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CROWNTURN COLLEGE, LAGOS ANNEX: 5, CROWNTURN CLOSE, IJANIKIN, OJO, LAGOS., WEBSITE: www.crownturn.org E-, CHEMISTRY, E – LEARNING NOTE, 2019/2020 ACADEMIC SESSION, FIRST TERM, CLASS: SS2, NAME:…………………………..……………………………………………….……………………………………………, DATE: ………………………………………………….., NAME:…………………………..………………………………………………………………….., CROWNTURN COLLEGE, WEBSITE: www.crownturn.org E-, SSS2 CHEMISTRY FIRST TERM SCHEME OF WORK, 1. Revision/ periodic law: periodic classification into blocks and families, 2. The Periodic Trends: Down the group and across the period, Atomic size, Ionic size, Ionization energy, Electron affinity, electro negativity, Diagonal relationship in the properties of elements in the periodic table., 3. Oxidation and reduction (Redox Reaction): Definition of terms, Relationship between oxidation number and IUPAC naming, 4. Identification of oxidizing and reducing agents, Balancing of Redox equation in acidic and alkaline media, 5. Ionic Theory: Distinguish between: electrovalent and covalent compounds, Electrolytes and non-electrolytes, weak and strong electrolytes, conductors and non-conductors., 6. Electrolysis: meaning of electrolysis, terminologies/electrolysis and electrolytes, factors affecting the preferential discharge of ions, 7. Electrolysis of specified electrolytes: acidified water (H2SO4), Copper (II) tetraoxosulphate (VI) solution, Brine (NaCl) solution, Project work: construction of electrolytic cell and electrochemical cells and uses of electrolysis, 8. Electrochemical cells: standard electrode potential, drawing and writing of cell diagrams – metal/metal ion systems, e.m.f of cells – electrochemical cells as a combination of two half – cell, calculation of e.m.f of cell., 9. Application of electrochemical cells: Distinction between primary and secondary cells, Types of electrochemical cells., 10. Faraday’s laws of electrolysis – calculation based on the laws., 11. Revision, 12 & 13. Examination, WEEK 1, THE PERIODIC TABLE, Periodic table is the arrangement of chemical elements in order of their atomic numbers. Atomic number, Z, of an element is the number of protons in one atom of that element. It can also arrangement of elements in rows and columns in a table., Any systematic study must ba based on a good system of classification. In chemistry, classification of elements began with Anthony Lavoisier, whose classification was essentially into metals and non-metals. Many scientists have attempted to classify chemical elements based on their properties. They include Newland, Lothar Meyer, Dobereiner and Dmitri Mendeleev. Mandeleev periodic law state that the “properties of an elements are in periodic function of their relative atomic masses”. This law fulfilled two important functions., 1. It summarised the properties of elements and classified them into groups with similar properties., 2. It enable predictions to be made about the properties of known and unknown elements and led to considerable research activity., The use of relative atomic mass was somehow confusing to chemists, because relative atomic mass is not in whole numbers., The modern periodic table is based on Mendeleev’s original idea in 1869. The basic assumption behind the modern periodic table known as Modern Periodic Law which states that “the properties of the elements are a periodic function of their atomic number”, THE MODERN PERIODIC TABLE, The modern periodic table is divided into vertical columns of elements called Groups and horizontal rows of elements called Periods. There are seven (7) periods and eight (8) groups. The first twenty (20) elements are:, 1. Hydrogen 6. Carbon 11. Sodium 16. Sulphur, 2. Helium 7. Nitrogen 12. Magnesium 17. Chlorine, 3. Lithium 8. Oxygen 13. Aluminium 18. Argon, 4. Beryllium 9. Fluorine 14. Silicon 19. Potassium, 5. Boron 10. Neon 15. Phosphorus 20. Calcium, Periodic table of elements, PERIOD OF ELEMENTS, The horizontal or row arrangement of elements is called period, and are numbered from 1 – 7. All elements in the same period have the same number of shell., PERIOD 1: have Two (2) elements, PERIOD 2: have Eight (8) elements, PERIOD 3: have Eight (8) elements, PERIOD 4: have Eighteen (18) elements, PERIOD 5: have Eighteen (18) elements, PERIOD 6: have Thirty – Two (32) elements, PERIOD 7: have Thirty – Two (32) elements, GROUP OF ELEMENTS, The vertical or column arrangement of elements is called group, and are numbered from 1 – 7 and 0. There are Eight(8) groups in the periodic table. All elements in the same group have the same number of valent electrons, or same number of electron at the outer most shell., GROUP 1: Alkali metals, GROUP 2: Alkaline - earth metals, GROUP 3: Boron Family, GROUP 4: Carbon Family, GROUP 5: Nitrogen Family, GROUP 6: Oxygen Family, GROUP 7: Halogen Family, GROUP 0: Noble or Inert gases, GROUP 1:, 1. They are univalent elements because they have one valance electron., 2. They ionized readily by donating one electron to form positive ions., 3. They are good reducing agent, their reducing properties increases down the group and decreases across the period., 4. They form electrovalent compounds, 5. They are very good conductors of electricity and are very electro-positive, 6. They react vigorously with cold water, 7. They are known as alkali metal, e.g lithium, sodium, potassium, rubidium, caesium and francium., GROUP 2:, 1. They are divalent elements because they have two valance electrons., 2. They ionized readily by donating two electrons to form positive ions., 3. They are good reducing agent, their reducing properties increases down the group and decreases across the period., 4. They form electrovalent compounds, 5. They are very good conductors of electricity and are very electro-positive, 6. Some react vigorously with cold water while others with steam., 7. They are known as alkaline-earth metal, e.g beryllium, magnesium, calcium, strontium, barium and radium., GROUP 3:, 1. They are trivalent elements because they have three valance electron., 2. They ionized readily by donating three electron to form positive ions., 3. They are good reducing agent, their reducing properties increases down the group and decreases across the period., 4. They form electrovalent compounds, 5. They are very good conductors of electricity and are very electro-positive, 6. Aluminium react with steam, 7. They are known as boron family, e.g Boron, aluminium, gallium, indium, and thallum., GROUP 4:, 1. They are tetravalent elements because they have four valance electron., 2. They exhibit two oxidation