Estimate the `E^(@)` reduction for `Cu|CuS` electrode.
Given `: K_(sp) ` of `CuS=8.0 xx 10^(-36),E^(@)._((Cu|Cu^(2+)))=-0.34V`

Options

A. 1.034 V
B. 1.0 V
C. `-0.694V`
D. 0.694 V

Answer

To estimate the \( E^\circ \) reduction potential for the \( Cu|CuS \) electrode, we will follow these steps:

### Step 1: Write the half-reactions
The oxidation and reduction half-reactions for copper and copper sulfide can be written as follows:

- **Oxidation (anode)**:
\[ Cu \rightarrow Cu^{2+} + 2e^- \]

- **Reduction (cathode)**:
\[ CuS + 2e^- \rightarrow Cu + S^{2-} \]

### Step 2: Combine the half-reactions
The net reaction can be obtained by combining the oxidation and reduction half-reactions:

\[ Cu + CuS \rightarrow Cu^{2+} + S^{2-} \]

### Step 3: Use the Nernst equation
The Nernst equation for the cell potential is given by:

\[ E_{cell} = E^\circ_{cell} - \frac{0.0591}{n} \log Q \]

Where:
- \( E_{cell} \) is the cell potential,
- \( E^\circ_{cell} \) is the standard cell potential,
- \( n \) is the number of moles of electrons transferred,
- \( Q \) is the reaction quotient.

### Step 4: Determine \( n \) and \( Q \)
From the half-reactions, we see that \( n = 2 \) (2 electrons are transferred). At equilibrium, \( Q \) becomes the solubility product \( K_{sp} \) of \( CuS \):

\[ K_{sp} = [Cu^{2+}][S^{2-}] \]

Given \( K_{sp} = 8.0 \times 10^{-36} \).

### Step 5: Set \( E_{cell} = 0 \) for equilibrium
At equilibrium, \( E_{cell} = 0 \). Thus, we can write:

\[ 0 = E^\circ_{cell} - \frac{0.0591}{2} \log(8.0 \times 10^{-36}) \]

### Step 6: Calculate \( \log K_{sp} \)
Calculate \( \log(8.0 \times 10^{-36}) \):

\[ \log(8.0 \times 10^{-36}) = \log(8.0) + \log(10^{-36}) = 0.903 - 36 = -35.097 \]

### Step 7: Substitute into the Nernst equation
Substituting back into the equation:

\[ 0 = E^\circ_{cell} - \frac{0.0591}{2} \times (-35.097) \]

### Step 8: Solve for \( E^\circ_{cell} \)
Calculating the right side:

\[ E^\circ_{cell} = \frac{0.0591}{2} \times 35.097 \]

Calculating this gives:

\[ E^\circ_{cell} = 0.0591 \times 17.5485 \approx 1.037 \text{ V} \]

### Step 9: Calculate \( E^\circ \) for \( Cu|CuS \)
Now we need to find \( E^\circ \) for the reduction of \( Cu|CuS \). We know:

\[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

Where \( E^\circ_{anode} = -0.34 \text{ V} \) (for \( Cu^{2+}/Cu \)). Let \( E^\circ_{cathode} = x \):

\[ 1.037 = x - (-0.34) \]

Thus,

\[ x = 1.037 - 0.34 = 0.697 \text{ V} \]

### Final Step: Determine the sign
Since we are looking for the reduction potential of \( Cu|CuS \), we take the negative:

\[ E^\circ_{Cu|CuS} = -0.697 \text{ V} \]

### Conclusion
The estimated \( E^\circ \) reduction for the \( Cu|CuS \) electrode is approximately:

\[ \boxed{-0.697 \text{ V}} \]