During the electrolytic reduction of alumina, the reaction at cathode is
- $2 \mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{O}_2+4 \mathrm{H}^{+}+4 e^{-}$
- $3 \mathrm{~F}^{-} \longrightarrow 3 \mathrm{~F}+3 e^{-}$
- $\mathrm{Al}^{3+}+3 e^{-} \longrightarrow \mathrm{Al}$
- $2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_2$
Solution
Asked in: AP EAMCET 2001