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 effect of doubling the concentrations of Zn and Cu ions on the EMF of the cell `Zn|Zn^(2+) || Cu^(2+)|Cu`, we will use the Nernst equation. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

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

2. **Overall Reaction**:
- The overall cell reaction can be written as:
\[ \text{Zn (s)} + \text{Cu}^{2+} (aq) \rightarrow \text{Zn}^{2+} (aq) + \text{Cu (s)} \]

3. **Apply the Nernst Equation**:
- The Nernst equation is given by:
\[ E_{cell} = E^0_{cell} - \frac{RT}{nF} \ln Q \]
- Where:
- \(E^0_{cell}\) is the standard EMF of the cell.
- \(R\) is the universal gas constant.
- \(T\) is the temperature in Kelvin.
- \(n\) is the number of moles of electrons transferred (which is 2 in this case).
- \(F\) is Faraday's constant.
- \(Q\) is the reaction quotient.

4. **Determine the Reaction Quotient (Q)**:
- Initially, let the concentrations of \(\text{Zn}^{2+}\) and \(\text{Cu}^{2+}\) be \(x\) and \(y\) respectively.
- The reaction quotient \(Q\) is given by:
\[ Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{x}{y} \]

5. **Calculate Initial EMF**:
- The initial EMF can be expressed as:
\[ E_{cell, initial} = E^0_{cell} - \frac{RT}{2F} \ln \left(\frac{x}{y}\right) \]

6. **Doubling the Concentrations**:
- When the concentrations of both ions are doubled, we have:
- \([\text{Zn}^{2+}] = 2x\)
- \([\text{Cu}^{2+}] = 2y\)
- The new reaction quotient \(Q'\) becomes:
\[ Q' = \frac{2x}{2y} = \frac{x}{y} \]

7. **Calculate Final EMF**:
- The final EMF can be expressed as:
\[ E_{cell, final} = E^0_{cell} - \frac{RT}{2F} \ln \left(\frac{2x}{2y}\right) = E^0_{cell} - \frac{RT}{2F} \ln \left(\frac{x}{y}\right) \]

8. **Conclusion**:
- Since the expression for \(E_{cell, final}\) is the same as \(E_{cell, initial}\), we conclude that:
\[ E_{cell, final} = E_{cell, initial} \]
- Therefore, the EMF of the cell remains unchanged when the concentrations of Zn and Cu ions are doubled.

### Final Answer:
The EMF of the cell remains the same when the concentrations of Zn and Cu ions are doubled.

---