What is a disproportionation reaction? Use the following reduction potentials $$ \begin{aligned} \mathrm{ClO}_{3}^{-} &+3 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}_{2}+\mathrm{H}_{2} \mathrm{O} & & \mathscr{E}^{\circ}=1.21 \mathrm{~V} \\ \mathrm{HClO}_{2}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO}+\mathrm{H}_{2} \mathrm{O} & & \mathscr{E}^{\circ}=1.65 \mathrm{~V} \end{aligned} $$ to predict whether \(\mathrm{HClO}_{2}\) will disproportionate.

In this exercise, we can see that \(\mathrm{HClO}_{2}\) is the product of the first reaction and the reactant in the second reaction. Therefore, the possible oxidation half-reaction is: $$ \mathrm{HClO}_{2} \longrightarrow \mathrm{ClO}_{3}^{-} + 3\mathrm{H}^{+} + 2 \mathrm{e}^{-} $$ and the possible reduction half-reaction is: $$ \mathrm{HClO}_{2} + 2\mathrm{H}^{+} + 2 \mathrm{e}^{-} \longrightarrow \mathrm{HClO} + \mathrm{H}_{2}\mathrm{O} $$

To find the reduction potential for the oxidation half-reaction, we need to reverse the first given reaction and add together the potentials of the reversed reaction and the reduction half-reaction. Reverse the first reaction to get: $$ \mathrm{HClO}_{2} - 3\mathrm{H}^{+} - 2\mathrm{e}^{-} \longrightarrow \mathrm{ClO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} $$ with the potential \(\mathscr{E}_{1}^{\circ}= -1.21\mathrm{~V}\). Now, add this reversed reaction to the reduction half-reaction: $$ \mathrm{HClO}_{2} + 3\mathrm{H}^{+} + 2\mathrm{e}^{-} - (\mathrm{HClO}_{2} - 3\mathrm{H}^{+} - 2\mathrm{e}^{-}) \longrightarrow \mathrm{ClO}_{3}^{-} + \mathrm{H}_{2}\mathrm{O} - (\mathrm{HClO} + \mathrm{H}_{2}\mathrm{O}) $$ Combined, this results in: $$ \mathrm{HClO}_{2} - \mathrm{HClO} - \mathrm{e}^{-} \longrightarrow \mathrm{ClO}_{3}^{-} $$ with a potential \(\mathscr{E}_{2}^{\circ}= \mathscr{E}_{1}^{\circ} + \mathscr{E}_{3}^{\circ}= (-1.21 + 1.65) \mathrm{~V} = 0.44\mathrm{~V}\).

Since the overall potential of the reaction, \(\mathscr{E}_{2}^{\circ}= 0.44\mathrm{~V}\), is positive, the reaction is energetically favorable and will proceed spontaneously. Therefore, \(\mathrm{HClO}_{2}\) will disproportionate in this reaction.