state, +2 and +4. The stability of the +2 state compounds increase and that of the +4 state compounds decrease on moving down the group., 3. The group trend in changing from non-metal to metal on moving down the group., 4. They form covalent compounds, 5. They are non conductors of electricity., 6. Carbon does not react with water in any form, but silicon and tin react with steam at red heat., 7. They are known as carbon family, e.g carbon, silicon, germanium, tin, and lead., GROUP 5:, 1. They are non -metals, 2. They show two valences of 3 and 5., 3. They are oxidizing agent., 4. They ionized readily by accepting three electrons to form negative ions, 5. They are bad conductors of electricity and are electro-negative, 6. They electrons acceptor, 7. They are known as nitrogen family, e.g Nitrogen, phosphorus, arsenic, antimony, and bismuth., GROUP 6:, 1. They are non -metals, 2. They ionized readily by accepting two electrons to form negative ions., 3. They are oxidizing agent., 4. They do not attack water in any form. Oxygen combine directly with hydrogen to form water., 5. They are bad conductors of electricity and are electro-negative, 6. They electrons acceptor, 7. They are known as oxygen family, e.g Oxygen, sulphur, selenium, tellurium and polonium., GROUP 7:, 1. They exist as diatomic molecule, e.g F2, Cl2 Br2 , I2., 2. They ionized to form univalent negative ion., 3. They are good oxidizing agent., 4. They are coloured, 5. They are most reactive non-metals and are very electro-negative, 6. They are electron acceptor., 7. They are known as halogen family, and exist mainly as salt e.g Fluorine, chlorine, bromine, iodine, and astatine., GROUP 0:, 1. They form few compounds and exist freely as monoatomic molecules in the atmosphere, e.g He, Ne, Ar e.t.c., 2. They have no bonding electrons in the outermost shell., 3. They are non-reactive elements, because their have complete filled valence electron., 4. They show similar properties among themselves., 5. They bear no resemblance to the halogens that come before them and the alkali metals that come after them., 6. They exhibit a gradation in their physical properties which is related to the progressive increase in their atomic numbers., 7. Their melting and boiling points increase while their ionization energy decrease down the group from helium to radon., 8. They are known as noble or inert gases e.g helium, neon, argon, krypton, xenon, and radon., THE REACTIVE METALS, The elements in Group I and Group II form a block of reactive metals .e.g Na, Mg, K, Ca. They are sometimes referred to as the “S – block elements since the outermost electrons in these metals are in S – subshells. These metals are all high in the electrochemical series and have lower densities, melting and boiling point than most other metals. They form stable, involatile ionic compounds, THE TRANSITION METALS, These elements are located between GROUP II and GROUP III in the periodic table which show similar behaviour. Transition elements are all metals with typical metallic properties such as high tensile strength and high melting points. They are sometime called the “d – block” elements since electrons are added to d – subshells across this block of elements. Elements in this group include Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn. They are much less reactive than the metals in group I and II, hence they have properties different from that of the reactive metals., GENERAL CHARACTERISTICS OF FIRST TRANSITION METALS, 1. They have variable oxidation state, 2. They form coloured ions (Compounds), 3. They form complex ions, 4. They are Para magnetism., 5. They are use as catalyst., THE LANTHANIDES AND ACTINIDES, These elements are found at the bottom of the periodic table, there are metals with typical metallic properties. The lanthanides and actinides form a block of elements with the transition metals. They are sometimes called the inner transition elements or “f – block” elements, since electrons are being added to f – subshells across this block of elements., Lanthanide is the sixth period series and actinide is the seventh period series., THE NON - METALS, These elements form a triangular block in the periodic table which are located between Group III and Group VII. They are sometimes called the “p – block” elements since the outermost electrons in these elements are going into p – subshells. Non-metals include N, O, Si, P and the halogens which are Group VII elements. These include F, Cl, Br, I, At., THE NOBLE GASES, The elements in Group 0 are known as the inert or noble gases. They have completely filled shells and have outer S – and P – subshells and have outer S – and P – subshells of electrons. They are very unreactive. Elements in this group include He, Ne, Ar, Kr, Xe, Rn., METALLOIDS (SEMI – METALS), Some elements are very difficult to classify as either metals or non-metals, such elements are called METALLOIDS OR SEMI – METALS. They include B, C, Si, Ge, As, Te and are found as follows in the periodic table., Group III – B, Group IV – C, Si, Ge, Group V – As, Group VI – Te, They are poor conductors of electricity., WEEK 2, ATOMIC RADII: Atomic radius is defined as half the distance between the nuclei when the bond linking the atoms is a single covalent bond., In other words it is the distance of closest approach to another atom in a given bonding situation. The value of atomic radii decreases from left to right across a period. This is because as the nuclear charge increase, the electrons enter the same shell at about the same distance from the nucleus., In moving down a group of the periodic table, the atomic radii increase while decreases from left to right across the period., IONIC RADII: like the atomic radii are determined by X – ray analysis of ionic crystal. However, this depends on the distance between the centres of the adjacent ions. Hence, it is the distance between the centre of one ion and the centre of its nearest neighbour of opposite charge., Ionic radii deals with electrovalent compound and its varies periodically like the atomic radii, i.e it decreases across the period and increase down the group., IONIZATION ENERGY: This is the minimum amount of energy required to remove a valence electron from an atom of the element to form an ion. Generally, there is a tendency of ionization energy to increase across a period and decrease down a group. The ionization energy of an atom is affected by (i) the distance of the outermost electron from the nucleus. (ii) the size of the positive nuclear charge. (iii) the screening effect