The `E^(c-)` for `Cu^(2+)//Cu^(o+),Cu^(o+)//Cu,Cu^(2+)//Cu,` are `0.15V,0.50V` and `0.325V`, respectively. The redox cell showing redox reaction `2Cu^(o+) rarr Cu^(2+)+Cu` is made. `E^(c-)` of this cell reaction and `DeltaG^(c-)` may be

Options

A. `E^(c-)=0.175V or E^(c-)=0.350V`
B. `n=1 or 2`
C. `Delta G^(c-)=-33.775k J`
D. All of these

Explanation

To solve the question, we need to determine the standard cell potential (E°) and the Gibbs free energy change (ΔG°) for the redox reaction given:

**Redox Reaction:**
\[ 2 \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu} \]

### Step 1: Identify the Half-Reactions
From the given standard electrode potentials:
1. \( \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \) has \( E° = 0.50 \, V \)
2. \( \text{Cu}^+ \rightarrow \text{Cu}^{2+} + e^- \) has \( E° = 0.15 \, V \)

### Step 2: Determine the Oxidation and Reduction Reactions
- **Oxidation (Anode):**
\[ \text{Cu}^+ \rightarrow \text{Cu}^{2+} + e^- \]

- **Reduction (Cathode):**
\[ \text{Cu}^{2+} + 2e^- \rightarrow \text{Cu} \]

### Step 3: Write the Overall Reaction
To balance the electrons, we multiply the oxidation half-reaction by 2:
\[ 2 \text{Cu}^+ \rightarrow 2 \text{Cu}^{2+} + 2e^- \]
Then, we add the reduction half-reaction:
\[ 2 \text{Cu}^+ \rightarrow \text{Cu}^{2+} + \text{Cu} \]

### Step 4: Calculate the Standard Cell Potential (E°cell)
Using the formula:
\[ E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} \]

Substituting the values:
- \( E°_{\text{cathode}} = 0.50 \, V \) (for Cu²⁺ to Cu)
- \( E°_{\text{anode}} = 0.15 \, V \) (for Cu⁺ to Cu²⁺)

Thus,
\[ E°_{\text{cell}} = 0.50 \, V - 0.15 \, V = 0.35 \, V \]

### Step 5: Calculate Gibbs Free Energy Change (ΔG°)
Using the formula:
\[ \Delta G° = -nFE°_{\text{cell}} \]
Where:
- \( n = 2 \) (number of moles of electrons transferred)
- \( F = 96500 \, C/mol \) (Faraday's constant)
- \( E°_{\text{cell}} = 0.35 \, V \)

Calculating ΔG°:
\[ \Delta G° = -2 \times 96500 \, C/mol \times 0.35 \, V \]
\[ \Delta G° = -2 \times 96500 \times 0.35 = -67,975 \, J \]
\[ \Delta G° = -67.975 \, kJ \]

### Final Answers
- \( E°_{\text{cell}} = 0.35 \, V \)
- \( \Delta G° = -67.975 \, kJ \)