Complete the following reaction: 2H_2O + 5SO_2 + 2MnO_4^- longrightar…

The correct answer is Option (1) → \(5SO_4^{2-}, \, \ 4H^+, \, \ 2Mn^{2+}\)

Sulphur dioxide is a reducing agent. Acidified permanganate oxidises S(+4) in \( \ce{SO2} \) to S(+6) in sulphate, and is itself reduced to Mn²⁺ (Mn +7 → +2, gain of 5 electrons). Balancing the electrons: 5 \( \ce{SO2} \) lose 5 × 2 = 10 electrons and 2 \( \ce{MnO4^{-}} \) gain 2 × 5 = 10 electrons.

    \( \ce{2MnO4^{-}} \) + \( \ce{5SO2} \) + \( \ce{2H2O} \) → \( \ce{2Mn^{2+}} \) + \( \ce{5SO4^{2-}} \) + \( \ce{4H^{+}} \)

Check: O: 8 + 10 + 2 = 20 on both sides; H: 4 = 4; charge: −2 on the left and +4 − 10 + 4 = −2 on the right. So the blanks are 5SO₄²⁻, 4H⁺ and 2Mn²⁺.

Why the other options are incorrect

• Option (2) 5SO₃²⁻, 4OH⁻, 2Mn²⁺ — Incorrect. Sulphite still has S in the +4 state, so no oxidation would have taken place, and the reaction produces \( \ce{H^{+}} \), not \( \ce{OH^{-}} \).

• Option (3) 5SO₄²⁻, 4H⁺, 2Mn³⁺ — Incorrect. In acid solution permanganate is reduced to \( \ce{Mn^{2+}} \); with \( \ce{Mn^{3+}} \) the electrons (10 lost, 8 gained) and the charges do not balance.

• Option (4) 5S₂O₃²⁻, 4H⁺, 2MnO₂ — Incorrect. Thiosulphate has S in a lower oxidation state (+2), which would mean \( \ce{SO2} \) is reduced; \( \ce{MnO2} \) is the product in neutral or alkaline solution.

NCERT (2026-27 edition):
Oxides and Oxoanions of Metals – “… (e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid: 5SO₃²⁻ + 2MnO₄⁻ + 6H⁺ → 2Mn²⁺ + 3H₂O + 5SO₄²⁻”