Consider the reactions: `2S_(2)O_(3)^(2-)(aq)+I_(2)(s)toS_(4)O_(6)^(2-)(aq)+2I^(-)(aq)`
`S_(2)O_(3)^(2-)(aq)+2Br_(2)(l)+5H_(2)O(l)to2SO_(4)^(2-)(aq)+4Br^(-)(aq)+10H^(+)(aq)`
Why does the same reductant ,thiosulphate react differently with iodine and bromine ?

Explanation

The standard reductant potential for `Br_(2)//2Br^(-)` system is greater than that for `I_(2)//2I^(-)` system`Eoverset0Br//2Br=1.09Vand Eoverset0I_(2)//2I^(-)=0.54V)`indicating `Br_(2)` is a stonger oxidising agent than `I_(2)` .The average oxidation number of S in `S_(2)O_(3)^(2-)` is +2 and that in `S_(4)O_(6)^(2-)` is 2.5 while the oxidation number of S is `SO_(4)^(2-)` is +6 .The oxidation per S -atom change unit in the reaction `S_(2)O_(3)^(2-)toS_(4)O_(6)^(2-)` ,while in the `S_(2)O_(3)^(2-)toSO_(4)^(2-)` this change occurs by 4 unit Being a stonger oxidising agent `Br_(2)` in capable of increasing the oxidation number of S in `S_(2)O_(3)^(2-)` to the maximum oxidation numbe of 6, thereby leading to the formation of `SO_(4)^(2-)` ion On the other hand, `I_(2)` being a weaker oxidising agent ,increasing the oxidation number of S in `to S_(2)O_(3)^(2-)`to an oxidation number of 2.5 result in the formation of ` S_(4)O_(6)^(2-)` iom.