of the inner electrons. Moving from left to right across the period, there is a general rise in the first ionization energy. This is due to the fact that the nuclear charge is increasing across the period., ELECTRON AFFINITY: Electron affinity is the energy change that occurs when an isolated gaseous atom acquires an electron and so becomes a negatively charge ion. e.g. X + e— X—, If the electron affinity is negative, a lot of energy is released when the ion is formed. Generally, electron affinities decrease (i.e increases in negative value) from left to right across a period (except noble gases which have positive values) and increase (i.e decreases in negative value) from top to bottom within a group. This is because, as the nuclear charge increase (i.e smaller atomic radii) the extra electrons enter the same electron shell and so are attracted more strongly., ELECTRONEGATIVITY: This is the measure of ability of an atom in a molecule to attract electrons to itself. Electro negativity of elements increase across a period but decrease down a group. This is because as we move from one element to the next across a period, the nuclear charge increases and the atomic radius decreases., Electro negativity can be used to predict the type of bonding a compound will have. When two elements of widely different electro negativity combine, an ionic compound is formed. Covalent bonding occurs between non-metal when the electro negativity differences are not very large., DIAGONAL RELATIONSHIP IN THE PROPERTIES OF ELEMENTS IN THE PERIODIC TABLE: In the periodic table, particularly for the elements lithium and magnesium, beryllium and aluminium, boron and silicon, there is diagonal resemblance. This is to be expected, since in moving from left to right across the periodic table, the metallic properties gradually decrease, while on descending a group, they increase. Hence, elements diagonally below on another have similar properties., Li Be B C, Na Mg Al Si, The ions of lithium and magnesium have the following characteristics:, 1. They are extensively hydrated due to their small size., 2. Their trioxocarbonates (IV) and trioxonitrate (V) undergo ready decomposition on heating., 3. Their tetraoxophospate (V) and trioxocarbonates (IV) are insoluble in water., 4. Both metals combine with nitrogen at high temperature., WEEK 3, OXIDATION AND REDUCTION (REDOX REACTIONS): In term of electron transfer, oxidation is defined as the process of electron loss to give a higher oxidation state (more positive), and reduction is defined as the process of electron gain to give a lower oxidation state (more negative)., A redox reaction is one in which a transfer of electrons take place between the reactants. This requires as the name “REDOX” implies. The simultaneous involvement of a reducing agent and an oxidising agent. The reducing agent is the specie which is able to donate electrons while the oxidising agent is the specie which is able to accept electrons., Oxidation is a process of electron loss or a process in which the oxidation state of some elements increases while reduction is a process of electron gain or a process in which the oxidation state of some elements decreases., Fe2+ Fe3+ + e— (Oxidation), Sn4+ + 2e— Sn2+ (Reduction), DETERMINATION OF THE OXIDATION STATES OF ELEMENTS IN COMPOUND, RULE 1: The oxidation number of an element in the free state is zero. e.g Mg = 0, Fe = 0, K = 0, e.t.c, RULE 2: The algebraic sum of the oxidation numbers of all the atoms in a molecule is zero, e.g MgSO4 = 0, H2SO4 = 0, etc, RULE3: in Ion, the algebraic sum of the oxidation numbers equal to charge on the ion, e.g. Fe2+ = +2, Fe3+ = +3, MnO— = --1, RULE 4: In any substance, the more electronegative atom has the negative oxidation number, e.g O2—, F—, Cl— and so on., While the less electronegative atom has the positive oxidation number, e.g Al3+, Zn2+, K+ and so on., RULE 5: The oxidation number of hydrogen in all compounds, except in metal hydrides is +1., RULE 6: The oxidation number of oxygen in all compounds, except in peroxides and in OF2 is –2, RELATIONSHIP BETWEEN OXIDATION NUMBER AND IUPAC NAMING, 1. Find the oxidation number of Mn in KMnO4, SOLUTION, Applying Rule 2, KMnO4 = O, K+ + Mn + O42— = O, 1 + Mn + 4(-2) = O, Mn = 8 – 1, Mn = + 7, IUPAC NAMING: Potassium tetraoxomanganate(VIII), 2. Find the oxidation number of Mn in MnO4—, SOLUTION, Applying Rule 3, MnO4 = -1, Mn + O42— = -1, Mn + 4(-2) = -1, Mn - 8 = – 1, Mn = -1 + 8, Mn = + 7, IUPAC NAMING: Tetraoxomanganate(VIII), 3. Find the oxidation number of Chromium (Cr) in K2Cr2O4, SOLUTION, Applying Rule 2, K2Cr2O4 = O, 2K+ + Cr2 + 4O2— = O, 2 + 2Cr + 4(-2) = O, 2Cr = +8 – 2, 2Cr = + 6, Cr = +3, IUPAC NAMING: Potassium tetraoxodichromate(III), 4. Calculate the Oxidation Number of copper in Cu2O, SOLUTION, Let y represents the Oxidation Number of each copper atom., Oxidation Number of one hydrogen atom, O is -2., In a neutral compound, sum of all Oxidation Number is zero., Therefore, in Cu2O: Cu2O, 2y + (-2) = 0, 2y = +2, y = +2/2 = +1, Thus, Oxidation Number of a copper atom in Cu2O = +1, IUPAC NAMING: Copper(I)Oxide, Determine the Oxidation Number of X in X2O72-, SOLUTION, Oxidation Number of each O atom = -2, The net charge on the ion = -2, Therefore, in X2O72-, 2X + 7(-2) = -2, 2X – 14 = -2, 2X = -2 + 14 = + 12, X = + 12/2 = +6, Therefore, Oxidation Number of each X in X2O72- is +6, IUPAC NAMING: heptaoxodiX (VI), WEEK 4, OXIDATION AND REDUCTION (REDOX REACTIONS): In term of electron transfer, oxidation is defined as the process of electron loss to give a higher oxidation state (more positive), and reduction is defined as the process of electron gain to give a lower oxidation state (more negative)., A reduction – oxidation reaction (commonly called a redox reaction) is one that occurs between a reducing and oxidizing agent. In a redox reaction, reduction and oxidation processes occur simultaneously. The oxidising agent is reduced while the reducing agent is oxidised., OX1 + Red2 Red1 + OX2, OX1 is reduced to Red1 and Red2 is oxidized to OX2. OX1 is the oxidising agent, and Red2 is the reducing agent. The reducing or oxidizing tendency of a substance that has the tendency to take on an electron or electrons and be reduced to a lower oxidation state, e.g., Fe3+ + e— Fe2+, Common oxidizing reagent are:, 1. Potassium tetraoxomanganate (VII), KMnO4, 2. Potassium heptaoxodichromate (VI) K2Cr2O7, 3. Oxygen, O2, 4. Ozone, O3, 5. Hydrogen perioxide, H2O2, 6. Tetraxoxmanganate (VII), MnO4—, 7. Lead(IV)oxide, PbO2, 8. Manganate(IV) oxide, MnO2, 9. Chlorine, Cl—, 