Consider the reactions:
$2 S_{2} O_{3}^{2 -} (a q) + I_{2} (s) ⟶ S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)$
$S_{2} O_{3}^{2 -} (a q) + 2 {B r}_{2} (l) + 5 H_{2} O (l) ⟶ 2 {S O}_{4}^{2 -} (a q) + 4 {B r}^{-} (a q) + 10 H^{+} (a q)$
Why does the same reductant thiosulphate react differently with iodine and bromine?

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Solution

$B r_{2}$ is a stronger oxidizing agent than $I_{2}$ . Due to this, it oxidizes $S_{2} O_{3}^{2 -}$ from $+ 2$ to $+ 6$ oxidation state, while $I_{2}$ oxidizes $S_{2} O_{3}^{2 -}$ from $+ 2$ to $+ 2.5$ which is comparatively lower oxidation state. Due to this, same reductant behaves differently in different reactions.

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Similar questions

Consider the reactions:

(aq) + I₂(s) → (aq) + 2I^–(aq)

aq) + 2Br₂(l) + 5 H₂O(l) → (aq) + 4Br^–(aq) + 10H⁺(aq)

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

2 S_{2} O_{3}^{2 -} (a q) + I_{2} (s) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

S_{2} O_{3}^{2 -} (a q) + 2 B r_{2} (l) + 5 H_{2} O (l) \to 2 S O_{4}^{2 -} (a q) + 4 B r^{-} (a q) + 10 H^{+} (a q)

2 S_{2} O_{3}^{2 -} (a q) + I_{2} (s) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

S_{2} O_{3}^{2 -} (a q) + 2 B r_{2} (l) + 5 H_{2} O (l) \to 2 S O_{4}^{2 -} (a q) + 4 B r^{-} (a q) + 10 H^{+} (a q)

I_{2 (a q)} + 2 S_{2} O_{3}^{2 -} (a q) \to S_{4} O_{6}^{2 -} (a q) + 2 I^{-} (a q)

2 S_{2} O_{3}^{2 -} + I_{2} ⟶ S_{4} O_{6}^{2 -} + 2 I^{⊖}

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