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

Explanation

To find the electrode potential for the reaction \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \), we can use the given half-reactions and their electrode potentials. Here’s a step-by-step solution:

### Step 1: Identify the Given Half-Reactions
We have two half-reactions provided:
1. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) with \( E^\circ = 0.337 \, \text{V} \) (let's call this Reaction A)
2. \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \) with \( E^\circ = 0.153 \, \text{V} \) (let's call this Reaction B)

### Step 2: Reverse Reaction B
To find the potential for the reaction \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \), we need to reverse Reaction B:
\[ \text{Cu}^+ \rightarrow \text{Cu}^{2+} + e^- \]
When we reverse a reaction, the sign of the electrode potential also changes:
\[ E^\circ = -0.153 \, \text{V} \]
Let's call this Reaction C.

### Step 3: Combine Reactions A and C
Now, we will add Reaction A and Reaction C:
- Reaction A: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)
- Reaction C: \( \text{Cu}^+ \rightarrow \text{Cu}^{2+} + e^- \)

When we add these reactions, we need to ensure that the electrons cancel out. To do this, we can multiply Reaction C by 2:
\[ 2 \text{Cu}^+ \rightarrow 2 \text{Cu}^{2+} + 2e^- \]
Now, we can add this to Reaction A:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (A) \]
\[ 2 \text{Cu}^+ \rightarrow 2 \text{Cu}^{2+} + 2e^- \quad (C) \]

### Step 4: Write the Net Reaction
Adding these gives:
\[ \text{Cu}^+ + e^- \rightarrow \text{Cu} \]
The \( \text{Cu}^{2+} \) cancels out, and we are left with the desired reaction.

### Step 5: Calculate the Electrode Potential
Now we can calculate the overall electrode potential using the formula:
\[ E^\circ_{\text{cell}} = \frac{N_1E_1 + N_2E_2}{N_1 + N_2} \]
Where:
- \( N_1 \) is the number of electrons in Reaction A (which is 2).
- \( E_1 \) is the electrode potential of Reaction A (0.337 V).
- \( N_2 \) is the number of electrons in Reaction C (which is -1 because we reversed it).
- \( E_2 \) is the electrode potential of Reaction C (-0.153 V).

Substituting the values:
\[ E^\circ_{\text{cell}} = \frac{(2 \times 0.337) + (-1 \times 0.153)}{2 - 1} \]
\[ E^\circ_{\text{cell}} = \frac{0.674 - 0.153}{1} = 0.521 \, \text{V} \]

### Final Answer
Thus, the electrode potential for the reaction \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \) is approximately \( 0.52 \, \text{V} \).

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