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STUDY MATERIAL PREPARED BY, © LUWANG HOSTEL CUM TUTORIAL CENTER, , Chapter – 8, , •, REDOX REACTION, , 1, , BASIC CONCEPT, In the previous chapter, we have studied neutralization reactions in which H+ ions from an acid, , (HCl, H2SO4, CH3COOH etc.) combine with OH− ions of a base (NaOH, Ca(OH)2, NH4OH etc.) to form a, salt and weakly ionized molecules of water. In this chapter, we shall discuss another important class of, reactions called reduction−oxidation or simply redox reactions. All these reactions are always accompanied, by energy changes in the form of heat, light or electricity., A large number of chemical and biological reactions fall in this category. Burning of different, types of fuels for obtaining energy for domestic, transport or industrial purpose, electrochemical processes, like manufacturing of caustic soda, digestion of food in animals, photosynthesis by plants, corrosion of, metals and operations of dry and wet batteries are diverse examples of redox reactions., 2, , OXIDATION AND REDUCTION, , In redox reactions one reactant called reductant or reducing agent is oxidised while the other, reactant called oxidant or oxidising agent is reduced., CLASSICAL CONCEPT OF OXIDATION AND REDUCTION, Originally, oxidation of a substance was defined as the addition of oxygen or any other, electronegative element to it. Due to the presence of oxygen in the atmosphere, many elements combine, with it. In fact the most common occurrence of elements on the earth is in the form of their oxides. Later, on, removal of hydrogen or any other electropositive element from a substance was also considered, as oxidation., (i) 2Mg(s) + O2(g) → 2MgO(s), Magnesium is oxidised by addition of oxygen to it., (ii) 2H2S(g) + O2(g) → 2S(s) + 2H2O(l), Hydrogen sulphide is oxidised by the removal of hydrogen from it., (iii) Mg(s) + F2(g) → MgF2(s), Magnesium is oxidised by the addition of fluorine to it., (iv) Mg(s) + S(s) → MgS(s), Magnesium is oxidised by the addition of sulphur to it., Similarly reduction of a substance was initially introduced for the processes in which metal oxides, were converted into metals by their reactions with hydrogen, carbon or other metals. Carbon monoxide is, also used in the reduction of ferric oxide to iron., (i) CuO(s) + H2(g) → Cu(s) + H2O(l), (ii) 2CuO(s) + C(s) → 2Cu(s) + CO2(g), (iii) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g), Here, in all the three cases mentioned above metal oxide is reduced by hydrogen, carbon or carbon, monoxide., Page number, , 1, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , MODERN CONCEPT OF OXIDATION AND REDUCTION, The general definition of oxidation is loss of electrons. In the above examples of oxidation, magnesium is converted into Mg2+ ion and in the process it loses two electrons., Similarly, reduction in general means acceptance of electrons. When CuO reacts with hydrogen it, gets converted into copper. In copper oxide, each Cu2+ ion accepts two electrons and gets reduced to, copper., COMPLIMENTARY NATURE OF OXIDATION AND REDUCTION REACTIONS, Whenever any substance is oxidised, another substance is always reduced at the same time and, vice−versa. In other words, oxidation and reduction reactions are complimentary. This is illustrated by the, following examples, (i) H2S(g) + Cl2(g) → 2HCl(g) + S(s), Here H2S is oxidised to S while Cl2 is reduced to HCl., (ii) MnO2(s) + 4HCl(aq) → MnCl2(aq) + Cl2(g) + 2H2O(l), Here, HCl is oxidised to Cl2 while MnO2 is reduced to MnCl2., The occurrence of electron transfer is more apparent in some redox reactions than others. When, metallic zinc is added to a solution containing copper(II) sulphate, zinc reduces Cu2+ by donating two, electrons to it., Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s), In the process, solution loses the blue colour that characterizes the presence of Cu2+ ions. The, oxidation and reduction reactions are, Zn → Zn2+ + 2e−, [Oxidation half reaction], 2+, −, Cu + 2e → Cu, [Reduction half reaction], Similarly, metallic copper reduces silver ions in a solution of silver nitrate (AgNO3), Cu(s) + 2AgNO3(aq) → Cu(NO3)2(aq) + 2Ag(s), The oxidation and reduction reactions are, Cu → Cu2+ + 2e−, [Oxidation half reaction], +, −, 2Ag + 2e → 2Ag [Reduction half reaction], PRACTICE PROBLEM, PP1., , Identify the oxidant and the reductant in the following reactions., (a) Zn(s) + ½O2(g) → ZnO(s), (b) Zn(s) + 2H (aq) → Zn (aq) + H2(g), +, , 3, , 2+, , (c) 2Al + Fe2O3, , → Al2O3 + 2Fe, , (d) H2S + 2FeCl3, , → 2FeCl2 + 2HCl + S, , THE OXIDATION STATE OF AN ELEMENT, , To describe the changes that occur in oxidation−reduction reactions and to write the correctly, balanced equations for such reactions, it is helpful to know the concept of the oxidation state (or oxidation, number) of an atom., , Page number, , 2, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , For monoatomic ions, the oxidation state is simply the charge on the ion. For example,, oxidation states of Na+, Mg2+, Cl− and N3− are +1, +2, −1 and −3 respectively. For covalently bonded, substances, the charge on an atom would be so small that it is impossible to calculate the exact charge on, each atom in the molecule. Thus, oxidation state of an atom in a covalently bonded molecule is defined as, “the charge an atom would possess if all the bonds associated with that atom are broken, heterolytically considering them to be completely ionic”. For example, in NH3 there are three N−H, bonds. If we consider all three N−H bonds to be ionic, then each H will possess a charge of +1 while N, acquires a charge of −3 (because N is more electronegative than H). In H2O2, there are two O−H bonds and, one O−O bond as evident from the structural formula H−O−O−H. Assuming each O−H bond to be ionic,, each H atom would possess +1 charge while each O would possess −1 charge. The O−O bond is not ionic, as the bond is between atoms of similar electronegativity. Thus