Browse Questions

# In the electrolysis of alumina using cryolite, the reaction that takes place at cathode is

(1) $4H_2O+4e^{-} \longrightarrow 2H_2+4\: OH^{-}$

(2) $4AI^{3+}+12e^{-} \longrightarrow 4AI$

(3) $6F_2+2AI_2O_3 \longrightarrow 4AIF_3 + 3O_2$

(4) $12F \longrightarrow 6F_2+12e^{-}$

(2) $4AI^{3+}+12e^{-} \longrightarrow4AI$