Given: `E_(Sn^(2+)//Sn) ^0 = -0.14V` , `E_(Pb^(2+)//Pb) ^0 = -0.13V`. Determine `[(Sn^(2+))//(Pb^(2+))]` at equilibrium. For cell reaction `Sn |Sn^(2+) || Pb^(2+) | Pb` , Take `(2.303RT)//F = 0.06V`

Options

A. (a) 2.15
B. (b) 5.35
C. (c) 6.10
D. (d) 3.25

Explanation

To solve the problem, we need to determine the concentration ratio of \( \text{Sn}^{2+} \) to \( \text{Pb}^{2+} \) at equilibrium for the given electrochemical cell reaction.

### Step-by-Step Solution:

1. **Identify the Cell Reaction**:
The cell reaction is given as:
\[ \text{Sn} | \text{Sn}^{2+} || \text{Pb}^{2+} | \text{Pb} \]
Here, tin (Sn) is oxidized to \( \text{Sn}^{2+} \) and lead \( \text{Pb}^{2+} \) is reduced to lead (Pb).

2. **Determine the Anode and Cathode**:
- **Anode**: Oxidation occurs at the anode. Thus, \( \text{Sn} \) is oxidized to \( \text{Sn}^{2+} \).
- **Cathode**: Reduction occurs at the cathode. Thus, \( \text{Pb}^{2+} \) is reduced to \( \text{Pb} \).

3. **Write the Half-Reactions**:
- Oxidation (Anode):
\[ \text{Sn} \rightarrow \text{Sn}^{2+} + 2e^- \]
- Reduction (Cathode):
\[ \text{Pb}^{2+} + 2e^- \rightarrow \text{Pb} \]

4. **Calculate the Standard Cell Potential (\( E^\circ_{\text{cell}} \))**:
Using the standard reduction potentials:
\[ E^\circ_{\text{Sn}^{2+}/\text{Sn}} = -0.14 \, \text{V} \]
\[ E^\circ_{\text{Pb}^{2+}/\text{Pb}} = -0.13 \, \text{V} \]
The cell potential is calculated as:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{cathode}} - E^\circ_{\text{anode}} = (-0.13) - (-0.14) = 0.01 \, \text{V} \]

5. **Use the Nernst Equation**:
The Nernst equation is given by:
\[ E = E^\circ - \frac{2.303RT}{nF} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]} \right) \]
At equilibrium, \( E = 0 \), so:
\[ 0 = E^\circ - \frac{2.303RT}{nF} \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]} \right) \]

6. **Substitute Known Values**:
Given \( E^\circ = 0.01 \, \text{V} \), \( n = 2 \), and \( \frac{2.303RT}{F} = 0.06 \, \text{V} \):
\[ 0 = 0.01 - 0.06 \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]} \right) \]

7. **Rearranging the Equation**:
\[ 0.01 = 0.06 \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]} \right) \]
\[ \log \left( \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]} \right) = \frac{0.01}{0.06} = \frac{1}{6} \]

8. **Taking the Antilog**:
\[ \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]} = 10^{\frac{1}{6}} \approx 1.47 \]

9. **Final Calculation**:
To find the numerical value:
\[ 10^{\frac{1}{6}} \approx 1.47 \]

### Conclusion:
The concentration ratio of \( \text{Sn}^{2+} \) to \( \text{Pb}^{2+} \) at equilibrium is approximately **1.47**.