In Hall - Heroult process, the following is used for reducing Al₂ O₃ :-

Solution

Chemistry ma’am · 4 steps

Look at what gets used up: the graphite anode burns away as it takes the oxygen from Al₂O₃.

1. The set-up. Purified alumina is dissolved in molten cryolite, Na₃AlF₆, with a little fluorspar, CaF₂, and electrolysed. A steel vessel lined with carbon is the cathode, and graphite rods dipped into the melt are the anode.
2. Electrode reactions. At the cathode, Al³⁺ + 3e⁻ → Al. At the anode, oxide ions give up electrons and react with the carbon: C + O²⁻ → CO + 2e⁻ and C + 2O²⁻ → CO₂ + 4e⁻. The anode is slowly burnt away and has to be replaced from time to time.

   2Al₂O₃ + 3C → 4Al + 3CO₂

3. Who is the reductant. In the overall reaction, carbon removes the oxygen from Al₂O₃ and is itself oxidised to CO and CO₂. Cryolite and CaF₂ come out unchanged. They only lower the melting point and improve the conductivity of the melt.
4. So. Graphite, the anode material, acts as the reducing agent for Al₂O₃. The answer is (c).

The idea to keep. In an electrolytic extraction, separate the substance that is consumed from the additives that only help the melt. Here the carbon anode is oxidised to CO₂, while cryolite and fluorspar are just helpers.

Answer from the NTA final answer key. Solution by PrepWiser.