The emf of the cell
`Zn|Zn^(2+)(1M)||Cu^(2+)|Cu(1M)`
is 1.1V. If the standard reduction potential of `Zn^(2+)|Zn` is -0.78V, what is the oxidation potential of `Cu|Cu^(2+)`?

Options

A. `+1.86V`
B. `0.32V`
C. `-0.32V`
D. `-1.86V`

Explanation

To solve the problem, we need to find the oxidation potential of the half-reaction \( \text{Cu} | \text{Cu}^{2+} \). We are given the following information:

- EMF of the cell, \( E_{\text{cell}} = 1.1 \, \text{V} \)
- Standard reduction potential of \( \text{Zn}^{2+} | \text{Zn} \), \( E^\circ_{\text{Zn}} = -0.78 \, \text{V} \)

### Step-by-Step Solution:

1. **Identify the Cell Components**:
The cell is represented as \( \text{Zn} | \text{Zn}^{2+}(1M) || \text{Cu}^{2+} | \text{Cu}(1M) \). Here, zinc is oxidized (anode) and copper is reduced (cathode).

2. **Write the Cell Reaction**:
The overall cell reaction can be written as:
\[ \text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu} \]

3. **Use the Formula for Cell EMF**:
The EMF of the cell can be expressed as:
\[ E_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Here, the cathode is copper and the anode is zinc.

4. **Substitute Known Values**:
We know:
- \( E_{\text{cell}} = 1.1 \, \text{V} \)
- \( E^\circ_{\text{anode}} = E^\circ_{\text{Zn}} = -0.78 \, \text{V} \)

Therefore, we can rearrange the equation:
\[ 1.1 = E^\circ_{\text{Cu}} - (-0.78) \]
This simplifies to:
\[ 1.1 = E^\circ_{\text{Cu}} + 0.78 \]

5. **Solve for \( E^\circ_{\text{Cu}} \)**:
Rearranging gives:
\[ E^\circ_{\text{Cu}} = 1.1 - 0.78 = 0.32 \, \text{V} \]

6. **Determine the Oxidation Potential**:
The oxidation potential is the negative of the reduction potential:
\[ E_{\text{oxidation}} = -E^\circ_{\text{Cu}} = -0.32 \, \text{V} \]

### Final Answer:
The oxidation potential of \( \text{Cu} | \text{Cu}^{2+} \) is \( -0.32 \, \text{V} \).