How can you work out, using the changes in oxidation numbers, which compound out of KCl, KBr and KI has the greatest reducing power?

We can work this out by observing the changes in oxidation state of sulphur in concentrated sulphuric acid (H₂SO₄) when this compound reacts with KCl, KBr and KI individually.
As suggested by its name, a reducing agent reduces other reactants.When something is reduced, there is a reduction in its oxidation number.
KCl + H₂SO₄ --> KHSO₄ + HClThe oxidation number of S is +6 in H₂SO₄ and +6 also in KHSO₄ (i.e. it does not change). Thus the S is not reduced.
KBr + H₂SO₄ --> KHSO₄ + HBr + Br₂ + SO₂The oxidation number of S is +6 in H₂SO₄ and +4 in SO₂. (A change in oxidation number of -2.)
KI + H₂SO₄ --> KHSO₄ + HI + I₂ + H₂S + SThe oxidation number of S is +6 in H₂SO₄, changing to -2 in H₂S and 0 in S. (Changes of -6 and -8 respectively)
The reaction with KI produces the greatest reduction in the oxidation state of S, and is therefore the strongest reducing agent.