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?

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Solution

The average oxidation number (O.N.) of S in is +2. Being a stronger oxidising agent than I₂, Br₂ oxidises to , in which the O.N. of S is +6. However, I₂ is a weak oxidising agent. Therefore, it oxidises to , in which the average O.N. of S is only +2.5. As a result, reacts differently with iodine and bromine.

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Similar questions

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)

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)

2 S_{2} O_{3}^{2 -} + I_{2} \to S_{4} O_{6}^{2 -} + 2 I^{-}

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

2 C u^{2 +} (a q) + 4 I^{-} (a q) ⇌ 2 C u I (s) + I_{2} (a q)

24.5 c m^{3}

0.100 m o l e d m^{- 3}

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

[A t o m i c m a s s o f c o p p e r = 63.5]

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