Thiosulphate reacts differently with iodine and bromine in the reaction given below : 2 S 2 O 3 2- + I 2 rightarrow S 4 O 6 2- +2 I - S 2 O 3 2- +5 Br 2 +5 H 2 O rightarrow 2 SO 4 2- +4 Br - +10 H + Which of the following statement justifies the above dual behaviour of thiosulphate?

Bromine is a stronger oxidant than iodine

Bromine undergoes oxidation and iodine undergoes reduction by iodine in these reactions

Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reaction

Bromine is a weaker oxidant than iodine

Chemistry · Redox Reactions

JEE Main 2024 — 8 April, Shift 1 — Question 64

Thiosulphate reacts differently with iodine and bromine in the reaction given below :

$2 \mathrm{~S}_{2} \mathrm{O}_{3}^{2-}+\mathrm{I}_{2} \rightarrow \mathrm{~S}_{4} \mathrm{O}_{6}^{2-}+2 \mathrm{I}^{-}$

$\mathrm{S}_{2} \mathrm{O}_{3}^{2-}+5 \mathrm{Br}_{2}+5 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{SO}_{4}^{2-}+4 \mathrm{Br}^{-}+10 \mathrm{H}^{+}$

Which of the following statement justifies the above dual behaviour of thiosulphate?

Option A:
Bromine undergoes oxidation and iodine undergoes reduction by iodine in these reactions
Option B:
Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reaction
Option C:
Bromine is a stronger oxidant than iodine
Correct
Option D:
Bromine is a weaker oxidant than iodine

Answer: C

Step-by-step solution

In the reaction $2S_2O_3^{2-} + I_2 \rightarrow S_4O_6^{2-} + 2I^-$ , calculate oxidation state of S. In $S_2O_3^{2-}$ , let oxidation state of S be x: $2x + 3(-2) = -2 \Rightarrow 2x - 6 = -2 \Rightarrow 2x = 4 \Rightarrow x = +2$ . In $S_4O_6^{2-}$ , let oxidation state be y: $4y + 6(-2) = -2 \Rightarrow 4y - 12 = -2 \Rightarrow 4y = 10 \Rightarrow y = +2.5$ . S oxidation state changes from +2 to +2.5 (oxidation). In the reaction $S_2O_3^{2-} + 5Br_2 + 5H_2O \rightarrow 2SO_4^{2-} + 10Br^- + 10H^+$ , calculate oxidation state of S in products. In $SO_4^{2-}$ , let oxidation state be z: $z + 4(-2) = -2 \Rightarrow z - 8 = -2 \Rightarrow z = +6$ . S oxidation state changes from +2 to +6 (oxidation). Both $I_2$ and $Br_2$ act as oxidizing agents. Since S is oxidized more with $Br_2$ (to +6 vs +2.5 with $I_2$ ), $Br_2$ is a stronger oxidant. Thus, statement C is correct.

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Chemistry
Chapter: Redox Reactions
Topic: Types of Redox Reactions & Balancing of Redox Reactions