Question

The alumina obtained from Bayer process is converted to aluminium metal by:

• A. Carbon reduction in blast furnace
• B. Electrolysis of molten alumina-cryolite mixture
• C. Electrolysis in aqueous NaOH solution
• D. Distillation under vacuum

Hint:

Cryolite (\( \text{Na}_3\text{AlF}_6 \)) serves two primary roles in the Hall-Héroult process:
1. Solves the high-melting point problem by lowering the bath melting point from \( \gt 2000^{\circ}\text{C} \) to \( \sim 950^{\circ}\text{C} \).
2. Improves electrical conductivity of the electrolyte bath.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question asks for the chemical extraction process used to reduce pure alumina (\( \text{Al}_2\text{O}_3 \)) obtained from the Bayer process into metallic aluminum.

Step 2: Key Formula or Approach:
Aluminum is a highly reactive metal with a very stable oxide. The Gibbs free energy of formation of \( \text{Al}_2\text{O}_3 \) is extremely negative, meaning carbon reduction is thermodynamically impossible at standard industrial furnace temperatures. Therefore, electrometallurgy must be used. The process is known as the Hall-Héroult process, which operates via the electrolysis of alumina dissolved in a molten solvent.

Step 3: Detailed Explanation:

• The Hall-Héroult Electrolytic Process:
- Pure alumina (\( \text{Al}_2\text{O}_3 \)) has an extremely high melting point (above \( 2050^{\circ}\text{C} \)), making direct electrolysis of pure molten alumina energetically impractical.
- To lower the operating temperature, alumina is dissolved in a molten bath of synthetic cryolite (\( \text{Na}_3\text{AlF}_6 \)) and aluminum fluoride (\( \text{AlF}_3 \)).
- This cryolite-based mixture forms a eutectic with a much lower melting point, allowing the electrolysis to operate efficiently at approximately \( 950^{\circ}\text{C}-980^{\circ}\text{C} \).

• Electrochemical Reactions:
- Carbon anodes and a carbon-lined cell cathode are used.
- At the Cathode (reduction):
\[ \text{Al}^{3+} + 3e^{-} \rightarrow \text{Al(l)} \]
- At the Anode (oxidation):
\[ \text{C(s)} + 2\text{O}^{2-} \rightarrow \text{CO}_2\text{(g)} + 4e^{-} \]
- Molten aluminum, being denser than the molten electrolyte, sinks to the bottom of the cell and is tapped off.
- Electrolysis in aqueous NaOH solution is impossible because hydrogen ions would reduce preferentially over aluminum ions due to water's electrochemical window.


Step 4: Final Answer:
Alumina is reduced to aluminum metal via the electrolysis of a molten alumina-cryolite mixture.
Thus, the correct option is (B).