What is the free energy change per mole of Cu(II) ion formed in a cell consisting of Cu|Cu(II) ion half-cell suitably connected to a `Ag|Ag^(+)` ion half-cell ? (Given: `E^(@) = 0.46 V`)

Options

A. `-75 KJ`
B. `-89 KJ`
C. `-45 KJ`
D. `-25 kJ`

Answer

To find the free energy change per mole of Cu(II) ion formed in the given electrochemical cell, we can use the relationship between Gibbs free energy change (ΔG°) and the cell potential (E°). The formula to calculate ΔG° is:

\[ \Delta G° = -nFE° \]

Where:
- \( n \) = number of moles of electrons transferred in the reaction
- \( F \) = Faraday's constant (approximately \( 96500 \, \text{C/mol} \))
- \( E° \) = standard cell potential (given as \( 0.46 \, \text{V} \))

### Step-by-step Solution:

1. **Identify the half-reactions**:
- The half-reaction for copper is:
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \quad (Reduction) \]
- The half-reaction for silver is:
\[ \text{Ag} \rightarrow \text{Ag}^{+} + e^- \quad (Oxidation) \]

2. **Balance the half-reactions**:
- To balance the electrons transferred, we multiply the silver half-reaction by 2:
\[ 2\text{Ag} \rightarrow 2\text{Ag}^{+} + 2e^- \]

3. **Determine the total number of electrons (n)**:
- From the balanced equations, we see that 2 electrons are involved in the overall reaction:
\[ \text{Cu}^{2+} + 2\text{Ag} \rightarrow \text{Cu} + 2\text{Ag}^{+} \]
- Thus, \( n = 2 \).

4. **Substitute values into the ΔG° equation**:
- Using \( n = 2 \), \( F = 96500 \, \text{C/mol} \), and \( E° = 0.46 \, \text{V} \):
\[ \Delta G° = -nFE° = -2 \times 96500 \, \text{C/mol} \times 0.46 \, \text{V} \]

5. **Calculate ΔG°**:
- First, calculate the product:
\[ \Delta G° = -2 \times 96500 \times 0.46 = -88780 \, \text{J/mol} \]

6. **Convert to kilojoules**:
- To convert joules to kilojoules, divide by 1000:
\[ \Delta G° = -88.78 \, \text{kJ/mol} \approx -89 \, \text{kJ/mol} \]

### Final Answer:
The free energy change per mole of Cu(II) ion formed is approximately \(-89 \, \text{kJ/mol}\).