.6, it is called acid rain. The effects, of acid rain include damage to the limestone and marble in statues and, buildings; weakening of the exposed metal on bridges and cars; damage to, bodies of water, wildlife, plants, forests and crops; and the contamination of the, drinking water supply. When acid rain flows into the rivers, it lowers the pH of, the river water. Aquatic life cannot survive in acidic water., 3. Plants Survival, Acid rain has indirect effects on plants. It weakens the trees by damaging, their leaves, limiting the nutrients available to them, or poisoning them with toxic, substances slowly released from the soil. Trees affected by acid rain do not, grow as quickly as usual, their leaves and needles turn brown and fall off when, they should be green and healthy, 4. Soil Acidity, Plants require a specific pH range for their healthy growth. Soil pH, should be maintained at >5.5. Acidic soil prevents or limits root development., Plants grown in acidic soil cannot absorb water and nutrients, are stunted, and, exhibit nutrient deficiency symptoms (especially those for phosphorus)., , Activity 4: SHOW ME WHAT YOU’VE GOT!, Directions: Investigate and analyze the given situation. Provide an explanation, for your answer., Hair stylists recommend slightly acidic and near neutral shampoo for, smoother hair. You find 5 brands that you like. The first has a pH of 3.6,, the second is 13, the third is 8.2, the fourth is 6.8 and the fifth is 9.7., Which one should you buy? Explain why it is better., The fourth brand that has a pH of 6.8. It is close to being neutral, (pH=7), but is below pH 7 making it slightly acidic., , Prepared by:, Rosemarie C. Fernandez, Itawes National High School, , Practice Personal Hygiene protocols at all times., 136
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, BUFFERS, Background Information for the Learners (BIL), A buffer is a solution that can resist pH change upon the addition of an, acidic or basic component. It can neutralize small amounts of added acid or, base, thus maintaining the pH of the solution relatively stable. This is important, for processes and/or reactions which require specific and stable pH ranges., Buffer solutions have a working pH range and capacity which dictate how much, acid/base can be neutralized before pH changes, and the amount by which it, will change. Buffer solutions are used as a means of keeping pH at a nearly, constant value in a wide variety of chemical applications. For example, blood in, the human body is a buffer solution., Human blood has narrow limits in which its pH must be maintained. The, pH range is between 7.35 to 7.45. A pH outside this range results in many, diseases, including cancer. Many foods cause acidity. There are substances in, the body that neutralize acids or bases when the pH of the blood is increased, or decreased. For this reason, the pH of the blood is controlled., What is a buffer composed of?, To effectively maintain a pH range, a buffer must consist of a weak, conjugate acid-base pair, meaning either a. a weak acid and its conjugate base,, or b. a weak base and its conjugate acid. The use of one or the other will simply, depend upon the desired pH when preparing the buffer. For example, the, following could function as buffers when together in solution:, •, , •, , •, , Acetic acid (weak organic acid w/ formula CH3COOH) and a salt, containing its conjugate base, the acetate anion (CH3COO-), such as, sodium acetate (CH3COONa), Pyridine (weak base w/ formula C5H5N) and a salt containing its, conjugate acid, the pyridinium cation (C5H5NH+), such as Pyridinium, Chloride., Ammonia (weak base w/ formula NH3) and a salt containing its conjugate, acid, the ammonium cation, such as Ammonium Hydroxide (NH4OH), , Practice Personal Hygiene protocols at all times., 137
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How does a buffer work?, A buffer is able to resist pH change because the two components, (conjugate acid and conjugate base) are both present in appreciable amounts, at equilibrium and are able to neutralize small amounts of other acids and bases, (in the form of H3O+ and OH-) when the are added to the solution. To clarify, this effect, we can consider the simple example of a Hydrofluoric Acid (HF) and, Sodium Fluoride (NaF) buffer. Hydrofluoric acid is a weak acid due to the strong, attraction between the relatively small F- ion and solvated protons (H3O+),, which does not allow it to dissociate completely in water. Therefore, if we obtain, HF in an aqueous solution, we establish the following equilibrium with only slight, dissociation (Ka(HF) = 6.6x10-4, strongly favors reactants):, HF(aq) + H2O(l) ⇌ F−(aq) + H3O+(aq), , (1), , We can then add and dissolve sodium fluoride into the solution and mix, the two until we reach the desired volume and pH at which we want to buffer., When Sodium Fluoride dissolves in water, the reaction goes to completion, thus, we obtain:, NaF(aq) + H2O(l) → Na+ (aq) + F−(aq), , (2), , Since Na+ is the conjugate of a strong base, it will have no effect on the, pH or reactivity of the buffer. The addition of NaF to the solution will, however,, increase the concentration of F- in the buffer solution, and, consequently, by Le, Châtelier’s Principle, lead to slightly less dissociation of the HF in the previous, equilibrium, as well. The presence of significant amounts of both the conjugate, acid,HF, and the conjugate base, F-, allows the solution to function as a buffer., This buffering action can be seen in the titration curve of a buffer solution., , As we can see, over the working range of the buffer. pH changes very, little with the addition of acid or base. Once the buffering capacity is exceeded, the rate of pH change quickly jumps. This occurs because the conjugate acid, Practice Personal Hygiene protocols at all times., 138
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or base has been depleted through neutralization. This principle implies that a, larger amount of conjugate acid or base will have a greater buffering capacity., If acid were added:, F−(aq) + H3O+(aq) ⇌ HF(aq) + H2O(l), In this reaction, the conjugate base, F-, will neutralize the added acid,, H3O+, and this reaction goes to completion, because the reaction of F- with, H3O+ has an equilibrium constant much greater than one. (In fact, the, equilibrium constant the reaction as written is just the inverse of the Ka for HF:, 1/Ka(HF) = 1/(6.6x10-4) = 1.5x10+3.) So long as there is more F- than H3O+,, almost all of the H3O+ will be consumed and the equilibrium will shift to the, right, slightly increasing the concentration of HF and slightly decreasing the, concentration of F-, but resulting in hardly any change in the amount of H3O+, present once equilibrium is re-established., , If base were added:, HF(aq) + OH−(aq) ⇌ F−(aq) + H2O(l), , (4), , In this reaction, the conjugate acid, HF, will neutralize added amounts of, base, OH-, and the equilibrium will again shift to the right, slightly increasing the, concentration of F- in the solution and decreasing the amount of HF slightly., Again, since most of the OH- is neutralized, little pH change will occur., These two reactions can continue to alternate back and forth with little pH, change., How can a buffer maintain the pH of a substance even if small amounts, of acids or bases are added into the substance? If an acid is added to the, substance, the conjugate base of the buffer neutralizes the acid added. If a, base is added to the substance, the conjugate acid of the buffer neutralizes the, base added. For a buffer solution made up of HCH3COO and NaCH3COO, the, conjugate acid is HCH3COO and the conjugate base is CH3COO-, If HCl is, added into the solution containing the acetic acid – acetate buffer, the HCl, reacts with the conjugate base of the buffer, CH3COO-. HCl is strong acid and, is completely ionized., HCl → H+ +, , Cl--, , The H+ reacts with the CH3COO-1, the conjugate base of the buffer, forming acetic acid., H+ + CH3COO-1 → HCH3COO, , Practice Personal Hygiene protocols at all times., 139
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If NaOH is added into the solution containing the acetic acid – acetate, buffer, the NaOH reacts with conjugate acid of the buffer, HCH3COO. NaOH is, a strong acid and is completely ionized., NaOH → Na+ + OH-1, OH- + HCH3COO → HOH + CH3COOThe OH reacts with the HCH3COO, the conjugate acid of the buffer, forming H2O and CH3COO-., This explains why a buffer is able to maintain its pH., , Learning Competency:, Describe how a buffer solution maintain its pH (STEM_GC11ABIVf-g-160), , Activity 1: WHO AM I?, Part A, Directions: Choose the best buffer solution., You need a buffer with pH 4.5 and have four acids and their sodium salt, available. Which acid should you use to get the best buffer?, Your acids:, •, •, •, •, , Chloroacetic acid, Benzoic acid, Propanoic acid, Hypochlorous acid, , 1. Which acid should you use to get the best buffer?, ______________________________________________________________, _____, 2. What is the pKa of the acid?, ______________________________________________________________, _____, , Practice Personal Hygiene protocols at all times., 140
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3. Why do they give the most efficient buffer solution?, ______________________________________________________________, ______________________________________________________________, ________________________________________________________, , Part B, Directions: Identify what is being described below. Use the jumbled words as, your basis., 1. A buffer that is made up of a weak acid plus its conjugate base., CDAI EFBFRU, 2. It can neutralize small amounts of added acid or base, thus maintaining, the pH of the solution relatively stable., RBEFUF, 3. Buffer range is the pH range where a buffer effectively __________, added acids and bases, while maintaining a relatively constant pH., NTSLUEZAERI, 4. A buffer made of weak base and its conjugate acid., BSAE EFBFRU, 5. A buffer is a solution of _____________ and conjugate base used to, resist pH change with added solute., WAKE DCIA, , Activity 2: MIND POWER, Directions: Answer the following questions briefly but substantially., 1. How does a mixture of a weak acid and its conjugate base help buffer a, solution, against pH changes?, ______________________________________________________________, ______________________________________________________________, __________________________________________, , Practice Personal Hygiene protocols at all times., 141
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2. What happens if a strong base is added to a buffer?, ______________________________________________________________, ______________________________________________________________, __________________________________________, , 3. What happens if a strong acid is added to a buffer?, ______________________________________________________________, _____________________________________________________________, 4. Why pH does not change very much when a small amount of a strong acid, or a strong base are added to a buffer?, ______________________________________________________________, _____________________________________________________________, ____________________________________________, 5. How can a buffer maintain the pH of a substance even if small amounts of, acids or bases are added into the substance?, ______________________________________________________________, ______________________________________________________________, ______________________________________________________, 6. What is pH?, ______________________________________________________________, ______________________________________________________________, ______________________________________________, 7. What is buffer?, ______________________________________________________________, ______________________________________________________________, ______________________________________________, 8. Why are buffer solutions important in foods?, ______________________________________________________________, _____________________________________________________________, _______________________________________________, , Practice Personal Hygiene protocols at all times., 142
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ANSWER KEY:, Activity 1: WHO AM I?, Part A, Directions: Choose the best buffer solution., You need a buffer with pH 4.5 and have four acids and their sodium salt, available. Which acid should you use to get the best buffer?, Your acids:, •, •, •, •, , Chloroacetic acid, Benzoic acid, Propanoic acid, Hypochlorous acid, , 1. Which acid should you use to get the best buffer?, ➢ Benzoic acid, ➢ Propanoic acid, 2. What is the pKa of the acid?, ➢ Benzoic acid – 4.20, ➢ Propanoic acid – 4.87, 3. Why do they give the most efficient buffer solution?, ➢ They give the most efficient buffer solution because their pKas is closest, to the pH of 4.5, Part B, Directions: Identify what is being described below. Use the jumbled words as, your basis., 1. A buffer that is made up of a weak acid plus its conjugate base., ACID BUFFER, 2. It can neutralize small amounts of added acid or base, thus maintaining, the pH of the solution relatively stable., BUFFER, 3. Buffer range is the pH range where a buffer effectively __________, added acids and bases, while maintaining a relatively constant pH., NEUTRALIZERS, Practice Personal Hygiene protocols at all times., 146
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4. A buffer made of weak base and its conjugate acid., BASE BUFFER, 5. A buffer is a solution of _____________ and conjugate base used to, resist pH change with added solute., WEAK ACID, Activity 2: MIND POWER, Directions: Answer the following questions briefly but substantially., 1. How does a mixture of a weak acid and its conjugate base help buffer a, solution against pH changes?, ➢ If we mix a weak acid (HA) with its conjugate base (A -), both the acid, and base components remain present in the solution. This is because, they do not undergo any reactions that significantly alter their, concentrations. The acid and conjugate base may react with one, another, HA + A- → A- + HA, but when they do so, they simply trade, places and the concentrations [HA] and [A-] do not change. In addition,, HA and A- only rarely react with water., 2. What happens if a strong base is added to a buffer?, ➢ If a strong base is added to a buffer, the weak acid will give up its H+ in, order to transform the base (OH-) into water (H2O) and the conjugate, base: HA + OH- → A- + H2O. Since the added OH- is consumed by this, reaction, the pH will change only slightly., 3. What happens if a strong acid is added to a buffer?, ➢ If a strong acid is added to a buffer, the weak base will react with the H+, from the strong acid to form the weak acid HA: H+ + a- → HA. The H+, gets absorbed by the A- instead of reacting with water to form H3O+ (H+),, so the pH changes only slightly., 4. Why pH does not change very much when a small amount of a strong acid, or a strong base are added to a buffer?, ➢ The pH does not change very much because [A]/[HA] does not change, very much. This is true as long as the amount of strong acid or strong, base added is small compared to the amount of conjugate acid or, conjugate base in the buffer. If you add much strong acid or base, you, will exceed the buffering capacity., Practice Personal Hygiene protocols at all times., 147
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5. How can a buffer maintain the pH of a substance even if small amounts of, acids or bases are added into the substance?, ➢ If an acid is added to the substance, the conjugate base of the buffer, neutralizes the acid added. If a base is added to the substance, the, conjugate acid of the buffer neutralizes the base added., 6. What is pH?, ➢ pH is a measure of the concentration of H+ [H3O+] ions in a solution. Only, the concentration of H+ and OH- molecules determine the pH., 7. What is buffer?, ➢ Solutions that control the pH of a substance when small amounts of, acids or bases are added into the substance, ➢ Buffer is made up of a weak acid plus its conjugate base (an acid buffer),, or it can also be a weak base plus its conjugate acid (a base buffer), 8. Why are buffer solutions important in foods?, ➢ Buffering solutions in foods olay an important role in maintain specific, pH values for optimum activity of enzymes, protein stability, and, functionality. pH also modify the color and flavor of foods and it is a, critical factor in the preservation of many processed foods. Overall pH, control is a major factor in maintaining the physical, chemical, and, microbiological stability of foods., 9. State two (2) ways in which buffers play a crucial role in the pharmaceutical, industry., ➢ Buffers adjust the pH of aqueous solutions for applications that require, predictable stability and best clinical outcomes. It control the pH of a, solution to minimize drug degradation to improve the efficacy and, delivery, , Practice Personal Hygiene protocols at all times., 148
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Activity 3: GUESS ME!, PART A, Directions: Identify whether the following solutions is a weak acid or weak, base. Write your answer beside each item., 1. CH3COOH, , weak acid, , 2. HF, , weak acid, , 3. NH3, , weak base, , 4. CH3NH2, , weak base, , 5. H2CO3, , weak acid, , 6. HCOOH, , weak acid, , 7. (CH3CH2)2NH, , weak base, , 8. HCN, , weak acid, , 9. NaHCO3, , weak base, , 10. H3PO4, , weak acid, , PART B, Directions: Match the weak acid/base to its conjugate base/acid. Write your, answer before each number., Column A, , Column B, , C 1. HC6H7O7 (citric acid), , A. NH3, , A 2. NH4, , B. CO32-, , B 3. HCO3-, , C. C6H7O7-, , E 4. CH3COOH, , D. HPO42-, , D 5. H2PO4-, , E. CH3COO-, , Prepared by:, JOVELYN Q. BANGAYAN, Aparri School of Arts and Trades, , Practice Personal Hygiene protocols at all times., 149
