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

Step by step solution

01

Determine the overall reaction's standard cell potential (E°)

We will reverse and add the given half-reactions to get the disproportionation reaction: Reversed 1st half-reaction: \(In^+(aq) \longrightarrow In^{3+}(aq) + 2 e^-\) (E° = +0.444 V) 2 × 2nd half-reaction: \(2 In^+(aq) + 2 e^- \longrightarrow 2 In(s) \) (E° = 2 × -0.126 V) Now add the reversed 1st and 2 × 2nd half-reactions: \(3 In^+(aq) \longrightarrow 2 In(s) + In^{3+}(aq)\) (E° = 0.444 - 2 × 0.126 V) Calculating E° for the overall reaction: E° = 0.444 - 2 × 0.126 = 0.192 V

02

Determine the equilibrium constant (K) using Nernst equation

Using the Nernst equation for the equilibrium condition and under standard conditions (\(E = E°\), Q = 1), we get: \(E° = \frac{RT}{nF} \ln K\) Here, \(E° = 0.192\) V (calculated in step 1) R = 8.314 J/(mol·K) (gas constant) T = 298.15 K (standard temperature) n = 3 (moles of electrons involved in the overall reaction) F = 96,485 C/mol (Faraday's constant) Solving for K: \(K = e^{\frac{nFE°}{RT}}\) K = \(e^{\frac{3 \times 96,485 \times 0.192}{8.314 \times 298.15}}\) K = 3.54 × 10^3 So, the equilibrium constant (K) for the disproportionation reaction is 3.54 × 10^3. b. Finding the standard Gibbs free energy (ΔG°) for In+(aq):

03

Determine the standard Gibbs free energy (ΔG°) for the overall reaction

Using the relationship between ΔG° and E°: ΔG° = -nFE° Here, ΔG° = standard Gibbs free energy (J/mol) n = 3 (moles of electrons involved) F = 96,485 C/mol (Faraday's constant) E° = 0.192 V (calculated in part a's step 1) ΔG° = -3 × 96,485 × 0.192 ΔG° = -55,554 J/mol or -55.554 kJ/mol

04

Determine the standard Gibbs free energy (ΔG°) for In+(aq)

From the given information, the standard Gibbs free energy (ΔG°) for In3+(aq) is -97.9 kJ/mol. Let's consider ΔG°(In+(aq)) for In+(aq) and X for the overall reaction (ΔG° = -55.554 kJ/mol) Using Hess's law: ΔG°(In+(aq)) + ΔG°(In3+(aq)) = ΔG° Therefore, ΔG°(In+(aq)) = ΔG° - ΔG°(In3+(aq)) ΔG°(In+(aq)) = -55.554 - (-97.9) ΔG°(In+(aq)) = 42.346 kJ/mol So, the standard Gibbs free energy (ΔG°) for In+(aq) is 42.346 kJ/mol.

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