In electrolysis of ${\mathrm{Al}}_2{\mathrm{O}}_3$ by Hall- Heroult process:

(a)In electrolysis of ‎Aluminum oxide by the Hall-Heroult process cryolite ${\mathrm{Na}}_3\left[{\mathrm{AlF}}_6\right]$ lowers the melting point of ${\mathrm{Al}}_2{\mathrm{O}}_3$ and increases its electrical conductivity

Alumina,${\mathrm{Al}}_2{\mathrm{O}}_3$ , is mixed in molten cryolite, ${\mathrm{Na}}_3\left[{\mathrm{AlF}}_6\right]$, in the Hall-Héroult process to reduce its melting point and facilitate electrolysis. The solution's conductivity is also improved by it.

(b) In the Hall-Heroult process aluminum( $\mathrm{Al}$) is obtained at the cathode and probably ${\mathrm{CO}}_2$ at anode because the procedure uses a fused combination of cryolite, alumina, and small amounts of fluorspar and aluminum fluoride to electrolyze alumina. The electrical conductivity of alumina is increased and the fusion temperature is lowered by the addition of cryolite and fluorspar.

 Al(III) is converted to Al metal at the cathode. Alumina and fluorine react at the anode to release oxygen. Carbon dioxide is created when oxygen and carbon are reacted.

Both (a) and (b) are correct

Hence Correct option is C