For the process: Cu^2+ + 2e^– - Cu; log[Cu^2+] vs. E_red graph is shown in the figure, where OA = 0.34 V. The electrode potential of the half-cell of Cu\|Cu^2+ (0.1 M) will be [2.303RT/F = 0.06]

Question

For the process: Cu^2+ + 2e^– -> Cu; log[Cu^2+] vs. E_red graph is shown in the figure, where OA = 0.34 V. The electrode potential of the half-cell of Cu\|Cu^2+ (0.1 M) will be [2.303RT/F = 0.06]

Prashant Kotian · Accepted Answer

Correct answer: A. The y-intercept OA corresponds to log[Cu^2+] = 0, which is E°_red = 0.34 V. For the Cu\|Cu^2+ oxidation half-cell, E_ox = -E_red = -(E°_red + (0.06/2)*log[Cu^2+]). Substituting [Cu^2+] = 0.1 M gives E_ox = -(0.34 + 0.03*(-1)) = -(0.34 - 0.03) = -0.31 V. Therefore, correct answer is A.

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