Given that standard potential for the following half - cell reaction at 298 K,
`Cu^(+)(aq)+e^(-)rarr Cu(s), E^(@)=0.52V`
`Cu^(2+)(aq)+e^(-)rarrCu^(+)(aq), E^(@)=0.16V`
Calculate the `DeltaG^(@)(kJ)` for the eaction, `[2Cu^(+)(aq)rarr Cu(s)+Cu^(2+)]`

Options

A. `-34.740`
B. `-65.720`
C. `-69.720`
D. `-131.440`

Answer

To solve the problem, we need to calculate the standard Gibbs free energy change (ΔG°) for the reaction:

\[ 2 \text{Cu}^+ (aq) \rightarrow \text{Cu} (s) + \text{Cu}^{2+} (aq) \]

### Step 1: Identify the Half-Reactions
We have two half-reactions with their standard potentials:

1. Reduction of Cu⁺ to Cu:
\[ \text{Cu}^+ (aq) + e^- \rightarrow \text{Cu} (s), \quad E^\circ = 0.52 \, \text{V} \]

2. Reduction of Cu²⁺ to Cu⁺:
\[ \text{Cu}^{2+} (aq) + 2e^- \rightarrow \text{Cu}^+ (aq), \quad E^\circ = 0.16 \, \text{V} \]

### Step 2: Reverse the Second Half-Reaction
To obtain the desired reaction, we need to reverse the second half-reaction:

\[ \text{Cu}^+ (aq) \rightarrow \text{Cu}^{2+} (aq) + e^-, \quad E^\circ = -0.16 \, \text{V} \]

### Step 3: Combine the Half-Reactions
Now we can add the two half-reactions together:

1. From the first half-reaction:
\[ \text{Cu}^+ (aq) + e^- \rightarrow \text{Cu} (s) \]

2. From the reversed second half-reaction:
\[ \text{Cu}^+ (aq) \rightarrow \text{Cu}^{2+} (aq) + e^- \]

Adding these gives:
\[ 2 \text{Cu}^+ (aq) \rightarrow \text{Cu} (s) + \text{Cu}^{2+} (aq) \]

### Step 4: Calculate the Standard Cell Potential (E°)
The standard cell potential for the overall reaction is calculated as follows:
\[ E^\circ_{\text{cell}} = E^\circ_{\text{reduction}} + E^\circ_{\text{oxidation}} \]
\[ E^\circ_{\text{cell}} = 0.52 \, \text{V} + (-0.16 \, \text{V}) \]
\[ E^\circ_{\text{cell}} = 0.52 \, \text{V} - 0.16 \, \text{V} = 0.36 \, \text{V} \]

### Step 5: Calculate ΔG°
Using the formula:
\[ \Delta G^\circ = -nFE^\circ_{\text{cell}} \]
Where:
- \( n = 2 \) (since 2 moles of electrons are transferred in the overall reaction)
- \( F = 96500 \, \text{C/mol} \) (Faraday's constant)
- \( E^\circ_{\text{cell}} = 0.36 \, \text{V} \)

Calculating ΔG°:
\[ \Delta G^\circ = -2 \times 96500 \, \text{C/mol} \times 0.36 \, \text{V} \]
\[ \Delta G^\circ = -2 \times 96500 \times 0.36 \]
\[ \Delta G^\circ = -69360 \, \text{J/mol} \]

### Step 6: Convert to kJ
To convert from joules to kilojoules:
\[ \Delta G^\circ = -69.36 \, \text{kJ/mol} \]

### Final Answer
Thus, the final answer for ΔG° for the reaction is:
\[ \Delta G^\circ \approx -69.36 \, \text{kJ} \]