J.R. S. answered 11/11/23
Ph.D. University Professor with 10+ years Tutoring Experience
Oxidation takes place at the ANODE
Reduction takes place a the CATHODE
Oxidation half reaction:
Mg(s) ==> Mg2+ + 2e-
Reduction half reaction:
Al3+ + 3e- ==> Al(s)
In order to equalize the electrons, we must multiply oxidation rxn by 3 and reduction rxn by 2:
3Mg(s) ==> 3Mg2+(aq) + 6e- half cell at the anode
2Al3+(aq) + 6e- ==> 2Al(s) half cell at the cathode