oxidation state of each O-atom in H2O2 is –, 1., , It is not necessary to know the structure of a molecule in order to calculate the oxidation state of an, atom in a molecule. The following set of rules are used to assign oxidation state to each atom in an ion or, molecule., 1. Each pure element has an oxidation state of zero. This is true whether the element is a monoatomic, gas, a metallic solid or a polyatomic molecule. Thus, Fe(s), N2(g), P4(s) and S8(s) are all in the zero, oxidation state., 2. In monoatomic ions, the oxidation state of the element is equal to the charge on the ion. Thus,, oxidation state of all alkali metals is +1 and that of alkaline earth metals is +2 in all, their compounds. In ionic solid K2S, the oxidation states of potassium and sulfur are +1 and −2, respectively. In AgCl, the oxidation state of silver and chlorine are +1 and −1 respectively., 3. The oxidation state of hydrogen in any molecule in which it is combined with another element is, +1, except in the metallic hydrides such as LiH or CaH2, where the oxidation state of hydrogen is, −1., 4. The oxidation state of oxygen in any molecule or ion in which it is combined with, another element is −2, except in the peroxides, the superoxides and in OF2. The peroxides (H2O2,, Na2O2, BaO2 etc.) are compounds in which there is an O−O covalent bond and the peroxide ion is, , O 22− . The oxidation state of oxygen in peroxides is −1. The superoxides (like KO2, CsO2 etc.) are, ionic compounds involving the superoxide ion, O −2 . The oxidation state of oxygen is −½ in, superoxide ion. In OF2, the oxidation state of oxygen is +2 since fluorine is more electronegative, than oxygen., 5. The oxidation state of all the halogens is −1 in all their compounds except where they are, combined with an element of higher electronegativity or in interhalogen compounds.. Oxidation, state of fluorine is always −1, since it is the most electronegative element., 6. In covalent compounds not involving hydrogen or oxygen, the more electronegative element is, assigned negative oxidation state while a less electronegative element is assigned a positive, oxidation state., 7. The algebraic sum of the oxidation numbers of all the atoms combined in a molecule or complex, ion must equal the net charge on the molecule or ion., , Question:, , Calculate the oxidation state of the underlined atoms in the given species., (a) NO 2+, , (b) NO 3−, , (c) KMnO 4, , (d) Cr 2 O 27 −, , (e) Fe 2 O 3, , (f) Fe 3 O 4, , Solution:, Let the oxidation state of N in NO2+ be ‘x’., , (a), Page number, , 3, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, x + [2(−2)] = +1, x=+5, , Thus, oxidation state of N in NO 2+ is +5., Let ‘x’ be the oxidation state of N in NO 3− ., x + [3(−2)] = −1, x=+5, , (b), , Thus, oxidation state of N in NO 3− is + 5., (c), , Let the oxidation state of Mn in KMnO4 be ‘x’., [1(+1)] + x + [4(−2)] = 0, x = +7, Thus, oxidation state of Mn in KMnO4 is + 7., Let the oxidation state of Cr in Cr2 O 72− be ‘x’., , (d), , 2x + [7(−2)] = −2, x =+6, Thus, oxidation state of Cr in Cr2 O 72− is +6., (e), , Let ‘x’ be the oxidation state of Fe in Fe2O3., 2x + [3(−2)] = 0, x=+3, Thus, oxidation state of Fe in Fe2O3 is +3., Let the oxidation state of Fe in Fe3O4 be ‘x’., 3x + [4(−2)] = 0, x = + 8/3, Thus, oxidation state of Fe in Fe3O4 is +8/3. This is the average oxidation state of Fe in Fe3O4., Actually, Fe3O4 is made up of equimolar quantity of FeO and Fe2O3., , (f), , 4, , TYPES OF REDOX REACTION, There are four general types of redox reactions, namely, 1. Combination reactions., 2. Decomposition reactions., 3. Displacement reactions., 4. Disproportionation reactions., COMBINATION REACTIONS, , A combination reaction may be represented by, A + B → C, If either A or B is an element, then the reaction is usually redox in nature. Combination reactions, are reactions in which two or more substances combine to form a single product., For example,, 0, , +4 −2, , 0, , S(s) + O2(g) → SO2(g), 0, , 0, , +2, , −3, , 3Mg(s) + N2(g) → Mg3N2(s), DECOMPOSITION REACTIONS, Decomposition reactions are the opposite of combination reactions. Specifically, a decomposition, reaction is the breakdown of a compound into two or more components., , Page number, , 4, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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Redox Reaction, , ©Luwang Hostel, , C → A + B, +2 −2, , 0, , 0, , 2HgO(s) → 2Hg(l) + O2(g), +5 −2, , −1, , 0, , 2KClO3(s) → 2KCl(s) + 3O2(s), +1 −1, , 0, , 0, , 2NaH(s) → 2Na(s) + H2(g), Note that we show oxidation states only for the elements that are oxidized or reduced., DISPLACEMENT REACTIONS, , In a displacement reaction, an ion (or atom) in a compound is replaced by an ion, (or atom) of another element:, A + BC → AC + B, Most displacement reactions fit into one of three subcategories: hydrogen displacement, metal, displacement or halogen displacement., Hydrogen Displacement, All alkali metals and some alkaline earth metals (Ca, Sr and Ba), which are the most reactive of the, metallic elements, will displace hydrogen from cold water., 0, , +1, , +1, , +1, , 0, , 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g), 0, , +1, , +2, , +1, , 0, , Ca(s) + 2H2O(l) → Ca(OH)2(aq) + H2(g), Less reactive metals, such as aluminium and iron, react with steam to give hydrogen gas:, 0, , +1, , +3, , 0, , 2Al(s) + 3H2O(g) → Al2O3(s) + 3H2(g), 0, , +1, , +3, , 0, , 2Fe(s) + 3H2O(g) → Fe2O3(s) + 3H2(g), Many metals, including those that do not react with water, are capable of displacing hydrogen from, acids. For example, zinc (Zn) and magnesium (Mg) do not react with water but do react with hydrochloric, acid, as follows:, 0, , +1, , +2, , 0, , Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g), 0, , +1, , +2, , 0, , Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g), Metal Displacement, A metal in a compound can be displaced by another metal in the elemental state., In addition to the two