If the `Pb^(2+)` ion concentration is maintained at `1.0 M` , what is the `[Cu^(2+)]` concentration when the cell potential drops to zero? `E^(o) ""_(cell)= 0.473V`, and cell reaction is `Pb (s)abs(Pb^(2+ )(1.0M))abs(Cu^(2+)(1.0xx10^(-4)M)) Cu(s)`

Options

A. `1xx10^(-16)M`
B. `1xx10^(16)M`
C. `1xx10^(-14)M`
D. `1xx10^(14)M`

Answer

To solve the problem, we will use the Nernst equation, which relates the cell potential to the concentrations of the reactants and products in an electrochemical cell.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Standard cell potential, \( E^\circ_{\text{cell}} = 0.473 \, \text{V} \)
- Concentration of \( \text{Pb}^{2+} = 1.0 \, \text{M} \)
- Concentration of \( \text{Cu}^{2+} \) is what we need to find when \( E_{\text{cell}} = 0 \).

2. **Write the Nernst Equation:**
The Nernst equation is given by:
\[ E_{\text{cell}} = E^\circ_{\text{cell}} - \frac{0.0591}{n} \log \left( \frac{[\text{Ox}]}{[\text{Red}]} \right) \]
where:
- \( n \) = number of moles of electrons transferred in the reaction.
- \( [\text{Ox}] \) = concentration of the oxidized species.
- \( [\text{Red}] \) = concentration of the reduced species.

3. **Determine the Reaction and Number of Electrons:**
The cell reaction is:
\[ \text{Pb (s)} + \text{Cu}^{2+} \rightarrow \text{Pb}^{2+} + \text{Cu (s)} \]
Here, lead (Pb) is oxidized to \( \text{Pb}^{2+} \) and copper (\( \text{Cu}^{2+} \)) is reduced to copper metal (Cu). The number of electrons transferred \( n = 2 \).

4. **Set Up the Nernst Equation for \( E_{\text{cell}} = 0 \):**
Since we want to find the concentration of \( \text{Cu}^{2+} \) when \( E_{\text{cell}} = 0 \):
\[ 0 = 0.473 - \frac{0.0591}{2} \log \left( \frac{[1.0]}{[\text{Cu}^{2+}]} \right) \]

5. **Rearranging the Equation:**
Rearranging gives:
\[ 0.473 = \frac{0.0591}{2} \log \left( \frac{1.0}{[\text{Cu}^{2+}]} \right) \]
Multiply both sides by \( \frac{2}{0.0591} \):
\[ \frac{2 \times 0.473}{0.0591} = \log \left( \frac{1.0}{[\text{Cu}^{2+}]} \right) \]

6. **Calculate the Left Side:**
\[ \frac{2 \times 0.473}{0.0591} \approx 16.31 \]
Thus,
\[ \log \left( \frac{1.0}{[\text{Cu}^{2+}]} \right) = 16.31 \]

7. **Exponentiate to Solve for \( [\text{Cu}^{2+}] \):**
Taking the antilogarithm:
\[ \frac{1.0}{[\text{Cu}^{2+}]} = 10^{16.31} \]
Therefore,
\[ [\text{Cu}^{2+}] = \frac{1.0}{10^{16.31}} = 10^{-16.31} \, \text{M} \]

8. **Final Result:**
\[ [\text{Cu}^{2+}] \approx 1.0 \times 10^{-16} \, \text{M} \]

### Conclusion:
The concentration of \( \text{Cu}^{2+} \) when the cell potential drops to zero is approximately \( 1.0 \times 10^{-16} \, \text{M} \).