During electrolysis of aluminium, the reactions taking place at cathode and anode are as follows:
4Al3+ + 12e- -> 4Al (aluminium metal at the (-)cathode) reduction.
6O2- – 12e- -> 3O2 (oxygen gas at the (+)anode) oxidation.
It can be observed that overall number of electrons involved are 12 in both the cases.
The overall reaction is
aluminium oxide -> aluminium + oxygen.
2Al2O3(l) -> 4Al(l) + 3O2(g)
Hence, there is no situation as such of al2o3 aluminum give 3 electron while oxygen take 2 electron where 1 electron go.