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, BUFFER SOLUTIONS: HENDERSON HASSELBALCH EQUATION, Background Information for the Learners (BIL), The Henderson-Hasselbalch equation is the equation commonly used in, chemistry and biology to determine the pH of a solution., •, •, , •, , •, •, , This equation shows a relationship between the pH or pOH of the solution,, the pKa or pKb, and the concentration of the chemical species involved., This equation was developed independently by the American biological, chemist L. J. Henderson and the Swedish physiologist K. A. Hasselbalch, to determine the pH of the bicarbonate buffer system in blood., This type of kinetic analysis has enabled us for nearly a century to relate, theoretically the changes of the acidic intensity of dilute solutions to a, quantity of acid or base added or subtracted., This equation can be considered as the backbone of acid-base physiology., This equation is commonly used to determine the amount of acid and, conjugate base required to prepare a buffer of the desired pH., , The Henderson–Hasselbalch equation mathematically connects the, measurable pH of a solution with the pKa (which is equal to -log Ka) of the acid., The equation is also useful for estimating the pH of a buffer solution and finding, the equilibrium pH in an acid-base reaction. The equation can be derived from, the formula of pKa for a weak acid or buffer. The balanced equation for an acid, dissociation is:, HA ⇌ H+ + A−, The acid dissociation constant is:, Ka, After taking the log of the entire equation and rearranging it, the result is:, , Practice Personal Hygiene protocols at all times., 150
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This equation can be rewritten as:, [𝐴−], , -pKa = -pH + log ([𝐻𝐴]), Distributing the negative sign gives the final version of the HendersonHasselbalch equation:, pH = pKa, In an alternate application, the equation can be used to determine the, amount of acid and conjugate base needed to make a buffer of a certain pH., With a given pH and known pKa, the solution of the Henderson-Hasselbalch, equation gives the logarithm of a ratio which can be solved by performing the, antilogarithm of pH/pKa:, , EXAMPLE:, What is the pH of a buffer solution consisting of 0.0350 M NH 3 and, 0.0500 M NH4+ (Ka for NH4+ is 5.6 x 10-10)? The equation for the reaction, is:, NH4+ ⇌ H+ + NH3, Assuming that the change in concentrations is negligible in order for the, system to reach equilibrium, the Henderson-Hasselbalch equation will be:, [NH3], , pH = pKa + log [NH4], pH, pH = 9.095, Calculating Changes in a Buffer Solution, The changed pH of a buffer solution in response to the addition of an, acid or a base can be calculated., If the concentrations of a solution of a weak acid and its conjugate base, are reasonably high, then the solution is resistant to changes in hydrogen ion, concentration. These solutions are known as buffers. It is possible to calculate, how the pH of the solution will change in response to the addition of an acid or, a base to a buffer solution., , Practice Personal Hygiene protocols at all times., 151
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EXAMPLE:, A solution is 0.050 M in acetic acid (HC2H3O2) and 0.050 M NaC2H3O2., Calculate the change in pH when 0.001 mole of hydrochloric acid (HCl), is added to a liter of solution, assuming that the volume increase upon, adding the HCl is negligible. Compare this to the pH if the same amount, of HCl is added to a liter of pure water., Step 1:, HC2H3O2(aq) ⇋ H+(aq) + C2H3O2-(aq), Recall that sodium acetate, NaC2H3O2, dissociates into its component, ions, Na+ and C2H3O2– (the acetate ion) upon dissolution in water. Therefore,, the solution will contain both acetic acid and acetate ions., Before adding HCl, the acetic acid equilibrium constant is:, Ka = [H+] [C2H3O2-] = [x (0.050), [HC2H3O2], (0.050), (assuming that x is small compared to 0.050 M in the equilibrium, concentrations) Therefore:, X, , =, , [H+], , =, , Ka, , =, , 1.76×10−5 M pH = pKa =, 4.75, In this example, ignoring the x in the [C2H3O2–] and [HC2H3O2] terms was, justified because the value is small compared to 0.050., Step 2:, The added protons from HCl combine with the acetate ions to form more, acetic acid:, C2H3O2− + H+ (from HCl) → HC2H3O2, Since all of the H+ will be consumed, the new concentrations will be, [HC2H3O2] = 0.051M and [C2H3O2-] = 0.049M before the new equilibrium is to, be established. Then, we consider the equilibrium concentrations for the, dissociation of acetic acid, as in Step 1:, HC2H3O2(aq), , ⇋, , H+(aq) +, , C2H3O2−(aq) we have,, Ka =, , [ 𝑥 (0.049)], (0.051), , Practice Personal Hygiene protocols at all times., 152
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X, , M, , pH = −log([H+]) = 4.74, In the presence of the acetic acid-acetate buffer system, the pH only, drops from 4.75 to 4.74 upon addition of 0.001 mol of strong acid HCl, a, difference of only 0.01 pH unit., Step 3:, Adding 0.001 M HCl to pure water, the pH is:, pH = −log([H+]) = 3.00, In the absence of HC2H3O2 and C2H3O2–, the same concentration of HCl, would produce a pH of 3.00., Buffers Containing a Base and Conjugate Acid, An alkaline buffer can be made from a mixture of the base and its, conjugate acid, but the formulas for determining pH take a different form., A base is a substance that decreases the hydrogen ion (H+), concentration of a solution. In the more generalized Bronsted-Lowry definition,, the hydroxide ion (OH) is the base because it is the substance that combines, with the proton. Ammonia and some organic nitrogen compounds can combine, with protons in solution and act as Brønsted-Lowry bases. These compounds, are generally weaker bases than the hydroxide ion because they have less, attraction for protons. For example, when ammonia competes with OH – for, protons in an aqueous solution, it is only partially successful. It can combine, with only a portion of the H+ ions, so it will have a measurable equilibrium, constant. Reactions with weak bases result in a relatively low pH compared to, strong bases. Bases range from a pH of greater than 7 (7 is neutral like pure, water) to 14 (though some bases are greater than 14)., An alkaline buffer can be made from a mixture of a base and its, conjugate acid, similar to the way in which weak acids and their conjugate bases, can be used to make a buffer., Calculating the pH of a Base, The pH of bases is usually calculated using the OH– concentration to find, the pOH first. This is done because the H+ concentration is not a part of the, reaction, while the OH– concentration is. The formula for pOH is:, pOH = −log([OH−]), , Practice Personal Hygiene protocols at all times., 153
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By multiplying a conjugate acid (such as NH4+) and a conjugate base, (such as NH3) the following is given:, Ka × Kb = [H3O+][NH3] × [NH4+][OH−], [NH4+], [NH3], Ka × Kb = [H3O+] [OH−], = Kw log(Ka) + log(Kb) =, log(Kw) pKa + pKb = pKw, = 14.00, The pH can be calculated using the formula:, pH = 14 − pOH, Weak bases exist in chemical equilibrium much in the same way as weak, acids do. A base dissociation constant (Kb) indicates the strength of the base., For example, when ammonia is put in water, the following equilibrium is set up:, NH3 + H2O ⇌ NH4+ + OH−, Kb = [NH4+][OH−], [NH3], Bases that have a large Kb will ionize more completely, meaning they are, stronger bases. NaOH (sodium hydroxide) is a stronger base than, (CH3CH2)2NH (diethylamine) which is a stronger base than NH3 (ammonia). As, the bases get weaker, the Kb values get smaller., For a buffer made up of a weak base and its conjugate acid, the equation, for determining the pH of the buffer will be:, pOH = pKb + log, EXAMPLE:, Calculate the pH of a buffer solution consisting of 0.051 M NH3 and 0.037, M NH4+. The Kb for NH3 = 1.8 x 10-5., NH3 + H2O ⇌ NH4+ + OH−, Kb = [NH4+] [OH−], Practice Personal Hygiene protocols at all times., 154
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[NH3], , Assuming the change (x) is negligible to 0.051 M and 0.037 M solutions:, Kb =, , 1.8 x 10-5 = [0.037][x], [0.051], x = [OH–] = 2.48 x 10-5, pOH = 4.61, pH = 14 – 4.61, = 9.39, , Learning Competency:, Calculate the pH of a buffer solution using the Henderson Hasselbalch equation, (STEM_GC11AB-IVf-g-161), , Activity 1: YOU COMPLETE ME!, Directions: Fill in the blanks with words that correspond to the, statements below. Choose the word in the word bank., , Weak acid, , buffer solution, , alkaline buffer Conjugate base, , base, , 1. The Henderson–Hasselbalch equation is useful for estimating the pH of a, _____________ and finding the equilibrium pH in an acid-base reaction., 2. The Henderson–Hasselbalch equation can be used to determine the, amount of acid and _________________needed to make a buffer of a, certain pH., , Practice Personal Hygiene protocols at all times., 155
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3. If the concentrations of a solution of a ____________ and its conjugate, base are reasonably high, then the solution is resistant to changes in, hydrogen ion concentration., 4. A ______________ is a substance that decreases the hydrogen ion (H+), concentration of a solution., 5. An alkaline buffer can be made from a mixture of a __________ and its, conjugate acid, similar to the way in which weak acids and their conjugate, bases can be used to make a buffer., , Activity 2: GIVE ME THE SOLUTION!, Directions: Calculate the following problems below. Show your complete, solution., 1. What is the pH of the solution containing 0.20 M NH3 and 0.15 M NH4Cl?, 2. Calculate the pH of a buffer solution formed by dissolving 0.350 mole, HCH3COO, and 0.550 mole of NaCH3COO in 0.950 L solution. Ka, HCH3COO = 1.8 x 10-5. The buffer in the solution is acid buffer made up of :, HCH3COO-, , -, , NaCH3COO-, , Weak acid, acid, CA, , salt of the weak, CB, , 3. Determine the pH of the following combination of solutions:, A. 8 mL of 0.04 M HCl, 5 mL of 0.8 M ammonia, pKa of ammonium = 9.26, , B. 6 mL of 0.25 ammonium, 5 mL of 0.34 M ammonia, pKa of ammonium = 9.26, , Activity 3: CHANGES IN ME!, Directions: Calculate the changes in pH of the following problems below., 1. A formic acid buffer is prepared with 0.010 M each of formic acid (HCOOH), and, , Sodium formate (NaCOOH). The K a for formic acid is 1.8 x 10-4. What, , Practice Personal Hygiene protocols at all times., 156
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is the pH, , of the solution? What is the pH if 0.0020 M of solid sodium, , hydroxide (NaOH) is, , added to a liter of buffer? What would be the pH of, , the sodium hydroxide, , solution without the buffer? What would the pH have, , been after adding sodium, , hydroxide if the buffer concentrations had been, , 0.10 M instead of 0.010 M?, 2. A. Calculate the pH of a 0.500 L buffer solution composed of 0.700 M formic, acid (HCOOH, Ka = 1.77 x 10¯4) and 0.500 M sodium formate (HCOONa)., B. Calculate the pH after adding 50.0 mL of a 1.00 M NaOH solution., , Activity 4: CALCULATE THE UNKNOWN!, Directions: Read and analyze the following questions and compute for is, unknown in the given problem., 1. Calculate the pH of a buffer made up of 0.544M CH3NH2 and 0.678M, CH3NH3+Cl-. Ka of CH3NH2 is 1.8 x 10-5. The buffer is made up of CH3NH2, (weak base) and CH3NH3+Cl- (salt or conjugate acid), 2. Calculate the pH of buffer made of 0.500M NH3 and 0.456M NH4Cl,, pKb NH3 = 4.74., 3. Calculate the pH of a buffer solution consisting of 0.960M NH3 and 0.608M, NH4+. The Kb for NH3 = 1.8 x 10-5. The reaction is:, NH3 + H2O ⇌ NH4+ + OH−, , Practice Personal Hygiene protocols at all times., 157
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ANSWER KEY, , ACTIVITY 1: YOU COMPLETE ME!, Directions: Fill in the blanks with words that corresponds to the statements, below. Choose the word in the word bank., Weak acid, , buffer solution, , alkaline buffer Conjugate base, base, , 1. The Henderson–Hasselbalch equation is useful for estimating the pH of a, buffer solution and finding the equilibrium pH in an acid-base reaction., 2. The Henderson–Hasselbalch equation can be used to determine the, amount of acid and conjugate base needed to make a buffer of a certain, pH., 3. If the concentrations of a solution of a weak acid and its conjugate base, are reasonably high, then the solution is resistant to changes in hydrogen, ion concentration., 4. An alkaline buffer is a substance that decreases the hydrogen ion (H+), concentration of a solution., 5. An alkaline buffer can be made from a mixture of a base and its conjugate, acid, similar to the way in which weak acids and their conjugate bases can, be used to make a buffer., Activity 2: GIVE ME THE SOLUTION!, Directions: Calculate the following problems below. Show your complete, solution., 1. What is the pH of the solution containing 0.20 M NH3 and 0.15 M NH4Cl?, Given:, , [NH3], , =, , 0.20, , (conjugate, , base), , [NH4Cl]= 0.15 (weak acid), pKa = 9.25 (for NH4+, from the table), Equation: pH = pKa + log [B-], [HB], pH = 9.25 + log 0.20, 0.15, , Practice Personal Hygiene protocols at all times., 160
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pH = 9.37, 2. Calculate the pH of a buffer solution formed by dissolving 0.350 mole, HCH3COO, and 0.550 mole of NaCH3COO in 0.950 L solution. Ka, HCH3COO = 1.8 x 10-5., The buffer in the solution is acid buffer made up, of, HCH3COONaCH3COOWeak acid, CA, , salt of the weak acid, CB, , Given:, Concentration of HCH3COO = 0.350 mole, Concentration of NaCH3COO = 0.550 mole, Volume = 0.950 L, Solution:, CH3COO = 0.350 mole = 0.368 M, 0.950 L, NaCH3COO = 0.550 mole = 0.579 M, 0.950 L, pH = pKa + log [CB], [CA], pKa = -logKa = -log(1.8 x 10-5) = 4.74, pH = 4.74 + log (0.579 M), (0.368 M), pH = 4.74 + 0.1968, pH = 4.94, 3. Determine the pH of the following combination of solutions:, A. 8 mL of 0.04 M HCl, 5 mL of 0.8 M, ammonia, pKa of ammonium = 9.26, Practice Personal Hygiene protocols at all times., 161
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The first reaction that has to happen is the creation of NH+4, ammonium., H+ + NH3 → NH4+, We need to determine how many moles of hydrogen ions that we have, from the HCl, that is equal to the amount of NH+4 that is formed., moles = 0.04 moles x, 1L, x 8mL = 3.2x10−4 moles, L, 1000 mL, We need to determine the total amount of moles of ammonia, and then, how many are remaining., Total moles = 0.8 moles x, 1L, x 5mL = 0.004 moles, L, 1000 mL, Remaining moles = 0.004 moles − 3.2x10−4 moles = 0.00368 moles, Now we can use the Henderson-Hasselbalch equation to solve for the pH, pKa = pH + log base = 9.26 + log 3.2x10−4 =, 8.2, acid, 0.00368, B. 6 mL of 0.25 M ammonium, 5 mL of 0.34 M ammonia, pKa of ammonium = 9.26, The relevant chemical reaction here is:, NH4+ + H2O, , NH3 + H3O+, , First we need to determine how many moles we have of NH4+ and NH3., Moles ammonium = 0.25 mole x, 1L, x 6mL = 0.0015 moles, L, 1000 mL, , Moles ammonia = 0.34 moles x, L, , 1L, x 5mL = 0.0017 moles, 1000 mL, , Now we can use the Henderson-Hasselbalch equation to solve for the pH, pH = pKa + log base = 9.26 + log 0.00170 = 9.31, acid, 0.0015, , Practice Personal Hygiene protocols at all times., 162
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Activity 3: CHANGES IN ME!, Directions: Calculate the changes in pH of the following problems below., 1. A formic acid buffer is prepared with 0.010 M each of formic acid (HCOOH), Sodium formate (NaCOOH). The K a for formic acid is 1.8 x 10-4. What, , and, , is the pH, , of the solution? What is the pH if 0.0020 M of solid sodium, , hydroxide (NaOH) is, , added to a liter of buffer? What would be the pH of the, , sodium hydroxide solution without the buffer? What would the pH have been, after adding sodium hydroxide if the buffer concentrations had been 0.10 M, instead of 0.010 M?, Step 1:, Solving for the buffer pH:, HCOOH ⇋ H+ + HCOO−, Assuming x is negligible, the Ka expression looks like:, Ka = x(0.010), (0.010), 1.8 x 10-4 = x =, [H+] pH = -log, [H+] = 3.74, Buffer:, , pH, , =, , 3.74 STEP 2:, Solving for the buffer pH after 0.0020 M NaOH has been added:, OH− + HCOOH → H2O + HCOO−, The concentration of HCOOH would change from 0.010 M to 0.0080 M and, the concentration of HCOO– would change from 0.010 M to 0.0120 M., Ka = x(0.0120), (0.0080), , Practice Personal Hygiene protocols at all times., 163
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After adding NaOH, solving for x = [H+] and then calculating the pH = 3.92., The pH went up from 3.74 to 3.92 upon addition of 0.002 M of NaOH., Step 3:, Solving for the pH of a 0.0020 M solution of NaOH:, pOH = -log, (0.0020) pOH, = 2.70 pH =, 14 – pOH pH, = 11.30, Without, , buffer:, , pH, , =, , 11.30, Step 4:, Solving for the pH of the buffer solution if 0.1000 M solutions of the weak, acid and its conjugate base had been used and the same amount of NaOH, had been added:, The concentration of HCOOH would change from 0.1000 M to 0.0980 M, and the concentration of HCOO– would change from 0.1000 M to 0.1020, M., Ka = x(0.1020), (0.0980), pH if 0.1000 M concentrations had been used = 3.77, This shows the dramatic effect of the formic acid-formate buffer in, keeping the solution acidic in spite of the added base. It also shows the, importance of using high buffer component concentrations so that the, buffering capacity of the solution is not exceeded., 2. A. Calculate the pH of a 0.500 L buffer solution composed of 0.700 M formic, acid (HCOOH, Ka = 1.77 x 10¯4) and 0.500 M sodium formate (HCOONa)., We can use the given molarities in the Henderson-Hasselbalch Equation:, pH = pKa + log [base /, acid] pH = 3.752 + log, Practice Personal Hygiene protocols at all times., 164
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[0.5 / 0.7] pH = 3.752 +, (−0.146) pH = 3.606, B. Calculate the pH after adding 50.0 mL of a 1.00 M NaOH solution., Step 1:, We need to determine the moles of formic acid and sodium formate, after the NaOH was added. We first calculate the amounts before the, addition of the, NaOH:, HCOOH ---> (0.700mol/L)(0.500L) = 0.350 mol, HCOONa ---> (0.500 mol/L) (0.500 L) = 0.250 mol, Step 2:, Now, determine the moles of NaOH:, NaOH ---> (1.00 mol/L) (0.0500 L) = 0.0500 mol, , Step 3:, NaOH reacts in a 1:1 molar ratio with HCOOH:, HCOOH ---> 0.350 mol − 0.0500 mol = 0.300 mol, HCOONa ---> 0.250 mol + 0.0500 mol = 0.300 mol, Step 4:, Calculate the new pH:, pH = 3.752 + log [0.300 /, 0.300] pH = 3.752 + log 1, pH = 3.752, , Practice Personal Hygiene protocols at all times., 165
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Activity 4: CALCULATE THE UNKNOWN!, Directions: Read and analyze the following questions and compute for is, unknown in the given problem., 1. Calculate the pH of a buffer made up of 0.544M CH3NH2 and 0.678M, CH3NH3+Cl-. Ka of CH3NH2 is 1.8 x 10-5. The buffer is made up of CH3NH2 (weak, base) and CH3NH3+Cl- (salt or conjugate acid), Given: Concentration of CH3NH2 = [CB] = 0.544M, Concentration of CH3NH3+Cl- = [CA] = 0.678M, Solution: pOH = pKb + log [CA], [CB], = 4.74 + log 0.678M, 0.544M, = 4.74 + 0.0956, = 4.8356, pOH = 14 – 4.8356, = 9.1644, 2. Calculate the pH of buffer made of 0.500M NH3 and 0.456M NH4Cl, pKb, NH3 = 4.74., Given: Concentration of NH3 = [CB] = 0.500M, Concentration of NH4Cl- = [CA] = 0.456M, Solution: pOH = pKb + log [CA], [CB], = 4.74 + log 0.456M, 0.500M, = 4.74 + (- 0.040), = 4.34, pOH = 14 – 4.34, = 9.66, , Practice Personal Hygiene protocols at all times., 166
