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.

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Answered by Sarah P. Chemistry tutor

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