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 reduction of `Al_(2)O_(3)` at `500^(@)C` is at least:

Options

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

Answer

To find the potential difference needed for the electrolytic reduction of \( Al_2O_3 \) at \( 500^\circ C \), we can follow these steps:

### Step 1: Understand the Reaction
The given reaction is:
\[ \frac{2}{3} Al_2O_3 \rightarrow \frac{4}{3} Al + O_2 \]
The Gibbs free energy change (\( \Delta_r G \)) for this reaction is \( +966 \, \text{kJ/mol} \).

### Step 2: Convert Gibbs Energy to Joules
Since \( \Delta_r G \) is given in kilojoules, we need to convert it to joules:
\[ \Delta_r G = 966 \, \text{kJ/mol} = 966 \times 10^3 \, \text{J/mol} = 966000 \, \text{J/mol} \]

### Step 3: Identify the Number of Electrons Involved
From the reaction, we can deduce that:
- Each \( Al^{3+} \) ion requires 3 electrons to be reduced to aluminum (Al).
- The stoichiometry of the reaction shows that \( \frac{2}{3} Al_2O_3 \) produces \( \frac{4}{3} Al \) and \( O_2 \).

Calculating the number of moles of \( Al^{3+} \):
\[ \text{Moles of } Al^{3+} = \frac{2}{3} \times 2 = \frac{4}{3} \]
Thus, the total number of electrons needed for the reduction of \( Al^{3+} \) ions:
\[ \text{Electrons for } Al^{3+} = \frac{4}{3} \times 3 = 4 \, \text{electrons} \]

### Step 4: Calculate the Gibbs Free Energy Relation to Electrode Potential
Using the relation:
\[ \Delta_r G = -nFE \]
Where:
- \( n \) = number of moles of electrons = 4
- \( F \) = Faraday's constant = \( 96500 \, \text{C/mol} \)
- \( E \) = electrode potential (in volts)

Rearranging gives:
\[ E = -\frac{\Delta_r G}{nF} \]

### Step 5: Substitute Values
Substituting the values into the equation:
\[ E = -\frac{966000 \, \text{J/mol}}{4 \times 96500 \, \text{C/mol}} \]
Calculating the denominator:
\[ 4 \times 96500 = 386000 \, \text{C/mol} \]
Now substituting back:
\[ E = -\frac{966000}{386000} \approx -2.5 \, \text{V} \]

### Step 6: Determine the Required Potential Difference
Since the potential difference must be positive for the electrolytic process, we take the absolute value:
\[ \text{Potential Difference} = 2.5 \, \text{V} \]

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

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