Consider the reactions : 2S2O2−3(aq)+I2(s)→S4O2−6(aq)+2I−(aq) S2O2−3(aq)+2Br2(l)+5H2O(l)→2SO2−4(aq)+4Br−(aq)+10H+(aq) Why does the same reductant, thiosulphate react differently with iodine and bromine ?
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Solution
Br2 is better oxidising agent than I2 hence, Br2 oxidises sulphur to its maximum oxidation state that is +6 while I2 does not. This is because Br2 is more electromagnetic than I2. Hence it easily gets converted to Br.