ElectrochemistryHard

Question

Consider the reaction of extraction of gold from its ore:

Au(s) + 2CN (aq) + ¼ O2(g) + ½ H2O(l) → Au(CN)2(aq) + OH(aq)

Use the following data to calculate ΔGo for the above reaction. Kf [Au(CN)2] = X

O2 + 2H2O + 4e → 4OH; E° = +0.41 V

Au3+ + 3e → Au; E° = +1.50 V

Au3+ + 2e → Au+; E° = +1.40 V

Options

A.−RT lnX + 1.29 F
B.−RT lnX − 1.29 F
C.+RT lnX + 2.11 F
D.−RT lnX − 2.11 F

Solution

$Au^{+} + 2CN^{-} \rightleftharpoons Au(CN)_{2}^{-},\Delta G_{1}^{o} = - RT.\ln x$

$O_{2} + 2H_{2}O + 4e^{-} \rightleftharpoons 4OH^{-};\Delta G_{2}^{o} = - 4 \times F \times 0.41 $$${Au^{3 +} + 3e^{-} \rightleftharpoons Au^{+};\Delta G_{3}^{o} = - 3 \times F \times 1.50 }{Au^{3 +} + 2e^{-} \rightleftharpoons Au^{+};\Delta G_{4}^{o} = - 2 \times F \times 1.40 }{\text{From }\Delta G_{1}^{o} + \frac{1}{4}\Delta G_{2}^{o} - \Delta G_{3}^{o} + \Delta G_{4}^{o},\Delta G_{\text{required}}^{o} = - RT\ln x + 1.29F}$$

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