Question

Consider the disproportionation reaction $ 2 \, \text{Cu}^{+} \rightleftharpoons \text{Cu}^{2+} + \text{Cu}(s) $. At equilibrium $[ \text{Cu}^{+} ] / [ \text{Cu}^{2+} ] = 10^4 $. If $ E_{\text{Cu}^{+}/\text{Cu}} = 0.15 \, \text{V} $, then $ E_{\text{Cu}^{2+}/\text{Cu}} $ (in V) is:

• A. \( -0.386 \)
• B. \( +0.386 \)
• C. \( +0.536 \)
• D. \( -0.268 \)

Hint:

When dealing with standard electrode potentials, always apply the Nernst equation to calculate the potential at non-standard conditions.

Answer 1

The Correct Option is B

The reaction is a disproportionation reaction. The relation between \( E \) and \( E^0 \) for a half-reaction is given by the Nernst equation: \[ E = E^0 - \frac{0.0591}{n} \log \frac{[\text{Red}]}{[\text{Ox}]} \] For the given reaction: \[ E_{\text{Cu}^{2+}/\text{Cu}} = E_{\text{Cu}^{+}/\text{Cu}} - \frac{0.0591}{n} \log \left(\frac{[\text{Cu}^{+}]^2}{[\text{Cu}^{2+}]}\right) \] Given \( [\text{Cu}^{+}] / [\text{Cu}^{2+}] = 10^4 \) and \( E_{\text{Cu}^{+}/\text{Cu}} = 0.15 \, \text{V} \), the required \( E_{\text{Cu}^{2+}/\text{Cu}} \) is: \[ E_{\text{Cu}^{2+}/\text{Cu}} = 0.15 \, \text{V} - \frac{0.0591}{1} \log(10^4) \] Simplifying this gives \( E_{\text{Cu}^{2+}/\text{Cu}} = +0.386 \, \text{V} \).