examples given in section 2.3 some more examples are given as follows, V2O5(s) + 5Ca(l) → 2V(l) + 5CaO(s), TiCl4(g) + 2Mg(l) → Ti(s) + 2MgCl2(l), Halogen Displacement, Another activity series summarizes the halogens behaviour in halogen displacement reactions:, F2 > Cl2 > Br2 > I2, The power of these elements as oxidizing agents decreases as we move down the group VII A, from fluorine to iodine. So molecular fluorine can displace chloride, bromide and iodide ions in solution., Page number, , 5, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , In fact, molecular fluorine is so reactive that it also attacks water. Thus, these reactions cannot be carried, out in aqueous solutions. On the other hand molecular chlorine can displace bromide and iodide ions in, aqueous solution. The displacement equations are, −1, , 0, , −1, , 0, , Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(l), −1, , 0, , −1, , 0, , Cl2(g) + 2NaI(aq) → 2NaCl(aq) + I2(s), , DISPROPORTIONATION REACTIONS, A special type of redox reaction is the disproportionation reaction. In this type of reactions, an, element in one oxidation state is simultaneously oxidized and reduced. One reactant in a disproportionation, reaction always contains an element that shows increase in its oxidation state in one product and decrease, in its oxidation state in another product. The element itself is in an intermediate oxidation state, i.e., both, higher and lower oxidation states of the element exist on the product side. The decomposition of hydrogen, peroxide is an example of a disproportionation reaction, −1, , −2, , 0, , 2H2O2(aq) → 2H2O(l) + O2(g), Question:, , Classify the following redox reactions and indicate changes in the oxidation states of the, elements involved in each reaction., (a) 2N2O(g) → 2N2(g) + O2(g), (b) 6Li(s) + N2(g) → 2Li3N(s), (c) Ni(s) + Pb(NO3)2(aq) → Pb(s) + Ni(NO3)2(aq), (d) 2NO2(g) + H2O(l) → HNO2(aq) + HNO3(aq), , Solution:, (a), , This is a decomposition reaction because one reactant dissociates into two different products. The, oxidation number of N changes from +1 to 0 while that of O changes from −2 to 0., This is a combination reaction because two reactants combine to form a single product. The oxidation, number of Li changes from 0 to +1 while that of N changes from 0 to −3., This is a metal displacement reaction. The Ni metal displaces (reduces) the Pb2+ ion., The oxidation number of Ni increases from 0 to +2 while that of Pb decreases from +2 to 0., The oxidation number of N is +4 in NO2, +3 in HNO2 and +5 in HNO3. Since oxidation state of N, both increases and decreases, this is an example of a disproportionation reaction., , (b), (c), (d), , PRACTICE PROBLEM, PP2., , What is the type of redox reactions given below?, (a) Fe + H2SO4 → FeSO4 + H2, (b) S + 3F2 → SF6, (c) 2CuCl → Cu + CuCl2, (d) 2Ag + PtCl2 → 2AgCl + Pt, , 5, , BALNCING OF REDOX REACTION, , Redox reactions involve both oxidation and reduction simultaneously. Oxidation means loss of, electrons and reduction means gain of electrons. Thus redox reactions involve electron transfer and the, , Page number, , 6, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , number of electrons lost is same as the number of electrons gained during the reaction. This aspect of, redox reaction can serve as the basis of a pattern for balancing redox reactions., There are two common and useful methods to balance redox reactions. These are, (a) Oxidation state method and, (b) Ion−electron method., BALANCING OF REDOX REACTIONS BY OXIDATION STATE METHOD, For balancing a redox reaction by oxidation state method, follow the order of steps as listed below. In, balancing some reactions, all the steps may not be required., (i), , For each redox reaction, deduce the oxidation state of the elements that are undergoing, oxidation and reduction., (ii) Separate the reactants and products into two half−reactions involving the elements alone that, show change in their oxidation states. Write the skeletal equations for each half−reaction., (iii) For each half−reaction, first balance the number of atoms of the element undergoing change, in oxidation state., (iv) Now find the total change in oxidation state by determining the change per atom and, multiplying it by the total number of atoms that undergo change in oxidation state. Also,, decide whether electrons are lost or gained. An increase in oxidation state means loss of, electrons and a decrease in oxidation state means gain of electrons., (v) Add the electrons lost or gained to the half equation. Lost electrons are placed on the product, side and gained electrons are kept on the reactant side., (vi) Now add both the half reactions after multiplying by suitable integers to make the number of, electrons lost and gained same., (vii) Transfer the coefficients of each reactant and product to the main skeleton equation., (viii) If the coefficients developed are not correct, then change them by inspection method., Such coefficient changes are required when an element from a compound goes in, 2 different compounds, one with the same oxidation state and the other with different oxidation, state., (ix) Count the charges on both sides of the equation and balance the charges in the equation by adding, requisite H+ or OH− to the required side. If the reaction occurs in acidic solution, use H+ and if it, occurs in basic solution, use OH−. If the reaction occurs in neutral solution, use H+ or OH− on any, of the side as needed i.e. in a neutral solution, if negative charges are needed for balancing, use, OH− and if positive charges are needed, use H+., (x) Balance the hydrogens and oxygens by adding the appropriate number of H2O molecules on the, required side., Question:, , Balance the following oxidation−reduction equation,, KMnO4 + KCl + H2SO4 → MnSO4 + K2SO4 + H2O + Cl2, by oxidation state method., , Solution:, Identify the oxidation and reduction half equations in ionic form., , (i), , Reduction half: Mn7+ → Mn2+, Oxidation