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3. Calculate the pH of a buffer solution consisting of 0.960M NH 3 and 0.608M, NH4+. The Kb for NH3 = 1.8 x 10-5. The reaction is:, NH3 + H2O ⇌ NH4+ + OH−, Kb = [NH4+] [OH−], [NH3], Assuming the change (x) is negligible to 0.051 M and 0.037 M solutions:, Kb =, , 1.8 x 10-5 = [0.608][x], [0.960], x = [OH–] = 2.84 x 10-5, pOH = 4.55, pH = 14 – 4.55 = 9.45, , Practice Personal Hygiene protocols at all times., 167
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APPENDIX: Table of acids with Ka and pKa values., , Prepared by:, , JOVELYN Q. BANGAYAN, Aparri School of Arts and Trades, , Practice Personal Hygiene protocols at all times., 168
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, OXIDATION-REDUCTION REACTIONS, , Background Information for the Learners, Variety of oxidation-reduction reactions affect us amazingly every day., Our society runs on batteries – in our calculators, laptop computers, cars, toys,, radios, mobile phones, and more. There are some other encounters that seems, to be a puzzle that we observe but don’t have dare to care., Some questions like, why does an apple turn brown whenever sliced?, Why does an eggplant change in color when sliced? Why do we paint iron, railings, roofs, window grills, and others?, Apples and eggplants immediately turns brown after slicing due to, exposure to air. We paint iron railings, galvanized iron of roofs, window grills, and other materials made of iron to combat corrosion. We electroplate jewelry, and computer chips with very thin coatings of gold or silver. We bleach our, clothes and develop our photographs in solution using chemical reactions that, involve electron transfer. Medical experts test for glucose in urine or alcohol in, the breath with reactions that show vivid color changes. Plants turn energy into, chemical compounds through series of reactions called photosynthesis. These, reactions all involve the transfer of electrons between substances in a chemical, process called oxidation-reduction or redox reaction. Redox reaction is also, responsible for providing energy for our body to move, think and stay alive., Oxidation Number, The oxidation number of an atom (sometimes called its oxidation state), represents the number of electrons gained, lost, or unequally shared by an, atom. Oxidation numbers can be zero, positive, or negative. An oxidation, number of zero means the atom has the same number of electrons assigned to, it as there are in the free neutral atom. A positive oxidation number means the, atom has fewer electrons assigned to it than in the neutral atom, and a negative, oxidation number means the atom has more electrons assigned to it that in the, neutral atom., The oxidation number of an atom that lost or gained electrons to form an, ion is the same as the positive or negative charge of the ion. The table below, shows the oxidation numbers of common ions., , Practice Personal Hygiene protocols at all times., 169
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Ion, , Oxidation, Number, +, H, +1, +, Na, +1, K+, +1, +, Li, +1, Ag+, +1, 2+, Cu, +2, Ca2+, +2, 2+, Ba, +2, 2+, Fe, +2, Mg2+, +2, 2+, Zn, +2, Al3+, +3, 3+, Fe, +3, Cl-1, Br, -1, F, -1, I-1, 2S, -2, O2-2, Table 1: Common Ions and their, respective oxidation numbers., , In the ionic compound NaCl, the, oxidation numbers are clearly +1 for Na+ ion, and -1 for Cl- ion. The Na+ ion has one less, electron than the neutral Na atom, and the, Cl- ion has one more electron than the, neutral Cl atom., Assigning correct oxidation numbers, to elements is essential, oxidation-reduction, , in balancing, , equations., , Oxidation, , numbers have variety of uses in chemistry –, from writing formulas to predicting properties, of compounds and assisting in the balancing, of oxidation-reduction reactions in which, electrons are transferred., , To, , correctly, , assign, , oxidation, , numbers, the rules are summarized below., , Rules in assigning Oxidation Numbers, 1. All elements in their free state (uncombined with other elements) have, an oxidation number of zero. (e.g. Na, Cu, Mg, H2, O2, Cl2, N2), 2. H is +1, except in metal hydrides, where it is -1 (e.g. NaH, CaH2)., 3. O is -2, except in peroxides, where it is -1, and in OF2, where it is +2., 4. The metallic element in an ionic compound has a positive oxidation, number., 5. In covalent compounds, the negative oxidation number is assigned to, the most electronegative atom., 6. The algebraic sum of the oxidation numbers of the elements in a, compound is zero., 7. The algebraic sum of the oxidation numbers of the elements in a, polyatomic ion is equal to the charge of the ion., , Practice Personal Hygiene protocols at all times., 170
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Steps in finding the Oxidation Number of an element within a compound., 1. Write the oxidation number of each known atom below the atom in the, formula., 2. Multiply each oxidation number by the number of atoms of that element, in the compound., 3. Write the expression indicating the sum of all oxidation numbers in the, compound. Remember: The sum of the oxidation numbers in a, compound is zero and equal to the charge for polyatomic ion., , OXIDATION NUMBER IN COMPOUNDS, Example, Determine the oxidation number of Sulfur in Sulfuric Acid:, H2SO4, Step 1:, , +1, , -2, , Step 2: 2(+1) = +2, , 4(-2) = -8, , Step 3: +2 + S + (-8) = 0, Step 4: S = +6, The Oxidation number of Sulfur in Sulfuric acid is +6, OXIDATION NUMBER IN POLYATOMIC IONS, Example, Determine the oxidation number for Carbon in the Oxalate ion:, C2O42Step 1:, , -2, , Step 2:, , 4(-2) = -8, , Step 3:, , 2C + (-8) = -2, , Step 4:, , 2C = +6, C = +3, , The oxidation number of Carbon in Oxalate ion is +3., OXIDATION – REDUCTION, Oxidation – reduction, also known as redox, is a chemical process in, which oxidation number of an element is changed. The process may involve, , the complete transfer of electrons to form ionic bonds or only a partial transfer, shifts of electrons to form covalent bonds., Practice Personal Hygiene protocols at all times., 171
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Oxidation occurs whenever the oxidation number of an element, increases as a result of losing electrons. Conversely, Reduction occurs, whenever the oxidation numbers of an element decreases as a result of, gaining electrons. For example, a change in oxidation number from +2 to +3, or -1 to 0 is oxidation; a change in oxidation number from +5 to +2 or from -2 to, -4 is reduction. Oxidation and reduction occur simultaneously in a chemical, reaction., , In the given chemical reaction wherein oxidation numbers are written, above the elements, Magnesium changes oxidation number from 0 to + 2, while, H changes oxidation number from +1 to 0. Magnesium loss electrons causing, an increase in its oxidation number, magnesium has been oxidized. On the, other hand, Hydrogen gains electrons causing a decrease in its oxidation, number, hydrogen has been reduced. The substance being oxidized is called, , reducing agent while the substance being reduced is oxidizing agent., To summarize:, LEO – Losing Electron is Oxidation, Increase in Oxidation Number, GER – Gaining Electron is Reduction, Decrease in Oxidation Number, Reducing agent – the substance being oxidized. Source: Tro, Nivaldo (2011), Introductory Chemistry (4th ed.)., Oxidizing agent – the substance being reduced., , Learning Competency, The learner will be able to define Oxidation and Reduction Reactions, (STEM_GC11AB-IVf-g-169)., , Practice Personal Hygiene protocols at all times., 172
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Activity 1: FIND THE CORRECT NUMBER, Directions: Assign the correct oxidation number of the individual atom in the, compound or ion below., 1., 2., 3., 4., 5., , S in S8, Cl in CaCl2, I in IO3C in HCO3S in Fe2(SO4)3, , 6. S in SO427. S in Na2SO4, 8. As in K3AsO4, 9. Cr in Cr2O7210. N in NH4+, , Activity 2: EXAMINE ME, Directions: Examine the given chemical reactions and answer the given, questions in each item., 1. Label the following as oxidation or reduction., _______a. Conversion of Br- to Br2, _______b. Conversion of Fe2+ to Fe3+, _______c. Change in oxidation number in a negative direction., 2. In the balanced equation: I2 + 5Cl2 + 6H2O → 2HIO3 + 10HCl, _______a. What element was reduced?, _______b. What element was oxidized?, _______c. What substance is the oxidizing agent?, _______d. What substance is the reducing agent?, 3. The equation below shows the corrosion of Iron when exposed to, oxygen:, 4Fe + 3O2 → 2Fe2O3, ______a. What element was reduced?, ______b. What element was oxidized?, ______c. What substance is the oxidizing agent?, ______d. What substance is the reducing agent?, 4. Burning of methane is common among households that uses gas stove., The reaction below shows the combustion of methane:, CH4 + 2O2 →, CO2 + 2H2O, ______a. What element was reduced?, ______b. What element was oxidized?, ______c. What substance is the oxidizing agent?, ______d. What substance is the reducing agent?, , Practice Personal Hygiene protocols at all times., 173
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Activity 3: LEOGER, Directions: Read carefully the given statements. Determine whether oxidation, or reduction is described. Write your answer on the space provided., ________1. An element changes oxidation number +5 to +2., ________2. Oxygen attained oxidation number of 0 from -2., ________3. Hydrogen have an initial oxidation of +1 and later had 0 after the, reaction., ________4. An element losses 2 electrons after a reaction., ________5. Chlorine reaches -1 charge after gaining an electron., ________6. An element undergoes change of oxidation number from +3 to 0., ________7. Copper changes from +2 to 0., ________8. From Cl2 having 0 as oxidation number, Cl later had -1 after, combining with Na., ________9. Lithium from its free state to LiCl., ________10. Carbon in CH4 before combustion and Carbon in CO2 after, combustion., , Reflection:, 1.I learned that ________________________________________________, _____________________________________________________________, _____________________________________________________________, , 2.I enjoyed most on ____________________________________________, ______________________________________________________________, ______________________________________________________________, , 3.I want to learn more on __________________________________________, ______________________________________________________________, _____________________________________________________________, , Practice Personal Hygiene protocols at all times., 174
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ANSWER KEY, Activity 1: FIND THE CORRECT NUMBER, 1. S = 0, 2. Cl = -1, 3. I = +5, 4. C = +4, 5. S = +6, 6. S = +6, 7. S = +6, 8. As = +5, 9. Cr = +6, 10. N = -3, Activity 2: EXAMINE ME, 1., 2., 3., 4., , (a) oxidation, (a) Cl, (a) O, (a) O, , (b) oxidation, (b) I, (b) Fe, (b) C, , (c) reduction, (c) Cl2, (c) O2, (c) O2, , (d) I2, (d) Fe, (d) CH4, , Activity 3: LEOGER, 1. Reduction, 2. Reduction, 3. Reduction, 4. Oxidation, 5. Reduction, 6. Reduction, 7. Reduction, 8. Reduction, 9. Oxidation, 10. Oxidation, , Prepared by:, , CLETO D. ABBIDO, Don Severino Pagalilauan National High School, , Practice Personal Hygiene protocols at all times., 176
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, BALANCING REDOX REACTIONS, Background Information for the Learners, In previous lessons, we learned how to balance chemical equations, using inspection. Many redox reactions can be balanced readily through, inspection, or by the trial and error., In this lesson we will be dealing with more systematic way of balancing, complex reactions such as oxidation-reduction reactions., How to balance Oxidation – Reduction reactions?, One systematic method for balancing oxidation – reduction equations is, based on the transfer of electrons between the oxidizing and reducing agents., Consider the equation:, 0, , 0, , +, , Na + Cl2 → NaCl, , -, , Above each element is the oxidation number of each in the reaction. In, this reaction, sodium metal loses one electron per atom when it changes to, sodium ion. At the same time, chlorine gains one electron per atom. Because, chlorine is a diatomic, two electrons per molecule are needed to form a chloride, ion from each atom. These electrons are furnished by two sodium atoms., Stepwise, the reaction may be written as two half-reactions, the oxidation halfreaction and the reduction half reaction:, 2 Na0, , →, , Cl20, , +, , 2e-, , →, , Cl20, , +, , 2 Na0, , →, , 2Na+, , +, , 2e-, , 2Cl-, , oxidation half-reaction, reduction half-reaction, , 2Na+Cl-, , When the two half-reactions, each containing the same number of, electrons, are added together algebraically, the electrons cancel out. In this, reaction there are no excess electrons; the two electrons lost by the two sodium, atoms were utilized by chlorine. In all redox reactions the loss of electrons by, the reducing agent must equal the gain of electrons by the oxidizing agent. Here, sodium is oxidized and chlorine is reduced. Chlorine is the oxidizing agent;, odium is the reducing agent., , Practice Personal Hygiene protocols at all times., 177
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In the following examples, we use the change in oxidation number, method, a system for balancing more complicated redox reactions., Change in oxidation method, Rule, , Example 1, , 1. Assign oxidation states to all, 0, +2, +3, 0, atoms, and, identify, the, Al(s) + Cu2+(aq) → Al3+(aq) + Cu(s), substances being oxidized and, Oxidation, reduced., Reduction, 2. Separate the overall reaction Oxidation: Al(s) → Al3+(aq), into two half-reactions, one for Reduction: Cu2+(aq) → Cu(s), oxidation and one for reduction., 3. Balance each half reaction,, with respect to mass in the, following order:, All atoms other than O and H are, a. Balance all elements other balanced, so proceed to other, than H and O, elements., No O, proceed to the next step., No H, proceed to the next step, b. Balance O by adding H2O, , c. Balance H by adding H+, 4. Balance each half-reaction, with respect to charge by, adding electrons to the right, side of the oxidation halfreaction and the left side of the, reduction half-reaction. (The, sum of the charges on both, sides of each equation should, then be equal.), 5. Make the number of electrons, in both half-reactions equal by, multiplying one or both halfreactions by a small whole, number., 6. Add the two half-reactions, together, canceling electrons, and other species as necessary., 7. Verify that the reaction is, balanced with respect to both, mass and charge., , Al(s) → Al3+(aq) + 3e2e- + Cu2+(aq) → Cu(s), , 2 (Al(s) → Al3+(aq) + 3e-), 3 (2e- + Cu2+(aq) → Cu(s)), , 2Al(s) → 2Al3+(aq) + 6e6e- + 3Cu2+(aq) → 3Cu(s), 2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s), Reactants, 2 Al, 3 Cu, +6 charge, , Products, 2 Al, 3 Cu, +6 charge, , Practice Personal Hygiene protocols at all times., 178
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References:, Hein, M., Best, L. R., et.al,(2005), Introduction to general, organic and, biochemistry(8th ed.).Hoboken, NJ: John Wiley and Sons, Tro, Nivaldo (2011), Introductory Chemistry (4th ed.).USA:Pearson Prentice, Hall, Chang, Raymond,.(2010) General Chemistry(10th ed) NY: McGraw Hill, Modern Chemistry Study Guide/Copyright © by Holt, Rinehart and Winston, Zumdahl,.Zumdahl,. Chemistry (7th ed) © USA: Houghton Mifflin Company, , Practice Personal Hygiene protocols at all times., 184
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GENERAL CHEMISTRY 2, Name ________________________, , Grade Level: ____________, , Date: _________________________, , Score: _________________, , LEARNING ACTIVITY SHEET, ELECTROCHEMICAL CELL, Background Information for the Learners (BIL), When life gives you lemons, generate electricity! How? Understanding the, basic principle of transfer of electrons and electro-chemical reactions will show, us how electricity is produced in lemons. The juice of lemon is acidic in nature, and works as a powerful electrolyte. The lemon itself serves as a reservoir for, transfer of electrons to and from the electrodes. When the two electrodes,, copper and zinc, are suspended in the acidic lemon juice, the atomic structure, of the atoms of both the electrodes starts breaking, resulting in production of, individual electrons., , https://www.brighthubengineering.com/commercial-electricalapplications/64675-why-does-lemon-conducts-electricity/, , Refer on the links below to watch video lessons., https://betterlesson.com/lesson/598318/when-life-gives-you-lemons-make-a-battery, , https://www.youtube.com/watch?v=GhbuhT1GDpI, Ever wondered what are the contents of battery? The branch of chemistry, which studies the relationship between chemical change and electricity is called, electrochemistry., All chemical reactions are fundamentally electrical in nature since electrons, are involved (in various ways) in all types of chemical bonding., Electrochemistry, however, is primarily the study of oxidation-reduction, phenomena. It deals with the study of production of electricity from energy, , Practice Personal Hygiene protocols at all times., 186
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released during spontaneous chemical reactions and the use of electrical, energy to bring about non-spontaneous chemical transformations., The relationships between chemical change and electrical energy have, theoretical as well as practical importance. Chemical reactions can be used to, produce electrical energy (in cells that are either called voltaic or galvanic cells)., Electrical energy can be used to bring about chemical transformations (in, electrolytic cells). In addition, the study of electrochemical processes leads to, an understanding, as well as to the systematization, of oxidation-reduction, phenomena that take place outside cells., , Electrochemical Cell, An electrochemical cell typically consists of, ■ Two electronic conductors (also called electrodes), ■ An ionic conductor (called an electrolyte), ■ the electron conductor used to link the electrodes is often a metal wire, such, as, copper wiring., , https://www.google.com/search?q=electrochemical+cell+images&tbm=isch&source=iu&ictx=1&fir=tes5h0gSLZPlM%252CdRnTq9ASXzkomM%252C_&vet=1&usg=AI4_kRgb15xsswliMbguEPEEDdpyGB5mA&sa=X&ved=2ahUKEwi_zrX2tqbqAhUZPnAKHZbMDtoQ9QEwBnoECAoQMA&biw=1508&bih=687#imgrc=lYAUY4ZowQT, HjM, , Electolytic cell: electrochemical cell in, which a non-spontaneous reactions is, driven by an external source of current., , Galvanic cell: electrochemical cell in, which electricity is produced as a result, of a spontaneous reaction (e.g.,, batteries, fuel cells, electric fish), , Practice Personal Hygiene protocols at all times., 187
