If the standard electrode poten tial of `Cu^(2+)//Cu` electrode is 0.34V. What is the electrode potential of 0.01 M concentration of `Cu^(2+)`?

Options

A. 0.399V
B. 0.281V
C. 0.222V
D. 0.176V

Explanation

To find the electrode potential of a 0.01 M concentration of \( Cu^{2+} \), we can use the Nernst equation. The Nernst equation is given by:

\[ E = E^\circ - \frac{RT}{nF} \ln Q \]

Where:
- \( E \) is the electrode potential we want to find.
- \( E^\circ \) is the standard electrode potential (0.34 V for \( Cu^{2+}/Cu \)).
- \( R \) is the universal gas constant (8.314 J/(mol·K)).
- \( T \) is the temperature in Kelvin (assume 298 K for standard conditions).
- \( n \) is the number of moles of electrons transferred in the half-reaction (2 for \( Cu^{2+} + 2e^- \rightarrow Cu \)).
- \( F \) is Faraday's constant (96485 C/mol).
- \( Q \) is the reaction quotient, which can be expressed as \( \frac{[Cu]}{[Cu^{2+}]} \). Since \( Cu \) is a solid, its activity is 1, so \( Q = \frac{1}{[Cu^{2+}]} \).

### Step-by-Step Solution:

1. **Identify the standard electrode potential**:
\[ E^\circ = 0.34 \, \text{V} \]

2. **Identify the concentration of \( Cu^{2+} \)**:
\[ [Cu^{2+}] = 0.01 \, \text{M} \]

3. **Calculate the reaction quotient \( Q \)**:
\[ Q = \frac{1}{[Cu^{2+}]} = \frac{1}{0.01} = 100 \]

4. **Use the Nernst equation**:
Since we are using logarithm base 10, we can simplify the Nernst equation:
\[ E = E^\circ - \frac{0.0591}{n} \log Q \]
Here, \( n = 2 \):
\[ E = 0.34 - \frac{0.0591}{2} \log(100) \]

5. **Calculate \( \log(100) \)**:
\[ \log(100) = 2 \]

6. **Substitute back into the equation**:
\[ E = 0.34 - \frac{0.0591}{2} \times 2 \]
\[ E = 0.34 - 0.0591 \]

7. **Final calculation**:
\[ E = 0.34 - 0.0591 = 0.2809 \, \text{V} \approx 0.281 \, \text{V} \]

### Conclusion:
The electrode potential of a 0.01 M concentration of \( Cu^{2+} \) is approximately **0.281 V**.