half: Cl− → Cl2, Balance the atoms that undergo change in oxidation state., Reduction half: Mn7+ → Mn2+, Oxidation half: 2Cl− → Cl2, Add the electrons lost or gained to each half equation., Reduction half: 5e− + Mn7+ → Mn2+, ……………….. A, , (ii), , (iii), , Page number, , 7, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , Oxidation half: 2Cl− → Cl2 + 2e−, , (iv), , ……………….. B, , Multiply equation A with 2 and equation B with 5 and then add the two half reactions., , (v), (vi), (vii), , Question:, , 5e− + Mn7+ → Mn2+ ] × 2, 2 Cl− → Cl2 + 2e− ] × 5, 2Mn7+ + 10 Cl− → 5Cl2 + 2Mn2+, Transfer the coefficients to the main equation., 2KMnO4 + 10KCl + H2SO4 → 2MnSO4 + K2SO4 + H2O + 5Cl2, Balance H and O atoms, 2KMnO4 + 10KCl + 8H2SO4 → 5Cl2 + 2MnSO4 + 8H2O + 6K2SO4, Finally balance by inspection method., 2KMnO4 + 10KCl + 8H2SO4 → 2MnSO4 + 6K2SO4 + 8H2O + 5Cl2, Balance the following redox equation,, K2Cr2O7 + HCl → KCl + CrCl3 + Cl2 + H2O, using oxidation state method., , Solution:, Identify the oxidation and reduction half equations in ionic form., , (i), , Reduction half: Cr212 + → Cr3+, Oxidation half: Cl− → Cl2, Balance the atoms that undergo change in oxidation state., , (ii), , Reduction half: Cr212 + → 2Cr3+, Oxidation half: 2Cl− → Cl2, Add the electrons lost or gained to each half equation., , (iii), , Reduction half: 6e− + Cr212 + → 2Cr3+, , ……………….. A, , Oxidation half: 2Cl → Cl2 + 2e, , ……………….. B, , −, , −, , Multiply equation B by 3 and add it to equation A ., , (iv), , 6e− + Cr212 + → 2Cr3+, 2Cl− → Cl2 + 2e− ] × 3, , Cr212 + + 6Cl− → 2Cr3+ + 3Cl2, (v), , Transfer the coefficients to the main equation., K2Cr2O7 + 6HCl → KCl + 2CrCl3 + 3Cl2 + H2O, By inspection, it is seen that the coefficient of HCl developed is not correct, as the number of moles, of Cl atoms on the product side are 13 while on the reactant side are only 6. So, the coefficient is, changed appropriately, seeing that the net reaction is completely balanced., K2Cr2O7 + 14HCl → 2KCl + 2CrCl3 + 3Cl2 + 7H2O, , (vi), , BALANCING REDOX REACTIONS BY ION−ELECTRON METHOD, This method of balancing redox reaction involves following steps., (i) For each redox reaction, deduce the oxidation state of the elements that are undergoing, oxidation and reduction., (ii) Separate the reactants and products into two half−reactions involving the elements that show, change in their oxidation states. Write the skeleton equations for each half−reaction., (iii) Balance each half−reaction separately involving the given steps., 1. First balance the atoms of the element undergoing oxidation or reduction., 2. Then balance atoms of the elements other than hydrogen and oxygen., 3. For balancing oxygen atoms in acidic or neutral medium, add suitable number of H2O, molecules to the side deficient in O while in alkaline medium, add equal number of H2O, molecules as the excess of O on the side having excess of, O atoms and add double the number of OH− ions on the opposite side of the equation., Page number, , 8, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , 4. In order to balance the hydrogen atoms in acidic or neutral medium, add required, number of H+ to the side deficient in H while in alkaline medium, add equal number of, OH− ions as the excess number of H atoms on the side having excess, H and add equal number of H2O molecules on the opposite side of the equation., (iv) Multiply each half−reaction by suitable integer to make the number of electrons lost and, gained same and add both the half−equations to get a completely balanced reaction., Mostly, the medium in which a redox reaction is to be balanced is given in the problem but if, the problem does not state the medium explicitly, then the medium is decided by looking at the reactants or, products. If an acid or base is one of the reactants or products, then that reactant or product decides the, medium. If ammonia is present, the solution would be basic, and if ammonium chloride is present, it would, be acidic. If metals which form insoluble hydroxides are given in their ionic form in a redox reaction, the, solution is acidic., Although, both the oxidation state method and ion−electron method lead to the correct form of the, balanced redox reaction but ion−electron method is considered to be superior to the oxidation state method, due to following advantages:, , In ion−electron method, all reactants and products are completely balanced, , (i), , (i.e. coefficient is developed for each one of them) whether they participate in the redox, change or not, while in oxidation state method, the balancing of coefficients is developed, only for the species involved in the redox change., The balancing coefficients developed in ion−electron method are always correct and need no, amendment while such coefficients are to be changed sometimes in oxidation number, method., , (ii), , Question:, , Balance the redox equation,, HNO3 + H2S → NO + S, by ion−electron method (acidic medium)., (i) Identify the oxidation and reduction halves., Reduction half: HNO3 → NO, Oxidation half: H2S → S, Atoms of the element undergoing oxidation and reduction are already balanced., Balancing O atoms,, Reduction half: HNO3 → NO + 2H2O, Oxidation half: H2S → S, Balancing H atoms,, Reduction half: 3H+ + HNO3 → NO + 2H2O, Oxidation half: H2S → S + 2H+, Balancing charge,, Reduction half: 3e− + 3H+ + HNO3 → NO + 2H2O……………….. A, , Solution:, , (ii), (iii), , (iv), , (v), , Oxidation half: H2S → S + 2H+ + 2e−, (vi), , ……………….. B, , Multiplying equation A by 2 and equation B by 3 and then adding them., 3e− + 3H+ + HNO3 → NO + 2H2O ] × 2, ……………….. A, H2S → S + 2H+ + 2e− ] × 3, , ……………….. B, , 2HNO3 + 3H2S → 3S + 2NO + 4H2O, , Question:, , Balance the following redox equation,, , AsO 33 − + MnO −4 → AsO 34− + MnO 2, using ion−electron method (alkaline medium)., , Solution:, , Page number, , 9, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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(i), , ©Luwang