What is `E^(c)._(red)` for the reaction`: Cu^(2+)+2e^(-) rarr Cu` in the half cell `Pt_(S^(2)|CuS|Cu)`. if `E^(c-)._(Cu^(2+)|Cu)` is `0.34V` and `K_(sp)` of `CuS=10^(-35)`?

Options

A. `0.34V`
B. `-0.6925V`
C. `+0.6925V`
D. `-0.66V`

Explanation

To find the standard reduction potential \( E^{\circ}_{\text{red}} \) for the reaction \( \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \) in the half-cell \( \text{Pt}_{(S^{2}|CuS|Cu)} \), we can use the Nernst equation.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The standard reduction potential \( E^{\circ}_{\text{Cu}^{2+}/Cu} = 0.34 \, \text{V} \).
- The solubility product \( K_{sp} \) of \( \text{CuS} = 10^{-35} \).
- The reaction we are interested in is \( \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \).

2. **Set Up the Nernst Equation**:
The Nernst equation at equilibrium (where \( E_{\text{cell}} = 0 \)) can be expressed as:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.0591}{n} \log K \]
Here, \( n \) is the number of electrons transferred (which is 2 for this reaction), and \( K \) is the equilibrium constant, which in this case is related to \( K_{sp} \) of \( \text{CuS} \).

3. **Equilibrium Condition**:
At equilibrium, \( E_{\text{cell}} = 0 \):
\[ 0 = E^{\circ}_{\text{Cu}^{2+}/Cu} - \frac{0.0591}{2} \log K_{sp} \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ E^{\circ}_{\text{Cu}^{2+}/Cu} = \frac{0.0591}{2} \log K_{sp} \]

5. **Substituting Known Values**:
Substitute \( K_{sp} = 10^{-35} \):
\[ E^{\circ}_{\text{Cu}^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \log(10^{-35}) \]

6. **Calculating the Logarithm**:
The logarithm can be calculated:
\[ \log(10^{-35}) = -35 \]

7. **Final Calculation**:
Substitute this value back into the equation:
\[ E^{\circ}_{\text{Cu}^{2+}/Cu} = 0.34 - \frac{0.0591}{2} \times (-35) \]
\[ E^{\circ}_{\text{Cu}^{2+}/Cu} = 0.34 + \frac{0.0591 \times 35}{2} \]
\[ E^{\circ}_{\text{Cu}^{2+}/Cu} = 0.34 + 1.03425 \]
\[ E^{\circ}_{\text{Cu}^{2+}/Cu} = 0.34 - 0.6925 \]
\[ E^{\circ}_{\text{Cu}^{2+}/Cu} = -0.3525 \, \text{V} \]

### Final Answer:
The standard reduction potential \( E^{\circ}_{\text{red}} \) for the reaction \( \text{Cu}^{2+} + 2e^{-} \rightarrow \text{Cu} \) is approximately \( -0.3525 \, \text{V} \).