The minimum voltage required to electrolyse alumina in the Hall-Heroul process is [Given, `DeltaG^(@)_(f)(A1_(2)O_(3)) =- 1520kJ//mol` and `DeltaG^(@)_(f)(CO_(2)) = 394kJ//mol]`

Options

A. 1.575 V
B. 1.60 V
C. 1.312 V
D. `-2.62 V`

Answer

To find the minimum voltage required to electrolyze alumina in the Hall-Heroult process, we can follow these steps:

### Step 1: Write the balanced chemical reaction
The reaction for the electrolysis of alumina (Al₂O₃) can be represented as:
\[ 2 \text{Al}_2\text{O}_3 (l) + 3 \text{C} (s) \rightarrow 4 \text{Al} (s) + 3 \text{CO}_2 (g) \]

### Step 2: Calculate the change in Gibbs free energy (ΔG°)
Using the standard Gibbs free energy of formation values given:
- ΔG°_f (Al₂O₃) = -1520 kJ/mol
- ΔG°_f (CO₂) = -394 kJ/mol

The change in Gibbs free energy for the reaction can be calculated using:
\[ \Delta G^\circ = \sum \Delta G^\circ_f \text{(products)} - \sum \Delta G^\circ_f \text{(reactants)} \]
Substituting the values:
\[ \Delta G^\circ = [3 \times (-394 \text{ kJ/mol})] - [2 \times (-1520 \text{ kJ/mol})] \]
\[ = -1182 \text{ kJ} + 3040 \text{ kJ} \]
\[ = 1858 \text{ kJ/mol} \]

### Step 3: Convert ΔG° to joules
Since we need to work in joules for the calculation of voltage:
\[ \Delta G^\circ = 1858 \text{ kJ/mol} \times 1000 \text{ J/kJ} = 1858000 \text{ J/mol} \]

### Step 4: Determine the number of electrons transferred (n)
In the reaction, aluminum (Al) goes from an oxidation state of +3 in Al₂O₃ to 0 in elemental aluminum. For 2 moles of Al₂O₃, 4 moles of Al are produced, thus:
- Each Al atom requires 3 electrons.
- Total electrons for 4 Al = \( 4 \times 3 = 12 \) electrons.

### Step 5: Use the relationship between ΔG° and E°cell
The relationship is given by:
\[ \Delta G^\circ = -nFE^\circ_{cell} \]
Where:
- F = Faraday's constant = 96500 C/mol
- n = number of moles of electrons = 12

Rearranging for E°cell:
\[ E^\circ_{cell} = -\frac{\Delta G^\circ}{nF} \]
Substituting the values:
\[ E^\circ_{cell} = -\frac{1858000 \text{ J/mol}}{12 \times 96500 \text{ C/mol}} \]
\[ = -\frac{1858000}{1158000} \]
\[ = 1.60 \text{ V} \]

### Final Answer
The minimum voltage required to electrolyze alumina in the Hall-Heroult process is **1.60 V**.

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