Given:
(i) `Cu^(2+)+2e^(-) rarr Cu, E^(@) = 0.337 V`
(ii) `Cu^(2+)+e^(-) rarr Cu^(+), E^(@) = 0.153 V`
Electrode potential, `E^(@)` for the reaction, `Cu^(+)+e^(-) rarr Cu`, will be

Options

A. 0.52 V
B. 0.90 V
C. 0.30 V
D. 0.38 V

Explanation

To find the electrode potential \( E^\circ \) for the reaction \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \), we can use the provided half-reactions and their standard electrode potentials.

### Step-by-Step Solution:

1. **Identify the Given Reactions and Their Potentials:**
- Reaction 1: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \), \( E^\circ = 0.337 \, \text{V} \)
- Reaction 2: \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \), \( E^\circ = 0.153 \, \text{V} \)

2. **Reverse Reaction 2:**
- To find the potential for \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \), we need to reverse Reaction 2:
- Reversed Reaction 2: \( \text{Cu}^+ \rightarrow \text{Cu}^{2+} + e^- \)
- The potential for the reversed reaction will change sign: \( E^\circ = -0.153 \, \text{V} \)

3. **Combine the Reactions:**
- Now, we can add the reversed Reaction 2 to Reaction 1:
- Reaction 1: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)
- Reversed Reaction 2: \( \text{Cu}^+ \rightarrow \text{Cu}^{2+} + e^- \)

When we add these reactions, we get:
\[ \text{Cu}^+ + e^- \rightarrow \text{Cu} \]

4. **Calculate the Overall Cell Potential:**
- The overall cell potential \( E^\circ \) can be calculated by adding the potentials of the two reactions:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reaction 1}} + E^\circ_{\text{reversed reaction 2}} \]
\[ E^\circ_{\text{cell}} = 0.337 \, \text{V} + (-0.153 \, \text{V}) = 0.337 \, \text{V} - 0.153 \, \text{V} = 0.184 \, \text{V} \]

5. **Final Result:**
- The electrode potential \( E^\circ \) for the reaction \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \) is \( 0.184 \, \text{V} \).