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) → \(Mn^{2+}, H_2O, I_2\)
Potassium permanganate is a strong oxidising agent, and the medium decides how far Mn is reduced. In acidic solution, Mn goes from +7 to +2:
\( \displaystyle \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \)
Iodide ions are oxidised to iodine:
\( \displaystyle \ce{2I^- -> I2 + 2e^-} \)
To balance the electrons (10 on each side), the permanganate half-reaction is multiplied by 2 and the iodide half-reaction by 5. Adding them:
\( \displaystyle \ce{2MnO4^- + 10I^- + 16H^+ -> 2Mn^2+ + 5I2 + 8H2O} \)
So the products are \( \ce{Mn^2+} \), \( \ce{H2O} \) and \( \ce{I2} \); the \( \ce{H^{+}} \) ions are used up, not formed.
Why the other options are incorrect
Option (2) \(Mn^{4+}, H^+, {IO_3}^-\) — Incorrect. Manganese(IV) (as \( \ce{MnO2} \)) and iodate are the products in neutral or faintly alkaline solution: \( \ce{2MnO4^- + H2O + I^- -> 2MnO2 + 2OH^- + IO3^-} \). In acid, \( \ce{H^{+}} \) is consumed, not produced.
Option (3) \(Mn^{4+}, H_2O, I_2\) — Incorrect. In acidic solution Mn is reduced all the way to +2, not to +4.
Option (4) \(Mn^{2+}, H^+, {IO_3}^-\) — Incorrect. Iodate is formed only in the neutral medium; in acid, iodide is oxidised to iodine, and \( \ce{H^{+}} \) is a reactant.
Note: This question is from the NCERT book (2024-25). The topic is not there in the latest NCERT book (2026-27), but it is still in the CUET 2026 syllabus, so do not skip it.
NCERT (2024-25 edition):
Oxides and Oxoanions of Metals – …
