Question

Hall-Heroult’s process is given by
1. \(ZnO+C\xrightarrow[]{coke,1673K}Zn+CO\)
2. Cr₂O₃ + 2Al ⟶ Al₂O₃ + 2Cr
3. 2Al₂O₃ + 3C ⟶ 4Al + 3CO₂
4. Cu²⁺ (aq) + H₂ (g) ⟶ Cu (s) + 2H⁺ (aq)

Answer 1

Correct Answer - Option 3 : 2Al₂O₃ + 3C ⟶ 4Al + 3CO₂

Concept:

Hall-Heroult’s process is an electro-reduction process by which pure alumina (Al₂O₃) is reduced to crude Al.

In this process, electrolysis of a fused mixture of Al₂O₃, Na₃ [AlF₆] (cryolite) and CaF₂ (fluorspar) is carried out at carbon cathode and graphite anode.

The overall reaction is represented as:

 2Al₂O₃ + 3C ⟶ 4Al + 3CO₂