A disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following disproportionation reactions: (a) $\mathrm{Fe}^{2+}(a q) \longrightarrow \mathrm{Fe}(s)+\mathrm{Fe}^{3+}(a q)$ (b) $\mathrm{Br}_{2}(l) \longrightarrow \mathrm{Br}^{-}(a q)+\mathrm{BrO}_{3}^{-}(a q)$ (acidic solution) (c) $\mathrm{Cr}^{3+}(a q) \longrightarrow \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{Cr}(s)$ (acidic solution) (d) $\mathrm{NO}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{3}^{-}(a q)$ (acidic solution)

\(Fe^{2+}(aq) \longrightarrow Fe(s) + Fe^{3+}(aq)\) First, identify the changes in oxidation numbers for the iron atoms: \( Fe^{2+}(aq)\) is reduced to \( Fe(s)\): Oxidation number decreases from +2 to 0. \( Fe^{2+}(aq)\) is oxidized to \( Fe^{3+}(aq)\): Oxidation number increases from +2 to +3. To balance the reaction, we have to make sure that the number of electrons transferred is the same during oxidation and reduction. The difference in electrons here is 1: Reduction: \( Fe^{2+}(aq) \longrightarrow Fe(s) + 1e^{-}\). Oxidation: \( Fe^{2+}(aq) \longrightarrow Fe^{3+}(aq) + 1e^{-}\). Now we can write the balanced disproportionation reaction: \( 2Fe^{2+}(aq) \longrightarrow Fe(s) + 2Fe^{3+}(aq) \).

\(Br_{2}(l) \longrightarrow Br^{-}(aq) + BrO_3^{-}(aq)\) (acidic solution) First, identify the changes in oxidation numbers for the bromine atoms: \( Br_{2}(l)\) is reduced to \( Br^{-}(aq)\): Oxidation number decreases from 0 to -1. \( Br_{2}(l)\) is oxidized to \( BrO_3^{-}(aq)\): Oxidation number increases from 0 to +5. To balance the reaction: Reduction: \( Br_{2}(l) \rightarrow 2Br^{-}(aq) + 2e^{-} \). Oxidation: \( Br_{2}(l) + 10e^{-} \rightarrow 2BrO{_3}^{-}(aq) + 12H^{+}(aq) \) (Considering the acidic solution). To balance the electrons, we must multiply the reduction half-reaction by 5: \(5 (Br_{2}(l) \rightarrow 2Br^{-}(aq) + 2e^{-})\). Now, we can write the balanced disproportionation reaction in acidic solution: \(3Br_{2}(l) + 12H^{+}(aq) \longrightarrow 6Br^{-}(aq) + 2BrO_3^{-}(aq) + 6H_{2}O(l)\).

\(Cr^{3+}(aq) \longrightarrow Cr_{2}O_{7}^{2-}(aq) + Cr(s)\) (acidic solution) First, identify the changes in oxidation numbers for the chromium atoms: \( Cr^{3+}(aq)\) is reduced to \( Cr(s)\): Oxidation number decreases from +3 to 0. \( Cr^{3+}(aq)\) is oxidized to \( Cr_{2}O_{7}^{2-}(aq)\): Oxidation number increases from +3 to +6. To balance the reaction: Reduction: \(2Cr^{3+}(aq) + 6e^{-} \rightarrow 2Cr(s)\). Oxidation: \(2Cr^{3+}(aq) \rightarrow Cr_{2}O_{7}^{2-}(aq) + 6e^{-} + 14H^{+}(aq)\) (Considering the acidic solution). Now we can write the balanced disproportionation reaction in acidic solution: \( 3Cr^{3+}(aq) + 14H^{+}(aq) \longrightarrow Cr_{2}O_{7}^{2-}(aq) + 2Cr(s) + 7H_{2}O(l)\).

\(NO(g) \longrightarrow N_{2}(g) + NO_{3}^{-}(aq)\) (acidic solution) First, identify the changes in oxidation numbers for the nitrogen atoms: \( NO(g)\) is reduced to \( N_{2}(g)\): Oxidation number decreases from +2 to 0. \( NO(g)\) is oxidized to \( NO_{3}^{-}(aq)\): Oxidation number increases from +2 to +5. To balance the reaction: Reduction: \(2NO(g) + 2e^{-} \rightarrow N_{2}(g)\). Oxidation: \(4NO(g) + 6H_2O(l) \rightarrow 4NO_{3}^{-}(aq) + 12H^{+}(aq) + 6e^{-}\) (Considering the acidic solution). To balance the electrons, we must multiply the reduction half-reaction by 3: \(3 (2NO(g) + 2e^{-} \rightarrow N_{2}(g))\). Now, we can write the balanced disproportionation reaction in acidic solution: \(6NO(g) + 6H_{2}O(l) \longrightarrow 2N_{2}(g) + 6NO_{3}^{-}(aq) + 12H^{+}(aq)\).