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https://www.quora.com/What-is-the-difference-between-electrolytic-and-galvanic-cell, , Galvanic (Voltaic) Cells, The operation of a galvanic (or voltaic cell is opposite to that of an, electrolytic cell. In galvanic cell,electrical energy is produced by a chemical, redox reaction, instead of a chemical reaction being produced by electricity., The classic example of a redox reaction for a galvanic cell is the reaction, between aqueous solutions of zinc (Zn) and copper (Cu)., Zn(s) + Cu++ (aq), , Zn++ (aq) + Cu(s), , In this cell, the zinc is oxidized, and the copper is reduced. Initially, this, produces a flow of electrons across a wire connected to two separate electrode, solutions, but the zinc solution becomes positively charged from losing, electrons and the copper solution becomes negatively charged from gaining, electrons giving them, that flow stops. No more negatively charged electrons, want to flow toward the negatively charged copper solution., , Learning Competency:, Identify the reaction occurring, (STEM_GC11ABIVf-g-172), , in, , the, , different, , parts, , of, , the, , cell., , Practice Personal Hygiene protocols at all times., 188
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Salt Bridge, , Figure 1. Salt Bridge, https://www.google.com/search?source=univ&tbm=isch&q=electrochemical+cells+illustration&sa=X&ved=2ahUKEwj, ErvGP-5vqAhWXF4gKHWjvAaoQsAR6BAgKEAE&biw=1508&bih=647#imgrc=Ca8tFhFcfihKzM, , The salt bridge is usually a U shaped tube filled with a concentrated salt, solution. The solution in this tube provides a way for ions to travel between the, two electrode solutions so that they can remain electrically neutral in charge., This enables the continuous flow of electrons. Use the link to better understand, the concept of salt bridge. https://youtu.be/C26pH8kC_Wk, , Electrolytic Cells, Electrolysis is used to drive an oxidation-reduction reaction in a direction in, which it does not occur spontaneously. The concept of reversing the direction, of the spontaneous reaction in a galvanic cell through the input of electricity is, at the heart of the idea of electrolysis. Electrolytic cells, like galvanic cells, are, composed of two half-cells--one is a reduction half-cell, the other is an oxidation, half-cell. Though the direction of electron flow in electrolytic cells may be, reversed from the direction of spontaneous electron flow in galvanic cells, the, definition of both cathode and anode remain the same--reduction takes place, at the cathode and oxidation occurs at the anode., , Practice Personal Hygiene protocols at all times., 190
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https://www.google.com/search?q=electrochemical+cell+diagram&tbm=isch&ved=2ahUKEwjigsqS-5vqAhUOBJQKHe-WD7cQ2cCegQIABAA&oq=electrochemical+cell&gs_lcp=CgNpbWcQARgBMgIIADICCAAyAggAMgIIADICCAAyAggAMgIIADICCAAyAggAMgIIAFCyawBWOGKrQFg3aKtAWgAcAB4AIABlAKIAZcTkgEFMC40LjeYAQCgAQGqAQtnd3Mtd2l6LWltZw&sclient=img&ei=phH0XuKXHY6I0ATvrb64, Cw&bih=647&biw=1508#imgrc=d1cfNzM1GclDnM, , When comparing a galvanic cell to its electrolytic counterpart, occurs on, the right-hand half cell. Because the directions of both half-reactions have been, reversed, the sign, but not the magnitude of the potential has been reversed., Note that copper is spontaneously plated onto copper cathode in the galvanic, cell whereas it requires a voltage greater than 0.78V from the battery to plate, iron on its cathode in the electrolytic cell., , Anode and Cathode, The electrode at which electrons are lost is known as the anode, and the, one at which electrons are gained is the cathode., ■ At the anode:, *Oxidation, or loss of electrons, is the process which occurs at the anode., *Negative ions from the electrolyte are discharged if they are halide ions such, as I- , Br- , and Cl-., , Practice Personal Hygiene protocols at all times., 191
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*SO42- and NO3- ions are not discharged. Instead, OH- ions from water are, discharged and O2 gas is produced., ■ At the cathode:, *Reduction, or gain of electrons, occurs at the cathode., *Positive ions from the electrolyte are discharged if they are H+ ions or ions, less reactive metals such as Cu2+ , Pb2+ or Ag+., , *Positive ions of reactive metals such as Na+ , K+ and Ca2+ are not discharged, in the presence of water. Instead, H+ ions from water are discharged and H2, gas is produced., , Oxidation-Reduction Reaction, At each electrode, an electrochemical reaction occurs. This reaction is, called a half cell reaction (since there are two electrodes in a typical cell at, which reactions, occur) The overall chemical reaction of the cell is given by combining the two, individual half cell reactions, , There are two fundamental types of half cell reactions:, ■ Oxidation reactions, ■ Reduction reactions, ■ A reaction is classified as oxidation or reduction depending on the direction, of, electron transfer, , Oxidation, ■ Involves the loss of an electron, ■ Involves the transfer of electrons from the species to the electrode, The term oxidation was originally used to describe reactions in which an, element combines with oxygen., R = O + ne, , Practice Personal Hygiene protocols at all times., 192
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The reaction between magnesium metal and oxygen to form, magnesium oxide involves the oxidation of magnesium., , Reduction, ■ Involves the gain of an electron, ■ Involves the transfer of electrons from the electrode to the species, O + ne = R, The term reduction comes from the Latin stem meaning "to lead back." Anything, that that leads back to magnesium metal therefore involves reduction., The reaction between magnesium oxide and carbon at 2000C, to form magnesium metal and carbon monoxide is an example of the reduction, of magnesium oxide to magnesium metal., , After electrons were discovered, chemists became convinced that oxidationreduction reactions involved the transfer of electrons from one atom to another., From this perspective, the reaction between magnesium and oxygen is written, as follows., 2 [Mg2+][O2-], , 2 Mg + O2, , In the course of this reaction, each magnesium atom loses two electrons to form, an Mg2+ ion., Mg, , Mg2+ + 2 e-, , And, each O2 molecule gains four electrons to form a pair of O2- ions., O2 + 4 e -, , 2 O2-, , Because electrons are neither created nor destroyed in a chemical reaction,, oxidation and reduction are linked. It is impossible to have one without the, other, as shown in the figure below., , Practice Personal Hygiene protocols at all times., 193
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Example 3: Reaction Between Zinc and Copper, This is a type of metal displacement reaction in which copper metal is obtained, when zinc displaces the Cu2+ion in the copper sulfate solution as shown in the, reaction below., Zn (s) + CuSO4 (aq) → ZnSO4 (aq) + Cu (s), The oxidation half-reaction can be written as: Zn → Zn2+ + 2e–, The reduction half-reaction can be written as: Cu2+ + 2e– → Cu, , Example 3: Reaction between Iron and Hydrogen Peroxide, Fe2+ is oxidized to Fe3+ by hydrogen peroxide when an acid is present. This, reaction is provided below., 2Fe2+ + H2O2 + 2H+ → 2Fe3+ + 2H2O, Oxidation half-reaction: Fe2+ → Fe3+ + e–, Reduction half-reaction: H2O2 + 2e– → 2 OH–, Thus, the hydroxide ion formed from the reduction of hydrogen peroxide, combines with the proton donated by the acidic medium to form water., , Practice Personal Hygiene protocols at all times., 194
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Activity 4, 1. TRUE, 2. FALSE, 3. FALSE, 4. TRUE, 5. TRUE, , Activity 5, A., , 3S(s) + 2KClO3(s), 0, , B., , (+5), , 3SO2(g) + 2KCl (s), , (+4), , (-1), , S: Oxidation number of S increases from 0 to +4, , Cl: Oxidation of element Cl element atom in KClO3 decreases from +5 to -1, C. Oxidant - S; Reductant - KClO, , Prepared by:, , IVON A. ADDATU, TUAO VOCATIONAL AND TECHNICAL SCHOOL PATA ANNEX, , Practice Personal Hygiene protocols at all times., 200
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, REDUCTION POTENTIAL, OXIDATION POTENTIAL AND CELL, POTENTIAL, Background Information for the Learners (BIL), Reduction potential also called as redox potential, oxidation/reduction, potential, or Eh, that measures the tendency of a chemical species to obtain, electrons and thereby be reduced. Reduction potential is measured in volts (V), or millivolts (mV). Each kind or species has its own intrinsic reduction potential., The more positive the potential is, the greater the species’ affinity for electrons,, or the more the species tends to be reduced., , Source: https://courses.lumenlearning.com/introchem/chapter/standard-reduction-potentials/, , Oxidation-reduction in a galvanic cell, In this galvanic cell, zinc reduces copper cations. The reaction yields, zinc cations and neutral copper metal., The standard reduction potential (E0) is measured under standard, conditions:, •, , 25 °C, , •, , 1 M concentration for each ion participating in the reaction, , •, , Partial pressure of 1 atm for each gas that is part of the reaction, , •, , Metals in their pure states, , Practice Personal Hygiene protocols at all times., 201
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Standard Reduction Potential, The standard reduction potential is well-defined relative to a standard, hydrogen electrode (SHE) reference electrode, which results a potential of 0.00, volts. The given values below in parentheses are standard reduction potentials, for half-reactions measured at 25 °C, 1 atmosphere, and with a pH of 7 in, aqueous solution., •, , CH3COOH + 2H+ + 2e– → CH3CHO + H2O (-0.58), , •, , 2H+ + 2 e– → H2 (0.0), , •, , O2 + 2H+ + 2e– → H2O2 (+0.7), , •, , O2 + 4H+ + 4e– → 2H2O (+1.64), Since the reduction potential measures the central tendency for a, , species to undergo reduction, comparing standard reduction potential for two, processes can be useful for determining how a reaction will proceed., It is written in the form of a reduction half reaction. An example can be, seen below where "A" is a generic element and C is the charge. Standard, Reduction Potential, AC-+ + Ce- → A, For example, copper's Standard Reduction Potential of EO = +0.340V) is, for this reaction:, Cu2+ + 2e− → Cu, Standard reduction or oxidation potentials can be determined using a, SHE (standard hydrogen electrode)., Figure (1) - Standard Hydrogen Electrode, , https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electro, chemistry/Voltaic_Cells/The_Cell_Potential, , Practice Personal Hygiene protocols at all times., 202
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Standard Oxidation potentials, The, , energy, , change,, , measured, , in, , volts,, , need, , to, , add, , or, , remove electrons to or from an element or compound. The reference reaction, is the deletion or removal of electrons from hydrogen in a standard hydrogen, half-cell (i.e. H2(gas) at 1 atm pressure delivered to a 1.0 M solution of, H+ ions at 25°C, into which a platinum electrode has been inserted): H 2 →, 2H+ + 2e−, This energy change is given the value of zero. The oxidation, potential of other species is determined relatively by measuring the potential, difference between a half-cell containing an aqueous solution of the oxidized, and reduced forms of the test substance, and the standard hydrogen half-cell., , Cell Potential, The cell potential, Ecell, is the measure of the potential difference, between two half cells in an electrochemical cell. The potential difference is, caused by the ability of electrons to flow from one half cell to the other. Electrons, are able to move between electrodes because the chemical reaction is a redox, reaction. A redox reaction occurs when a certain substance is oxidized, while, another is reduced. During oxidation, the substance loses one or more, electrons, and thus becomes positively charged. Conversely, during reduction,, the substance gains electrons and becomes negatively charged. This relates, to the measurement of the cell potential because the difference between the, potential for the reducing agent to become oxidized and the oxidizing agent to, become reduced will determine the cell potential. The cell potential (Ecell) is, measured in voltage (V), which allows us to give a certain value to the cell, potential., , How are Standard Reduction Potentials Applied, Standard reduction potentials are used to determine the standard cell, potential. The standard reduction cell potential and the standard oxidation cell, potential can be combined to determine the overall Cell Potentials of a galvanic, cell. The equations that relate these three potentials are shown below:, Eocell = Eoreduction of reaction at cathode + Eooxidation of reaction at anode, or alternatively, Practice Personal Hygiene protocols at all times., 203
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Predicting Spontaneity, 1. A spontaneous reaction only occurs when the oxidizing agent is above the, reducing agent in the Table of Reduction Potentials., 2. For any functioning galvanic cell, the measured cell potential has a positive, value., Eocell = positive, the reaction will occur and is spontaneous, Eocell = negative, the reaction will not occur and is not spontaneous, , The Activity Series, When solving for the standard cell potential, the species oxidized, and, the species reduced must be identified. This can be done using an activity, series. Below is a table of standard reduction potentials., , https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Supplemental_Modules_(Analytical_Chemistry)/Electro, chemistry/Voltaic_Cells/The_Cell_Potential, , Learning Competency:, •, , Define reduction potential, oxidation potential and cell potential, (STEM_GC11AB-IVf-g-176), , Practice Personal Hygiene protocols at all times., 205
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Activity 1: GIVE ME THE SOLUTION, Directions: Solve the standard cell potential of the following problem., 1. Calculate the standard cell potential produced by a galvanic cell consisting, of a sodium electrode in contact with a solution of Na+ ions and a copper, electrode in contact with a solution of Cu2+ ions. Which is anode and which is, the cathode?, Na+(aq) + e- ⇌ Na(s) E° = -2.71 V (must be flipped), Cu2+(aq) + 2e- ⇌ Cu(s) E° = 0.34 V, 2. What is the voltage produced by a voltaic cell consisting of a calcium, electrode in contact with a solution of Cu2+ ions. Which is anode and which is, the cathode?, Ca2+(aq) + 2e- ⇌ Ca(s) E° = -2.87 V (must be flipped), Cu2+(aq) + 2e- ⇌ Cu(s) E° = 0.34 V, 3. Calculate the standard cell potential produced by a galvanic cell consisting, of a nickel electrode in contact with a solution of Ni2+ ions and a silver electrode, in contact with a solution of Ag+ ions. Which is anode and which is the cathode?, Ni2+(aq) + 2e- ⇌ Ni(s) E° = -0.26 V (must be flipped), Ag+(aq) + e- ⇌ Ag(s) E° = 0.80 V, 4. What is the voltage produced by a galvanic ell consisting of an aluminum, electrode in contact with a solution of Al3+ ions and an iron electrode in contact, with a solution of Fe2+ ions. Which is anode and which is the cathode?, Al3+(aq) + 3e- ⇌ Al(s) E° = -1.66 V (must be flipped), Fe2+(aq) + 2e- ⇌ Fe(s) E° = -0.44 V, , 5. A voltaic cell is constructed using electrodes based on the following half, reactions:, Mg2+(aq) + 2e-, , Mg(s), , Cu2+(aq) + 2e-, , Cu(s), , a) Which is the anode and which is the cathode in this cell?, ANODE: ________, , CATHODE: _________, , b) What is the standard cell potential?, Mg2+ (aq) + 2e- ⇌ Mg(s) E° = -2.37 V (must be flipped), Cu2+ (aq) + 2e- ⇌ Cu(s) E° = 0.34 V, , Practice Personal Hygiene protocols at all times., 206
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Activity 4: THINK ABOUT IT!, Directions: Explain briefly the following questions below., , 1. What does the standard reduction potential measure?, _________________________________________________________, _________________________________________________________, 2. The standard reduction potential of Fe3+ is +0.77V. What is its standard, oxidation potential?, ______________________________________________________________, ______________________________________________________________, 3. What are the differences between the standard reduction potential and, standard oxidation potential, and how are the two related?, ______________________________________________________________, ______________________________________________________________, 4. Explain how standard reduction potentials or standard oxidation potentials, are applied., ______________________________________________________________, ______________________________________________________________, 5. What conditions must be met for a potential to be standard?, ______________________________________________________________, ______________________________________________________________, 6. Based on the activity series, which species will be oxidized and reduced:, Zn2+ or H+., ______________________________________________________________, ______________________________________________________________, 7. When standard reduction potentials are measured, what are the potentials, relative to?, ______________________________________________________________, ______________________________________________________________, , 8. Explain how the activity series is used., ______________________________________________________________, ______________________________________________________________, 9. How is a standard reduction potential measured?, , Practice Personal Hygiene protocols at all times., 208
