Standard cell voltage for the cell `Pb//Pb^(2+)||Sn^(2+)//Sn " is " -0.01V`. If the cell is the exhibit `E_("cell") = 0`, the value of log `[Sn^(2+)]//[Pb^(2+)]` should be

Options

A. 0.33
B. 0.5
C. 1.5
D. `-0.5`

Answer

To solve the problem, we need to find the value of \(\log \left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right)\) given that the standard cell voltage \(E_{cell} = 0\) and the standard cell voltage for the cell \(Pb//Pb^{2+}||Sn^{2+}//Sn\) is \(-0.01V\).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Standard cell voltage \(E_{cell} = 0\)
- Standard cell voltage \(E^{\circ}_{cell} = -0.01V\)

2. **Use the Nernst Equation:**
The Nernst equation relates the cell potential to the concentrations of the reactants and products:
\[ E_{cell} = E^{\circ}_{cell} - \frac{RT}{nF} \log \left(\frac{[Products]}{[Reactants]}\right) \]
Here, \(R\) is the universal gas constant, \(T\) is the temperature in Kelvin, \(n\) is the number of moles of electrons transferred, and \(F\) is Faraday's constant.

3. **Determine the Number of Electrons Transferred (\(n\)):**
The half-reaction for the oxidation of lead (\(Pb\)) and the reduction of tin (\(Sn^{2+}\)) involves 2 electrons:
\[ Pb \rightarrow Pb^{2+} + 2e^{-} \quad (Oxidation) \]
\[ Sn^{2+} + 2e^{-} \rightarrow Sn \quad (Reduction) \]
Thus, \(n = 2\).

4. **Substituting Values into the Nernst Equation:**
We can substitute the known values into the Nernst equation:
\[ 0 = -0.01 - \frac{(8.314)(298)}{(2)(96500)} \log \left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right) \]

5. **Calculate \(\frac{RT}{nF}\):**
First, calculate \(\frac{RT}{nF}\):
\[ \frac{RT}{nF} = \frac{(8.314)(298)}{(2)(96500)} \approx 0.0591 \]

6. **Rearranging the Nernst Equation:**
Rearranging the equation gives:
\[ 0.01 = 0.0591 \log \left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right) \]

7. **Solving for the Logarithm:**
Now, isolate the logarithm:
\[ \log \left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right) = \frac{0.01}{0.0591} \]
Calculate the right side:
\[ \log \left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right) \approx 0.169 \]

8. **Final Expression:**
Therefore, the final expression for the logarithm is:
\[ \log \left(\frac{[Sn^{2+}]}{[Pb^{2+}]}\right) \approx 0.169 \]