Electron balance in electrolysis half-reactions

Steps:

• Molten Al2O3 dissociates into Al^{3+} and O^{2-} ions.
• Cathode (reduction): Al^{3+} + 3e^- → Al (3 electrons per Al).
• Anode (oxidation): 2O^{2-} → O_2 + 4e^- (4 electrons per O_2).
• Balance electrons using LCM of 3 and 4 (12): 4 Al from cathode, 3 O_2 from anode.
• Overall: 2Al_2O_3 → 4Al + 3O_2, so Al : O_2 = 4:3.

Why C is correct:

• The balanced equation 2Al_2O_3 → 4Al + 3O_2 (from electron balance) gives molar ratio of Al to O_2 as 4:3.

Why the others are wrong:

• A: Incorrect ratio; balanced equation shows 4:3, not 1:2.
• B: In industrial process with carbon anode, CO_2 forms, not O_2 gas.
• D: Reduction occurs at cathode by definition; oxidation at anode.

Final answer: C