Hostel, , Redox Reaction, , Identify the oxidation and reduction halves., , Reduction half: MnO −4 → MnO2, Oxidation half: AsO 33 − → AsO 34−, , (ii), (iii), , Atoms of the element undergoing oxidation and reduction are already balanced., Balancing O atoms,, Reduction half: 2H2O + MnO −4 → MnO2 + 4OH−, Oxidation half: 2OH− + AsO 33 − → AsO 34− + H2O, , (iv), , Balancing H atoms,, H atoms are already balanced in both the half−reactions., Balancing charge,, , (v), , Reduction half: 3e− + 2H2O + MnO −4 → MnO2 + 4OH−………….. A, Oxidation half: 2OH− + AsO 33 − → AsO 34− + H2O + 2e−………….. B, Multiply equation A by 2 and equation B by 3 and then add A and B ., 3e− + 2H2O + MnO4− → MnO2 + 4OH− ] × 2, , (vi), , 2OH− + AsO 33 − → AsO 34− + H2O + 2e− ] × 3, 3 AsO 33 − + 2 MnO −4 + H2O → 3 AsO 34− + 2MnO2 + 2OH−, 6, , ELECTROCHEMICAL CELLS, , Let us take a zinc rod and immerse it in a solution of CuSO4 taken in a beaker as shown in the, figure 1. After some time, we see that the Zn rod starts dissolving and Cu starts depositing on its surface., This happens because Zn gets oxidised to Zn2+ ions, which passes into the solution and 2 electrons remain, on the Zn rod for each Zn2+ ion going in solution. Then, Cu2+ ion from the solution takes up these 2, electrons to form Cu, which deposits on the Zn rod., Zn rod, , Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s), , CuSO4, solution, Figure 1, , In this reaction, Zn acts as reducing agent, which is able to reduce Cu2+ to Cu by transferring 2, electrons and Cu2+ ion acts as an oxidising agent, which oxidises Zn to Zn2+ ions and itself gets reduced to, Cu. In this case, there is a direct transfer of electrons from Zn rod to Cu2+ ion and some heat is also, evolved., The above experiment is slightly modified in a manner that for the same redox reaction, transfer of, electrons takes place indirectly and the heat of the reaction is converted into electrical energy. This, necessitates the separation of Zn rod from CuSO4 solution. Such cells in which the oxidised and reduced, species are separated and connected through electrical wires are called electrochemical cells., Electrochemical cells are the cells in which chemical energy is transformed into electrical energy., This means that chemical reactions produce electric current., An electrochemical cell consists of two half−cells or electrodes. The electrodes are metallic, conductors dipped in an electrolyte, which is an ionic conductor dissolved in water. A conducting rod and a, suitable electrolyte comprise an electrode or half−cell compartment. The two electrodes may share the, same electrolyte or have different electrolytes., , In a given electrochemical cell, combination of any of the two electrodes can be used., The cell may even contain same type of electrodes with different concentration of electrolytes. When an, Page number, , 10, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , ‘inert metal’ is part of the electrode, it acts as a source of electrons, but does not take part in the reaction. If, the electrolytes are different, the two compartments may be joined by a salt bridge, which is a concentrated, solution of KCl in agar−agar jelly that completes the electrical circuit and enables the cell to function., Thus, salt bridge provides an electrical contact between the two solutions without allowing them to mix, with each other., The simplest electrochemical cell to study is Daniel cell, which is shown in figure 2. This is, prepared by dipping Zn rod in a solution of ZnSO4 in one beaker and by dipping Cu strip in a solution of, CuSO4 in another beaker. It consists of two redox couples, Zn2+ | Zn at one end and Cu2+ | Cu at the other, end. Since Zn has a higher oxidation potential than Cu, so it has a higher tendency to get oxidised than Cu, and conversely Cu2+ has higher tendency to get reduced than Zn2+. At this stage, no reaction takes place in, either of the two beakers. Now, the two half−cells are connected by connecting wire through an ammeter., As soon as the connection is made, Zn rod starts dissolving i.e. Zn atom changes to Zn2+ by losing 2, electrons. The Zn2+ ion passes into the solution, thereby increasing its concentration and 2 electrons remain, on the Zn rod. Thus, ZnSO4 solution now has +2 unit extra charge and Zn rod has −2 unit extra charge., Thus, there is a charge separation of 4 units between the Zn rod and ZnSO4 solution. This charge, separation develops a potential referred to as oxidation potential; since oxidation takes place on this, electrode. The potential is called standard oxidation potential, if the concentration of Zn2+ in the solution is, 1 M. Similar potential is developed at Cu2+ | Cu half−cell also, when Cu2+ discharges and deposits on Cu, rod. This potential is referred to as reduction potential; as reduction takes place on this electrode. The, potential is called standard reduction potential, if the concentration of Cu2+ in the solution in 1 M., Current flow, , Flow of e−, , Ammeter, Anode (−), , Cathode (+), , Zn strip, , Salt bridge, (KCl or KNO3), , ZnSO4, solution, , Cu strip, , CuSO4, Solution, Figure 2. Setup of a Daniel cell, , The zinc rod which has the electrons left by the zinc (that got oxidized) becomes negatively, charged while the Cu rod which lost electrons to Cu2+ becomes positively charged. Thus, electric current, flows through the connected wire, which is indicated by a deflection in ammeter showing that a chemical, reaction is occurring in the cell. During the course of reaction, zinc rod gets dissolved and copper gets, deposited on the copper rod. Thus, the concentration of the anode solution increases while that of cathode, solution decreases. The flow of electrons occurs from the zinc rod to copper rod in the external circuit and, indirectly from cathode to anode in the internal circuit., , The current flow is in the direction opposite to electron flow. The reactions occurring at the two, electrodes are, At anode:, At