An aqueous solution of \(\mathrm{SO}_{2}\) reduces (a) aqueous \(\mathrm{KMnO}_{4}\) to \(\mathrm{MnSO}_{4}(a q),(\mathbf{b})\) acidic aqueous \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to aqueous \(\mathrm{Cr}^{3+},(\mathrm{c})\) aqueous \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}\) to mercury metal. Write balanced equations for these reactions.

Step 1: Write the half reactions Oxidation half-reaction (SO2 to SO4^2-): \(SO_{2} \to SO_{4}^{2-}\) Reduction half-reaction (MnO4^- to Mn^2+): \(MnO_{4}^{-} \to Mn^{2+}\) Step 2: Balance the atoms Oxidation half-reaction: \(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+}\) Reduction half-reaction: \(5e^{-} + 8H^{+} + MnO_{4}^{-} \to Mn^{2+} + 4H_{2}O\) Step 3: Balance the charges In this case, charges are balanced in the half-reactions. Step 4: Combine the half-reactions Multiply the oxidation half-reaction by 5 and reduction half-reaction by 2, and then add. \(5(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+})\) \(2(5e^{-} + 8H^{+} + MnO_{4}^{-} \to Mn^{2+} + 4H_{2}O)\) Reaction 1 Balanced Equation: \(5SO_{2} + 2KMnO_{4} + 2H_{2}O \to 5SO_{4}^{2-} + 2Mn^{2+} + 12H^{+} + 2K^{+}\) \(5SO_{2} + 2KMnO_{4} + 2H_{2}O \to K_{2}Mn_{2}(SO_{4})_{5} + 6H_{2}O\)

Step 1: Write the half reactions Oxidation half-reaction (SO2 to SO4^2-): \(SO_{2} \to SO_{4}^{2-}\) Reduction half-reaction (Cr2O7^2- to Cr^3+): \(Cr_{2}O_{7}^{2-} \to 2Cr^{3+}\) Step 2: Balance the atoms Oxidation half-reaction: \(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+}\) Reduction half-reaction: \(6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \to 2Cr^{3+} + 7H_{2}O\) Step 3: Balance the charges In this case, charges are balanced in the half-reactions. Step 4: Combine the half-reactions Multiply the oxidation half-reaction by 6 and reduction half-reaction by 1, and then add. \(6(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+})\) \(6e^{-} + 14H^{+} + Cr_{2}O_{7}^{2-} \to 2Cr^{3+} + 7H_{2}O\) Reaction 2 Balanced Equation: \(6SO_{2} + K_{2}Cr_{2}O_{7} + 14H^{+} \to 6SO_{4}^{2-} + 4H_{2}O + 2Cr^{3+} + 2K^{+}\) \(6SO_{2} + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} \to 6H_{2}SO_{4} + 4H_{2}O + 2Cr^{3+} + 2K^{+}\) \(6SO_{2} + K_{2}Cr_{2}O_{7} + 7H_{2}SO_{4} \to 6H_{2}SO_{4} + Cr_{2}(SO_{4})_{3} + K_{2}SO_{4} + 4H_{2}O\)

Step 1: Write the half reactions Oxidation half-reaction (SO2 to SO4^2-): \(SO_{2} \to SO_{4}^{2-}\) Reduction half-reaction (Hg^2+ to Hg): \(Hg^{2+} \to Hg\) Step 2: Balance the atoms Oxidation half-reaction: \(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+}\) Reduction half-reaction: \(2e^{-} + Hg^{2+} \to Hg\) Step 3: Balance the charges In this case, charges are balanced in the half-reactions. Step 4: Combine the half-reactions Multiply the oxidation half-reaction by 1 and reduction half-reaction by 4, and then add. \(SO_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+})\) \(4(2e^{-} + Hg^{2+} \to Hg)\) Reaction 3 Balanced Equation: \(SO_{2} + 4Hg^{2+} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+} + 4Hg\) \(SO_{2} + Hg_{2}(NO_{3})_{2} + 2H_{2}O \to SO_{4}^{2-} + 4H^{+} + 2Hg + 2NO_{3}^{-}\) \(SO_{2} + Hg_{2}(NO_{3})_{2} + 2H_{2}O \to H_{2}SO_{4} + 2Hg + 2HNO_{3}\)