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in Redox Reactions by (30.5k points)
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Consider the reactions:

2S2O32-(aq) + I2(s) ⟶ S4O62-(aq) + 2I-(aq)

S2O32-(aq) + 2Br2(I) ⟶ 5H2O(I) ⟶ 2SO42-(aq) + 4Br-(aq) + 10H(aq)

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

1 Answer

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The average O.N. of S in S2O32- is + 2 while in S4O62- it is + 2.5. The O.N. of S in SO42- is+6. Since Br2 is a stronger oxidising agent that l2 , it oxidises S of S2O32- to a higher oxidation state of + 6 and hence forms SO42- ion. l2 , however, being a weaker oxidising agent oxidises S of S2O32- ion to a lower oxidation of + 2.5 in S4O62- ion.

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