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 \(\mathrm{Cr}^{3+}\) (\mathbf{c} ) \text { aqueous } \mathrm { Hg } _ { 2 } ( \mathrm { NO } _ { 3 } ) _ { 2 } \text { to mercury metal. Write balanced } equations for these reactions.

First, we have to write the unbalanced chemical equations representing the reactions. a) \(\mathrm{KMnO}_4 + \mathrm{SO}_2 \rightarrow \mathrm{MnSO}_4\) b) \(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + \mathrm{SO}_2 \rightarrow \mathrm{Cr}^{3+} + \mathrm{K}^{+}\) c) \(\mathrm{Hg}_2(\mathrm{NO}_3)_2 + \mathrm{SO}_2 \rightarrow \mathrm{Hg} + \mathrm{NO}_3^-\)

Next, we will balance the atoms in the chemical equations using the oxidation-reduction method. a) Oxidation half-reaction: \(\mathrm{SO}_{2} \rightarrow \mathrm{SO}_{4}^{2-}\) Reduction half-reaction: \(\mathrm{MnO}_{4}^{-} \rightarrow \mathrm{Mn}^{2+}\) Multiply the oxidation half-reaction by 2: \(2 \mathrm{SO}_{2} \rightarrow 2 \mathrm{SO}_{4}^{2-}\) Multiply the reduction half-reaction by 5: \(5 \mathrm{MnO}_{4}^{-} \rightarrow 5 \mathrm{Mn}^{2+}\) Now add the half-reactions and balance the charges by adding appropriate numbers of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions, if needed: \(5 \mathrm{KMnO}_{4} + 2 \mathrm{SO}_{2} + 2 \cdot 2 \mathrm{H}^{+} \rightarrow 5 \mathrm{Mn}^{2+} + 2 \mathrm{SO}_{4}^{2-} + 5 \mathrm{K}^{+} + 8-5 \, \mathrm{H}^{+}\) \(5 \mathrm{KMnO}_{4} + 2 \mathrm{SO}_{2} + 6 \mathrm{H}^{+} \rightarrow 5 \mathrm{MnSO}_{4} + 8 \mathrm{H}_{2}\mathrm{O}\) b) Oxidation half-reaction: \(\mathrm{SO}_2 \rightarrow \mathrm{SO}_4^{2-}\) Reduction half-reaction: \(\mathrm{Cr}_2\mathrm{O}_7^{2-} \rightarrow \mathrm{Cr}^{3+}\) Multiply the oxidation half-reaction by 3: \(3 \mathrm{SO}_{2} \rightarrow 3 \mathrm{SO}_{4}^{2-}\) No multiplication needed for the reduction half-reaction. Now add the half-reactions and balance the charges by adding appropriate numbers of \(\mathrm{H}^{+}\): \(1 \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 3 \mathrm{SO}_2 + 6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{K}^{+} + 3 \mathrm{SO}_{4}^{2-} + 7 \, \mathrm{H_2O}\) c) Oxidation half-reaction: \(\mathrm{SO}_2 \rightarrow \mathrm{SO}_4^{2-}\) Reduction half-reaction: \(\mathrm{Hg}^{2+} \rightarrow \mathrm{Hg}\) No multiplication needed for the oxidation half-reaction. Multiply the reduction half-reaction by 2: \(2 \mathrm{Hg}^{2+} \rightarrow 2 \mathrm{Hg}\) Now add the half-reactions and balance the charges by adding appropriate numbers of \(\mathrm{H}^{+}\) and \(\mathrm{OH}^{-}\) ions, if needed: \(\mathrm{Hg}_2(\mathrm{NO}_3)_2 + \mathrm{SO}_2 + 2 \mathrm{H}^{+} \rightarrow 2 \mathrm{Hg} + \mathrm{SO}_4^{2-} + 4 \mathrm{NO}_3^-\) The three balanced equations are: a) \(5 \mathrm{KMnO}_{4} + 2 \mathrm{SO}_{2} + 6 \mathrm{H}^{+} \rightarrow 5 \mathrm{MnSO}_{4} + 8 \mathrm{H}_{2}\mathrm{O}\) b) \(1 \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 + 3 \mathrm{SO}_2 + 6 \mathrm{H}^{+} \rightarrow 2 \mathrm{Cr}^{3+} + 6 \mathrm{K}^{+} + 3 \mathrm{SO}_{4}^{2-} + 7 \, \mathrm{H_2O}\) c) \(\mathrm{Hg}_2(\mathrm{NO}_3)_2 + \mathrm{SO}_2 + 2 \mathrm{H}^{+} \rightarrow 2 \mathrm{Hg} + \mathrm{SO}_4^{2-} + 4 \mathrm{NO}_3^-\)