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Question
Consider the reactions:
\[\ce{2S_2O_3^{(2-)}(aq) + l_2(S) -> S_4O_6^{(2-)}(aq) + 2l-(aq)}\]
\[\ce{S_2O_3^{(2-)}(aq) + 2Br_2(l) + 5H_2O(l) -> 2SO_4^{2-} (aq) + 4Br-(aq) + 10H+ (aq)}\]
Why does the same reductant, thiosulphate react differently with iodine and bromine?
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Solution
The average oxidation number (O.N.) of `"S"_2"O"_3^(2-)` in is +2. Being a stronger oxidising agent than I2, Br2 oxidises `"S"_2"O"_3^(2-)` to `"SO"_4^(2-)`, in which the O.N. of S is +6. However, I2 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.
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