cathode:, , Page number, , 11, , Zn(s) → Zn2+(aq) + 2e−, Cu2+(aq) + 2e− → Cu(s), , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , This current flow stops after some time. Think, why it happens? As zinc rod dissolves, let one, mole of Zn atoms dissolve to form Zn2+ ions, which pass into the solution and 2 mole of electrons remain, on the rod. Thus, rod becomes negatively charged and the solution becomes positively charged. With the, passage of time, the solution becomes so much positively charged that any Zn atom getting oxidised and, trying to get into the solution, would be repelled by the solution and thus the oxidation of Zn stops. Same, phenomenon occurs at cathode and the reduction of Cu2+ at cathode ceases. As the flow of electrons stop,, so does the flow of current. This problem can be removed with the use of salt bridge. The salt bridge, contains a highly soluble electrolyte (like KCl, NH4NO3, NH4Cl, KNO3 etc) in which ionic mobilities of, cation and anion are of comparable order., Now, let us examine the function of salt bridge. Since zinc ions are produced as electrons leave the, zinc electrode, this tends to produce a net positive charge in the left compartment. The salt bridge then, throws ions having equivalent opposite charge into the solution to maintain electrical neutrality. Thus, the, salt bridge keeps the solution neutral by passing appropriate amounts of cations or anions to the two, half−cells (compartments). Thus, the purpose of the salt bridge is to prevent any net charge accumulation, in either of the compartments. Salt bridge does not participate chemically in the cell reaction but it is, essential for the cell to operate. With the use of salt bridge, the flow of electrons becomes continuous and, cell continues to operate. But after some more time, cell ceases to operate. Can you guess why it stops, operating now?, CELL NOTATION OF AN ELECTROCHEMICAL CELL, (i) Anode is written on the left side and cathode is written on the right side., (ii) Phase boundaries are indicated by vertical bar or slash., (iii) Concentration of the electrolytes in the anode and cathode must be written in parenthesis., (iv) In case of a gas, the partial pressure is to be mentioned in atm or mm Hg., (v) A comma is used to separate two chemical species present in the same solution., (vi) A double vertical line i.e. || denotes that a salt bridge is present., (vii), EMF of the cell is written on the extreme right of the representation., For example,, (i) Zn(s) | ZnSO4(c1 M) || CuSO4(c2 M) | Cu(s), , ;, , Ecell, , (ii) Pt | H2(P1 atm) | HCl (c M) | AgCl(s) | Ag, , ;, , ', E cell, , (iii) Pt | Fe2+ (c1 M), Fe3+(c2 M) || Ag+ (c M) | Ag, , ;, , E "cell, , Note: In some cell representations (as in (ii) above), the salt bridge is not indicated which implies that, the electrolyte is common to both anode and cathode compartments., 7, , STANDARD REDUCTION, , When the concentrations of the Cu2+ and Zn2+ ions are both 1.0 M, we find that the voltage or emf, of the Daniell cell is 1.10 V at 25°C. This voltage must be related directly to the redox reaction, but how?, just as the overall cell reaction can be thought of as the sum of two half−cell reactions, the measured emf, of the cell can be treated as the sum of the electrical potentials at the Zn and Cu electrodes. Knowing one, of these electrode potentials, we could obtain the other by subtraction. It is impossible to measure the, potential of just a single electrode, but if we arbitrarily set the potential value of a particular electrode at, zero, we can use it to determine the relative potentials of other electrodes. The hydrogen electrode, serves, as the reference for this purpose. Hydrogen gas is bubbled into a hydrochloric acid solution at 25°C. The, platinum electrode has two functions. First, it provides a surface on which the dissociation of hydrogen, molecules can take place, H2 → 2H+ + 2e−, , Page number, , 12, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , Second, it serves as an electrical conductor to the external circuit. Under standard state conditions, (when the pressure of H2 is 1 atm and the concentration of the HCl solution is 1 M), the potential for the, reduction of H+ at 25°C is taken to be exactly zero:, 2H+ (1 M) + 2e− → H2(1 atm), ; E° = 0 V, The superscript “º” denotes standard−state conditions and E° is standard reduction potential, or, the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at, 1 atm. Thus, the standard reduction potential of the hydrogen electrode is defined as zero. The hydrogen, electrode is called the standard hydrogen electrode (SHE)., We can use the SHE to measure the potentials of other kinds of electrodes. For example, consider, an electrochemical cell with a zinc electrode and a SHE. In this case the zinc electrode is anode and the, SHE is the cathode. We deduce this fact from the decrease in mass of the zinc electrode during the, operation of the cell, which is consistent with the loss of zinc to the solution caused by the oxidation, reaction:, Zn(s) → Zn2+(aq) + 2e−, The cell diagram is, Zn(s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt(s), As mentioned earlier, the Pt electrode provides the surface on which the reduction takes place., When all the reactants are in their standard states (that is, H2 at 1 atm, H+ and Zn2+ ions at, 1 M each, the emf of the cell is 0.76 V at 25°C., We can write the half−cell reactions as follows:, Zn(s) → Zn2+ (1 M) + 2e−, 2H+ (1 M) + 2e− → H2 (1 atm), Zn(s) + 2H+ (1 M) → Zn2+ (1 M) + H2 (1 atm), , Anode (oxidation):, Cathode (reduction):, Overall:, , o, , By convention, the standard emf of the cell, E cell , which is composed of a contribution from the, anode and a contribution from the cathode, is given by, o, , o, , o, , E cell = E cathode − E anode, o, , o, , where E cathode and E anode are the standard reduction potential of the cathode and anode, respectively. For the Zn−SHE cell, we write, o, , o, , E cell = E H+ | H − E Zn2 + | Zn, o, , 2, , o, , 0.76 V = 0 − E Zn2 + | Zn, where the subscript H+ | H2 means 2H+ + 2e− → H2 and the subscript Zn2+ | Zn means Zn2+ +, o, , 2e− → Zn. Thus the standard reduction potential of zinc, E Zn2 + | Zn , is −0.76 V., The standard electrode potential of copper can be obtained in a similar fashion, by using a cell with, a copper electrode and a SHE. In this case, the copper electrode is the cathode because its mass increases, during the operation of the cell, as is consistent with the reduction reaction:, Cu2+(aq) + 2e− → Cu(s), The cell diagram is, Pt(s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu(s), and the half−cell reactions are, Anode (oxidation):, H2 (1 atm) → 2H+ (1 M) + 2e−, 2+, Cathode (reduction):, Cu (1 M) + 2e− → Cu(s), Overall: H2 (1 atm) + Cu2+ (1 M) → 2H+ (1 M) + Cu(s), Under standard−state conditions and at 25°C, the emf of the cell is 0.34 V, so we write, Page number, , 13, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , + 0.15, Cu2+ + e− → Cu+, 4+, −, 2+, + 0.13, Sn + 2e → Sn, +, −, 0.00, 2H + 2e → H2, 3+, −, Fe + 3e → Fe, − 0.036, 2+, −, Pb + 2e → Pb, − 0.126, 2+, −, Sn + 2e → Sn, − 0.14, −, −, AgI + e → Ag + I, − 0.151, 2+, −, Ni + 2e → Ni, − 0.25, 2+, −, Co + 2e → Co, − 0.28, 2+, −, Cd + 2e → Cd, − 0.403, 3+, −, 2+, Cr + e → Cr, − 0.41, 2+, −, Fe + 2e → Fe, − 0.44, 3+, −, Cr + 3e → Cr, − 0.74, 2+, −, Zn + 2e → Zn, − 0.762, −, −, 2H2O + 2e → H2 + 2OH, − 0.828, 2+, −, Mn + 2e → Mn, − 1.18, 3+, −, Al + 3e → Al, − 1.66, −, −, H2 + 2e → 2H, − 2.25, 2+, −, Mg + 2e → Mg, − 2.37, +, −, Na + e → Na, − 2.71, 2+, −, Ca + 2e → Ca, − 2.87, 2+, −, Ba + 2e → Ba, − 2.90, +, −, Cs + e → Cs, − 2.92, +, −, K + e → K, − 2.93, +, −, Li + e → Li, − 3.05, The above table lists standard reduction potentials for a number of half−cell reactions., By definition, the SHE has an E° value of 0.00 V. Below the SHE the negative standard reduction, potentials increase and above it the positive standard reduction potentials increase. It is important to know, the following points about the table in calculations:, 22., 23., 24., 25., 26., 27., 28., 29., 30., 31., 32., 33., 34., 35., 36., 37., 38., 39., 40., 41., 42., 43., 44., 45., 46., , 1. The E° values apply to the half−cell reactions as read in the forward (left to right) direction., 2. The more positive E° is, the greater the tendency for the substance to be reduced. For example,, the half−cell reaction, F2 (1 atm) + 2e− → 2F− (1 M), ; E° = 2.87 V, has the highest positive E° value among all the half−cell reactions. Thus, F2 is the strongest, oxidizing agent because it has the greatest tendency to be reduced. At the other extreme is the, reaction, Li+ (1 M) + e− → Li (s), ; E° = −3.05 V, +, which has the most negative E° value. Thus Li is the weakest oxidizing agent because it is the, most difficult species to reduce. Conversely, we say that F− is the weakest reducing agent and, Li metal is the strongest reducing agent. Under standard−state conditions, the oxidizing agents, (the species on the left−hand side of the half−reactions in the given table) increase in strength, from bottom to top and the reducing agents (the species on the right−hand side of the, half−reactions) increase in strength from top to bottom., 3. The half−cell reactions are reversible. Depending on the conditions, any electrode can act, either as an anode or as a cathode. Earlier we saw that the SHE is the cathode (H+ is reduced to, H2) when coupled with zinc in a cell and that it becomes the anode (H2 is oxidized to H+) when, used in a cell with copper., Page number, , 15, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , 4. Under standard−state conditions, any species on the left of a given half−cell reaction will react, spontaneously with a species that appears on the right of any half−cell reaction located below, it in the given table. This principle is sometimes called the diagonal rule. In the case of the, Daniel cell, Cu2+ (1 M) + 2e− → Cu(s), ; E° = 0.34 V, 2+, −, Zn (1 M) + 2e → Zn(s), ; E° = −0.76 V, We see that the substance on the left of the first half−cell reaction is Cu2+ and the substance on, the right in the second half−cell reaction is Zn. Therefore, as we saw earlier, Zn spontaneously, reduces Cu2+ to form Zn2+ and Cu., 5. Changing the stoichiometric coefficients of a half−cell reaction does not affect the value of E°, because electrode potentials are intensive properties. This means that the value of E° is, unaffected by the size of the electrodes or the amount of solutions present., For example,, I2(g) + 2e− → 2I− ( 1 M), ; E° = 0.53 V, but E° does not change if we multiply the half−reaction by 2 :, 2I2(s) + 4e− → 4I− (1 M), ; E° = 0.53 V, Question:, , Predict what will happen if molecular bromine (Br2) is added to a solution containing NaCl, and NaI at 25°C. Assume all species are in their standard states. [Use, Eo, , Cl2 |Cl −, , o, = 1.36V, E Br, , 2 |B r, , = 1.07V and E o, , I2 | I −, , = 0.53V], , Solution:, To predict what redox reaction(s) will take place, we need to compare the standard reduction, potentials for the following half−reactions., Cl2 (1 atm) + 2e− → 2Cl− (1 M), ; E° = 1.36 V, −, −, Br2(l) + 2e → 2Br (1 M), ; E° = 1.07 V, I2(s) + 2e− → 2I− (1 M), ; E° = 0.53 V, Applying the diagonal rule, we see that Br2 will oxidize I− but will not oxidize Cl−. Therefore, the, only redox reaction that will occur appreciably under standard state conditions is, Oxidation:, 2I− (1 M) → I2(s) + 2e−, Reduction: Br2(l) + 2e− → 2Br− (1 M), Overall:, 2I− (1 M) + Br2(l) → I2(s) + 2Br− (1 M), , PRACTICE PROBLEMS, 2+, , PP3. Can Sn reduce Zn (aq) under standard state conditions?, PP4. What is the standard emf of an electrochemical cell made of a Cd−electrode in a 1.0 M, Cd(NO3)2 solution and a Cr electrode in a 1.0 M Cr(NO3)3 solution? (E oCd2 + |Cd = –0.403 V ;, , E oCr + 3 |Cr = –0.74 V), , 8, , REDOX REACTIONS AS THE BASIS FOR TITRATION, , In acid-base systems we come across with a titration method for finding out the