`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. `0.343+(0.0591)/(2)`
B. `-0.34+(0.0591)/(2)`
C. 0.34
D. `(0.34)`

Explanation

To solve the problem, we need to determine the electrode oxidation potential of the half-cell reaction given by \( \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \) when the concentration of \( \text{Cu}^{2+} \) is 0.1 M. We are given that the intercept of the graph of \( E_{\text{red}} \) versus \( \ln[\text{Cu}^{2+}] \) is 0.34 V.

### Step-by-step Solution:

1. **Identify the Reaction and Given Information**:
The half-cell reaction is:
\[ \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \]
The intercept of the graph \( E_{\text{red}} \) versus \( \ln[\text{Cu}^{2+}] \) is given as 0.34 V. This value represents the standard reduction potential \( E^{\circ} \) for the reaction.

2. **Determine the Standard Reduction Potential**:
From the information given, we have:
\[ E^{\circ}_{\text{red}} = 0.34 \, \text{V} \]

3. **Convert to Oxidation Potential**:
The oxidation potential \( E^{\circ}_{\text{ox}} \) is the negative of the reduction potential:
\[ E^{\circ}_{\text{ox}} = -E^{\circ}_{\text{red}} = -0.34 \, \text{V} \]

4. **Apply the Nernst Equation**:
The Nernst equation for the half-cell reaction is given by:
\[ E = E^{\circ} - \frac{0.0591}{n} \log \left( \frac{[\text{products}]}{[\text{reactants}]} \right) \]
For our reaction, the products are \( \text{Cu} \) (which is a solid and has an activity of 1) and the reactants are \( \text{Cu}^{2+} \) with a concentration of 0.1 M. Here, \( n = 2 \) (the number of electrons transferred).

5. **Substituting Values into the Nernst Equation**:
Substitute \( E^{\circ}_{\text{ox}} \), \( n \), and the concentration into the Nernst equation:
\[ E = -0.34 - \frac{0.0591}{2} \log \left( \frac{1}{0.1} \right) \]
Since \( \frac{1}{0.1} = 10 \), we have:
\[ E = -0.34 - \frac{0.0591}{2} \cdot 1 \]

6. **Calculating the Final Value**:
Calculate \( \frac{0.0591}{2} \):
\[ \frac{0.0591}{2} = 0.02955 \]
Thus:
\[ E = -0.34 - 0.02955 = -0.36955 \, \text{V} \]

7. **Final Answer**:
Rounding to two decimal places, the electrode oxidation potential of the half-cell \( \text{Cu} | \text{Cu}^{2+} (0.1M) \) is:
\[ E \approx -0.37 \, \text{V} \]