10. Heptaoxodichromate (VI), Cr2O72—, 11. Trioxonitate(V), NO3—, 12. Iron(III), Fe3+, A reducing agent is the substance that has the substance that has the tendency to give up an electron or electrons and be oxidised. e.g. Fe2+ Fe3+ + e—, Common reducing reagents are:, 1. Iron(II), Fe2+, 2. Sodium thiosulphate, Na2S2O3., 3. Hydrogen, H2, 4. Carbon, C, 5. Carbon(II)oxide, CO, 6. Hydrogen sulphide, H2S, 7. Sulphur (IV) oxide, SO2, 8. Ammonia, NH3, 9. Tin, Sn2+, 10. Potassium iodide, KI, Metallic elements being electropositive tend to lose their electrons more easily than non-metallic elements which are electro negative. The redox potential of a system is a measure of its oxidizing and reducing capacity. Thus, powerful oxidizing agent has a large and positive potential., 1. Addition of oxygen to substance is referred to as oxidation while removal of oxygen from substance is reduction., 2. Addition of hydrogen to substance is referred to as oxidation while removal of hydrogen from substance is reduction., 3. Addition of electronegative elements and removal of electropositive elements, while reduction include the removal of electronegative elements and addition of electropositive elements., 4. Transfer of electron, TEST FOR OXIDISING AGENTS, 1. Pass hydrogen sulphide gas, H2S, through a solution of the unknown. A yellow sulphur deposited if formed., S2—(aq) S(s) + 2e—, 2. To a little of the unknown add a little dilute tetraoxosulphate(VI) acid and potassium iodide solution. A brown colouration is formed, 2I— I2 + 2e—, Brown, The brown colouration is due to the formation of iodine. If a drop of starch solution is then added, a blue black colouration forms., 3. To a solution of the unknown add iron(II) chloride solution. A brown solution is formed, Fe2+ Fe3+ + e—, Green brown, TEST FOR REDUCING AGENTS, To a little of the unknown, add a little tetraoxosulphate(VI) acid, then either potassium tetraoxomanganate(VII) solution or potassium heptaoxodichromate(VI) solution drop by drop. Warm if necessary., 1. If potassium tetraoxomanganate(VII) solution is used, the purple colour disappears., MnO4— + 8H+ +5e— Mn2+ + 4H2O, (Purple) (Colourless), 2. If potassium heptaoxodichromate(VI) solution is used, the orange colour changes to green., Cr2O72— + 14H+ +6e— 2Cr3+ + 7H2O, (Orange) (Green), 3. Add iron(III) chloride solution to a little of the unknown. The brown colour change to pale green., Fe3+ + e— Fe2+, (Brown) (Green), BALANCING OF REDOX EQUATIONS, Redox equations can be balanced by two method: the oxidation state method and the ion-electron method. The one of interest to us is the ion – electron method. This method involves writing two half – equation. One for the oxidation, the other for reduction., In balance redox equation, the following rules should be followed., 1. Breaking down the equation into an oxidation half – equation and a reduction half – equations., 2. The two half – equations are then balanced separately., 3. combine the two balanced half – equations in such a way that overall equation does not show any change in a number of electrons., EXAMPLE 1, Balance the redox reaction between iron(II) tetraoxosulphate(VI) and potassium tetraoxomanganate(VII) in acid solution may be represented by the ionic equation., SOLUTION, Fe2+ + MnO4— Fe3+ + Mn2+ (in acid solution), Oxidation half equation is Fe2+ Fe3+, Reduction half equation is MnO4— Mn2+, In acid solution, H+ ions and H2O molecules may be added while in alkaline solution, OH— ions and H2O molecule may be added., MnO4— + 8H+ Mn2+ + 4H2O -, Fe2+ Fe3+ + e— ………..(i) oxidation half equation is balanced thus, MnO4— + 8H+ + 5e— Mn2+ + 4H2O ……….(ii) - Reduction half equation is balanced thus, Multiply the equation (i) by the numbers of electrons in equation(ii) and vice versa., 5Fe2+ Fe3+ + 5e— …….. (iii), MnO4— + 8H+ + 5e— Mn2+ + 4H2O ……..(iv), Add up equation (iii) and (iv), 5Fe2+ + MnO4— + 8H+ + 5e— 5Fe3+ + 5e— + Mn2+ + 4H2O, 5Fe2+ + MnO4— + 8H+ 5Fe3+ + Mn2+ + 4H2O (Balanced), EXAMPLE 2, Balance the redox equation: Fe3+(aq) + Cl—(aq) Fe + 1/2Cl2, SOLUTION, Oxidation half equation is 2Cl— Cl—, Reduction half equation is Fe3+ Fe, 2Cl— Cl2 + 2e— ………..(i) oxidation half equation is balanced thus, Fe3+ + 3e— Fe ……….(ii) - Reduction half equation is balanced thus, Multiply the equation (i) by the numbers of electrons in equation(ii) and vice versa., 6Cl— 3Cl2 + 6e— …….. (iii), 2Fe3+ + 6e— 2Fe ……..(iv), Add up equation (iii) and (iv), 2Fe3+ + 6Cl— + 6e— 2Fe + 6e— + 3Cl2, 2Fe3+ + 6Cl— 2Fe + 3Cl2 (Balanced), EXAMPLE 3, Balance the redox equation: MnO4—(aq) + Cl—(aq) + H+(aq) Mn2+(aq) + Cl2(aq) + H2O, SOLUTION, Oxidation half equation is 2Cl— Cl2, Reduction half equation is MnO4— Mn2+, MnO4— + 8H+ Mn2+ + 4H2O -, 2Cl— Cl2 + 2e— ………..(i) oxidation half equation is balanced thus, MnO4— + 8H+ + 5e— Mn2+ + 4H2O ……….(ii) - Reduction half equation is balanced thus, Multiply the equation (i) by 5 and equation(ii) by 2., 10Cl— 5Cl2 + 10e— …….. (iii), 2MnO4— + 16H+ + 10e— 2Mn2+ + 8H2O ……..(iv), Add up equation (iii) and (iv), 10Cl— + 2MnO4— + 16H+ + 10e— 5Cl2 + 10e— + 2Mn2+ + 8H2O, 10Cl— + 2MnO4— + 16H+ 5Cl2 + 2Mn2+ + 8H2O (Balanced), EXAMPLE 4, Balance the redox equation: 2I— + MnO4— + H+ I2 + MnO2 + 2H2O, SOLUTION, Oxidation half equation is 2I— I2, Reduction half equation is MnO4— MnO2, In acid solution, H+ ions and H2O molecules may be added while in alkaline solution, OH— ions and H2O molecule may be added., MnO4— + 4H+ MnO2 + 2H2O, MnO4— + 4H+ + 3e— MnO2 + 2H2O, 2I— I2 + 2e— ………..(i) oxidation half equation is balanced thus, MnO4— + 4H + 3e— MnO2 + 2H2O ……….(ii) - Reduction half equation is balanced thus, Multiply the equation (i) by the numbers of electrons in equation(ii) and vice versa., 6I— 3I2 + 6e— …….. (iii), 2MnO4— + 4H+ + 6e— 2MnO2 + 2H2O ……..