In a neutral or faintly alkaline solution, permanganate oxidises thiosulphate almost quantitatively to sulphate:
\( x\,\ce{MnO4-} + \ce{3S2O3^2- + H2O} \longrightarrow x\,\ce{MnO2} + \ce{6SO4^2- + 2OH-} \)
What is the value of x in the balanced equation?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 8
In neutral or faintly alkaline solution permanganate is reduced only to MnO2. The number of permanganate ions follows from the electrons given up by thiosulphate.
Mn: +7 in MnO4− → +4 in MnO2, a gain of 3 electrons per Mn.
S in S2O32−: 2y + 3(−2) = −2, so y = +2; in SO42− S is +6. Each S loses 4 electrons.
3 S2O32− contain 6 S atoms: electrons lost = 6 × 4 = 24.
\( x = \frac{24}{3} = 8 \).
Check, charge: left 8(−1) + 3(−2) = −14; right 6(−2) + 2(−1) = −14.
NCERT (2026-27 edition):
4.4.1 Oxides and Oxoanions of Metals – “2. In neutral or faintly alkaline solutions:” … “(b) Thiosulphate is oxidised almost quantitatively to sulphate: 8MnO4– + 3S2O32– + H2O → 8MnO2 + 6SO42– + 2OH–”