Identify the correct products formed during the oxidizing reaction of permanganate ion with sulphite ion in acidic solution
A. \(2OH^-\)
B. \(2Mn^{2+}\)
C. \(3H_2O\)
D. \(MnO_2\)
E. \(5SO_4^{2-}\)
Choose the correct answer from the option given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → B, C and E only
In acidic solution permanganate is reduced to \( \ce{Mn^2+} \) (Mn: +7 → +2, 5 electrons) and sulphite is oxidised to sulphate (S: +4 → +6, 2 electrons). The two half-reactions are:
\( \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \) (× 2)
\( \ce{SO3^2- + H2O -> SO4^2- + 2H^+ + 2e^-} \) (× 5)
Multiplying to balance the electrons (10 each) and adding, 16 H+ − 10 H+ = 6 H+ and 8 H2O − 5 H2O = 3 H2O remain:
\( \ce{5SO3^2- + 2MnO4^- + 6H^+ -> 2Mn^2+ + 3H2O + 5SO4^2-} \)
(A) \( \ce{2OH^-} \) — False. In acid solution \( \ce{H^+} \) ions are consumed and water is formed; \( \ce{OH^-} \) ions are formed only in neutral or alkaline medium.
(B) \( \ce{2Mn^2+} \) — True. In acid, each \( \ce{MnO4^-} \) gains 5 electrons and becomes \( \ce{Mn^2+} \); two permanganate ions give two \( \ce{Mn^2+} \) ions.
(C) \( \ce{3H2O} \) — True. The oxygen removed from permanganate combines with \( \ce{H^+} \) to form water; after balancing, three water molecules remain on the product side.
(D) \( \ce{MnO2} \) — False. Permanganate is reduced only to \( \ce{MnO2} \) in neutral or faintly alkaline solution.
(E) \( \ce{5SO4^2-} \) — True. Each sulphite ion loses 2 electrons (S: +4 → +6) and becomes sulphate; five sulphite ions give five sulphate ions.
NCERT (2026-27 edition):
Potassium Permanganate – …
