Hall heroult process:
In the Hall-Héroult process the following simplified reactions take place at the carbon electrodes:
Cathode: $\mathrm{A}{\mathrm{l}}^{3+}+3{\mathrm{e}}^{-1}\to \mathrm{A}\mathrm{l}$
Anode: ${\mathrm{O}}^{2-}+\mathrm{C}\to \mathrm{C}\mathrm{O}+2{\mathrm{e}}^{-}\mathrm{ }$

Overall:
$\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3+3\mathrm{C}\to 2\mathrm{A}\mathrm{l}+3\mathrm{C}\mathrm{O}$
And $2\mathrm{ }\mathrm{A}{\mathrm{l}}_2{\mathrm{O}}_3+3\mathrm{C}\to 4\mathrm{A}\mathrm{l}+3\mathrm{C}{\mathrm{O}}_2$

In reality, much more $\mathrm{C}{\mathrm{O}}_2$ is formed at the anode than $\mathrm{C}\mathrm{O}:$
Cryolite ($\mathrm{N}{\mathrm{a}}_3\mathrm{A}\mathrm{l}{\mathrm{F}}_6$) lowers the melting point of the mixture and make it more conductive in nature so as to carry out electrolysis of the ore in Hall-Heroult process.