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

Options

A. 1.86V
B. 0.34V
C. 0.31V
D. `-1.86V`

Answer

To solve the problem, we need to find the standard reduction potential of the half-reaction \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) given the EMF of the cell and the standard reduction potential of the zinc half-reaction.

### Step-by-Step Solution:

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

2. **Write the Half-Reactions**:
- At the anode (oxidation):
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]
- At the cathode (reduction):
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

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

4. **Use the Nernst Equation**:
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
Here, \( n \) is the number of moles of electrons transferred (which is 2 in this case), and \( Q \) is the reaction quotient.

5. **Calculate the Reaction Quotient \( Q \)**:
Since both zinc and copper ions are at 1 M concentration:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1}{1} = 1 \]

6. **Substituting Values into the Nernst Equation**:
Since \( \log(1) = 0 \):
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - 0 \]
Thus,
\[ E_{\text{cell}} = E^\circ_{\text{cell}} \]

7. **Calculate \( E^\circ_{\text{cell}} \)**:
The standard cell potential can be expressed as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} \]
Given that \( E^\circ_{\text{anode}} \) (for zinc) is -0.78 V, we can express it as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{Cu}^{2+}/\text{Cu}} - (-0.78) \]
Rearranging gives:
\[ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = E^\circ_{\text{cell}} + 0.78 \]

8. **Substituting the Given EMF**:
We know \( E_{\text{cell}} = 1.1 \, \text{V} \):
\[ E^\circ_{\text{Cu}^{2+}/\text{Cu}} = 1.1 + 0.78 = 1.1 - (-0.78) = 1.1 + 0.78 = 0.32 \, \text{V} \]

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

---