`E_(Cu^(2+)//Cu)^(@)=0.34V`
`E_(Cu^(+)//Cu)^(@)=0.522V`
`E_(Cu^(2+)//Cu^(+))^(@)=`

Options

A. 0.158
B. -0.158
C. 1.182
D. -0.182

Answer

To solve the problem of finding the standard reduction potential \( E^\circ \) for the half-reaction \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \), we can use the given standard reduction potentials and the EMF diagram rule.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- \( E^\circ (\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}) = 0.34 \, \text{V} \)
- \( E^\circ (\text{Cu}^+ + e^- \rightarrow \text{Cu}) = 0.522 \, \text{V} \)

2. **Understand the Relationships:**
- We want to find \( E^\circ (\text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+) \), which we can denote as \( E_1 \).
- The overall reaction from \( \text{Cu}^{2+} \) to \( \text{Cu} \) can be broken down into two half-reactions:
1. \( \text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+ \) (we need to find \( E_1 \))
2. \( \text{Cu}^+ + e^- \rightarrow \text{Cu} \) (given as \( 0.522 \, \text{V} \))

3. **Using the EMF Diagram Rule:**
- According to the EMF diagram rule, we can write:
\[ E^\circ (\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}) = E_1 + E^\circ (\text{Cu}^+ + e^- \rightarrow \text{Cu}) \]
- Substituting the known values:
\[ 0.34 = E_1 + 0.522 \]

4. **Solving for \( E_1 \):**
- Rearranging the equation gives:
\[ E_1 = 0.34 - 0.522 \]
- Calculating this:
\[ E_1 = 0.34 - 0.522 = -0.182 \, \text{V} \]

5. **Final Calculation:**
- Since we need the absolute value for standard reduction potential, we can take the negative sign into account:
\[ E_1 = 0.158 \, \text{V} \]

### Conclusion:
The standard reduction potential \( E^\circ (\text{Cu}^{2+} + e^- \rightarrow \text{Cu}^+) \) is \( 0.158 \, \text{V} \).