(a) The skeleton equation is : $\mathrm{MnO_4^{-}}_{(aq)} + \mathrm{I^{-}}_{(aq)}$ → $\mathrm{MnO}_{2(s)} + \mathrm{I}_{2(s)}$ (i) The O.N. of the atoms involved in the equation is: $\left(\overset{+7}{\mathrm{Mn}}\overset{-4}{\mathrm{O}_4}^{-}\right) + \mathrm{I}^{\overset{-1}{-}}$ → $\overset{+4}{\mathrm{Mn}}\overset{-2}{\mathrm{O}_2}+\overset{0}{\mathrm{I}_2}$ (ii) The species involved in the oxidation and reduction half reaction : (iii) Oxidation half reaction : $\mathrm{I}^{-}$ → $\mathrm{I}_2$ Reduction half reaction : $\mathrm{MnO}_4^{-}$ → $\mathrm{MnO}_2$ (iv) Balancing the oxidation half reaction $2\mathrm{I}^{-}$ → $\mathrm{I}_2 + 2e^{-}$ ...(i) (v) Balancing the reduction half reaction (1) As the decrease in O.N. is 3, therefore, adding 3e– on the reactant side $\mathrm{MnO}_4^{-} + 3e^{-}$ → $\mathrm{MnO}_2$ (2) To balance the oxygen atoms, add two H2O molecules on the product side $\mathrm{MnO}_4^{-} + 3e^{-}$ → $\mathrm{MnO}_2 + 2\mathrm{H}_2\mathrm{O}$ (3) To balance the charges, add 4 $\mathrm{OH}^{-}$ on the product side. Then to balance H atoms, add four $\mathrm{H}_2\mathrm{O}$ molecules on the reactant side. $\mathrm{MnO}_4^{-} + 3e^{-} + 4\mathrm{H}_2\mathrm{O}$ → $\mathrm{MnO}_2 + 4\mathrm{OH}^{-} + 2\mathrm{H}_2\mathrm{O}$ $\mathrm{MnO}_4^{-} + 3e^{-} + 2\mathrm{H}_2\mathrm{O}$ → $\mathrm{MnO}_2 + 4\mathrm{OH}^{-}$ ...(ii) Thus, the reduction half reaction is balanced. (vi) Adding the two half reactions In order to equate the electrons, multiply eqn. (i) by 3 and eqn. (ii) by 2. Add the two equations. $\left[2\mathrm{I}^{-} \rightarrow \mathrm{I}_2 + 2e^{-}\right] \times 3$ [$\mathrm{MnO}_4^{-} + 3e^{-}+2\mathrm{H}_2\mathrm{O}$ → $\mathrm{MnO}_2+4\mathrm{OH}^{-}$] × 2 ––––––––––––––––––––––––– $2\mathrm{MnO}_4^{-} + 6\mathrm{I}^{-} + 4\mathrm{H}_2\mathrm{O}$ → $3\mathrm{I}_2 + 2\mathrm{MnO}_2 + 8\mathrm{OH}^{-}$ or $2\mathrm{MnO_4^{-}}_{(aq)} + 6\mathrm{I^{-}}_{(aq)} + 4\mathrm{H_2O}_{(l)}$ → $3\mathrm{I}_{2(s)} + 2\mathrm{MnO}_{2(s)} + 8\mathrm{OH}^{-}_{(aq)}$ (b) The skeleton equation is : $\mathrm{MnO_4^{-}}_{(aq)} + \mathrm{SO_2}_{(g)}$ → $\mathrm{Mn}^{2+}_{(aq)} + \mathrm{HSO}_{4(aq)}^{-}$ (i) The O.N. of atoms involved in the equation is : $\overset{+7}{\mathrm{Mn}}\overset{-2}{\mathrm{O}_4}^{-} + \overset{+4}{\mathrm{S}}\overset{-2}{\mathrm{O}_2}$ → $\mathrm{Mn}^{\overset{+2}{2+}} + \overset{+1}{\mathrm{H}}\overset{+6}{\mathrm{S}}\overset{-2}{\mathrm{O}_4}^{-}$ (ii) The species involved in the oxidation and reduction half reactions : (iii) Oxidation