Answer
a) Oxidation: 6Cl$^{-}$(l) --> 3Cl$_{2}$(g) + 6e$^{-}$ (anode)
Reduction: 2Al$^{3+}$(l) + 6e$^{-}$ --> 2Al(l) (cathode)
b) 2AlCl$_{3}$(l) --> 2Al(l) + 3Cl$_{2}$(g)
c) Chlorine gas is produced at anode. Liquid aluminum is produced at cathode.
Work Step by Step
a) Oxidation: 6Cl$^{-}$(l) --> 3Cl$_{2}$(g) + 6e$^{-}$ (anode)
Reduction: 2Al$^{3+}$(l) + 6e$^{-}$ --> 2Al(l) (cathode)
b) 2AlCl$_{3}$(l) --> 2Al(l) + 3Cl$_{2}$(g)
c) Chlorine gas is produced at anode. Liquid aluminum is produced at cathode.