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ANSWER KEY, , Activity 1: GIVE ME THE SOLUTION, 1., , ANODE: Na(s) ⇌ Na+(aq) + e- E° = 2.71 V, CATHODE: Cu2+(aq) + 2e- ⇌ Cu(s) E° = 0.34 V, E° = 3.05 V, , 2., , ANODE: Ca(s) ⇌ Ca+(aq) + e- E° = 2.87 V, CATHODE: Cu2+(aq) (aq) + 2e- ⇌ Cu(s) E° = 0.34 V, E° = 3.21 V, , 3., , ANODE: Ni(s) ⇌ Ni2+(aq) + 2e- E° = 0.26 V, CATHODE: Ag+(aq) + e- ⇌ Ag(s) E° = 0.80 V, E° = 1.06 V, , 4., , ANODE: Al(s) ⇌ Al3+ (aq) + 3e- E° = 1.66 V, CATHODE: Fe2+(aq) + 2e- ⇌ Fe(s) E° = -0.44 V, E° = 1.22 V, , 5. a. ANODE: Mg, , CATHODE: Cu, , b. ANODE: Mg(s) ⇌ Mg2+ (aq) + 2e- E° = 2.37 V, CATHODE: Cu2+ (aq) + 2e- ⇌ Cu(s) E° = 0.34 V, E° = 2.71 V, , Activity 2: SPONTANEOUS OR NON-SPONTANEOUS, 1., , E° = -2.38 V, E° is negative, therefore the cell is non-spontaneous., , 2., , E° = 0.56 V, E° is positive, therefore the cell is spontaneous., , 3., , E° = -1.07 V, E° is negative, therefore the cell is non-spontaneous., , 4., , E° = 1.09 V, E° is positive, therefore the cell is spontaneous., , 5., , E° = -2.33 V, E° is negative, therefore the cell is non-spontaneous., , Practice Personal Hygiene protocols at all times., 211
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Activity 3: TRUE OR FALSE, 1. True, , 2. much like, 3. can be combined, 4. True, 5. True, , Activity 4: THINK ABOUT IT!, 1. Standard reduction potential measures the tendency for a given chemical, species to be reduced., 2. The standard oxidation potential and standard reduction potential are, always opposite in sign for the same species. The oxidation potential is 0.77V., 3. The standard oxidation potential measures the tendency for a given chemical, species to be oxidized as opposed to be reduced. For the same chemical, species the standard reduction potential and standard oxidation potential are, opposite in sign., 4. Standard reduction and oxidation potentials can be applied to solve for the, standard cell potential of two different non hydrogen species. Examples can be, seen in Cell Potentials., 5. The cell must be at 298K, 1atm, and all solutions must be at 1M., 6. H+ is farther up on the activity series then Zn2+ so H+ is reduced while Zn2+ is, oxidized., 7. Standard reduction potentials are measured with relativity to hydrogen which, has be universally set to have a potential of zero., 8. The activity series is a list of standard reduction potentials in descending, order of the tendency for chemical species to be reduced. Species at the top, are more likely to be reduced while species at the bottom are more likely to be, oxidized., 9. A standard reduction potential is measured using a galvanic cell which, contains a SHE on one side and an unknown chemical half cell on the other, , Practice Personal Hygiene protocols at all times., 212
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side. The amount of charge that passes between the cells is measured using a, voltmeter., 10., , Prepared by:, SHAROLYN T. GALURA, Licerio Antiporda Sr. National High School-Dalaya Annex, , Practice Personal Hygiene protocols at all times., 213
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, STANDARD REDUCTION POTENTIAL, Background Information for the Learners (BIL), Reduction potential is a measure of the tendency of a chemical species, to be reduced or to gain electrons while, oxidation potential is the tendency of, a chemical species to be oxidized or to give up electrons. In order to compare, the potential of one – half cell with another, it is necessary to measure all cell, potentials under standard condition. This is due to the fact that several factors, affect potential such as the nature of reaction in each half – cell, concentration, of reactants and products in solution, pressure of gaseous substances, and, temperature. Hence, measurement is done under these standard conditions:, ● solids and liquids are present in their standard states, ● solution concentration is 1M, ● temperature is 25oC or 298oK, ● pressure is 1 atm or 1 bar, A cell potential measured under these conditions is called standard, potential. Standard reduction potential, denoted EO red, is the reduction, potential measured at standard conditions, and the units is volt, V. The, superscript o indicates standard conditions. For the same chemical species, the, standard reduction potential has the same magnitude as the standard oxidation, potential, but opposite in sign. For example, the standard reduction potential at, 25oC, listed in the table is:, Zn+2 (aq, 1M) + 2e- → Zn (s), , Eo red = -0.762 V, , To obtain the standard oxidation potential, consider the reverse reaction, and change the sign., Zn (s) → Zn+2 (aq, 1M) + 2e-, , Eo ox = +0.762 V, , Therefore:, Eo oxidation = - Eoreduction, , Practice Personal Hygiene protocols at all times., 214
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The cell potential is the difference between two electrode potentials. The, potential difference between the two electrodes of a galvanic cell provides the, driving force that pushes the electrons to the external circuit. When the cell, potentials are measured under standard conditions, they are called standard, cell potentials, denoted Eo cell. Since Eo cell is measured in volts, it may also, be called cell voltage or cell emf. Note, however, that cell voltage has a, magnitude but no sign., The cell potential, Eo is given by the standard reduction potential of the, cathode reaction, Eo red (cathode) minus the standard reduction potential at the, anode, Eo red (anode)., Eocell = Eo red (cathode) - Eo red (anode), Eocell can also be obtained by using the equation,, Eocell = Eo red (cathode) + Eo ox (anode), Another way of finding the Eocell, using the cell notation as basis, is the, following formula:, Eocell = Eo (right) - Eo (left), For example, in the galvanic cell represented by the cell notation,, Zn(s)│Zn+2ǁFe+3, Fe+2(aq)│Pt(s), , Where Zn is the anode and Fe is the cathode. E ocell is, therefore,, obtained by using the equation., Eocell = Eo red (Fe+3 , Fe+2/Pt) - Eo red (Zn/Zn+2), , Practice Personal Hygiene protocols at all times., 215
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Standard Hydrogen Electrode, Every galvanic cell involves two half – cells. One cannot measure, individual half – cell potential directly since potential is only evident when it is, connected to a second electrode. It is, therefore, necessary to have another, electrode against which the potential of other electrodes can be measured. The, standard electrode usually adopted as reference point is the hydrogen, electrode. This electrode consists of platinum metal surrounded by hydrogen, gas in equilibrium with the solution of H+1 ions. The half – cell reaction is, represented by the following cell notation:, Pt│H2(g)(1atm)│H+1(aq), 1M, , Fig. 1: Diagram of a Standard Hydrogen Electrode, Source:http://www.mikeblaber.org/oldwine/chm1046/notes/Electro/CellEMF/C, ellEMF.htm, The reference half – reaction is the reduction of H+1(aq) to H2(g) under, stadards conditions. It is assigned a standard reduction potential value of zero, volt., 2H+1(aq) + 2e- → H2(g), , Eo = 0.0 V, , This value serves as the basis in calculating standard cell potentials, using different electrodes as shown below., , Practice Personal Hygiene protocols at all times., 216
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Fig. 2 A Galvanic cell using a SHE and Standard Cu/Cu+2 Electrode, Source: http://kec.edu.np/wp-content/uploads/2016/08/Electrochemical-cellsshrawan.pdf, For this particular galvanic cell, note that Cu is the cathode and the cell, voltage is +0.34 V. The cell notation for this cell is:, Pt│H2(g)(1 atm)│H+1(aq), 1MǁCu+2(aq),1M│Cu(s), Using the defined standard reduction potential of H+1 which is zero volt, and equation 1.1, the standard reduction potential of Cu +2/Cu half – reaction, can be determined., Eocell = Eo(cathode) - Eo(anode), + 0.34 V = Eo(cathode) - 0 V, Eo(cathode) = + 0.34 V + 0 V, Eo(cathode) = +0.34 V, Therefore, a standard reduction potential of +0.34 V can be assigned to, the reduction of Cu+2 to Cu., Cu+2(aq, 1M) + 2e- → Cu(s), , Eo (reduction) = + 0.34 V, , The standard reduction potentials for other half-reactions are also, determined in the same fashion the standard reduction potential of Cu+2/Cu, half-reaction was obtained. These standard reduction potentials as shown in, Table 1.1 can be combined to calculate the emf or standard potentials of other, cell., Practice Personal Hygiene protocols at all times., 217
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Source: https://opentextbc.ca/chemistry/chapter/17-3-standard-reduction-potentials/, An important point that can be gathered from the table of standard, reduction potential is that all reactions are written as reduction. The more, positive the standard reduction potential is, the greater the possibility that it will, be reduced and the stronger its oxidizing property. The more negative the value, of the reduction potential, the less likely the half-reaction will occur as reduction, and the greater the tendency to act as a reducing agent., Consider for example, galvanic cell where Zn/Zn+2 has an Eored = -0.762, V, while Cu/Cu+2 has an Eored = + 0.34 V. This indicates that since Cu/Cu+2 is, more positive, it will be reduced and act an oxidizing agent, while Zn/Zn+2 which, is less positive will be oxidized and act as a reducing agent. In calculating the, standard cell potential, the half – reaction which is more positive is always the, cathode., SAMPLE PROBLEM 1:, Using the standard reduction potential table, calculate the standard, potential of the cell consisting of Al and Ag immersed in 1M Ag + ions and, Al+3 ions., SOLUTION:, Given: Al+3/Al,, Ag+/Ag, , Eored = - 1.66 V, Eored = + 0.80 V, , Required: Eocell, Strategy: Assign the half – cell with the more positive value as, the cathode., o, Hence, E cell = Eo(cathode) - Eo(anode), = + 0.80 V – (-1.66 V), = 2.46 V, SAMPLE PROBLEM 2:, Using the standard reduction potential table, calculate the standard, potential of the galvanic cell that uses the redox reaction:, Cr2O7-2(aq) + 14H+1(aq) + 6I-1, , → 2Cr+3(aq) + 3I2(s) + 7H2O(l), , SOLUTION:, Given: Cr2O7-2(aq) + 6e- → 2Cr+3, (reduction), -1, 6I → 3I2(s) + 6e- (oxidation), From the table 1.1: Cr2O7-2/Cr+3 : Eored = 1.33 V, I-1/I2 :, Eored = 0.54 V, o, Required: E cell, Strategy: The half – cell with the more positive value is the, cathode, , Practice Personal Hygiene protocols at all times., 220
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Substituting in the formula:, Eocell = Eo(cathode) - Eo(anode), = 1.33 V – 0.54 V, = 0.79 V, , Learning Competency:, Calculate the standard cell potential (STEM_GC11AB-IVf-g-178), , Activity 1: CRITICAL THINKING, Directions: Answer the following briefly but substantially., 1. What does the standard reduction potential measure?, ______________________________________________________________, ______________________________________________________________, ___________________________________________, , 2. What are the differences between the standard reduction potential and, standard, oxidation potential, and how are the two related?, ______________________________________________________________, _____________________________________________________________, , 3. What conditions must be met for a potential to be standard?, ______________________________________________________________, ______________________________________________________________, ___________________________________________, , 4. When standard reduction potentials are measured, what are the potentials, relative to?, ______________________________________________________________, ______________________________________________________________, __________________________________________, Practice Personal Hygiene protocols at all times., 221
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5. How is a standard reduction potential measured?, ______________________________________________________________, ______________________________________________________________, ___________________________________________, , 6. What are the factors that affect the cell potentials?, ______________________________________________________________, ______________________________________________________________, ____________________________________________, , 7. What does hydrogen electrode consists?, ______________________________________________________________, ______________________________________________________________, ____________________________________________, , Activity 2: CALCULATE THE UNKNOWN!, Directions: Calculate the following problems below. Refer to Table 1 for the, standard reduction potential value at 25oC. Show your complete, solution., 1. Using the standard reduction potential table, calculate the standard potential, of, the a galvanic cell based on the two standard half – reactios:, Cd+2(aq) + 2e- →, , Cd(s), , Pb+2(aq) + 2e- → Pb(s), 2. What is the standard cell potential for a galvanic cell that consists of Au +3/Au, and Ni+2/Ni half – cell?, 3. The oxidation half – cell of the redox equation is:, Cu(s) → Cu+2 (aq) + 2e-, , Eoox = -0.340V, , Ag+ e- → Ag(s) x 2, , Eored = 0.800V, , Calculate for the standard cell potential., Practice Personal Hygiene protocols at all times., 222
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4. Determine the Eocell between Cu and Zn., 5. Determine the Eocell for the reaction:, Al(s) + Mn+2 aq) → Zn(s) + Al+3 (aq), Q6: Calculate the Eocell of Cd+2/Cd half cell and Ag+1/Ag half cell, Q7: Determine the Eocell between Aluminum and Iodine., Q8: For the Zn-Cu+2 galvanic cell, Eocell is 1.10 V and Zn is the anode. Given, that, the standard reduction potential of the anode is -0.76 V, calculate the Eored for, the, reduction of Cu+2 to Cu., , Activity 3: WHO AM I?, Directions:, A. Identify what is asked in the problem., Q1: Given the following,, Cu+2 + 2e- → Cu, , +0.34, , Fe+2 + 2e- → Fe, , -0.44V, , Al+3 + 3e- → Al, , -1.66V, , Mg+2 + 2e- → Mg, , -2.38V, , A. Based on the Eo values, which metal is the most easily oxidized?, B. Which is the weakest reducing agent?, C. Calculate the Eo cell of Aluminum and Copper, D. Calculate the Eo cell of Magnesium and Aluminum., , Practice Personal Hygiene protocols at all times., 223
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ANSWER KEY:, Activity 1: CRITICAL THINKING, 1. What does the standard reduction potential measure?, Standard reduction potential measures the tendency for a given chemical, species to be reduced., 2. What are the differences between the standard reduction potential and, standard oxidation potential, and how are the two related?, The standard oxidation potential measures the tendency for a given, chemical species to be oxidized as opposed to be reduced. For the same, chemical species the standard reduction potential and standard oxidation, potential are opposite in sign., 3. What conditions must be met for a potential to be standard?, Solids and liquids are present in their standard states, solution, concentration is 1M, temperature is 25oC or 298oK, pressure is 1 atm or 1, bar, 4. When standard reduction potentials are measured, what are the potentials, relative to?, Standard reduction potentials are measured with relativity to hydrogen, which has be universally set to have a potential of zero., 5. How is a standard reduction potential measured?, A standard reduction potential is measured using a galvanic cell which, contains a SHE on one side and an unknown chemical half - cell on the, other side. The amount of charge that passes between the cells is, measured using a voltmeter., 6. What are the factors that affect cell potentials?, The nature of reaction in each half – cell, concentration of reactants and, products in solution, pressure of gaseous substances, and temperature, 7. What does hydrogen electrode consist of?, This electrode consists of platinum metal surrounded by hydrogen gas in, equilibrium with the solution of H+1 ions., , Activity 2: CALCULATE THE UNKNOWN!, 1. Using the standard reduction potential table, calculate the standard potential, of the a galvanic cell based on the two standard half – reactios:, Cd+2(aq) + 2e- →, , Cd(s), , Pb+2(aq) + 2e- → Pb(s), Practice Personal Hygiene protocols at all times., 226
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, SPONTANEOUS REDOX REACTIONS, Background Information for the Learners (BIL), To determine whether a redox reaction is spontaneous or not, it is, possible to use half – cell potentials to calculate the cell potential. If the, calculated cell potential is a positive quantity, the reaction is spontaneous. If the, calculated cell potential is negative, the reaction is nonspontaneous. Ordinarily,, this principle is applied at a standard condition, hence, standard cell potential,, Eocell, is likewise a measure of spontaneity of a redox reaction., Eocell > 0; spontaneous, Eocell ˂ 0; nonspontaneous, Consequently, it is possible to determine the feasibility of a cell to, generate electric current from calculated cell potentials. Any reaction that can, occur in a galvanic cell to produce a positive cell voltage must be spontaneous., If cell voltage is positive, then the cell can generate electricity and if negative, it, is not capable of applying electric current. Consider the following examples to, show how spontaneity of a redox reaction is determined., SAMPLE PROBLEM:, A cell consists of iron and silver in a solution of 1M silver ions and 1M, iron (II) ions. Determine the direction of spontaneous reaction and calculate the, standard cell potential., SOLUTION:, Given: From the table, the half reactions and the corresponding standard, reduction potential are:, Oxidation:, , Fe+2(aq) + 2e- → Fe(s), , EOred = - 0.45 V, , Reduction:, , Ag+1(aq) + e-, , → Ag(s), , EOred = + 0.80 V, , Required:, , Eocell, , Strategy: The half – reaction with more positive reduction potential, oxidizes the half – reaction with the lesser potential. Ag+/Ag is more positive,, which means it has a greater tendency to be reduced and can oxidized Fe, and,, , Practice Personal Hygiene protocols at all times., 230
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Activity 3: SOLVE AND HUNT ME!, Directions: Solve the following redox reactions. Choose the correct Eocell in, the box. Write only the letter of your correct answer., a. – 1.259 V, , b. + 0.12 V, , c. + 1.08 V, , d. - 0.26 V, , e. – 1.38 V, , 1. Determine the standard cell potential when iodine oxidize silver to silver ions., 2. Determine the overall reaction and its standard cell potential at 25 °C for the, reaction involving the galvanic cell in which cadmium metal is oxidized to, 1 M cadmium(II) ion and a half-cell consisting of an aluminum electrode in, 1 M aluminum nitrate solution., 3. Determine the overall reaction and its standard cell potential at 25 °C for this, reaction., Cu(s)∣Cu2+(aq)∥Au3+(aq)∣Au(s), 4. Balance this redox reaction and determine its voltage., Li+ + Al → Li + Al3+, 5. Balance this redox reaction and determine its voltage., Cu2+ + Ag + Cl− → Cu + AgCl, , Practice Personal Hygiene protocols at all times., 235