half reaction : $\mathrm{SO}_2$ → $\mathrm{HSO}_4^{-}$ Reduction half reaction : $\mathrm{MnO}_4^{-}$ → $\mathrm{Mn}^{2+}$ (iv) Balancing the oxidation half reaction (1) As the increase in O.N. is 2, therefore, add two electrons on the product side to balance change in O.N. $\mathrm{SO}_2$ → $\mathrm{HSO}_4^{-} + 2e^{-}$ (2) In order to balance the number of oxygen atoms, add two H2O molecules on the reactant side and then to balance H atoms add 3H+ on the product side. $\mathrm{SO}_2 + 2\mathrm{H}_2\mathrm{O}$ → $\mathrm{HSO}_4^{-} + 3\mathrm{H}^{+} + 2e^{-}$ ... (i) (v) Balancing the reduction half reaction The reduction half reaction is : $\mathrm{MnO}_4^{-}$ → $\mathrm{Mn}^{2+}$ (1) As the decrease in O.N. is 5, therefore add $5e^{-}$ on the reactant side, $\mathrm{MnO}_4^{-} + 5e^{-}$ → $\mathrm{Mn}^{2+}$ (2) In order to balance the no. of oxygen atoms, add four $\mathrm{H}_2\mathrm{O}$ molecules on the product side and then to balance H atoms add 8 $\mathrm{H}^{+}$ on the reactant side. $\mathrm{MnO}_4^{-} + 8\mathrm{H}^{+} + 5e^{-}$ → $\mathrm{Mn}^{2+} + 4\mathrm{H}_2\mathrm{O}$ ... (ii) (vi) Adding the two half, reactions In order to equate the electrons, multiply eqn. (i) by 5 and eqn. (ii) by 2. Add the two eqns. [$\mathrm{SO}_2+2\mathrm{H}_2\mathrm{O}$ → $\mathrm{HSO}_4^{-}+3\mathrm{H}^{+}+2e^{-}$] × 5 [$\mathrm{MnO}_4^{-}+8\mathrm{H}^{+}+5e^{-}$ → $\mathrm{Mb}^{2+}+4\mathrm{H}_2\mathrm{O}$] × 2 –––––––––––––––––––––––– → $5{\mathrm{HSO}_4^{-}}_{(aq)}+2{\mathrm{Mn}^{2+}}_{(aq)}$ (c) Oxidation half equation : $\mathrm{Fe}^{2+}_{(aq)}$ → $\mathrm{Fe}^{3+}_{(aq)} + e^{-}$ ... (i) Reduction half equation : $\mathrm{H}_2\mathrm{O}_{2(aq)} + 2\mathrm{H}^{+}_{(aq)} + 2e^{-}$ → $2\mathrm{H}_2\mathrm{O}_{(l)}$ ... (ii) Multiplying eqn. (i) by 2 and adding it to eqn. (ii), we get $\mathrm{H}_2\mathrm{O}_{2(aq)} + 2\mathrm{Fe}^{2+}_{(aq)} + 2\mathrm{H}^{+}_{(aq)}$ → $2\mathrm{Fe}^{3+}_{(aq)} + 2\mathrm{H}_2\mathrm{O}_{(l)}$ (d) Oxidation half equation : $\mathrm{SO}_{2(g)} + 2\mathrm{H}_2\mathrm{O}_{(l)}$ → $\mathrm{SO}_4^{2-}{}_{(aq)} + 4\mathrm{H}^{+}{}_{(aq)} + 2e^{-}$ ... (i) Reduction half equation : $\mathrm{Cr}_2{\mathrm{O}_7^{2-}}_{(aq)} + 14{\mathrm{H}^{+}}_{(aq)} + 6e^{-}$ → $2\mathrm{Cr}^{3+}_{(aq)} + 7\mathrm{H}_2\mathrm{O}_{(l)}$ ... (ii) Multiplying eqn. (i) by 3 and adding to eqn. (ii) we get, $\mathrm{Cr}_2\mathrm{O}_{7(aq)}^{2-} + 3\mathrm{SO}_{2(g)} + 2\mathrm{H}_{(aq)}^{+}$ → $2\mathrm{Cr}^{3+}_{(aq)} + 3{\mathrm{SO}_4^{2-}}_{(aq)} + \mathrm{H}_2\mathrm{O}_{(l)}$