strength of one, solution against the other using a pH sensitive indicator. Similarly, in redox systems, the titration method, can be adopted to determine the strength of a reductant/ oxidant using a redox sensitive indicator. The, usage of indicators in redox titration is illustrated below:, (i), , In one situation, the reagent itself is intensely coloured, e.g. permanganate ion, MnO 4− . Here MnO −4 acts, as the self indicator. The visible end point in this case is achieved after the last of the reductant (Fe+2 or, , Page number, , 16, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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Redox Reaction, Cr2 O 72− + H2O, , ©Luwang Hostel, 2CrO 24− + 2H+, , be considered a redox reaction?, 5., , Nitric acid acts as an oxidising agent while nitrous acid acts both as an oxidising as well as a, reducing agent. Why?, , 6., , Write a balanced ionic equation to represent the oxidation of iodide ion (I−) by permanganate ion, , (MnO 4− ) in basic solution to yield molecular iodine (I2) and manganese (IV) dioxide (MnO2)., 7., , What is the function of salt bridge? What kind of electrolytes should be used in a salt bridge?, , 8., , What is a cell diagram? Write the cell diagram for an electrochemical cell consisting of an, aluminium electrode placed in a 1 M Al(NO3)3 solution and a silver electrode placed in 1 M, AgNO3 solution. Determine standard emf of the cell if E oAl+ 3 |Al = −1.66 V, E oAg+ |Ag = 0.799 V., , 9., , Determine the oxidation state of bold lettered atoms in the following reactions:, (a) 3KClO3 + 24 HCl → 8 KCl + 12H2O + 9Cl2 + 6ClO2, (b) 3I2 + 6NaOH → NaIO3 + 5NaI + 3H2O, , 10., , Identify the reactants getting oxidised or reduced in each of the following reactions., (a) CuSO4 + KI → 2CuI + I2 + 2K2SO4, (b) 2NaBr + Cl2 → 2NaCl + Br2, , 11., , State which of the following changes are oxidation, reduction, both or none?, (a) Na → NaOH, (b) Cl2 → Cl– + ClO 3−, , 12., , Six mole of Cl2 undergoes a loss and gain of 10 mole of electrons to form two oxidation states of, Cl (viz. Cl+5, Cl–1). Write down the complete reaction involved., , 13., , Arrange the following compounds in the order of, (a) increasing oxidation state of Mn: MnCl2, MnO2, KMnO4., (b) increasing oxidation state of N: NH3, N3H, N2O, NO, N2O5., (c) decreasing oxidation state of X: HXO, HXO3, HXO2, HXO4., , 14., , Name one compound each in which oxidation number of, (a) Oxygen is +2, (b) Oxygen is −1, (c) Hydrogen is −1, (d) Nitrogen is +1, (e) Cl is +4, (f) Oxygen is −2, , 15., , Balance the following redox reactions by ion electron method, (a) H2O2 + Fe2+ → Fe3+ + H2O, (in acidic medium), (b) Br2 → BrO 3− + Br−, , (in basic medium), , (c) S 2 O 32− + I2 → I− + S 4 O 26−, , (in acidic medium), , 16. State true or false, (a), A redox change involves oxidation and reduction both occurring simultaneously., (b), Oxidation number of Ni in Ni(CO)4 is +2., 17. Predict the reductant in the reaction C 2 O 24− + MnO 4− + H + → Mn +2 + CO 2 ,, , Page number, , 18, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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©Luwang Hostel, , Redox Reaction, , ANSWERS TO PRACTICE PROBLEMS, , PP1. Oxidant (reduces itself), (a) O2(g), (b) H+(aq), (c) Fe2O3, (d) FeCl3, PP2. (a) Hydrogen displacement reaction, (b) Combination reaction, (c) Disproportionation reaction, (d) Metal displacement reaction, , (, , o, , o, , Reductant (oxidizes itself), Zn(s), Zn(s), Al, H2S, , ), , PP3., , No E Sn2 + | Sn > E Zn2 + | Zn, , PP4., , E cell = E Cd2 + | Cd − E Cr 3 + |Cr = −0.403 −(−0.74) = 0.337 V., , o, , o, , o, , ANSWERS TO EXERCISE, CBSE PROBLEMS, 1., , 2., , Oxidant (reduces itself), Reductant (oxidizes itself), (a) H+, Zn, (b) I2, Na2S2O3, (c) Cl2, CH4, 3−, (d) [Fe(CN)6], H2O2, Si = −4 ; B = +3, O = +2, I = +7, Mn = +7, Cr = +6., , 3., , Oxidation state of Cl in HCl, HClO, ClO −4 , ClO2 are −1, +1, +7 and +4 respectively., , 4., , Oxidation state of Cr in Cr2 O 72− = +6., Oxidation state of Cr in CrO 24− = +6., So, the above reaction cannot be regarded as a redox reaction., , 5., , In HNO3, the oxidation state of N is maximum +5; while in HNO2, the oxidation state of, N is +3 and hence, HNO2 can act both as an oxidising as well as a reducing agent., , 6., , 6I− + 2MnO −4 + 4H2O → 3I2 + 2MnO2 + 8OH−, , 7., , The function of salt bridge is to maintain the solution neutral by passing appropriate amount of, cations or anions to the two half cells. Any one of the four electrolytes, namely. KCl, NH4Cl,, NH4NO3 or KNO3 can be used in a salt bridge., , 8., , Al | Al+3 (1M) || Ag+ (1M) | Ag ; E cell = E Ag+ | Ag – E Al3 + | Al = 0.799 – (–1.66) = 2.459 V., , 9., , 11., 12., , (a) Oxidation state of Cl in KClO3, HCl, KCl, Cl2 and ClO2 is +5, –1, –1, 0, and +4 respectively., (b) Oxidation state of I in I2, NaIO3 and NaI is 0, +5, and –1, (a) Oxidised: KI ; Reduced: CuSO4, (b) Oxidised: NaBr ; Reduced: Cl2, (a) Oxidation (b) Both oxidation and reduction., 6Cl2 → 2Cl+5 + 10 Cl–, , 13., , (a), , 10., , o, , +2, , Page number, , +4, , o, , o, , ++, , Mn Cl 2 , Mn O 2 , KM nO 4, , 20, , For any queries, , Whatsapp us at, , 8132835854, 8132827573

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Redox Reaction, –, , –3, , (b), , NH3 ,, +7, , 14., 15., , 1, +1, 3, N3 H, N 2, , ©Luwang Hostel, +2, , +5, , O, NO, N 2 O 5, , +5, , +3, , +1, , (c), (a), (a), , H XO 4 , H XO 3 , H XO 2 , H XO, OF2 (b) Na2O2 (c) NaH (d) N2O (e) ClO2 (f) H2O, H2O2 + 2Fe2+ + 2H+ → 2Fe3+ + 2H2O, , (b), , 3Br2 + 6OH− → BrO 3− + 5Br− + 3H2O, , (c), , 2S 2 O 32− + I2 → 2I − + S 4 O 26−, , 16., , (a) True, , (b) False. It is zero., , 17., , C 2 O 24−, , 18., , Number of moles of the Cl– ions oxidised =, Number of moles of electrons lost =, , 2, 35.5, , 2, 35.5, , 2, × 6.023 × 10 23 = 3.39 × 1022, 35.5, The possible values of x may 1, 3 or 5, zero to +5, , Number of electrons lost =, 19., 20., , Page number, , 21, , For any queries, , Whatsapp us at, , 8132835854, 8132827573