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Question:

Consider the reactions :

2 S2O2– 3 (aq) + I2(s) → S4 O2– 6(aq) + 2I – (aq)

S2O2– 3(aq) + 2Br2(l) + 5 H2O(l) → 2SO2–4(aq) + 4Br–(aq) + 10H+(aq)

Why does the same reductant, thiosulphate react differently with iodine and bromine ?

Answer:

The average oxidation number (O.N.) of S in  S2O2– 3 is +2. Being a stronger oxidising agent than I2, Br2 oxidises S2O2– 3 to SO2-4 , in which the O.N. of S is +6. However, I2 is a weak oxidising agent.

Therefore, it oxidises S2O2– 3 to S4O2– 6 , in which the average O.N. of S is only +2.5. As a result, S2O2– 3 reacts differently with iodine and bromine.


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