Consider the reaction:
`2S_(2)O_(3)^(2-)(aq)+I_(2)(s) rarr S_(4)O_(6)^(2-)(aq) + 2I^(Θ)(aq)`
`2S_(2)O_(3)^(2-)(aq) + 2Br_(2)(l) + 5H_(2)O(l) rarr 2SO_(4)^(2-)(aq) + 4Br^(Θ)(aq)+10H^(o+)(aq)`
Why does the same reductant, thiosulphate, react differently with iodine and bromine?

Explanation

The average O.N of S in `S_(2)O_(3)^(2-)` is +2 while in `S_(4)O_(6)^(2-)` it is +2.5 The O.N of S in `SO_(4)^(2-)` is +6 since `Br_(2)` is a stronger oxidising agent that `I_(2)` it oxidises s of `S_(2)O_(3)^(2-)` to higher oxidation state of +6 and hence forms `SO_(4)^(2-)` ion `I_(2)` howevr being a weaker oxidising agent oxidises S of `S_(2)O_(3)^(2-)` ion to a lower oxidation of +2.5 in `S_(4)O_(6)^(2-)` ion it is because of this reason that thiosuplhate reacts differently with `Br_(2)` and `I_(2)`