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ANSWER KEY:, Activity 1: MIND POWER!, 1. What makes a redox reaction spontaneous?, A redox reaction is spontaneous if the standard electrode potential for the, redox reaction, Eo (redox reaction), is positive. If Eo (redox reaction) is, positive, the reaction in the forward direction is spontaneous., 2. Which is a stronger reducing agent Cr2+ or Fe2+ and why?, Cr2+ is a stronger reducing agent than Fe2+. This can be explained on the, basis of the standard electrode potential values Eo (Cr3+/Cr2+ = -0.41 V), and Eo Fe3+/Fe2+ = +0.77 V). Thus Cr2+ is easily oxidized to Cr3+ but, Fe2+ cannot be as easily oxidized to Fe3+., 3. Is the decomposition of water a spontaneous redox? Support your answer., No, the electrolysis of water is an example of a nonspontaneous redox, reaction that occurs in the presence of electricity., 4. Differentiate spontaneous and non – spontaneous redox reaction., If the calculated cell potential is a positive quantity and it can generate, electric current, the reaction is spontaneous. If the calculated cell, potential is negative and cannot generate electric current, the reaction is, nonspontaneous., , Activity 2: WHO AM I?, 1. spontaneous, 2. spontaneous, 3. spontaneous, 4. non – spontaneous, 5. spontaneous, 6. spontaneous, 7. spontaneous, 8. non – spontaneous, 9. non – spontaneous, 10. non – spontaneous, , Practice Personal Hygiene protocols at all times., 238
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, ELECTROCHEMISTRY, Background Information for the Learners (BIL), Electrochemistry is a branch of chemistry that deals with the, interconversion of chemical energy and electrical energy. Electrochemistry has, many common applications in everyday life. All sorts of batteries, from those, used to power a flashlight to a calculator to an automobile, rely on chemical, reactions to generate electricity. Electricity is used to plate objects with, decorative metals like gold or chromium. Electrochemistry is important in the, transmission of nerve impulses in biological systems. Redox chemistry, the, transfer of electrons, is behind all electrochemical processes., An electrochemical cell is any device that converts chemical energy into, electrical energy or electrical energy into chemical energy. There are three, components that make up an electrochemical reaction. There must be a, solution where redox reactions can occur. These reactions generally take place, in water to facilitate electron and ion movement. A conductor must exist for, electrons to be transferred. This conductor is usually some kind of wire so that, electrons can move from one site to another. Ions also must be able to move, through some form of salt bridge that facilitates ion migration., Batteries consist of one or more electrochemical cells that store chemical, energy for later conversion to electrical energy. Batteries are used in many dayto-day devices such as cellular phones, laptop computers, clocks, and cars., Batteries are composed of at least one electrochemical cell which is used for, the storage and generation of electricity. Though a variety of electrochemical, cells exist, batteries generally consist of at least one voltaic cell. Voltaic cells, are also sometimes referred to as galvanic cells. Chemical reactions and the, generation of electrical energy is spontaneous within a voltaic cell, as opposed, to the reactions electrolytic cells and fuel cells., , Practice Personal Hygiene protocols at all times., 240
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ELECTROCHEMISTRY INVOLVED IN LECLANCHE DRY CELL, , Photo courtesy of Gerhard H Wrodnigg, Images used with permission from, Wikipedia, , Dry cell or Leclanche cell is a primary cell, handy for sporadic use, with, positive anode of zinc encompassed by a mixture of manganese dioxide and, powdered carbon in a pot, which is porous. The pot and the negative zinc, terminal remained in a container holding ammonium chloride solution. The, electromotive force (emf) is nearly 1 -4 volt., Types of Leclanche’s cell include:, a), zinc, (Carbon, cathode), b) zinc chloride (Ammonium chloride electrolyte reinstated by zinc chloride), c) alkaline manganese (Ammonium chloride terminal displaced by potassium, hydroxide), How does Lechanche works?, How does it work?, The process which generates power in a Leclanché cell starts when zinc, particles on the surface of the anode oxidize, i.e. when zinc atoms surrender, their valence electrons to end up becoming the positively charged particles. The, zinc particles move far from the anode while leaving their electrons on its, surface that makes the anode more negatively charged than the cathode. At, the point when the cell is associated in an outer electrical circuit, the excess, electrons on the zinc anode gush through the circuit to the cathode made up of, carbon. This flow of electrons frames the electric current., , Practice Personal Hygiene protocols at all times., 241
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What are the parts of dry cell or Lechanche cell?, A standard dry cell comprises a zinc anode, usually in the form of a, cylindrical pot, with a carbon cathode in the form of a central rod. The electrolyte, is ammonium chloride in the form of a paste next to the zinc anode., Practical Use, The Leclanche cell was utilized widely for telegraphy, electric bell and, signaling work; and for work where intermittent and low current was needed., The battery cell by Georges Leclanche proved out to be extremely, advantageous in the early years of the telephones., ELECTROCHEMISTRY INVOLVED IN BUTTON BATTERIES?, , Button batteries. Photo courtesy of Gerhard H Wrodnigg, Images used with, permission from Wikipedia, , A watch battery or button cell is a small single cell battery shaped as a, squat cylinder typically 5 to 25 mm (0.197 to 0.984 in) in diameter and 1 to 6, mm (0.039 to 0.236 in) high — resembling a button. A metal can forms the, bottom body and positive terminal of the cell. An insulated top cap is the, negative terminal., , •, , •, , What are button cell batteries made of?, Button cells are single cells, usually disposable primary cells. Common anode, materials are zinc or lithium. Common cathode materials are manganese, dioxide, silver oxide, carbon monofluoride, cupric oxide or oxygen from the air., What is battery in electronics?, Batteries consist of one or more electrochemical cells that store chemical, energy for later conversion to electrical energy. Batteries are used in many, day-to-day devices such as cellular phones, laptop computers, clocks, and, cars., Practice Personal Hygiene protocols at all times., 242
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•, , What kinds of devices use a button cell battery?, Button cells are used to power small portable electronics devices such as wrist, watches, pocket calculators, artificial cardiac pacemakers, implantable cardiac, defibrillators, automobile keyless entry transmitters, and hearing aids. Wider, variants are usually called coin cells., ELECTROCHEMISTRY INVOLVED IN FUEL CELLS, , Wikipedia en.wikipedia.org., , A fuel cell can be defined as an electrochemical cell that generates electrical, energy from fuel via an electrochemical reaction. These cells require a, continuous input of fuel and an oxidizing agent (generally oxygen) in order to, sustain the reactions that generate the electricity., Fuel cell, any of a class of devices that convert the chemical energy of a fuel, directly into electricity by electrochemical reactions. A fuel cell resembles, a battery in many respects, but it can supply electrical energy over a much, longer period of time. This is because a fuel cell is continuously supplied with, fuel and air (or oxygen) from an external source, whereas a battery contains, only a limited amount of fuel material and oxidant that are depleted with use., For this reason fuel cells have been used for decades in space probes,, satellites, and manned spacecraft., , Practice Personal Hygiene protocols at all times., 243
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Around the world thousands of stationary fuel cell systems have been installed, in utility power plants, hospitals, schools, hotels, and office buildings for both, primary and backup power; many waste-treatment plants use fuel, cell technology to generate power from the methane gas produced by, decomposing garbage., Types of Fuel Cells, Various types of fuel cells have been developed. They are generally classified, on the basis of the electrolyte used, because the electrolyte determines the, operating temperature of a system and in part the kind of fuel that can be, employed., 1. The Polymer Electrolyte Membrane (PEM) Fuel Cell, •, •, •, •, •, , These cells are also known as proton exchange membrane fuel cells (or, PEMFCs)., The temperature range that these cells operate in is between 50oC to, 100o, The electrolyte used in PEMFCs is a polymer which has the ability to, conduct protons., A typical PEM fuel cell consists of bipolar plates, a catalyst, electrodes,, and the polymer membrane., Despite having eco-friendly applications in transportation, PEMFCs can, also be used for the stationary and portable generation of power., , 2. Solid Acid Fuel Cell, •, •, •, •, , A solid acid material is used as the electrolyte in these fuel cells., The molecular structures of these solid acids are ordered at low, temperatures., At higher temperatures, a phase transition can occur which leads to a, huge increase in conductivity., Examples, of, solid, acids, include, CsHSO4 and, CsH2PO4 (cesium hydrogen sulfate and cesium dihydrogen phosphate, respectively), , 3. Alkaline Fuel Cell, •, •, •, •, , This was the fuel cell which was used as the primary source of electricity, in the Apollo space program., In these cells, an aqueous alkaline solution is used to saturate a porous, matrix, which is in turn used to separate the electrodes., The operating temperatures of these cells are quite low (approximately, 90oC)., These cells are highly efficient. They also produce heat and water along, with electricity., , Practice Personal Hygiene protocols at all times., 244
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4. Solid Oxide Fuel Cell, •, •, •, •, , These cells involve the use of a solid oxide or a ceramic electrolyte (such, as yttria-stabilized zirconia)., These fuel cells are highly efficient and have a relatively low cost, (theoretical efficiency can even approach 85%), The operating temperatures of these cells are very high (lower limit of, 600oC, standard operating temperatures lie between 800 and 1000 oC)., Solid oxide fuel cells are limited to stationary applications due to their, high operating temperatures., , 5. Molten Carbonate Fuel Cell, •, , •, •, , •, , The electrolyte used in these cells is lithium potassium carbonate salt., This salt becomes liquid at high temperatures, enabling the movement, of carbonate ions., Similar to SOFCs, these fuel cells also have a relatively high operating, temperature of 650o, The anode and the cathode of this cell are vulnerable to corrosion due, to the high operating temperature and the presence of the carbonate, electrolyte., These cells can be powered by carbon-based fuels such as natural, gas and biogas., , Applications of fuel cell, Fuel cell technology has a wide range of applications. Currently, heavy, research is being conducted in order to manufacture a cost-efficient automobile, which is powered by a fuel cell. A few applications of this technology are listed, below., •, •, •, •, •, •, , Fuel cell electric vehicles, or FCEVs, use clean fuels and are therefore, more eco-friendly than internal combustion engine-based vehicles., They have been used to power many space expeditions including the, Appolo space program., Generally, the byproducts produced from these cells are heat and water., The portability of some fuel cells is extremely useful in some military, applications., These electrochemical cells can also be used to power several electronic, devices., Fuel cells are also used as primary or backup sources of electricity in, many remote areas., , Practice Personal Hygiene protocols at all times., 245
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Thus, the different types of fuel cells and the working of an alkaline fuel cell, are briefly discussed in this article along with some of the applications of these, electrochemical cells. To learn more about this technology and other related, topics, register with BYJU’S and download the mobile application on your, smartphone., ELECTROCHEMISTRY INVOLVED IN LEAD BATTERIES, , Lead Storage Battery I chem.wisc.edu, A lead storage battery, also known as a lead-acid battery, is the oldest, type of rechargeable battery and one of the most common energy storage, devices. These batteries were invented in 1859 by French physicist Gaston, Planté, and they are still used in a variety of applications. Most people are, accustomed to using them in vehicles, where they have the ability to provide, high currents for cranking power., Although the batteries are reliable, they have a limited life, are heavy to ship,, and contain toxic materials that require specific removal methods at the end of, their useful life. Lead-acid batteries have moderate power density and good, response time. Depending on the power conversion technology incorporated,, batteries can go from accepting energy to supplying energy instantaneously., Lead-acid batteries are affected by temperature and must be maintained in, order to achieve maximum life expectancy., , Practice Personal Hygiene protocols at all times., 246
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•, •, •, •, , Lead- acid batteries, also known as lead storage batteries, can store a, lot of charge and provide high current for short periods of time., The basic design of lead-acid batteries has not changed significantly, since 1859 when Planté designed them, although some improvements, were made by Faure., Lead-acid batteries are capable of being recharged, which is important, for their use in cars., Discharging the stored energy relies on both the positive and negative, plates becoming lead(II) sulfate and the electrolyte losing much of its, dissolved sulfuric acid., , Learning Competency, Describe the electrochemistry involved in some common batteries:, a. Leclanche dry cell, b. Button batteries, c. Fuel cells, d. Lead storage batteries (STEM_GC11ABIVf-g180), , Activity I. MIND POWER, Directions: Multiple-choice questions on batteries and alternative sources of, energy, 1.A battery consists of:, (a) a cell (b) a circuit (c) a generator (d) a number of cells, 2.Which of the following statements is false?, a. A Leclanché cell is suitable for use in torches, b. A nickel–cadmium cell is an example of a primary cell, c. When a cell is being charged its terminal p.d. exceeds the cell, e.m.f., d. (d) A secondary cell may be recharged after use., 3.Which of the following statements is false? When two metal electrodes are, used in a simple cell, the one that is higher in the electrochemical series:, a. tends to dissolve in the electrolyte, b. is always the negative electrode, c. reacts most readily with oxygen, d. acts as an anode, , Practice Personal Hygiene protocols at all times., 247
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4.The negative pole of a dry cell is made of:, a.carbon, b. copper, c. zinc, d. mercury, , 5.The energy of a secondary cell is usually renewed:, a. by passing a current through it, b. it cannot be renewed at all, c. by renewing its chemicals, d. by heating it., , 6.A fuel cell is used to convert chemical energy to _________________., a.Mechanical energy, , b. electrical energy, , c. Solar energy, , d. potential, , 7.Which of the following statements is true?, a. A zinc–carbon battery is rechargeable and is not classified as, hazardous, b. A nickel–cadmium battery is not rechargeable and is classified as, hazardous, c. A lithium battery is used in watches and is not rechargeable, d. An alkaline manganese battery is used in torches and is classified as, hazardous, , 8. Which of the following is NOT example of a fuel cell?, a. hydrogen-oxygen cell, , b. methyl-oxygen cell, , c. propane-oxygen cell, , d. hexanone-oxygen cell, , Practice Personal Hygiene protocols at all times., 248
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Activity 2 MIND POWER, Directions: Answer the following questions briefly but substantially, 1.What are the chemical reactions in a battery?, , ______________________________________________________________, _____., , 2. What is the importance of electrochemistry?, , ______________________________________________________________, ______., 3. What is a battery in electrochemistry?, , ______________________________________________________________, _____.., , 4.What is lead storage battery in chemistry?, , ______________________________________________________________, _________., , Practice Personal Hygiene protocols at all times., 249
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References:, Harwood, William, Herring, Geoffrey, Madura, Jeffry, and Petrucci,, Ralph. General Chemistry: Principles and Modern Applications. Ninth Edition., Upper Saddle River, New Jersey: Pearson Prentice Hall, 2007., Frank Brescia, John Arents, Herbert Meislich, amos Turk, Fundamentals of, Chemistry 4th Edition, Ma. Christina Padolina, PhD, Laboratory Manual and Workbook in Chemistry, Marasinghe, B.Dr. 2010. Upper Secondary Chemistry. A textbook of chemistry, for Grades 11 &12., Bettelheim, Brown, Campbell, Farrell, Introduction to General, Organic and, Biochemistry, 8th Edition, Sackheim, George I, Lehman, Dennis D., Chemistry for Health Sciences, 8th, Edition, , Practice Personal Hygiene protocols at all times., 251
