The standard electrode potential of `Cu|Cu^(+2)` is - 0.34 Volt. At what concentration of `Cu^(+2)` ions, will this electrode potential be zero?

Options

A. `1.98 xx 10^(-12)M`
B. `2.98 xx 10^(-12)M`
C. `1.49 xx 10^(-12)M`
D. `5.96 xx 10^(-12)M`

Answer

To determine the concentration of \( \text{Cu}^{2+} \) ions at which the electrode potential becomes zero, we can use the Nernst equation. The Nernst equation relates the cell potential to the standard electrode potential and the concentrations of the reactants and products.

### Step-by-Step Solution:

1. **Write the Nernst Equation**:
The Nernst equation for the half-cell reaction is given by:
\[ E_{\text{cell}} = E^{\circ} + \frac{0.0591}{n} \log \left( \frac{1}{[\text{Cu}^{2+}]} \right) \]
where:
- \( E_{\text{cell}} \) is the cell potential,
- \( E^{\circ} \) is the standard electrode potential,
- \( n \) is the number of moles of electrons transferred in the reaction (for \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \), \( n = 2 \)),
- \( [\text{Cu}^{2+}] \) is the concentration of copper ions.

2. **Substitute Known Values**:
We know from the problem that \( E^{\circ} = -0.34 \, \text{V} \) and we want to find the concentration when \( E_{\text{cell}} = 0 \, \text{V} \). Substituting these values into the Nernst equation gives:
\[ 0 = -0.34 + \frac{0.0591}{2} \log \left( \frac{1}{[\text{Cu}^{2+}]} \right) \]

3. **Rearranging the Equation**:
Rearranging the equation to isolate the logarithmic term:
\[ 0.34 = \frac{0.0591}{2} \log \left( \frac{1}{[\text{Cu}^{2+}]} \right) \]
Multiply both sides by \( 2 \):
\[ 0.68 = 0.0591 \log \left( \frac{1}{[\text{Cu}^{2+}]} \right) \]

4. **Solving for the Logarithm**:
Divide both sides by \( 0.0591 \):
\[ \log \left( \frac{1}{[\text{Cu}^{2+}]} \right) = \frac{0.68}{0.0591} \]
Calculate the right-hand side:
\[ \log \left( \frac{1}{[\text{Cu}^{2+}]} \right) \approx 11.48 \]

5. **Finding the Concentration**:
To find \( [\text{Cu}^{2+}] \), we take the antilogarithm:
\[ \frac{1}{[\text{Cu}^{2+}]} = 10^{11.48} \]
Therefore:
\[ [\text{Cu}^{2+}] = \frac{1}{10^{11.48}} \approx 2.98 \times 10^{-12} \, \text{M} \]

### Final Answer:
The concentration of \( \text{Cu}^{2+} \) ions at which the electrode potential becomes zero is approximately \( 2.98 \times 10^{-12} \, \text{M} \).