The following standard reduction potentials have been determined for the aqueous chemistry of indium: $$\begin{array}{ll}\mathrm{In}^{3+}(a q)+2 \mathrm{e}^{-} \longrightarrow \operatorname{In}^{+}(a q) & \mathscr{E}^{\circ}=-0.444 \mathrm{~V} \\ \mathrm{In}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \operatorname{In}(s) & \mathscr{E}^{\circ}=-0.126 \mathrm{~V}\end{array}$$ 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 \(\mathrm{In}^{+}(a q)\) if \(\Delta G_{\mathrm{f}}^{\circ}=-97.9 \mathrm{~kJ} / \mathrm{mol}\) for \(\mathrm{In}^{3+}(a q)\) ?

Since indium is both oxidized and reduced in the disproportionation reaction, we need to divide the reaction into two half-reactions. For the disproportionation reaction: \(3 \operatorname{In}^{+}(a q) \longrightarrow 2 \operatorname{In}(s)+\operatorname{In}^{3+}(a q)\) We can split it into half-reactions as: Reduction half-reaction: \(2\operatorname{In}^{+}(a q) + e^{-} \longrightarrow 2\operatorname{In}(s)\) Oxidation half-reaction: \(\operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}^{3+}(a q) + e^{-}\)

First, multiply the oxidation half-reaction by 2 to double the number of electrons. This gives: \(\operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}^{3+}(a q) + e^{-}\) Then multiply by 2: \(2\operatorname{In}^{+}(a q) \longrightarrow 2\operatorname{In}^{3+}(a q) + 2e^{-}\) Next, subtract the reduction half-reaction from the oxidation half-reaction: \((2\operatorname{In}^{+}(a q) \longrightarrow 2\operatorname{In}^{3+}(a q) + 2e^{-}) - (2\operatorname{In}^{+}(a q) + e^{-} \longrightarrow 2\operatorname{In}(s))\) Which results in the whole reaction: \(3 \operatorname{In}^{+}(a q) \longrightarrow 2\operatorname{In}(s)+\operatorname{In}^{3+}(a q)\) Now, the standard potential for the disproportionation reaction, \(\mathscr{E}_{dis}^{\circ}\), can be calculated by subtracting the reduction half-reaction potential, \(\mathscr{E}_1^{\circ}\), from the oxidation half-reaction potential, \(\mathscr{E}_2^{\circ}\), obtained from the given reduction potentials. Therefore: \(\mathscr{E}_{dis}^{\circ} = \mathscr{E}_2^{\circ} - \mathscr{E}_1^{\circ} = (-0.444 V) - (-0.126 V) = -0.318 V\)

Now, we will use the Nernst Equation to calculate the equilibrium constant, K, for the disproportionation reaction: \(\mathscr{E}_{dis}^{\circ} = - \frac{RT}{2F} \ln K\) In this equation, R is the ideal gas constant (8.314 J/(mol·K)), T is the temperature in Kelvin (assume 298 K as the standard temperature), and F is the Faraday constant (96485 C/mol). Then we can solve for K: \(-0.318 V = - \frac{8.314 J \cdot 298 K}{2 \cdot 96485 C/mol} \ln K\) Solve for K: \(K \approx 8.247 \times 10^{11}\) So, the equilibrium constant for the disproportionation reaction is approximately \(8.25 \times 10^{11}\). b. Calculate the standard Gibbs free energy of formation for In+(aq):

We will use the relationship between Gibbs free energy and the standard reduction potential: \(\Delta G_{red}^{\circ} = -nF\mathscr{E}_{red}^{\circ}\) Where \(\Delta G_{red}^{\circ}\) is the Gibbs free energy change for the reduction of In3+(aq) to In+(aq), n is the number of electrons involved in the reaction (in this case, n=2), F is the Faraday constant (96485 C/mol), and \(\mathscr{E}_{red}^{\circ} = -0.444V\) (the potential given for the reduction of In3+(aq) to In+(aq)): \(\Delta G_{red}^{\circ} = -(2)(96485 C/mol)(-0.444 V) = 85764 J/mol = 85.8 kJ/mol\)

Since we are given \(\Delta G_{f}^{\circ}\) for In3+(aq) to be -97.9 kJ/mol, and we have just calculated the Gibbs free energy change for the reduction process as 85.8 kJ/mol, we can find the standard Gibbs free energy of formation for In+(aq), \(\Delta G_{f, \rm{In}^+}^{\circ}\), using the following equation: \(\Delta G_{f, \rm{In}^+}^{\circ} = \Delta G_{f, \rm{In}^{3+}}^{\circ} - \Delta G_{red}^{\circ} = -97.9\, \mathrm{kJ/mol} - 85.8\, \mathrm{kJ/mol} = -183.7\, \mathrm{kJ/mol}\) So, the standard Gibbs free energy of formation, \(\Delta G_{f}^{\circ}\), for In+(aq) is approximately -183.7 kJ/mol.