Question

What is a disproportionation reaction? Use the following reduction potentials:

$\mathbf{Cl}{{\mathbf{O}}_{\mathbf{3}}}^{\mathbf{-}}\mathbf{+}\mathbf{3}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }{\mathbf{HClO}}_{\mathbf{2}}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$$\mathit{\epsilon }\mathbf{°}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{90}\mathbf{}\mathbf{V}$

${\mathbf{HClO}}_{\mathbf{2}}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{+}\mathbf{2}{\mathbf{e}}^{\mathbf{-}}\mathbf{\to }\mathbf{HClO}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$$\mathit{\epsilon }\mathbf{°}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{65}\mathbf{}\mathbf{V}$

To predict whether${\mathbf{HClO}}_{\mathbf{2}}$will be disproportionate.

Short answer

The overall reaction of disproportionation of${\mathrm{HClO}}_2$ is said to be a spontaneous reaction.

Step-by-step solution

Define spontaneous reaction

A reaction that favours the formation of products at the conditions under which the reaction is occurring is called aspontaneous reaction.

Evaluating sum for standard voltage

Chlorous acid is an inorganic compound having the formula${\mathrm{HClO}}_2$. It is said to be a weak acid.

We then can calculate the likelihood of${\mathrm{HClO}}_2$to undergo a disproportionation reaction based on the given half-reactions in the question.

Redox reactions are chemical reactions that are expressed in terms of reduction and oxidation of the half-reactions. They involve the transfer of electrons between two species.

Oxidation reaction involves the loss of electrons, whereas reduction happens by gaining electrons.

The standard voltage ($\mathrm{E}°$) for the cell is the sum of the standard voltages of the reduction and oxidation half-reactions.

$$\begin{gathered} {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{red}}-{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{oxd}} \\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cathode}}-{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{anode}} \end{gathered}$$

Evaluating disproportion reaction

A disproportionation reaction is said to be a redox reaction in which the same compound gets oxidized and reduced into two separate half-reactions.

Based on the reaction potentials, the half-reaction with a higher (positive)$\mathrm{E}°$serves as the cathode.

On the other hand, the one with lower$\mathrm{E}°$serves as the anode.

Then the half-reactions are:

The cathode (reduction) is: ${\mathrm{HClO}}_2+2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to \mathrm{HClO}+{\mathrm{H}}_2\mathrm{O}$

The anode (oxidation) is: ${\mathrm{HClO}}_2+{\mathrm{H}}_2\mathrm{O}\to +3{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}$

The overall reaction is: $2{\mathrm{HClO}}_2\to {\mathrm{ClO}}_3^{-}+{\mathrm{H}}^{+}+\mathrm{HClO}$

Based on the overall reaction,${\mathrm{HClO}}_2$gets oxidized to ${\mathrm{HClO}}_2$, and then reduced to $\mathrm{HClO}$.

If we want to know whether this reaction is spontaneous or not.

Then, we need to calculate the standard voltage of the cell.

Here ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cathode}}=1.65\mathrm{V}$and ${{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{anode}}=1.21\mathrm{V}$.

Substitute the values as:

role="math" localid="1663831284313" $$\begin{gathered} {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}={{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cathode}}-{{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{anode}} \\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}=1.65\mathrm{V}-(1.21\mathrm{V}) \\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}=1.65\mathrm{V}-1.21\mathrm{V} \\ {{\mathrm{E}}^{\mathrm{o}}}_{\mathrm{cell}}=0.44\mathrm{V} \end{gathered}$$

A positive voltage that forms across the electrodes of a voltaic cell indicates that the oxidation-reduction reaction is said to be a spontaneous reaction for reduction at the cathode and oxidation at the anode.

Therefore, the disproportionation of${\mathrm{HClO}}_2$is a spontaneous reaction.