The oxidation numbers of all the atoms are: In SO_{4}^{2-}, S = 6+, O = 2-; Al_{(s)} = 0; In SO_{3}^{2-}, S = 4+, O = 2-, and Al^{3+} = 3+

Al --> Al^{3+} + 3 e^{-} (x 2) gives 2 Al --> 2 Al^{3+} + 6 e^{-}

2 e^{-} + SO_{4}^{2-} + 2 H^{+} --> SO_{3}^{2-} + H_{2}O (x 3) gives

2 Al --> 2 Al^{3+} + 6 e^{-}

6 e^{-} + 3 SO_{4}^{2-} + 6 H^{+} --> 3 SO_{3}^{2-} + 3 H_{2}O

Which, when combined, gives the net ionic REDOX equation as:

2 Al + 3 SO_{4}^{2-} + 6 H^{+} ---> 2 Al^{3+} + 3 SO_{3}^{2-} + 3 H_{2}O

It is now balanced in acid (H^{+}). To balance it now in base, "neutralize" the acid by adding OH^{-} to each side: So we add 6 OH^{-} to both the reactants side & the products side. This results in 6 HOH (water) molecules on the left and 6 OH^{-} on the right. We have 3 H_{2}O molecules on the right from the acid balance, so we can subtract those three from the 6 on the left, and the resulting net ionic REDOX equation becomes:

2 Al + 3 SO_{4}^{2-} + 3 H_{2}O --> 2 Al^{3+} + 3 SO_{3}^{2-} + 6 OH^{-}