What are the products of the following reaction in neutral solution?
$MnO^{4-}+ {S_2O_3}^{2-} + H_2O → ?$
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $MnO_2, {SO_4}^{2-}, HO^-$
Core Concept — Behavior of $\text{KMnO}_4$ in Different Media
|
Medium |
Reduction product of $MnO_4^−$ |
|
Acidic |
$\text{Mn}^{2+}$ |
|
Neutral |
$\mathbf{MnO_2}$ (brown precipitate) |
|
Alkaline |
$\text{MnO}_4^{2-}$ |
So in neutral medium $\rightarrow \text{MnO}_4^-$ reduces to $\text{MnO}_2$.
Oxidation of Thiosulfate ($\text{S}_2\text{O}_3^{2-}$)
Thiosulfate acts as a reducing agent and gets oxidized to sulfate ($\mathbf{SO_4^{2-}}$) in strong oxidation conditions.
Why each option is considered
Option 1: $\text{MnO}_2, \text{SO}_3^{2-}, \text{H}^+$: $\text{SO}_3^{2-}$ is sulfite, not the final oxidation product here. Also $\text{H}^+$ is not expected in neutral medium.
Option 2: $\text{MnO}_2, \text{SO}_2, \text{OH}^-$: $\text{SO}_2$ formation occurs in acidic breakdown, not here. Not the typical redox product with permanganate.
Option 3: $\text{MnO}_2, \text{SO}_4^{2-}, \text{OH}^-$: Correct medium behavior: $\text{MnO}_4^- \rightarrow \text{MnO}_2$. $\text{S}_2\text{O}_3^{2-} \rightarrow \text{SO}_4^{2-}$. Neutral/basic tendency gives $\text{OH}^-$.
Option 4: $\text{Mn}^{2+}, \text{SO}_4^{2-}, \text{H}^+$: $\text{Mn}^{2+}$ forms only in acidic medium, not neutral.