(iv), Add up equation (iii) and (iv), 6I— + 2MnO4— + 8H+ + 6e— 3I2 + 6e— + 2MnO2 + 4H2O, 6I— + 2MnO4— + 8H+ 3I2 + 2MnO2 + 4H2O (Balanced), WEEK 5, IONIC THEORY, Ionic theory is of the view that when an electrolyte is melted or dissolved in water, some, if not all the substance breaks up (dissociate) into freely moving charged particles called ions. This process of dissociation into ions is known as ionization. Ionization occurs before an electric current is passed into the electrolyte and the ions move about at random in all direction., DISTINGUISH BETWEEN ELECTROVALENT AND COVALENT COMPOUNDS, ELECTROVALENT BOND: is the transfer of electrons from metallic atoms to non-metallic atoms during a chemical reaction, i.e donor – acceptor principle. Both ions now posses stable outer duplet or octet structure of noble gases, and are held together by strong electrostatic force of attraction which constitute the electrovalent or ionic bond that form electrovalent compound., PROPERTIES OF ELECTROVALENT (IONIC) COMPOUND, 1. Ionic compounds do not contain molecules, instead they consist of aggregates of positive and negative ions arranged in an orderly pattern of three-dimension crystal lattices, 2. They are solids at room temperature and do not vaporise easily, 3. They have high melting and boiling point because of the strong electrovalent bond between the ions., 4. Ionic compounds readily dissolve in water and other polar solvent., 5. They are good conductors of electricity when in molten state or solution, i.e, they are electrolytes. This is because the ions are free to move about when the compound is a liquid state., 6. They do not dissolve in non-polar solvents such as toluene, benzene, ether, trichloromethane etc., COVALENT BOND: This is not transfer of electrons but shared of electrons. A covalent bond consists of a pair of electrons shared by two atoms. The shared electrons are each contributed by the reacting atoms. Covalent bond can also be regarded as resulting from the overlap of electron clouds., PROPERTIES OF COVALENT COMPOUND, 1. Covalent compounds consist of molecules which have a definite shape., 2. They are gases or volatile liquid because their molecules being electrically neutral are not bound by strong attractive forces., 3. They have low melting and boiling points. This is because the weak intermolecular forces (hydrogen bond in liquid and vander waals in crystals) that bond the molecules can easily be broken by a small amount of heat energy., 4. Covalent compounds readily dissolve in non-polar organic solvents such as benzene, toluene and carbon disulphide., 5. Covalent compounds do not conduct electricity, i.e they are non-electrolytes. This is because the molecules do not contain charged particle., DISTINGUISH BETWEEN ELECTROLYTES AND NON-ELECTROLYTES, ELECTROLYTES: This is a compound either in solution or molten form which conducts electric current and is decomposed at the electrodes in the process. e.g Acids, Alkalis and soluble salts., NON-ELECTROLYTES: These are compounds which in solution or molten form do not conduct electricity or become decomposed by it. The non-electrolytes exist only in the form of molecules and are incapable of ionization. Examples includes alcohols, sugar, tetrachloromethane, etc., DISTINGUISH BETWEEN WEAK AND STRONG ELECTOLYTES, WEAK ELECTROLYTES: Weak electrolytes are those which dissociate partially (or slightly) into its ions in dilute solutions. Example include organic acids, aqueous ammonia, etc., CH3COOH(aq) CH3COO—(aq) + H+, STRONG ELECTROLYTES: Strong electrolytes are those which dissociated completely in dilute solutions. Example are salts, minerals acids, caustic alkalis., NaCl(aq) Na+(aq) + Cl—(aq), DISTINGUISH BETWEEN CONDUCTORS AND NON-CONDUCTORS, CONDUCTORS: These are substances which allow electricity to flow through them. Examples are metals, organic solution, salt (Aqueous), acid solution. The best conductor is silver, and copper is next to it., NON-CONDUCTORS: These are substance that does not allow electricity to flow through them. It is also known as Insulators. Examples are wood, paper, air, rubber, plastic., WEEK 6, ELECTROLYSIS, This is the effect of electricity on chemical substance. Electrolysis is the chemical decomposition of compound which takes place when an electric current is passed through either a solution or the molten form of the compound., TERMINOLOGIES USED IN ELECTROLYSIS, 1. ELECTRODES: these are conductors, (positive and negative poles of carbons or metal) through which an electric current enter or leaves the electrolytes., 2. ANODES: Anode is a positive (+ve) electrode at which the electron leaves and current enters the electrolytes., 3. CATHODE: Cathode is a negative (-ve) electrode at which the electron enters and current leaves the electrolytes., 4. IONS: Ion can be referred to as an atom or group of atoms that possesses electric charge. They are charged particles which exist in electrolytes and takes part in electrolysis. There are the immediate product of decomposition of an electrolytes. The ions which go to the anode are called anions, those ions which go to the cathode are called cation., 5. ELECTROLYTIC CELL: This consists of a container of electrolyte with two electrodes connected to a suitable direct current supply., FACTORS AFFECTING THE PREFERENTIAL DISCHARGE OF IONS, 1. Position of ions in the electrochemical series., 2. Concentration of ions, 3. Nature of the electrodes., 1. POSITION OF IONS IN THE ELECTROCHEMICAL SERIES: according to electrochemical series, the more electropositive a metals is, the more it has greater tendency to give up electron(s). In the same manner, the more electronegative an ion will have, the greater tendency to accept electrons., 2. CONCENTRATION: Increase in concentration of a given ion tends to promote its discharge from solution. The influence of concentration, however, is effective only when the two competing ions are closely positioned in the electrochemical series, while it is minimal if the ions are widely separated in their positions., 3. NATURE OF ELECTRODES: Some electrodes do not take part in electrolytic reaction. Hence. They are described as inert electrodes. Platinum and carbon (graphite) usually behave as inert electrode. Some electrode influence the discharge of ions because they have strong affinity for certain ions., WEEK 7, ELECTROLYSIS OF SPECIFIED ELECTROLYTES, 1. ELECTROLYSIS OF DILUTE TETRAOXOSULPHATE (VI) ACID (ACIDIFIED WATER):, The electrolytic cell used is shown above. It is designed for collection of gaseous products at the two electrodes and it is known as Hofman Voltameter. The electrodes used are platinum foil. The ions present in the electrolyte are:, Cations Anions, H2SO4 2H+(g) SO42—(aq), H2O H+(aq) OH—(aq), At the Cathode: H+ ion migrates to the cathode and takes up electrodes to form neutral hydrogen atoms., H+(aq) + e— H(g), The hydrogen atoms then combine in pairs to form diatomic hydrogen gas molecule, H+(g) + H+(g) H2(g)., At the anode: SO42— and OH—ions migrates to the anode. OH—ions being lower in the electrochemical series is preferentially discharged and lose its electrons to the anode to become a neutral - OH group., OH—(aq) OH + e—, The neutral –OH groups combine in pairs to form one molecule of water and one atom of oxygen., OH + OH H2O(l) + O(g), The oxygen atoms then react to form the diatomic