`Cu^(2+)+2e^(-) rarr Cu`. For this, graph between `E_(red)` versus `ln[Cu^(2+)]` is a straight line of intercept `0.34V`, then the electrode oxidation potential of the half cell `Cu|Cu^(2+)(0.1M)` will be

Options

A. `0.34+(0.0591)/(2)`
B. `-0.34-(0.0591)/(2)`
C. `0.34`
D. `-0.34+(0.0591)/(2)`

Explanation

To solve the problem, we need to determine the electrode oxidation potential (EOP) of the half-cell reaction where copper (Cu) is oxidized to copper ions (Cu²⁺) at a concentration of 0.1 M. The reduction potential (E_red) for the reaction \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) is given, and we can use the Nernst equation to find the required oxidation potential.

### Step-by-Step Solution:

1. **Identify the Reduction Reaction**:
The reduction reaction is given as:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

2. **Use the Nernst Equation for Reduction**:
The Nernst equation for the reduction potential is:
\[ E_{\text{red}} = E^\circ_{\text{red}} - \frac{0.0591}{n} \log \left( \frac{1}{[\text{Cu}^{2+}]} \right) \]
where \( n = 2 \) (the number of electrons transferred).

3. **Convert Logarithm to Natural Logarithm**:
To convert from logarithm (base 10) to natural logarithm (base e), we use the relationship:
\[ \log x = \frac{\ln x}{2.303} \]
Thus, the equation becomes:
\[ E_{\text{red}} = E^\circ_{\text{red}} + \frac{0.0591}{n \cdot 2.303} \ln [\text{Cu}^{2+}] \]

4. **Substitute Known Values**:
Given that the intercept \( E^\circ_{\text{red}} = 0.34 \, V \) and the concentration \( [\text{Cu}^{2+}] = 0.1 \, M \), we can substitute these values into the Nernst equation:
\[ E_{\text{red}} = 0.34 - \frac{0.0591}{2} \log \left( \frac{1}{0.1} \right) \]

5. **Calculate Logarithm**:
\[ \log(0.1) = -1 \]
Therefore:
\[ E_{\text{red}} = 0.34 - \frac{0.0591}{2} \cdot (-1) \]
\[ E_{\text{red}} = 0.34 + 0.02955 \]
\[ E_{\text{red}} = 0.36955 \, V \]

6. **Determine Oxidation Potential**:
The oxidation potential \( E_{\text{OP}} \) is related to the reduction potential by:
\[ E_{\text{OP}} = -E_{\text{red}} \]
Thus:
\[ E_{\text{OP}} = -0.36955 \, V \]

7. **Final Calculation**:
The oxidation potential can be expressed as:
\[ E_{\text{OP}} = -0.34 + 0.02955 \approx -0.31045 \, V \]

### Conclusion:
The electrode oxidation potential of the half-cell \( \text{Cu} | \text{Cu}^{2+} (0.1M) \) is approximately:
\[ E_{\text{OP}} \approx -0.310 \, V \]