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

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

We are 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

### Step 1: Write down the relevant half-reactions
We know the half-reactions and their standard potentials:

1. \( \text{Cu}^{+} + e^- \rightarrow \text{Cu} \) (1 electron transfer)
2. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) (2 electrons transfer)

### Step 2: Use the Nernst equation
To find the potential for the reaction \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^{+} \), we can use the relationship between the potentials of the half-reactions.

From the first half-reaction, we can see that:

\[ E^\circ(\text{Cu}^{+}/\text{Cu}) = X_1 \]

From the second half-reaction, we can express it in terms of the reaction we want to find:

\[ E^\circ(\text{Cu}^{2+}/\text{Cu}) = X_2 \]

### Step 3: Relate the potentials
The reaction \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^{+} \) can be derived from the other two half-reactions. We can write:

\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E^\circ = X_2) \]
\[ \text{Cu}^{+} + e^- \rightarrow \text{Cu} \quad (E^\circ = X_1) \]

If we reverse the first reaction, we get:

\[ \text{Cu} \rightarrow \text{Cu}^{+} + e^- \quad (E^\circ = -X_1) \]

Now, we can add this to the second reaction:

\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (E^\circ = X_2) \]
\[ \text{Cu} \rightarrow \text{Cu}^{+} + e^- \quad (E^\circ = -X_1) \]

### Step 4: Combine the reactions
Adding these two reactions gives:

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

The potential for this combined reaction is:

\[ E^\circ(\text{Cu}^{2+}/\text{Cu}^{+}) = E^\circ(\text{Cu}^{2+}/\text{Cu}) + E^\circ(\text{Cu}/\text{Cu}^{+}) \]

Substituting the values:

\[ E^\circ(\text{Cu}^{2+}/\text{Cu}^{+}) = X_2 - X_1 \]

### Final Answer
Thus, the standard electrode potential for the half-reaction \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^{+} \) is:

\[ E^\circ = X_2 - X_1 \]