In acidic solution, potassium permanganate oxidises sulphite ions to sulphate ions. How many moles of sulphite ions are oxidised by 0.2 mol of KMnO4?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.5 mol
In a redox reaction the electrons lost by the reducing agent must equal the electrons gained by the oxidising agent. In acid solution, permanganate is reduced to Mn2+.
Mn: +7 in MnO4− → +2 in Mn2+; each MnO4− ion gains 5 electrons.
S: +4 in SO32− → +6 in SO42−; each SO32− ion loses 2 electrons.
Electrons balance when 2 MnO4− ions (2 × 5 = 10 electrons) react with 5 SO32− ions (5 × 2 = 10 electrons).
\( \displaystyle \ce{MnO4^- + 8H+ + 5e^- -> Mn^2+ + 4H2O} \)
\( \displaystyle \ce{SO3^2- + H2O -> SO4^2- + 2H+ + 2e^-} \)
\( \displaystyle \ce{5SO3^2- + 2MnO4^- + 6H+ -> 2Mn^2+ + 3H2O + 5SO4^2-} \)
\( \displaystyle n(\ce{SO3^2-}) = 0.2\ \mathrm{mol} \times \frac{5}{2} = 0.5\ \mathrm{mol} \)
NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “1. In acid solutions:” … “(e) Sulphurous acid or sulphite is oxidised to a sulphate or sulphuric acid: 5SO32– + 2MnO4– + 6H+ → 2Mn2+ + 3H2O + 5SO42–”