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 calculate the standard Gibbs free energy change (ΔG°) for the reaction:

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

we will follow these steps:

### Step 1: Identify the half-reactions and their standard potentials

We have the following half-reactions with their standard potentials:

1. \( \text{Cu}^+(aq) + e^- \rightarrow \text{Cu}(s) \) with \( E^\circ = 0.52 \, \text{V} \)
2. \( \text{Cu}^{2+}(aq) + e^- \rightarrow \text{Cu}^+(aq) \) with \( E^\circ = 0.16 \, \text{V} \)

### Step 2: Reverse the second half-reaction

Since we need \( \text{Cu}^+ \) to be oxidized to \( \text{Cu}^{2+} \), we reverse the second half-reaction:

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

When we reverse a reaction, the sign of the standard potential changes:

\[ E^\circ = -0.16 \, \text{V} \]

### Step 3: Combine the half-reactions

Now we can combine the two half-reactions:

1. \( \text{Cu}^+(aq) + e^- \rightarrow \text{Cu}(s) \) (oxidation)
2. \( 2 \text{Cu}^+(aq) \rightarrow \text{Cu}^{2+}(aq) + e^- \) (reduction)

Adding these reactions gives:

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

### Step 4: Calculate the overall standard potential

The overall standard potential \( E^\circ \) for the reaction is calculated as follows:

\[ E^\circ = E^\circ_{\text{oxidation}} + E^\circ_{\text{reduction}} \]

Substituting the values we have:

\[ E^\circ = 0.52 \, \text{V} + (-0.16 \, \text{V}) = 0.34 \, \text{V} \]

### Step 5: Calculate ΔG°

The relationship between Gibbs free energy change and standard potential is given by:

\[ \Delta G^\circ = -nFE^\circ \]

Where:
- \( n \) = number of moles of electrons transferred (which is 1 for this reaction)
- \( F \) = Faraday's constant \( \approx 96500 \, \text{C/mol} \)
- \( E^\circ \) = standard potential calculated above

Substituting the values:

\[ \Delta G^\circ = -1 \times 96500 \, \text{C/mol} \times 0.34 \, \text{V} \]

Calculating this gives:

\[ \Delta G^\circ = -32910 \, \text{J} \]

### Step 6: Convert to kJ

To convert Joules to kilojoules, we divide by 1000:

\[ \Delta G^\circ = -32.91 \, \text{kJ} \]

### Final Answer

The value of \( \Delta G^\circ \) for the reaction is approximately:

\[ \Delta G^\circ \approx -32.91 \, \text{kJ} \]

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