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. 3.99V
B. 0.3105V
C. 0.222V
D. 0.176V

Answer

To find the electrode potential of a 0.01 M concentration of \( Cu^{2+} \), we can use the Nernst equation. The Nernst equation relates the standard electrode potential to the concentration of the ions involved in the half-reaction.

### Step-by-Step Solution:

1. **Identify the Standard Electrode Potential**:
The standard electrode potential (\( E^\circ \)) for the \( Cu^{2+}/Cu \) half-reaction is given as \( 0.34 \, V \).

2. **Write the Nernst Equation**:
The Nernst equation for the reduction of \( Cu^{2+} \) to \( Cu \) is:
\[ E = E^\circ - \frac{0.0591}{n} \log \left( \frac{1}{[Cu^{2+}]} \right) \]
where:
- \( E \) is the electrode potential we want to find,
- \( E^\circ \) is the standard electrode potential,
- \( n \) is the number of electrons transferred in the half-reaction (which is 2 for \( Cu^{2+} + 2e^- \rightarrow Cu \)),
- \( [Cu^{2+}] \) is the concentration of \( Cu^{2+} \).

3. **Substitute the Values**:
Here, \( n = 2 \) and \( [Cu^{2+}] = 0.01 \, M \).
\[ E = 0.34 - \frac{0.0591}{2} \log \left( \frac{1}{0.01} \right) \]

4. **Calculate the Logarithm**:
\[ \log \left( \frac{1}{0.01} \right) = \log(100) = 2 \]

5. **Substitute the Logarithm into the Equation**:
\[ E = 0.34 - \frac{0.0591}{2} \times 2 \]

6. **Simplify the Equation**:
\[ E = 0.34 - 0.0591 \]
\[ E = 0.2809 \, V \]

7. **Final Calculation**:
Rounding to three decimal places, we find:
\[ E \approx 0.281 \, V \]

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