Question

In a cell, a copper electrode was used as a cathode. What is the electrode potential (in V) of the copper electrode dipped in 0.1 M Cu$^{2+}$ solution at 298 K? $$ E^{\circ} = 0.34\, V, \quad \frac{2.303RT}{F} = 0.059\, V $$

• A. 0.34
• B. 0.31
• C. 0.37
• D. 0.40

Hint:

Use the Nernst equation to calculate electrode potential when ion concentration deviates from standard conditions.

Answer 1

The Correct Option is B

Using the Nernst equation: \[ E = E^{\circ} - \frac{2.303RT}{nF} \log \left( \frac{[Cu^{2+}]}{[Cu]} \right) \] For a concentration of 0.1 M Cu\(^{2+}\), the electrode potential becomes: \[ E = 0.34 - 0.059 \log (1/0.1) = 0.34 - 0.059 \times 1 = 0.31 \, V \]