Question

What is the electrode potential (in V) of copper electrode dipped in \( 10^{-2} \, \text{M} \, \text{Cu}^{2+} \) solution?

• A. 0.399
• B. 0.1405
• C. 0.199
• D. 0.281

Hint:

The Nernst equation allows us to calculate electrode potentials at non-standard conditions.

Answer 1

The Correct Option is D

Step 1:
The Nernst equation is used to calculate the electrode potential: \[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{[\text{Cu}^{2+}]}{[\text{Cu}]}\right) \] where \( E^\circ \) is the standard electrode potential (0.34 V for Cu), \( n \) is the number of electrons transferred (2 for Cu), and \( [\text{Cu}^{2+}] \) is the concentration of \( \text{Cu}^{2+} \).
Step 2:
Substituting the values: \[ E = 0.34 - \frac{0.0591}{2} \log \left( \frac{1}{10^{-2}} \right) \] \[ E = 0.34 - \frac{0.0591}{2} \log (100) \] \[ E = 0.34 - \frac{0.0591}{2} \times 2 \] \[ E = 0.34 - 0.0591 = 0.281 \] Thus, the electrode potential is 0.281 V.