Consider the reaction of extraction of gold from its ore: Au(s) + 2CN^– (aq) + 1/4 O2(g) + 1/2 H2O(l — Electrochemistry Chemistry Question
Question
Consider the reaction of extraction of gold from its ore: Au(s) + 2CN^– (aq) + 1/4 O2(g) + 1/2 H2O(l) -> Au(CN)2^–(aq) + OH^–(aq). Use the following data to calculate ΔG° for the above reaction. K_f[Au(CN)2^–] = X; O2 + 2H2O + 4e^– -> 4OH^–; E° = +0.41 V; Au^3+ + 3e^– -> Au; E° = +1.50 V; Au^3+ + 2e^– -> Au^+; E° = +1.40 V
💡 Solution & Explanation
First find E° for Au^+ + e^- -> Au: ΔG = 3F(1.50) - 2F(1.40) = 4.5F - 2.8F = 1.7F, so E° = 1.70 V. The overall cell consists of the reduction of O2 (E_cathode = 0.41 V) and oxidation of Au in presence of CN^-. E°_anode for Au + 2CN^- -> Au(CN)2^- + e^- is -E°_Au+/Au + (RT/F)ln(Kf) = -1.70 + (RT/F)ln(X). E°_cell = 0.41 - 1.70 + (RT/F)ln(X) = -1.29 + (RT/F)ln(X). ΔG° = -nFE°_cell = -F * (-1.29 + (RT/F)ln X) = 1.29 F - RT ln X. Therefore, correct answer is A.