Consider the cell `Zn|Zn^(2+) || Cu^(2+)|Cu. ` If the concentration of Zn and Cu ions are doubled, the emf of the cell.

Options

A. Doubles
B. Reduces of haif
C. Remains same
D. Becomes zero

Explanation

To solve the problem regarding the electrochemical cell \( \text{Zn} | \text{Zn}^{2+} || \text{Cu}^{2+} | \text{Cu} \) and the effect of doubling the concentrations of Zn and Cu ions on the emf of the cell, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Cell Components**:
- The cell consists of zinc (Zn) and copper (Cu) electrodes.
- The half-reactions are:
- At the anode (oxidation): \( \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \)
- At the cathode (reduction): \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \)

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

3. **Determine the Reaction Quotient (Q)**:
- The reaction quotient \( Q \) for the cell reaction is given by:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]
- Here, we only consider the aqueous species, ignoring the solids.

4. **Use the Nernst Equation**:
- The Nernst equation for the cell is:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log Q \]
- For the given cell, \( n = 2 \) (since 2 electrons are transferred).

5. **Calculate the Initial EMF**:
- Plugging in the expression for \( Q \):
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log \left( \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \right) \]

6. **Consider the Effect of Doubling Concentrations**:
- If the concentrations of both \( \text{Zn}^{2+} \) and \( \text{Cu}^{2+} \) are doubled, then:
\[ Q' = \frac{2[\text{Zn}^{2+}]}{2[\text{Cu}^{2+}]} = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} \]
- Thus, \( Q' = Q \).

7. **Calculate the New EMF**:
- The new EMF will be:
\[ E_{\text{cell}}' = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log Q' \]
- Since \( Q' = Q \), we have:
\[ E_{\text{cell}}' = E^\circ_{\text{cell}} - \frac{0.0591}{2} \log Q \]
- Therefore, \( E_{\text{cell}}' = E_{\text{cell}} \).

8. **Conclusion**:
- The emf of the cell remains the same even after doubling the concentrations of Zn and Cu ions.

### Final Answer:
The emf of the cell remains the same.