Given that `K_(sp)` of `CuS=10^(-35)` and `E_Cu//Cu^(2+)= (-0.34V)`.
The standard oxidation potential of `Cu|CuS|S^(2-)` Half- cell is :

Options

A. `(1.0V)`
B. `(0.693V)`
C. `(-0.690V)`
D. (1.0V)`

Explanation

To find the standard oxidation potential of the half-cell reaction \( \text{Cu} | \text{CuS} | \text{S}^{2-} \), we can use the Nernst equation and the relationship between solubility product \( K_{sp} \) and standard potentials.

### Step-by-Step Solution:

1. **Identify the Half-Cell Reaction**:
The relevant half-cell reaction for the formation of copper sulfide (CuS) from copper ions (Cu²⁺) and sulfide ions (S²⁻) can be written as:
\[ \text{Cu}^{2+} + \text{S}^{2-} \rightarrow \text{CuS} + 2e^- \]

2. **Write the Nernst Equation**:
The Nernst equation for this half-cell reaction is:
\[ E^\circ = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - \frac{0.059}{n} \log K_{sp} \]
where \( n \) is the number of electrons transferred (which is 2 in this case), and \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} \) is given as -0.34 V.

3. **Substitute the Values**:
Given:
- \( K_{sp} = 10^{-35} \)
- \( E^\circ_{\text{Cu}^{2+}/\text{Cu}} = -0.34 \, \text{V} \)
- \( n = 2 \)

Substitute these values into the Nernst equation:
\[ E^\circ = -0.34 - \frac{0.059}{2} \log(10^{-35}) \]

4. **Calculate the Logarithm**:
The logarithm can be calculated as:
\[ \log(10^{-35}) = -35 \]
Therefore, substituting this back into the equation gives:
\[ E^\circ = -0.34 - \frac{0.059}{2} \times (-35) \]

5. **Simplify the Equation**:
Calculate \( \frac{0.059}{2} \times (-35) \):
\[ \frac{0.059}{2} = 0.0295 \]
\[ 0.0295 \times (-35) = -1.0325 \]
So, the equation becomes:
\[ E^\circ = -0.34 + 1.0325 \]

6. **Final Calculation**:
Now, calculate the final value:
\[ E^\circ = 0.6925 \, \text{V} \approx 0.693 \, \text{V} \]

### Conclusion:
The standard oxidation potential of the half-cell \( \text{Cu} | \text{CuS} | \text{S}^{2-} \) is approximately \( 0.693 \, \text{V} \).