8.18. Balance the following redox reactions by ion-electron method. (a) MnO4–(aq) +I–(aq) → MnO2 (s) + I2 (s) (in basic medium) (b) MnO4–(aq) + SO2(g) → Mn2+(aq) +H2SO4–(in acidic solution) (c) H2O2 (aq) + Fe2+(aq) → Fe3+(aq) + H2O(l) (in acidic solution) (d) Cr2O72- (aq) + SO2 (g) → Cr3+ (aq) + SO42-(aq) (in acidic solution)

(a) The balanced half reaction equations are: Oxidation half equation: I− (aq) → I2 (s) - (i) Reduction half reaction equation: MnO4− (aq) → MnO2 (aq) - (ii) Balance I atoms and charges in the oxidation half reaction. 2I− (aq) → I2 (s) + 2e− In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction. MnO4− (aq) + 3e−→ MnO2 (aq) Balance charge in the reduction half reaction by adding 4 hydroxide ions to product side. MnO4− (aq) + 3e− → MnO2 (aq)+4OH− To balance O atoms, add 2 water molecules to reactant side. MnO4− (aq)+ 3e− +2H2O→MnO2 (aq) + 4OH− To equalize the number of electrons, multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2. 6I− (aq) → 3I2 (s)+6e− 2MnO4− (aq) + 6e−+4H2O → 2MnO2 (aq)+8OH− Add two half-cell reactions to obtain the balanced equation. 2MnO4−? (aq) + 6I− (aq) + 4H2? O2? (l) → 2MnO2? (s) + 3I2? (s) + 8OH− (b) The balanced half reaction equations are: Oxidation half equation: SO2 (g) + 2H2O (l) → HS04– (aq) + 3H+ (aq) +2e– … (i) Reduction half equation: MnO4– (aq) + 8H+ (aq) + 5e– → Mn2+ (aq) + 4H2O (l) ………. (ii) Multiply Eq. (i) by 3 and Eq. (ii) by 2 and add, we have, 2MnO4– (aq) + 5S02 (g) + 2H20 (l) + H+ (aq) → 2Mn2+ (aq) + 5HSO4– (aq) (c) Oxidation half equation: Fe2+ (aq) → Fe3+ (aq) + e– … (i) Reduction half equation: H2O2 (aq) + 2H+ (aq) + 2e– → 2H2O (l) … (ii) Multiply Eq. (i) by 2 and add it to Eq. (ii), we have, H2O2 (aq) +2Fe2+ (aq) +2H+ (aq) → 2Fe3+ (aq) + 2H2O (l) (d)Oxidation half equation: SO2 (g) + 2H2O (l) → SO42- (aq) + 4H+ (aq) + 2 e– … (i) Reduction half equation: Cr2O72– (aq) + 14H+ (aq) + 6e– → 2Cr3+ (aq) + 7H2O (l) … (ii) Multiply Eq. (i) by 3 and add it to Eq. (ii), we have, Cr2O72– (aq) + 3SO2 (q) + 2H+ (aq) → 2Cr3+ (aq) + 3SO42- (aq) + H2O (l)