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 oxidation number (O.N.) of S in `S_(2)O_(3)^(2-)` is +2 . Being a stronger oxidising agent than `I_(2) , Br_(2)` oxidises `S_(2)O_(3)^(2-)` to `SO_(4)^(2-)` in which the O.N. of S is +6 . However , `I_(2)` is a weak oxidising agent . Therefore , it oxidises `S_(2)O_(3)^(2-)` to
`S_(4)O_(6)^(2-)` in which the average O.N. of S is only `+ 2.5` . As a result `S_(2)O_(3)^(2-)` reacts differently with iodine and bromine .