oxygen gas molecule., O(g) + O(g) O2(g), Summary of Reactions, Cathodic half reaction – 4H+(aq) + 4e— 2H2(g), Anodic half reaction – 4OH—(aq) 2H2O(i) + O2(g) + 4e—, Over all reaction – 4H+(aq) + 4OH—(aq) 2H2(g) + 2H2O(l) + O2(g), 2. ELECTROLYSIS OF CONCENTRATED SODIUM CHLORIDE SOLUTION (BRINE SOLUTION): The apparatus is the same as for dilute tetraoxosulphate (vi) acids. The cathode may be platinum or carbon but the anode must be carbon to resist attack by chlorine. The ion present are:, Cations Anions, NaCl(aq) Na+(aq) Cl—(aq), H2O H+(aq) OH—(aq), At the Cathode: Na+ and H+ ion migrates to the cathode. Although Na+ ion is higher concentration, H+ ion is preferentially discharged because of its lower position in the electrochemical series. H+ takes up electrodes to form neutral hydrogen atoms., H+(aq) + e— H(g), The hydrogen atoms then combine in pairs to form diatomic hydrogen gas molecule, H+(g) + H+(g) H2(g)., Thus, hydrogen gas is obtained at the cathode., At the anode: Cl—and OH—ions migrate to the anode. Cl—ions is preferentially discharged because it is in higher concentration than OH—ion. The effect of concentration is importance here because Cl—and OH—are close to each other in the series. Cl—(aq) Cl(g) + e--, The chlorine atoms then combine in pairs to give chlorine gas molecules. Cl(g) + Cl(g) Cl2, Chlorine gas is obtained at the anode., Summary of Reactions, Cathodic half reaction – 2H+(aq) + 2e— H2(g), Anodic half reaction – 2Cl—(aq) Cl2(g) + 2e—, Over all reaction – 2H+(aq) + 2Cl—(aq) H2(g) + Cl2(g), 3. ELECTROLYSIS OF COPPER (II) TETRAOXOSULPHATE (VI) SOLUTION USING DIFFERENT ANODES: With carbon or platinum electrode., Cations Anions, CuSO4 Cu2+(g) SO42—(aq), H2O H+(aq) OH—(aq), At the Cathode: Cu2+ and H+ ion migrates to the cathode. Cu2+ being lower than H+ in the electrochemical series is preferentially discharged to acquired two electrons to be deposited as metallic copper on the cathode., Cu2+(aq) + 2e— Cu(s), At the anode: SO42— and OH—ions migrates to the anode. OH—ions being lower in the electrochemical series is preferentially discharged and lose its electrons to the anode to become a neutral - OH group., OH—(aq) OH + e—, The neutral –OH groups combine in pairs to form one molecule of water and one atom of oxygen., OH + OH H2O(l) + O(g), The oxygen atoms then react to form the diatomic oxygen gas molecule., O(g) + O(g) O2(g), Summary of Reactions, Cathodic half reaction – 2Cu2+(aq) + 4e— 2Cu(s), Anodic half reaction – 4OH—(aq) 2H2O(i) + O2(g) + 4e—, Over all reaction – 2Cu2+(aq) + 4OH—(aq) 2Cu(s) + 2H2O(l) + O2(g), USES OF ELECTROLYSIS, 1. Extraction of highly electropositive metals such as Na, K, Mg, Al, Zn and highly electronegative non-metals such as F2, Cl2, O2, etc., 2. Purification of less electropositive metals such as copper, mercury, silver, gold, etc., 3. Electroplating; that is, coating the surface of one metal with another metal to achieve improved appearance or prevent corrosion or give thickness to worn parts of machinery or combination of these., 4. Preparation of certain important substances such as sodium hydroxide, hydrogen and chlorine from electrolysis of brine using mercury cathode., WEEK 8, ELECTROCHEMICAL CELLS: This is a set-up which convert chemical energy to electrical energy. It consists of two half cells – an oxidation half –cell reaction and a reduction half cell reaction. The overall redox reaction results in a flow of electrons i.e. an electric current. An example of an electrochemical cell is a zinc electrode dipping into a solution of zinc tetraoxosulphate (vi), connected to a copper (ii) tetraoxosulphate (vi). The two solutions are separated by a porous partition. The porous partition allows electrical contact but prevents excessive mixing of the electrolytes by inter diffusion. In both electrochemical and electrolytic cells, it is important to note that oxidation always occurs at the anode and reduction at the cathode. Electrons always flow from the anode to the cathode. In an electrochemical cell the negative is the anode while the positive electrode is the cathode. In an electrolytic cell, however, the negative electrode is the cathode while the positive electrode is the anode. In place of a porous partition, the two half – cells of an electrochemical cell can be linked by a salt bridge., ELECTRODE POTENTIAL: this is the potential different set-up between an element and a solution of it ions. It is the measure of the tendency of an element to form ions. When a metal plate is dipped into a solution containing its ions, some of the atoms from the metal plate will ionize and go into solution as positively charged ions, leaving behind their valence electron on the surface of the plate., Metal(s) Metal ion(aq) + electrons, Zn(s) Zn2+(aq) + 2e—, Electrode potential vary from one metal ion/metal system to another and the value depends on:, 1. The concentration of ions in the solution, 2. The temperature at which the measurement is made, 3. The overall energy change., STANDARD ELECTRODE POTENTIAL: This is the potential difference set-up between the metal and one – molar solution of it ions at 250C or 298K using hydrogen as arbitrary standard., DRAWING AND WRITING OF CELL DIAGRAMS: When the Cu2+(aq) / Cu(s) system or half cell is connected to the 2H+(aq) / H2(g) or half cell by a salt bridge, the voltameter shows a reading of 0.34 volts. This is the potential difference between the two half cells and is known as electromotive force or e.m.f of the cell., The half cell reactions at the copper and hydrogen electrodes show that electrons flow from the hydrogen electrodes to the copper electrode. Hence, the electrode potential of copper is +0.34V., At the platinum in hydrogen electrode H2(g) 2H+(aq) + 2e— (oxidation), At the copper electrode Cu2+(aq) + 2e— Cu(s) (Reduction), Overall reaction Cu2+(aq) + H2(g) Cu(s) + 2H+, Standard electrode potential (E0) of some common metal ion/metal system., It is found that the potential difference between the two electrodes increases as the resistance in the external circuit is increased., The e.m.f of a cell: when two half cells are joined together through a salt bridge, the e.m.f of the cell formed is the algebraic difference between two electrode potential. Hence, the e.m.f of the cell (E0 total) formed by the system, Zn(s)/Zn2+(aq) // Cu2+(aq) / Cu(s)., The e.m.f of a cell is defined as the standard electrode potential of the right – hand electrode minus the standard electrode potential of the left – hand electrode., CALCULATION OF E.M.F OF A CELL:, Zn(s)/Zn2+(aq) // Cu2+(aq) / Cu(s)., E0total = E0(R.H.E) -- E0(L.H.E), = +0.34V – (-0.76V), = +1.10V, When this method is used, then a positive e.m.f indicates that the left – hand electrode (zinc) is capable of reducing copper ions to copper., Suppose the cell had been written as:, Calculation of e.m.f of a cell:, Cu(s)/Cu2+(aq) // Zn2+(aq) / Zn(s)., E0total = E0(R.H.E) -- E0(L.H.E), = – 0.76V – (+0.43V), = -1.10V, The negative value of e.m.f indicates that the left – hand electrode (copper) is not capable of or cannot reducing zinc ions to zinc., A positive e.m.f indicates that the reaction is thermodynamically feasible, as written down from left to right. While a negative e.m.f implies that the reaction is thermodynamically not feasible as written., WEEK 9, APPLICATION OF ELECTROCHEMICAL CELLS, 1. DANIEL CELL: This types of cell consist of copper vessel filled with a saturated solution of copper (II) tetraoxosulphate (VI) and a porous pot filled with a zinc tetraoxosulphate (VI) solution is a zinc rod., The initial e.m.f of denial cell is about 1.1 volts. This cells has to be replaced once the chemicals in it are depleted. In Daniel cell, the zinc goes into solution as Zn2+ ions – Zn Zn2+ + 2e— and the electrodes produced flow in the external circuit to the copper electrode where Cu2+ ions in solution accept them and deposit. Cu2+ + 2e— Cu, 2. LECLANCHE’ CELL: Leclanche’ cell consists of a zinc metal container as the anode and a graphite rod surrounded by manganes (IV) oxide as the cathode. The electrolytes is a pastes of ammonia chloride, At the anode: The zinc atoms lose two electrons each to dissolve in the electrolyte as positive charged zinc ions., Zn(s) Zn2+(aq) + 2e—, At the cathode: The ammonium ion accept the electrons to become reduced to ammonia and hydrogen gas. The hydrogen gas is removed by the manganese (IV) oxide to prevent it from adhering to the cathode., 2NH4+(aq) + 2e— 2NH3(g) + H2(g) (Reduction), 3. SECONDARY CELL: Cell which can be recharge are known as secondary cells. They can be recharged by passing a direct current through them. An example of a secondary cell is the lead accumulator, used in motor cars. The cathode and anode are lead (IV) oxide (PbO2) and metallic lead respectively. The electrolytes is a dilute tetraoxosulphate (VI) acid solution., At the anode: the lead atoms releases two electrons each to become oxidized to lead (II) ions Pb2+ which then combine with the tetraoxosulphate (VI) acid., Pb(S) Pb2+(aq) + 2e—, Pb2+(aq) + SO42-(aq) PbSO4(s), At the cathode: The electrons from the anode are accepted at the cathode where the lead (IV) oxide and the hydrogen ions H+, from the electrolytes undergo reduction to produce lead (II) ions and water., PbO2 + 4H+(aq) + 2e— Pb2+(aq) + 2H2O(l)., The lead (II) ions then combine with the tetraoxosulphate (VI) ions from the electrolytes to become deposited at the cathode as lead (II) tetraoxosulphate (VI)., Pb2+(aq) + SO42-(aq) PbSO4(s)., DISTINCTION BETWEEN PRIMARY AND SECONDARY CELLS, PRIMARY CELLS, 1. They are not rechargeable, 2. They are affected by defects such as local – action and polarization, 3. They have a very high internal resistance and hence can give a small current with very high drop of terminal p.d, Example of primary cells are: simple (voltaic) cell, Daniel cell, Leclanche cell e.t.c, SECONDARY CELLS, 1. They are rechargeable, 2. They are not affected by defects such as local – action and polarization, 3. They have a very low internal resistance and hence can give a large current with very little drop of terminal p.d, Example of secondary cells are: lead – acid accumulator, alkaline cell, e.t.c, WEEK 10, FARADAY’S LAWS OF ELECTROLYSIS, FARADAY’S FIRST LAW OF ELECTROLYSIS: State that the mass (m) of an element liberated at (or dissolved from) an electrode during electrolysis is directly proportional to the quantity of electricity (Q) that passed through it., Mathematically, M α Q, Q = It, M α It, M = Zit, Where Z is a constant for any given substance, and is known as the electrochemical equivalent of the substance., M = Mass of substance in gram, Q = Quantity of electricity in coulombs, I = Current in ampere, t = Time in seconds, VERIFICATION OF FARADAY’S FIRST LAW OF ELECTROLYSIS, Method, 1. The beaker if filled to two – thirds of its volume with 0.1 mol per dm3 copper (II) tetraoxosulphate (VI) solution., 2. The copper plate (about 5cm by 3cm) are cleaned with emery paper and then weighed, 3. The circuit is connected as shown above, and the variable resistor adjusted to maintain a current of 1A for 15 minutes., 4. The cathode is removed, washed with water and ethanol; dried and then reweighed, 5. The cathode is replaced and the same current passed for another 15 minutes, 6. The process of cleaning, drying and reweighing is repeated. In this way three more readings are obtained., The results are tabulated as shown in the table below, A graph of mass of copper deposited (g) against the quanitity of electricity (C) is plotted, Conclusion: The mass of copper deposited at the cathode is directly proportional to the quantity of electricity passed., FARADAY’S SECOND LAW OF ELECTROLYSIS: Faraday’s second law of electrolysis state that when the same quantity of electricity is passed through different electrolytes, the relative number of moles of the elements discharged are inversely proportional to the charges on the ions of each of the elements respectively., 1 Faraday’s = 96500 coulombs = the minimum quantity of electricity required to liberate one mole of single – charge ions during electrolysis., Mass = Quantity of Electricity, Molar mass Faraday, M Q, Mm F, VERIFICATION OF FARADAY’S SEOND LAW OF ELECTROLYSIS, Method, 1. The two beakers are respectively filled to two – thirds of their volumes with 0.5 mol per dm3 copper (II) tetraoxosulphate (VI) solution and silver trioxonitrate(V) solution., 2. The copper and silver plate (about 5cm by 3cm) are cleaned with emery paper and then weighed, 3. The circuit is connected as shown above, and the variable resistor adjusted to maintain a steady current of 0.5A passed through the solution for 20 minutes., 4. The cathode is removed, washed with water and ethanol; dried and then reweighed to obtain the masses of copper and silver deposited, 5. The ratio of the number of moles of copper and silver deposited is calculated., 6. The process of cleaning, drying and reweighing is repeated. In this way three more readings are obtained., Amounts n (Number of moles) = Mass of element deposited, Its relative atomic mass., Observation: On passing the same quantity of electricity through the solutions, the ratio of the number of moles of copper and silver deposited is 1:2., This ratio is inversely proportional to the ratio of the charges on the ions, Cu2+ and Ag+ , or the number of moles of electrons required to liberate 1 mole each of the ions., Cu2+(aq) + 2e— Cu(S) and Ag+(aq) + e— Ag(s), 2 moles 63.5g 1 mole 108g, Conclusion: When the same quantity of electricity is passed through solution of different electrolytes the relative number of moles of the elements deposited are inversely proportional to the charges on the ions of each of the elements respectively., CALCULATION BASED ON THE FIRST LAW OF ELECTROLYSIS, 1. In an electrolysis experiment, the ammeter records a steady current of 1A. The mass of copper deposited in 30mins is 0.66g. Calculate the error in the ammeter reading.