In the extraction of aluminium from \(Al_2O_3\) we generally add \(Na_3AlF_6\) or \(CaF_2\):
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → To lower the melting point of the mix and bring conductivity
Aluminium is obtained by the electrolysis of fused alumina (Hall-Héroult process). Pure \( \ce{Al2O3} \) melts above 2300 K and is a poor conductor when fused. It is therefore mixed with cryolite (Na3AlF6) or CaF2, which lowers the melting point of the mix (to about 1240 K) and increases its electrical conductivity. The fused mixture is electrolysed with a steel cathode and graphite anodes:
Cathode: Al3+(melt) + 3e− → Al(l)
Anode: C(s) + O2−(melt) → CO(g) + 2e−; C(s) + 2O2−(melt) → CO2(g) + 4e−
Net reaction: \( \ce{2Al2O3} \) + 3C → 4Al + \( \ce{3CO2} \)
Why the other options are incorrect
Option (1) To reduce it — Incorrect. \( \ce{Al^{3+}} \) is reduced by the electric current at the cathode; cryolite and \( \ce{CaF2} \) are not reducing agents.
Option (2) To oxidize the mixture — Incorrect. No oxidation of the alumina is wanted; aluminium has to be reduced, and the oxygen is removed at the carbon anode.
Option (3) To increase the melting point of the mix and bring conductivity — Incorrect. The additives lower the melting point; a higher melting point would make the process more costly.
NCERT (Pre-rationalised edition, 2022-23):
Electrochemical Principles of Metallurgy – “… In the metallurgy of aluminium, purified Al2O3 is mixed with Na3AlF6 or CaF2 which lowers the melting point of the mix and brings conductivity.”