Identify the oxidation numbers on all atoms in this reaction. Balance the reaction in acidic solution and basic solution.

Question

SO₄²⁻(aq) + Al(s) ® SO₃²⁻ (g) + Al³⁺(aq)

Answer 1

The oxidation numbers of all the atoms are: In SO₄^2-, S = 6+, O = 2-; Al_(s) = 0; In SO₃^2-, S = 4+, O = 2-, and Al³⁺ = 3+

Al --> Al³⁺ + 3 e^- (x 2) gives 2 Al --> 2 Al³⁺ + 6 e^-

2 e^- + SO₄^2- + 2 H⁺ --> SO₃^2- + H₂O (x 3) gives

2 Al --> 2 Al³⁺ + 6 e^-

6 e^- + 3 SO₄^2- + 6 H⁺ --> 3 SO₃^2- + 3 H₂O

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

2 Al + 3 SO₄^2- + 6 H⁺ ---> 2 Al³⁺ + 3 SO₃^2- + 3 H₂O

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 H₂O 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 SO₄^2- + 3 H₂O --> 2 Al³⁺ + 3 SO₃^2- + 6 OH^-