[electrochemical equivalent of copper =0.00033gC-1], Solution, M = ZIt, M = 0.66g, Z = 0.00033gC-1, t = 30mins = 1800 seconds, I = M/Zt, I = 0.66/0.00033 x 1800, I = 0.66/0.594, I = 1.11A, The error in the ammeter reading is 1.11 – 1 = 0.11A, 2. Calculated the time in minutes, required to plate a substance of total surface area 300cm2, a layer of copper 0.6mm thick, if a constant current of 2A is maintained. Assuming the density of copper is 8.8g/cm3 and one coulombs liberates 0.00033g copper., Solution, Given that area = 300cm2 , thickness = 0.6mm = 0.06cm, Mass = 0.00033g, density = 8.8g/cm3, Density = mass/volume, Mass = density x volume, Mass = 8.8 x 300 x 0.06, Mass = 158.4g, From M = Zit, t = m/ZI, t = 158.4/2 x 0.00033, t = 240000secs, t= 4000mins, 3. At what time must a current of 5Amp pass through a solution of zinc sulphate to deposited 1g of zinc. Electrochemical equivalent (e.c.e) = 0.0003387, Solution, Given current = 5Amp, e.c.e = 0.0003387g/c, Mass = 1g, M = ZIt, t = m/ZI, time = 1/0.0003387 x 25, time = 1,180secs, time = 19mins 40 seconds, 4. In an electrolysis experiment, a cathode of mass 5g is found to weigh 5.01g, after a current of 5A flows for 50 seconds. What is the electrochemical equivalent for the deposited substance?, Solution, M = 5.01 – 5 = 0.01g, t = 50sec, I = 5A, M = ZIt, Z = m/It, Z = 0.01/50 x 5, Z = 0.01/250, Z = 4 x 10-5g/c, 5. The electrochemical equivalent of silver is 0.0012g/c. if 0.36g of silver is to be deposited by electrolysis on a surface by passing a steady current for 5.0 minutes. Calculate the value of the current., Solution, Z= 0.0012g/c, m = 0.36g, t = 5mins = 300secs,, M = ZIt, I = m/Zt, I = 0.36/0.0012 x 300, I = 0.36/0.36, I = 1A., CALCULATION BASED ON THE SECOND LAW OF ELECTROLYSIS, 6. A current of 4.5A is passed through a solution of gold salt for 1 hour 45 minutes. Calculate, (i) The mass of gold deposited, (ii) The number of moles of gold deposited, (iii) If the same current is used, find the time taken for 5.5g of gold to be deposited (Au = 197, 1 Faraday = 96500c), Solution, (i) Au+ + e— Au, 197g 1F 197g, Mass = Quantity of Electricity, Molar mass Faraday, M Q, Mm F, Mass = ?, Molar mass = 197g, Quantity of electricity = I x t =, I = 4.5A, t = 1 hour 45minutes = 105 minutes = 105 x 60 = 6300 seconds, Quantity of electricity = I x t = 4.5 x 6300 = 28,350C, Faraday = 96500F, Mass = 28,350, 197 96500, Mass = 0.29378, 197, Mass of Gold deposited = 57.88g, (ii) Number of mole = Mass, Molar mass, Number of mole = 57.88, 197, The number of mole of Gold deposited = 0.30mol, (iii) Mass = Quantity of Electricity, Molar mass Faraday, M Q, Mm F, 5.5 = Quantity of Electricity, 197 96500, 0.02792 = Quantity of Electricity, 96500, Quantity of Electricity = 2694.28C. This is the quantity of electricity (Q) required for 5.5g of Au to be deposited., Q = It, t = Q/I, t = 2694, 4.5, t = 598.7, The time taken is 9.98 minutes, 7. Calculate the current that must be passed into a solution of aluminium salt for 1hr.30minutes in order to deposited 1.5g of Aluminium (Al = 27), Solution, Al3+ + 3e— Al(s), 27g 3F 27g, Mass = Quantity of Electricity, Molar mass Faraday, M Q, Mm F, Mass = 1.5g, Molar mass = 27g/mol, Quantity of electricity = ?, Faraday = 3 x 96500 = 289500, 1.5 = Quantity of electricity, 27 289500, 0.0556 = Quantity of electricity, 289500, Quantity of electricity = 16096.2C, Quantity of electricity = I x t, I = Q/t, t = 1 hr.30mins = 90mins = 90 x 60 = 5400seconds, I = 16096.2, 5400, The current (I) = 2.98Amperes, 8. 0.222g of a divalent metals is deposited when a current of 0.45A is passed thought a solution of its salt for 25 minutes. Calculate the relative atomic mass of the metal. (1 Faraday = 96500 coulombs), Solution, M2+ + 2e— M, 1mole 2F 1 mole of atoms., Mass = Quantity of Electricity, Molar mass Faraday, M Q, Mm F, Mass = 0.222g, molar mass = ?, Quantity of electricity = It = 0.45 x 25 x 60 = 675C, Faraday = 2 x 96500 = 193000F, 0.222 = 675, Mm 193000, 0.222 = 0.003497, Mm, Molar mass = 0.222, 0.0003497, The relative atomic mass of metal M is 63.5g/mol, Therefore metal M is copper., 9. A given quantity of electricity was passed through three electrolytic cells connected in series containing solutions of Silver trioxonitrate (V), Copper(II) tetraoxosulphate (VI) and Sodium Chloride respectively. If 10.5g of Copper are deposited in the second electrolytic cell. Calculate, (a) The mass of Silver deposited in the first cell., (b) The Volume of Chloride liberated in the third cell at 18oC and 760mmHg pressure. (Ag=108, Cu=63.5, 1Faraday=96500C, molar volume of gases at s.t.p =22.4dm3.), Solution, In the second cell, Cu2+(aq) + 2e— Cu(s), 63.5g 2F 63.5g, Mass = Quantity of Electricity, Molar mass Faraday, M Q, Mm F, mass of copper = 10.5g, molar mass of copper 63.5g/mol, Faraday = 2 x 96500 = 193000C, Quantity of electricity = ?, 10.5 = Quantity of electricity, 63.5 193000, 0.16535 = Quantity of electricity, 193000, Quantity of electricity = 31913.4 Coulombs, (a) The mass of silver deposited in the first cell, Ag+(aq) + e— Ag(s), 108g 1F 108g, Mass = Quantity of Electricity, Molar mass Faraday, M Q, Mm F, M = 31913.4, 108 96500, M = 0.3307, 108, Mass = 108 x 0.3307, Mass of silver deposited = 35.7g, (b) Volume of chlorine liberated in the third cell, 2Cl— Cl2 + 2e—, 2F, Mass = Quantity of Electricity, Molar mass Faraday, M 31913.4, Mm 2 x 96500, Mole = 31913.4, 193000, Mole = 0.1654mol, 1 molar volume of gas = 22.4dm3 x mole, Volume of chlorine = 22400cm3 x 0.1654, Volume of chlorine at s.t.p = 3704cm3, Convert the volume at s.t.p to its volume at 180C and 740mmHg, P1V1 = P2V2, T1 T2, Where P1 = 760mmHg, P2 = 740mmHg, T1 = 273 K T2 = 18 + 273 = 291K, V1 = 3704cm3 V2 = ?, V2 = P1V1T2 760 x 3704 x 291, P2T1 740 x 273 Volume of chlorine = 4055cm3.