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

Disproportionation reactions are redox reactions in which a single reactant takes part both as the reducing agent and the oxidizing agent. In these reactions, the reactant is simultaneously oxidized and reduced, leading to the formation of two different products. The species undergoing oxidation loses electrons, whereas the species undergoing reduction gains electrons.

We are given the reduction potentials of two reactions involving the species HClO2: Reaction 1: ClO3- + 3H+ + 2e- → HClO2 + H2O E°₁ = 1.21 V Reaction 2: HClO2 + 2H+ + 2e- → HClO + H2O E°₂ = 1.65 V We can use these reduction potentials to determine whether the disproportionation of HClO2 into ClO3- and HClO is possible by calculating the overall spontaneous reaction direction.

First, we need to determine the half-reactions for the disproportionation of HClO2: Oxidation half-reaction (reverse of Reaction 2): HClO2 → HClO + H+ + e- E°₃ = -1.65 V Reduction half-reaction (Reaction 1): ClO3- + 6H+ + 2e- → 2HClO2 + 2H2O E°₄ = 1.21 V Now, we need to balance the two half-reactions so that the number of electrons is equal: Oxidation half-reaction (balanced): 2HClO2 → 2HClO + 2H+ + 2e- 2xE°₃ = -3.30 V Reduction half-reaction (unchanged): ClO3- + 6H+ + 2e- → 2HClO2 + 2H2O E°₄ = 1.21 V Next, we find the potential for the disproportionation reaction by adding the oxidation and reduction half-reactions: Overall reaction: 2HClO2 + ClO3- + 6H+ → 2HClO + 2H+ + 2H2O E° (overall) = E°₄ + 2xE°₃ = 1.21 - 3.30 = -2.09 V

As the potential difference for the disproportionation reaction (E° (overall)) is negative (-2.09 V), it indicates that the disproportionation of HClO2 is non-spontaneous and unfavorable. Therefore, HClO2 will not disproportionate under standard conditions.