The standard reduction potential of `Cu^(2+)//Cu` and `Cu^(2+)//Cu^(+)` are `0.337` and `0.153` respectively. The standard electrode potential of `Cu^(+)//Cu` half `-` cell is

Options

A. `0.184 V`
B. `0.827V`
C. `0.521V`
D. `0.490V`

Explanation

To find the standard electrode potential of the half-cell reaction \( Cu^+ // Cu \), we can use the given standard reduction potentials of the half-cells \( Cu^{2+} // Cu \) and \( Cu^{2+} // Cu^+ \).

### Step-by-step solution:

1. **Identify the given standard reduction potentials**:
- \( E^\circ (Cu^{2+} + 2e^- \rightarrow Cu) = 0.337 \, V \)
- \( E^\circ (Cu^{2+} + e^- \rightarrow Cu^+) = 0.153 \, V \)

2. **Write the reactions corresponding to the given potentials**:
- For \( Cu^{2+} + 2e^- \rightarrow Cu \) (Reaction 1)
- For \( Cu^{2+} + e^- \rightarrow Cu^+ \) (Reaction 2)

3. **Determine the reaction for \( Cu^+ \rightarrow Cu \)**:
- To find the standard electrode potential for the reaction \( Cu^+ + e^- \rightarrow Cu \), we can manipulate the above reactions.
- We can express \( Cu^+ \) in terms of \( Cu^{2+} \) using Reaction 2:
\[ Cu^{2+} + e^- \rightarrow Cu^+ \quad \text{(reverse this reaction)} \]
This gives:
\[ Cu^+ \rightarrow Cu^{2+} + e^- \]
- Now, we can subtract this from Reaction 1:
\[ Cu^{2+} + 2e^- \rightarrow Cu \quad \text{(keep this as is)} \]

4. **Combine the reactions**:
- By subtracting the reversed Reaction 2 from Reaction 1, we get:
\[ (Cu^{2+} + 2e^- \rightarrow Cu) - (Cu^+ \rightarrow Cu^{2+} + e^-) \]
- This simplifies to:
\[ Cu^+ + e^- \rightarrow Cu \]

5. **Calculate the standard electrode potential**:
- The standard electrode potential for the reaction \( Cu^+ + e^- \rightarrow Cu \) can be calculated using the equation:
\[ E^\circ (Cu^+ // Cu) = E^\circ (Cu^{2+} // Cu) - E^\circ (Cu^{2+} // Cu^+) \]
- Substituting the values:
\[ E^\circ (Cu^+ // Cu) = 0.337 \, V - 0.153 \, V = 0.184 \, V \]

### Final Answer:
The standard electrode potential of the half-cell \( Cu^+ // Cu \) is \( 0.184 \, V \).