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ANSWER KEY, ACTIVITY 1, 1. D, , 2. B, , 6. C, , 7. D, , 3. D, , 4. C, , 5. A, , ACTIVITY 2, 1. The, , battery, , operates, , through, , electrochemical, , reactions, , called oxidation and reduction. These reactions involve the exchange, of electrons between chemical species. If a chemical species loses one, or more electrons, this is called oxidation. The opposite process, the, gain of electrons, is called reduction., 2. Electrochemistry is also vital in a wide range of important technological, applications. For example, batteries are important not only in storing, energy for mobile devices and vehicles, but also for load leveling to, enable the use of renewable energy conversion technologies., 3. A battery is an electrochemical cell or series of cells that produces an, electric current. In principle, any galvanic cell could be used as a battery., An ideal battery would never run down, produce an unchanging voltage,, and be capable of withstanding environmental extremes of heat and, humidity., 4. Lead-acid batteries, also known as lead storage batteries, can store a, lot of charge and provide high current for short periods of time. ..., Discharging the stored energy relies on both the positive and negative, plates becoming lead(II) sulfate and the electrolyte losing much of its, dissolved sulfuric acid., , Prepared by:, DOLORES ARAGON-LIBAN, Magalalag National High School, , Practice Personal Hygiene protocols at all times., 252
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, CORROSION, Background Information for the Learners (BIL), , Corrosion – Keystagewiki.com, , Corrosion is one of the most common phenomena that we observe in, our daily lives. You must have noticed that some objects made of iron are, covered with some orange or reddish-brown coloured layer at some point in, time. The formation of this layer is the result of a chemical process known as, rusting, which is a form of corrosion., For example, formation of a layer of a reddish scale or hydrated ferric, oxide (Fe3O4) on the surface of iron, also known as “rusting of iron”. Corrosion, can be defined as the process through which refined metals are converted into, more stable compounds such as metal oxides, metal sulfides, or metal, hydroxides. The rusting of iron involves the formation of iron oxides via the, action of atmospheric moisture and oxygen. Corrosion is usually an undesirable, phenomenon since it negatively affects the desirable properties of the metal., For example, iron is known to have good tensile strength and rigidity (especially, alloyed with a few other elements). However, when subjected to rusting, iron, objects become brittle, flaky, and structurally unsound. Corrosion can be, Practice Personal Hygiene protocols at all times., 253
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classified as an electrochemical process since it usually involves redox, reactions between the metal and certain atmospheric agents such as water,, oxygen, sulphur dioxide, etc., Electrochemical corrosion of metals occurs when electrons from atoms at, the surface of the metal are transferred to a suitable electron acceptor or, depolarizer. Water must be present to serve as a medium for the transport of, ions. The most common depolarizers are oxygen, acids, and the cations of less, active metals., , Factors Affecting Corrosion, 1. Exposure of the metals to air containing gases like CO2, SO2, SO3 etc., 2. Exposure of metals to moisture especially salt water (which increases the, rate of corrosion)., 3. Presence of impurities like salt (eg. NaCl)., 4. Temperature: An increase in temperature increases corrosion., 5. Nature of the first layer of oxide formed: some oxides like Al 2O3 forms an, insoluble protecting layer which can prevent further corrosion. Others like rust, easily crumble and expose the rest of the metal., 6. Presence of acid in the atmosphere: acids can easily accelerate the process, of, corrosion., Types of Corrosion, Some of the corrosion types include;, 1. Crevice Corrosion, Whenever there is a difference in ionic concentration between any two local, areas of a metal, a localized form of corrosion know as crevice corrosion can, occur. Examples of areas where crevice corrosion can occur are gaskets, the, undersurface of washers, and bolt heads., Example: All grades of aluminium alloys and stainless steels undergo crevice, corrosion., 2. Stress Corrosion Cracking, Stress Corrosion Cracking can be abbreviated to ‘SCC’ and refers to the, cracking of the metal as a result of the corrosive environment and the tensile, tress placed on the metal. It often occurs at high temperatures., Example: Stress corrosion cracking of austenitic stainless steel in chloride, solution., , Practice Personal Hygiene protocols at all times., 254
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3. Intergranular Corrosion, Intergranular corrosion occurs due to the presence of impurities in the grain, boundaries that separate the grain formed during the solidification of the metal, alloy. It can also occur via the depletion or enrichment of the alloy at these grain, boundaries., Example: Aluminum-base alloys are affected by IGC., 4. Galvanic Corrosion, When there exists an electric contact between two metals that are, electrochemically dissimilar and are in an electrolytic environment, galvanic, corrosion can arise. It refers to the degradation of one of these metals at a joint, or at a junction. A good example of this type of corrosion would be the, degradation that occurs when copper, in a salt-water environment, comes in, contact with steel., Example: When aluminium and carbon steel are connected and immersed in, seawater, aluminum corrodes faster and steel is protected., 5. Pitting Corrosion, Pitting Corrosion is very unpredictable and therefore is difficult to detect. It is, considered one of the most dangerous types of corrosion. It occurs at a local, point and proceeds with the formation of a corrosion cell surrounded by the, normal metallic surface. Once this ‘Pit’ is formed, it continues to grow and can, take various shapes. The pit slowly penetrates metal from the surface in a, vertical direction, eventually leading to structural failure it left unchecked., , Example: Consider a droplet of water on a steel surface, pitting will initiate at t, he centre of the water droplet (anodic site)., 6. Uniform Corrosion, This is considered the most common form of corrosion wherein an attack on, the surface of the metal is executed by the atmosphere. The extent of the, corrosion is easily discernible. This type of corrosion has a relatively low impact, on the performance of the material., Example: A piece of zinc and steel immersed in diluted sulphuric acid would, usually dissolve over its entire surface at a constant rate., , PRINCIPLES OF CORROSION, 2.1 HALF CELL REACTIONS, Corrosion is an electrochemical process in which metals and alloys undergo, transformation into predominantly oxides, hydroxides, and aqueous salts., , Practice Personal Hygiene protocols at all times., 255
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In the corrosion process, two reactions take place. In one, the anodic reaction,, metal atoms are ionized and pass into solution leaving their electrons within the, original metal surface. In the second, the cathodic reaction, the free electrons, within the metal are taken up by chemical species such as O2 and H2O in, reduction reactions., Consider a simplified version of the corrosion reaction between iron and, water. The overall reaction proceeds as follows:, Fe + 2H2O → Fe (OH)2 + H2 (1), The overall reaction can be broken down into the oxidizing ANODIC reaction, Fe → Fe2+ + 2e-, , (2), , and the reducing CATHODIC reaction, 2H2O + 2e- → H2 + 2(OH)-, , (3), , Figure 2.1 depicts this corrosion process., The reaction 2 and 3 are called ‘half cell’ reactions. Reaction 2 is the half of the, process which is responsible for the damage during corrosion. The speed at, which this reaction proceeds is directly related to the corrosion rate., 2.2 ELECTRODE POTENTIALS, Values of ELECTRODE POTENTIAL are associated with each of the half cell, reactions and give a measure of the likelihood for the reaction to occur. Figure, 2.2 depicts standard electrode potentials as measured on the Standard, Hydrogen Electrode Scale for some selected half cell reactions. This scale sets, as datum a value of zero volts for the reduction of Hydrogen., , The more reactive the metal the more negative is its standard potential. In the, case of Iron, it is -0.440V, whereas, for the more inert engineering metals, such, as Platinum, the standard electrode potential is +1.20V., This information allows us to determine whether a metal will corrode in a given, environment., , Practice Personal Hygiene protocols at all times., 256
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FIGURE 2.1 – Schematic of the Corrosion Process, Write the charges of the ions properly and remove the highlights, Half Cell, Reactions, Zn → Zn2+ + 2e-, , Standard Electrode, Potential (Volts), -0.76, , Fe → Fe2+ + 2e-, , -0.44, , 2H+ + 2e- → H2, , 0.00, , 2H2O + 2e- → H2 + 2(OH)-, , 0.40, , O2 + 4H+ + 4e- → 2H2O, , 1.23, , Cl2 + 2e- → 2Cl-, , 1.36, , FIGURE 2.2 – Selected Standard Half-Cell Potentials, , Under standard conditions, the potential of the corrosion cell is the, difference between the cathodic reaction half cell potential and the anodic half, cell potential. For reaction 1 the potential difference between the cathodic and, anodic half cell reactions is:, E = +0.401 – (-0.440) = +0.841V, The positive value of the potential indicates that the reaction is possible, and that corrosion will occur under these conditions. However, theory does not, permit the calculation of the corrosion rate; this has to be measured, preferably, Practice Personal Hygiene protocols at all times., 257
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in-situ. Thus a calculation of the corrosion cell potential is not able to predict, the magnitude of damage likely to occur by the corrosion reaction., , 2.3 CORROSION RATES AND POLARISATION, The existence of a potential difference between the anodic and cathodic half, cell regions generates a current flow. This has the effect of reducing the, potential difference between the half cells, phenomena known as, polarization. The shape of the polarization curve is affected by many factors,, but in its simplest form is depicted in Figure 2.3., At the point of intersection of the two polarization curves, a metal will be freely, corroding. The potential at which this occurs is called the Free Corrosion, Potential (Ecorr) which generates a Corrosion Current (Icorr). From the, corrosion current, it is possible to calculate corrosion rates., , Control of Corrosion, Since both the cathodic and anodic steps must take place for corrosion, to occur, prevention of either one will stop corrosion. The most obvious strategy, is to stop both processes by coating the object with a paint or other protective, coating. Even if this is done, there are likely to be places where the coating is, broken or does not penetrate, particularly if there are holes or screw threads. A, more sophisticated approach is to apply a slight negative charge to the metal,, thus making it more difficult for the reaction to take place:, M⟶M2++2e−.(16.8.5)(16.8.5)M⟶M2++2e−., Practice Personal Hygiene protocols at all times., 258
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Protection Method 1: Sacrificial Coatings, One way of supplying this negative charge is to apply a coating of a more, active metal. Thus a very common way of protecting steel from corrosion is to, coat it with a thin layer of zinc; this process is known as galvanizing. The zinc, coating, being less noble than iron, tends to corrode selectively. Dissolution of, this sacrificial coating leaves behind electrons which concentrate in the iron,, making it cathodic and thus inhibiting its dissolution., Protection Method 2: Cathodic Protection, A more sophisticated strategy is to maintain a continual negative, electrical charge on a metal, so that its dissolution as positive ions is inhibited., Since the entire surface is forced into the cathodic condition, this method is, known as cathodic protection. The source of electrons can be an external direct, current power supply (commonly used to protect oil pipelines and other buried, structures), or it can be the corrosion of another, more active metal such as a, piece of zinc or aluminum buried in the ground nearby, as is shown in the, illustration of the buried propane., , Corrosion Activities, Note: This activity is intended to follow the Corrosion Investigation &, Presentation activity where students investigate and summarize factors, affecting corrosion and corrosion prevention. However, it can be done on its, own if students have been introduced to the concept of corrosion., Introduction:, What does corrosion look like? Compare samples of corroded and un-corroded, metals such as iron, aluminum, copper, zinc, silver and magnesium. Do all, corroded samples look the same? What is corrosion a result of? What are, some common objects that are subject to corrosion? Do all metals corrode at, the same rate? What are some methods used to prevent corrosion?, Corrosion is a major destructive process that results in costly and unsightly, gaping holes or cracks in aircraft, automobiles, boats, gutters, screens,, plumbing, and many other items constructed of every metal except gold., Corrosion results from the overwhelming tendency of metals to react, electrochemically with oxygen, water, and other substances in the aqueous, environment. Fortunately, most useful metals react with the environment to, form protective films of corrosion reaction products that prevent the metals from, going into solution as ions. For example, aluminum forms a protective coating, of aluminum oxide on its surface that prevents further corrosion. This is, because many metals are not stable in their elemental state; they prefer to, combine with elements such as oxygen to form metal oxides that are lower in, energy and thus more stable., , Practice Personal Hygiene protocols at all times., 259
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Corrosion occurs via two opposing electrochemical reactions. The first is the, anodic reaction, in which a metal is oxidized, releasing electrons into the metal., The other is the cathodic reaction, in which a solution species (often O 2 or H+), is reduced, removing electrons from the metal. The two reactions can take, place on one metal or on two dissimilar metals (or metal sites) that are, electrically connected., There are several different reactions that describe the process of corrosion and, most are fairly complex, depending on the type of corrosion and the conditions., Visit the following website to view some typical anode and cathode reactions, involved in corrosion:http://www.arvanitakis.com/en/cl/corrosion_reactions.htm, , Learning Competency:, Apply electrochemical principles to explain corrosion (STEM_GC11ABIVf-g181), , Activity 1: MULTIPLE CHOICE, Directions: Read each item carefully. Encircle the letter that corresponds to, the correct answer., 1. Corrosion of metals involves, a. Physical reactions, , b. Chemical reactions, , c. Both, , d. None, , 2. The following factors play vital role in corrosion process, a. Temperature, , b. Solute concentration, , c. Both d. None, , 3. Which of the following can be used for cathodic protection:, a. Al, , b. Cd, , c. Cu, , d. Either, , 4. The rusting of iron is the __________________, a. oxidation corrosion, b. liquid metal corrosion, c. wet corrosion, d. corrosion by other gases, 5. Corrosion is uniform in ____________________, a. dry corrosion, b. wet corrosion, c. pitting corrosion, d. water line corrosion, , Practice Personal Hygiene protocols at all times., 260
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Extension Activities:, Research answers/explanations for the following:, 1. Is corrosion reversible?, 2. Explain what is meant by the term “anodizing” and describe where it is, used., 3. Explain why aluminum doesn’t corrode as quickly as many other, metals despite its high reactivity., 4. Describe the process of sacrificial protection. What is it and why is it, used?, 5. How does corrosion affect the economy?, 6. Why do gold, platinum and other precious metals not corrode?, , Reflection, 1.I learned that ____________________________________________, _______________________________________________________, _______________________________________________________, , 2.I enjoyed most on ________________________________________, _______________________________________________________, _______________________________________________________, , 3.I want to learn more on ____________________________________, _______________________________________________________, _______________________________________________________, , Practice Personal Hygiene protocols at all times., 261
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References:, , Frank Brescia, John Arents, Herbert Meislich, amos Turk, Fundamentals of, Chemistry 4th Edition, Ma. Christina Padolina, PhD, Laboratory Manual and Workbook in Chemistry, Marasinghe, B.Dr. 2010. Upper Secondary Chemistry. A textbook of, chemistry for Grades 11 &12., Bettelheim, Brown, Campbell, Farrell, Introduction to General, Organic and, Biochemistry, 8th Edition, Sackheim, George I, Lehman, Dennis D., Chemistry for Health Sciences, 8th, Edition, , Practice Personal Hygiene protocols at all times., 262
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ANSWER KEY, ACTIVITY 1., 1. B, , 2. C, , 3. A, , 4. A, , 5A, , EXTENSION ACTIVITIES, 1. NO, Corrosion is an irreversible interfacial reaction of a material (metal,, ceramic, polymer) with its environment which results in consumption of, the material or in dissolution into the material of a component of the, environment., 2. Anodizing is an electrolytic passivation process used to increase the, thickness of the natural oxide layer on the surface of metal parts. The, process is called anodizing because the part to be treated forms the, anode electrode of an electrolytic cell., 3., , Aluminum unlike iron and, conditions., , Its, , surface, , steel,does, is, , not rust or corrode in, , protected, , by, , a, , natural, , moist, layer, , of aluminium oxide. This prevents the metal below from coming into, contact with air and oxygen., 4. Sacrificial protection is the protection of iron or steel against corrosion, by using a more reactive metal. ... The iron pipe will be connected to a, more reactive metal such as magnesium through cooper wires, the, magnesium will donate its electrons to the iron preventing it from, rusting., 5. Corrosion of metals costs the U.S. economy almost $300 billion per, year at current prices. Approximately one-third of these costs could be, reduced by broader application of corrosion-resistant materials and the, application of best corrosion-related technical practices., 6., , Gold and gold alloys: Gold, , is an, , extremely, , inert precious, , metal and will not oxidize., , Prepared by:, DOLORES ARAGON-LIBAN, Magalalag National High School, , Practice Personal Hygiene protocols at all times., 263
