The following standard reduction potentials have been det mined for the aqueous chemistry of indium: $$ \begin{array}{cl} \mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V} \end{array} $$ a. What is the equilibrium constant for the disproportionation reaction, where a species is both oxidized and reduced, shown below? $$ 3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ b. What is \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\operatorname{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q) ?\)

We are given the disproportionation reaction: $$ 3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$ In this reaction, In+ ions are both oxidized and reduced.

To determine the overall redox reaction, we need to find an oxidation and reduction reaction that will give the disproportionation reaction when both are combined. Based on the given reduction reactions: 1. Reduction reaction: $\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q)$ 2. Oxidation reaction (reverse of the second given reaction): \(\mathrm{In}^{+}(a q) \longrightarrow \operatorname{In}(s) + \mathrm{e}^{-}\) Now, we need to balance these two half-reactions to combine them properly. To do this, we multiply the reduction reaction by 2 and the oxidation reaction by 3: 1. Balanced reduction reaction: $2[\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q)]$ 2. Balanced oxidation reaction: \(3[\mathrm{In}^{+}(a q) \longrightarrow \operatorname{In}(s) + \mathrm{e}^{-}]\) When we combine the two half-reactions, we get the disproportionation reaction: $$ 3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q) $$

To calculate the standard potential E°, subtract the oxidation reaction's standard reduction potential from the reduction reaction's standard reduction potential: $$ \mathscr{E}^{\circ}_{cell}=\mathscr{E}^{\circ}_{red} - \mathscr{E}^{\circ}_{ox} $$ Insert the values of the standard reduction potential: $$ \mathscr{E}^{\circ}_{cell}=(-0.444 \mathrm{~V}) - (-0.126 \mathrm{~V})=-0.318 \mathrm{~V} $$

Now, we will use the Nernst equation to find the equilibrium constant, K: $$ \mathscr{E}^{\circ}= \frac{-RT}{nF} \ln K $$ Where: - \(\mathscr{E}^{\circ}\) is the standard cell potential - R is the gas constant, 8.314 J/(mol·K) - T is the temperature in Kelvin, 298K - n is the total number of moles of electrons transferred. In this case, n = 6. - F is the Faraday constant, 96485 C/mol Plugging the values into the equation: $$ -0.318V = \frac{-8.314 J/(mol \cdot K) \cdot 298K}{6 \cdot 96485 C/mol} \ln K $$ Now, we can solve for K: $$ K = e^{\frac{-0.318V \cdot 6 \cdot 96485 C/mol}{8.314 J/(mol \cdot K) \cdot 298K}}=9.7 \times 10^7 $$ So, the equilibrium constant, K, for the disproportionation reaction is \(9.7 \times 10^7\).

We are given the standard Gibbs free energy of formation for In³+(aq), ΔG°\(_\text{f}\) = -97.9 kJ/mol. We can use the following relationship between ΔG°, standard cell potential, and Faraday constant: $$ \Delta G^{\circ}=-n F \mathscr{E}^{\circ} $$ First, we calculate the overall ΔG° of the disproportionation reaction: $$ \Delta G^{\circ}_{disproportionation} = -6 \cdot 96485 C/mol \cdot (-0.318V) = 184.2 \mathrm{~kJ/mol} $$ Now, let's write the ΔG°\(_\text{f}\) for the overall reaction: $$ \Delta G^{\circ}_{disproportionation} = 2 \Delta G^{\circ}_{\text{f}}(\mathrm{In}^{3+}(aq)) - 3 \Delta G^{\circ}_{\text{f}} (\mathrm{In}^{+}(aq)) $$ Plug in the value of ΔG°\(_\text{f}\)(In³+(aq)) and ΔG°\(_\text{f}\): $$ 184.2 \mathrm{~kJ/mol} = 2(-97.9 \mathrm{~kJ/mol}) - 3 \Delta G^{\circ}_{\text{f}} (\mathrm{In}^{+}(aq)) $$ Now, solve for ΔG°\(_\text{f}\)(In+(aq)): $$ \Delta G^{\circ}_{\text{f}} (\mathrm{In}^{+}(aq)) = \frac{2(-97.9 \mathrm{~kJ/mol}) - 184.2 \mathrm{~kJ/mol}}{-3} = 102.6 \mathrm{~kJ/mol} $$ So, the standard Gibbs free energy of formation, ΔG°\(_\text{f}\), for In+(aq) is 102.6 kJ/mol.