The Gibbs energy for the decomposition of `Al_(2)O_(3)` at `500^(@)C` is as follows:
`(2)/(3)Al_(2)O_(3) rarr (4)/(3)Al + O_(2), Delta_(r)G = +966kJ mol^(-1)`
The potential difference needed for electrolytic reeduction of `Al_(2)O_(3)` at `500^(@)C` is at least:

Options

A. 5.0 V
B. 4.5 V
C. 3.0 V
D. 2.5 V

Explanation

To find the potential difference needed for the electrolytic reduction of \( Al_2O_3 \) at \( 500^\circ C \), we can use the relationship between Gibbs free energy change (\( \Delta_r G \)), the number of electrons transferred (\( n \)), Faraday's constant (\( F \)), and the cell potential (\( E_{cell} \)).

### Step-by-Step Solution:

1. **Identify the Reaction and Gibbs Free Energy Change**:
The decomposition reaction given is:
\[ \frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2 \]
The Gibbs energy change for this reaction is:
\[ \Delta_r G = +966 \, \text{kJ/mol} \]

2. **Convert Gibbs Free Energy to Joules**:
Since we need to work in standard units, convert \( \Delta_r G \) from kJ to J:
\[ \Delta_r G = 966 \, \text{kJ/mol} \times 1000 \, \text{J/kJ} = 966000 \, \text{J/mol} \]

3. **Determine the Number of Electrons Transferred (\( n \))**:
From the half-reactions:
- Reduction:
\[ \frac{2}{3} Al^{3+} + 4e^- \rightarrow \frac{4}{3} Al \]
- Oxidation:
\[ \frac{2}{3} O^{2-} \rightarrow O_2 + 4e^- \]
We see that \( n = 4 \) electrons are transferred.

4. **Use the Gibbs Energy Equation**:
The relationship between Gibbs free energy change and cell potential is given by:
\[ \Delta_r G = -nFE_{cell} \]
Rearranging this gives:
\[ E_{cell} = -\frac{\Delta_r G}{nF} \]

5. **Substitute the Known Values**:
Using Faraday's constant \( F = 96500 \, \text{C/mol} \):
\[ E_{cell} = -\frac{966000 \, \text{J/mol}}{4 \times 96500 \, \text{C/mol}} \]

6. **Calculate the Cell Potential**:
\[ E_{cell} = -\frac{966000}{386000} \approx -2.5 \, \text{V} \]

7. **Determine the Required Potential Difference**:
Since the potential difference needed for the electrolytic reduction must be positive, we take the absolute value:
\[ \text{Required Potential Difference} = 2.5 \, \text{V} \]

### Final Answer:
The potential difference needed for the electrolytic reduction of \( Al_2O_3 \) at \( 500^\circ C \) is at least **2.5 V**.

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