Choose the correct products formed in acidic medium.
\( 5{SO_3}^{2-} + 2{MnO_4}^- + 6H^+→? \)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2Mn^{2+} + 3H_2O + 5{SO_4}^{2-}\)
Permanganate is a strong oxidising agent whose product depends on the medium: in acidic medium \( \ce{MnO4-} \) (Mn +7) is reduced to \( \ce{Mn^2+} \), gaining five electrons, while in neutral or faintly alkaline medium it gives \( \ce{MnO2} \). Here it oxidises sulphite (S +4) to sulphate (S +6).
Reduction half-reaction:
\( \displaystyle \ce{MnO4^- + 8H+ + 5e- -> Mn^2+ + 4H2O} \)
Oxidation half-reaction:
\( \displaystyle \ce{SO3^2- + H2O -> SO4^2- + 2H+ + 2e-} \)
To balance the electrons, the first is multiplied by 2 and the second by 5 (10 electrons each). Adding them and cancelling the water and \( \ce{H+} \) that appear on both sides:
\( \displaystyle \ce{2MnO4^- + 16H+ + 5SO3^2- + 5H2O -> 2Mn^2+ + 8H2O + 5SO4^2- + 10H+} \)
\( \displaystyle \ce{5SO3^2- + 2MnO4^- + 6H+ -> 2Mn^2+ + 3H2O + 5SO4^2-} \)
So the products are \( \ce{2Mn^2+} \), \( \ce{3H2O} \) and \( \ce{5SO4^2-} \).
Why the other options are incorrect
Option (1) \(2MnO_2 + 6H^+ + 5{SO_4}^{2-}\) — Incorrect. \( \ce{MnO2} \) is the product in neutral or faintly alkaline solution, and the six \( \ce{H+} \) taken on the left cannot reappear unchanged on the right.
Option (3) \(2Mn^{2+}+ 5{SO_4}^{2-} + 6OH^-\) — Incorrect. Hydroxide ions cannot be formed in an acidic solution; the \( \ce{H+} \) is used up to make water.
Option (4) \(2MnO_2 + 6H_2O +5/2{S_2O_3}^{2-}\) — Incorrect. In thiosulphate sulphur is +2, so sulphite would be reduced, not oxidised, while permanganate is also reduced; one species has to be oxidised.
NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – “… 1. In acid solutions: … (e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid:
\( \ce{5SO3^2- + 2MnO4^- + 6H^+ -> 2Mn^2+ + 3H2O + 5SO4^2-} \)”