What is the equilibrium constant of the reaction: 2Fe^3+ + Au^+ - 2Fe^2+ + Au^3+? Given E°_Au+\|Au = 1.68 V, E°_Au3+\|Au = 1.50 V, E°_Fe3+\|Fe2+ = 0.75 V and 2.303RT/F = 0.06.

Question

What is the equilibrium constant of the reaction: 2Fe^3+ + Au^+ -> 2Fe^2+ + Au^3+? Given E°_Au+\|Au = 1.68 V, E°_Au3+\|Au = 1.50 V, E°_Fe3+\|Fe2+ = 0.75 V and 2.303RT/F = 0.06.

Prashant Kotian · Accepted Answer

Correct answer: B. Calculate E°_Au3+\|Au+ oxidation: Au^3+ + 3e^- -> Au (1.50V, ΔG = -4.5F). Au^+ + e^- -> Au (1.68V, ΔG = -1.68F). Au^+ -> Au^3+ + 2e^- (ΔG = 4.5F - 1.68F = 2.82F). E°_ox = -1.41 V. Total E°_cell = 0.75 - 1.41 = -0.66 V. log(K) = (n * E°_cell) / 0.06 = (2 * -0.66) / 0.06 = -22. K = 10^-22. Therefore, correct answer is B.

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