`Cu^(+) + e rarr Cu, E^(@) = X_(1)` volt,
`Cu^(2+) + 2e rarr Cu, E^(@) = X_(2)`X_(2) volt
For `Cu^(2+) + e rarr Cu^(+), E^(@)` will be :

Options

A. `x_(1)-2x_(2)`
B. `x_(1)+2x_(2)`
C. `x_(1)+x_(2)`
D. `2x_(2)-x_(1)`

Explanation

To solve the problem, we need to find the standard electrode potential \( E^\circ \) for the half-reaction:

\[ \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \]

Given the following standard electrode potentials:

1. \( \text{Cu}^+ + e^- \rightarrow \text{Cu}, \quad E^\circ = X_1 \) volts
2. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}, \quad E^\circ = X_2 \) volts

We can derive the potential for the third half-reaction using the relationships between these half-reactions.

### Step-by-Step Solution:

1. **Write down the half-reactions:**
- Reaction 1: \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \) (with \( E^\circ = X_1 \))
- Reaction 2: \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) (with \( E^\circ = X_2 \))
- Reaction 3: \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \) (we want to find \( E^\circ \) for this reaction)

2. **Relate the reactions:**
- We can express Reaction 3 as a combination of Reaction 1 and Reaction 2. Specifically, we can manipulate the equations:
- Start with Reaction 2 and reverse Reaction 1:
\[ \text{Cu} \rightarrow \text{Cu}^+ + e^- \quad (E^\circ = -X_1) \]
- Now, add this to Reaction 2:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E^\circ = X_2) \]
- The combined reaction becomes:
\[ \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \]

3. **Calculate the standard potential for Reaction 3:**
- The overall potential for the combined reaction can be expressed as:
\[ E^\circ_{\text{total}} = E^\circ_{\text{Reaction 2}} + E^\circ_{\text{reversed Reaction 1}} \]
- Substitute the potentials:
\[ E^\circ = X_2 - X_1 \]

4. **Adjust for the electron transfer:**
- Since we are looking for the potential of the half-reaction \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \), we need to account for the fact that we are effectively taking one electron from the second reaction:
\[ E^\circ = X_2 - X_1 \]

5. **Final expression:**
- Rearranging gives us:
\[ E^\circ = 2X_2 - X_1 \]

### Conclusion:
The standard electrode potential for the half-reaction \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \) is given by:
\[ E^\circ = 2X_2 - X_1 \]