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 solve the problem of determining the concentration of \( \text{Cu}^{2+} \) ions at which the electrode potential becomes zero, we will use the Nernst equation. Here are the steps:

### Step 1: Write the Nernst Equation
The Nernst equation is given by:
\[ E_{\text{cell}} = E^0_{\text{cell}} - \frac{0.0591}{n} \log Q \]
where:
- \( E_{\text{cell}} \) is the cell potential,
- \( E^0_{\text{cell}} \) is the standard electrode potential,
- \( n \) is the number of moles of electrons transferred in the reaction,
- \( Q \) is the reaction quotient.

### Step 2: Identify the Reaction
For the half-reaction:
\[ \text{Cu} \rightarrow \text{Cu}^{2+} + 2e^- \]
Here, \( n = 2 \) because 2 electrons are transferred.

### Step 3: Define the Reaction Quotient \( Q \)
The reaction quotient \( Q \) for this reaction is:
\[ Q = \frac{[\text{Cu}^{2+}]}{1} = [\text{Cu}^{2+}] \]
since the concentration of solid copper is taken as 1.

### Step 4: Set the Cell Potential to Zero
We want to find the concentration of \( \text{Cu}^{2+} \) when \( E_{\text{cell}} = 0 \):
\[ 0 = E^0_{\text{cell}} - \frac{0.0591}{n} \log Q \]
Substituting \( E^0_{\text{cell}} = -0.34 \, \text{V} \) and \( n = 2 \):
\[ 0 = -0.34 - \frac{0.0591}{2} \log [\text{Cu}^{2+}] \]

### Step 5: Rearrange the Equation
Rearranging gives:
\[ 0.34 = \frac{0.0591}{2} \log [\text{Cu}^{2+}] \]

### Step 6: Solve for \( \log [\text{Cu}^{2+}] \)
Multiply both sides by \( \frac{2}{0.0591} \):
\[ \log [\text{Cu}^{2+}] = \frac{0.34 \times 2}{0.0591} \]
Calculating the right side:
\[ \log [\text{Cu}^{2+}] = \frac{0.68}{0.0591} \approx 11.50 \]

### Step 7: Find the Concentration \( [\text{Cu}^{2+}] \)
To find \( [\text{Cu}^{2+}] \), we take the antilog:
\[ [\text{Cu}^{2+}] = 10^{-11.50} \]
This can be expressed as:
\[ [\text{Cu}^{2+}] = 3.16 \times 10^{-12} \, \text{M} \]

### Step 8: Match with Given Options
The closest option to \( 3.16 \times 10^{-12} \, \text{M} \) is \( 2.98 \times 10^{-12} \, \text{M} \).

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