Standard cell voltage for the cell `Pb//Pb^(2+)||Sn^(2+)//Sn " is " -0.01V`. If the cell 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 follow these steps:

### Step 1: Write the half-reactions
For the cell `Pb//Pb^(2+)||Sn^(2+)//Sn`, we identify the half-reactions:
- At the anode (oxidation):
\[ \text{Pb} \rightarrow \text{Pb}^{2+} + 2e^- \]
- At the cathode (reduction):
\[ \text{Sn}^{2+} + 2e^- \rightarrow \text{Sn} \]

### Step 2: Write the overall cell reaction
Combining the half-reactions, we get the overall reaction:
\[ \text{Pb} + \text{Sn}^{2+} \rightarrow \text{Pb}^{2+} + \text{Sn} \]

### Step 3: Use the Nernst equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^{\circ}_{\text{cell}} - \frac{0.059}{n} \log \frac{[\text{Products}]}{[\text{Reactants}]} \]
Where:
- \(E^{\circ}_{\text{cell}} = -0.01 \, \text{V}\)
- \(n = 2\) (number of electrons transferred)

### Step 4: Set \(E_{\text{cell}} = 0\)
According to the problem, \(E_{\text{cell}} = 0\). Plugging this into the Nernst equation gives:
\[ 0 = -0.01 - \frac{0.059}{2} \log \frac{[\text{Pb}^{2+}]}{[\text{Sn}^{2+}]} \]

### Step 5: Rearranging the equation
Rearranging the equation to isolate the logarithmic term:
\[ 0.01 = -\frac{0.059}{2} \log \frac{[\text{Pb}^{2+}]}{[\text{Sn}^{2+}]} \]

### Step 6: Solve for the logarithm
Multiply both sides by \(-\frac{2}{0.059}\):
\[ \log \frac{[\text{Pb}^{2+}]}{[\text{Sn}^{2+}]} = \frac{0.01 \times 2}{0.059} \]
Calculating the right side:
\[ \log \frac{[\text{Pb}^{2+}]}{[\text{Sn}^{2+}]} \approx \frac{0.02}{0.059} \approx 0.338 \]

### Step 7: Find the desired logarithm
We need to find \(\log \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]}\). Using the property of logarithms:
\[ \log \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]} = -\log \frac{[\text{Pb}^{2+}]}{[\text{Sn}^{2+}]} \]
Thus:
\[ \log \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]} = -0.338 \]

### Final Answer
The value of \(\log \frac{[\text{Sn}^{2+}]}{[\text{Pb}^{2+}]}\) is approximately \(-0.338\).

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