The products for the reaction of $KMnO_4$ with potassium iodide in acidic solutions are
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $Mn^{2+}, H_2O, I_2$
In an acidic solution, KMnO₄ is a strong oxidising agent.
- $\mathrm{MnO_4^-}$ is reduced to $\mathrm{Mn^{2+}}$
- $\mathrm{I^-}$ is oxidised to $\mathrm{I_2}$
- Water is formed
The balanced reaction is:
$\mathrm{MnO_4^-} + 10\mathrm{I^-} + 16\mathrm{H^+} \rightarrow 2\mathrm{Mn^{2+}} + 5\mathrm{I_2} + 8\mathrm{H_2O}$
Therefore, the products are Mn²⁺, H₂O, and I₂