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, ELECTRODE REACTIONS DURING ELECTROLYSIS, Background Information for the Learners (BIL), Electrodes and Electrode Reactions, An electrode reaction refers to the net oxidation and reduction process, that takes place at an electrode. This reaction may take place in a single, electron-transfer step, or as a succession of two or more steps. The substances, that receive and lose electrons are called the electroactive species., , Fig. 1: Electron transfer at an anode, Source: Silverio,Angelina.”Exploring Life Through Science Series: General Chemistry 2.” In, Teachers Wraparound Edition. Quezon City, Phoenix Pulishing House, Inc., 2017, , This process takes place within the very thin interfacial region at the, electrode surface, and involves quantum mechanical tunneling of electrons, between the electrode and the electroactive species. The work required to, displace the H2O molecules in the hydration spheres of the ions constitutes part, of the activation energy of the process., , Practice Personal Hygiene protocols at all times., 264
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In Zn/Cu cell, the electrode reaction involves a metal and its hydrated, cation; we call such electrodes metal-metal ion electrodes. These are a number, of other kinds of electrodes which are widely encountered in electrochemistry, and analytical chemistry., , Ion-Ion Electrodes, Many electrode reactions involve only ionic species, such as Fe 2+ and, Fe3+. If neither of the electroactive species is a metal, some other metal must, serve as a conduit for the supply or removal of electrons from the system. In, order to avoid complications that would arise from electrode reactions involving, this metal, a relatively inert substance such as platinum is commonly used., Such a half cell would be represented as:, Pt(s) II Fe3+(aq), Fe2+(aq) II ..., and the half-cell reaction would be:, Fe2+ (aq) → Fe3+ (aq) + eThe reaction occurs at the surface of the electrode (fFig. 1 above). The, electroactive ion diffuses to the electrode surface and adsorbs (attaches) to it, by van der waals and coulombic forces. In doing so, the waters of hydration, that are normally attached to any ionic species must be displaced. This process, is always endothermic, sometimes to such an extent that only a small fraction, of the ions be able to contact the surface closely enough to undergo electron, transfer, and the reaction will be slow. The actual electron-transfer occurs by, quantum-mechanical tunnelling., , Gas Electrodes, Some electrode reactions involve a gaseous species such as H2, O2, or, Cl2. Such reactions must also be carried out on the surface of an, electrochemically inert conductor such as platinum. A typical reaction of, considerable commercial importance is:, Cl- (aq) → ½ Cl2 (g) + eSimilar reactions involving the oxidation of Br2 or I2 also take place at, platinum surfaces., , Practice Personal Hygiene protocols at all times., 265
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Insoluble-salt Electrodes, A typical electrode of this kind consists of a silver wire covered with a, thin coating of silver chloride, which is insoluble in water. The electrode reaction, consists in the oxidation and reduction of the silver:, AgCl(s) + e- → Ag (s) + Cl- (aq), The half cell would be represented as:, . . . II Cl- (aq) I AgCl(s) I Ag(s), Although the usefulness of such an electrode may not be immediately, apparent, this kind of electrode finds very wide appliction in electrochemical, measurements., , Reference Electrodes, In most electrochemical experiments our interest is concentrated on only, one of the electrode reactions. Since all measurements must be on a complete, cell involving two electrode systems, it is common practice to employ a, reference electrode as the other half of the cell. The major requirements of a, reference electrode are that it be easy to prepare and maintain, and that its, potential be stable. The last requirement essentially means that the, concentration of any ionic species involved in the electrode reaction must be, held at a fixed value. The most common way of accomplishing this is to use an, electrode reaction involving a saturated solution of an insoluble salt of the ion., One such system, the silver-silver chloride electrode has already been, mentioned:, Ag I AgCl(s) I Cl-(aq) II . . ., Ag(s) + Cl(aq) → AgCl(s) + eThis electrode usually takes the form of a piece of silver wire coated with, AgCl. The coating is done by making the silver the anode in an electrolytic cell, containing HCl; the Ag+ ions combine with Cl- ions as fast as they are formed, at the silver surface., , Practice Personal Hygiene protocols at all times., 266
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Figure 2, Source: Silverio,Angelina.”Exploring Life Through Science Series: General Chemistry 2.” In, Teachers Wraparound Edition. Quezon City, Phoenix Pulishing House, Inc., 2017, , The other common reference electrode is the calomel electrode, calomel, is the common name for mercury(I) chloride. Such a half cell would be, represented as, Hg I Hg2+(aq) I KCl ll . . ., and the half cell reaction would be:, Hg(l) + Cl- → ½ HgCl2(s) + eThe potentials of both of these electrodes have been very accurately, determined against the hydrogen electrode. The latter is seldom used in routine, electrochemical measurements because it is more difficult to prepare; the, platinum surface has to be specially treated by preliminary electrolysis. Also,, there is need for a supply of hydrogen gas which makes it somewhat, cumbersome and hazardous., , Learning Competency:, Explain the electrode reactions during electrolysis (STEM_GC11ABIVf-g-182), , Practice Personal Hygiene protocols at all times., 267
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Activity 3: Fact or Bluff, Directions: Write FACT if the statement is true, otherwise write BLUFF if it is, false. Write your answer on the space provided., ___________1. An electrode reaction refers to the reduction process that takes, place at an electrode., ___________2. Metal and its hydrated cation as electrodes are called metalmetal ion electrodes., ___________3. The actual electron transfer occurs by quantum mechanical, tunnelling., ___________4. H2, O2 and Cl2 are examples of non-gaseous species., ___________5. Caronel is the common name for mercury (I) chloride., , Activity 4: Time To Think!, Directions: Answer the following questions logically. Write your answer on the, space provided., 1. Why platinum is commonly used in electrode reactions involving ionic only, species?, Explain._____________________________________________________, ___________________________________________________________, ___________________________________________________________, ___________________________________________________________, ____________________, 2. What are the major requirements of a reference electrode? Enumerate., ___________________________________________________________, ___________________________________________________________, ___________________________________________________________, ___________________________________________________________, ____________________, , Practice Personal Hygiene protocols at all times., 269
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ANSWER KEY, ACTIVITY 1:, Possible answers:, 1. Ion-ion electrode - The electroactive ion diffuses to the electrode surface, and adsorbs (attaches) to it by van der waals and coulombic forces. This, process is always endothermic. The actual electron-transfer occurs by, quantum-mechanical tunnelling., 2. Gas electrode - electrode reactions that involves a gaseous species such, as H2, O2, or Cl2. It must also be carried out on the surface of an, electrochemically inert conductor such as platinum., 3. Insoluble-salt Electrodes- The electrode reaction consists in the oxidation, and reduction of the silver., 4. Reference Electrodes – commonly uses an electrode reaction involving a, saturated solution of an insoluble salt of the ion., , ACTIVITY 2:, 1. Gas electrode, 2. Ion-ion electrode, 3. Insoluble-salt electrode, 4. Reference electrode, 5. Ion-ion electrode, , ACTIVITY 3:, 1. BLUFF, 2. FACT, 3. FACT, 4. BLUFF, 5. BLUFF, , Practice Personal Hygiene protocols at all times., 272
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ACTIVITY 4:, Possible answers:, 1. In order to avoid complications that would arise from electrode reactions, involving a metal, a relatively inert substance such as platinum is commonly, used., 2. The major requirements of a reference electrode are that it be easy to, prepare and maintain, and that its potential be stable. The last requirement, essentially means that the concentration of any ionic species involved in the, electrode reaction must be held at a fixed value., , Prepared by:, , KIMBERLY PAGDANGANAN, Licerio Antiporda Sr. National High School- Dalaya Annex, , Practice Personal Hygiene protocols at all times., 273
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GENERAL CHEMISTRY 2, Name: ____________________________, , Grade Level: _________, , Date: _____________________________, , Score: ______________, , LEARNING ACTIVITY SHEET, Reactions in Some Commercial Electrolytic Processes, Background Information for the Learners (BIL), Electrolysis has various applications in our day-to-day living. Below are, some of the commercial applications of it, including how these electrolytic, processes react:, Electrorefining of copper, Electrolysis can be used for purposes other than preparing elements., One example is the refining of copper. Very pure copper is often required in, the manufacture of electrical equipment. (A purity of 99.999 percent is not, unusual.) The easiest way to produce a product of this purity is with electrolysis., An electrolytic cell for refining copper contains very pure copper at the, cathode, impure copper at the anode, and copper sulfate as the electrolyte., When the anode and cathode are connected to a battery, electrons flow into, the cathode, where they combine with copper ions (Cu2+) in the electrolyte:, Cu2+ + 2e- → Cu0, Pure copper metal (Cu0 in the above equation) is formed on the cathode., At the anode, copper atoms (Cu0) lose electrons and become copper ions, (Cu2+) in the electrolyte:, Cu0 – 2e- → Cu 2+, Overall, the only change that occurs in the cell is that copper atoms from, the impure anode become copper ions in the electrolyte . Those copper ions, are then plated out on the cathode. Any impurities in the anode are just left, behind, and nearly 100 percent pure copper builds up on the cathode., Practice Personal Hygiene protocols at all times., 274
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Electroplating, Another important use of electrolytic cells is in the electroplating silver,, gold, chromium, and nickel. Electroplating produces a very thin coating of these, expensive metals on the surfaces of cheaper metals, giving them the, appearance and the chemical resistance of the expensive ones., In silver plating, the object to be plated (a spoon, for example) is used, as the cathode. A bar of silver metal is used as the anode. And the electrolyte, is a solution of silver cyanide (AgCN). When this arrangement is connected to, a battery, electrons flow into the cathode where they combine with silver ions, (Ag+) from the electrolyte to form silver atoms (Ag0):, Ag+ + 1e- → Ag0, These silver atoms plate out as a thin coating on the cathode – in this, case, the spoon. At the anode, silver atoms give up electrons and become silver, ions in the electrolyte:, Ag0 – 1e- → Ag0, Silver is cycled, therefore, from the anode to the ectrolyte to the cathode,, where it is plated out., Aluminum Production, In 1886, Charles Hall and Paul L.T. Heroult discovered that molten, cryolite (a sodium aluminum fluoride mineral), , could be used to dissolve, , alumina and that the resulting chemical reaction would produce metallic, aluminum. They found that by passing an electrical current through the cryolite, /alumina mixture they could bring about the chemical reaction that converts the, alumina to metallic aluminum. The Hall-Heroult process, as it is known, remains, in use today., The cathode for the electrolytic process is a carbon lining made up of, pre-formed carbon-cathode blocks. Other carbon materials line the sides of the, pot and refractory materials generally sit between the carbon cathode and the, steel potshell. Over the typical three to ten year life of a pot, materials such as, alumina, molten aluminum, calcium, fluorides and sodium infiltrate the cathode, Practice Personal Hygiene protocols at all times., 275
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lining and cause it deteriorate. After a defined period, or when indicators, determine that the lining is at risk, the lining material is replaced and is removed, from the pot is called Spent Pontliner or SPL. Regain recovers and processes, the Spent Potliner., The anode for the electrolytic process is a large carbon block held by a, metal frame and suspended in the molten cryolite/alumina mixture to conduct, the current. Most of the anode is consumed in the production process. At the, end of its service, the remaining carbon anode “butt” is removed and cleaned, of residues and a large proportion of the butt is recycled to make new carbon, anode blocks. Some of the anode carbon and cleaning residues are however, not re-usable. Regain recovers and processes this otherwise unusable anode, carbon., , Learning Competency, Describe the reactions in some commercial electrolytic processes, (STEM_GC11ABIVf-g-183), , Activity 1: Copper Refining, Directions: Read carefully the statement below then answer the questions that, follow. Write your answers on the space provided., , 1. Blister copper is an impure sample, of copper containing small amounts, of zinc and gold. Blister copper is, purified using electrolysis., a) What does the cathode contains?, _____________________________________, b) What does the anode contains?, _____________________________________, , Practice Personal Hygiene protocols at all times., 276
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c) What will happen when the anode and, cathode are connected to a power supply?, ________________________________________________________, ________________________________________________________, ____________, d) What will happen to the impurities in the anode? Explain., ________________________________________________________, ________________________________________________________, ____________, , Activity 2: Let’s Electroplates Things!, , Directions: Consider the following diagram and answer logically the question, that follow. Write your answer on the space provided., , 1. A copper spoon is plated with silver in an, electrolytic cell., a) What is the anode?, ______________________________________________, b) What is the cathode?, ______________________________________, c) Write the equation for the half-reaction, that occurs in the copper spoon., ________________________________________________________, ________________________________________________________, ____________, , Practice Personal Hygiene protocols at all times., 277
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d) Write the equation for the half-reaction, that occurs in the silver metal bar., ________________________________________________________, ________________________________________________________, ____________, , 2. A trophy manufacturer electroplates an iron trophy with gold., a) What is the anode?, ________________________________________________________, __________________________________________________, b) What is the cathode?, ________________________________________________________, ________________________________________________, , Activity 3: What a Wonderful Ring!, Directions: Draw an electrolytic cell that could be used to plate an iron ring, with silver. Be sure to include all of the necessary parts. In addition, label the, anode, the cathode, and the solution used., , Practice Personal Hygiene protocols at all times., 278
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Activity 4: What to Use for My Aluminum?, Directions: Below is a diagram of a carbon-lined container that is used to, convert alumina into metal during electrolysis process. Choose the word from, the box that correctly label its part., Steel Potshell Carbon lining, Molten Bath, , Solid Bath Surface, , Refractory lining, , Molten Aluminum, , Activity 4: Fact or Bluff?, Directions: Write FACT if the statement is true. Otherwise, write BLUFF if it is, false. Write your answer on the space provided., _______________1. Electroplating uses an electrolytic cell in which the object, to be plated is immeresed in a solution of the metal to be deposited., _______________2. Electrorefining is only applicable in coppers., _______________3. Copper atoms from pure anode become copper ions in, the electrolyte., _______________4. Hall-Heroult process involves electrolytic reduction of, alumina in cells or pots., _______________5. The cathode for the electrolytic process is a refractory, lining., , Practice Personal Hygiene protocols at all times., 279
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ANSWER KEY, ACTIVITY 1:, 1. a. The pure copper, b. The blister/impure copper, c. When the anode and cathode are connected to a battery, electrons flow, into the, cathode, where they combine with copper ions (Cu2+) in the electrolyte., d. The impurities in the anode are just left behind, and nearly 100 percent, pure, copper builds up on the cathode, , ACTIVITY 2:, 1. a. The silver metal bar, b. The copper spoon, c. Ag+ + 1e- → Ag0, d. Ag0 – 1e- → Ag0, 2. a. The gold bar, b. The iron trophy, , ACTIVITY 3:, , Practice Personal Hygiene protocols at all times., 282
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ACTIVITY 4:, , ACTIVITY 5:, 1. FACT, 2. BLUFF, 3. BLUFF, 4. FACT, 5. BLUFF, , Prepared by:, , KIMBERLY PAGDANGANAN, Licerio Antiporda Sr National High School- Dalaya Annex, , Practice Personal Hygiene protocols at all times., 283
