In a faintly alkaline medium, the thiosulphate ion is oxidized to a s…

The correct answer is Option (3) → 41/3

The coefficients of a redox equation are fixed by one rule: the total number of electrons lost must equal the total number gained. Find the change in oxidation number of each element first, then balance the charge, and only then check the atoms.

Manganese goes from +7 in \( \ce{MnO4^{-}} \) to +4 in \( \ce{MnO2} \), so each Mn gains 3 electrons. Sulphur goes from +2 (average) in \( \ce{S2O3^{2-}} \) to +6 in \( \ce{SO4^{2-}} \). Each thiosulphate ion holds two sulphur atoms, so it loses 2 × 4 = 8 electrons. In faintly alkaline solution the two half-reactions are:

    \( \displaystyle \ce{MnO4^{-} + 2H2O + 3e^{-} -> MnO2 + 4OH^{-}} \)

    \( \displaystyle \ce{S2O3^{2-} + 10OH^{-} -> 2SO4^{2-} + 5H2O + 8e^{-}} \)

    \( \displaystyle \text{electrons gained}=3a,\quad \text{electrons lost}=8b \)

    \( \displaystyle 3a=8b\quad\Rightarrow\quad a=8,\ b=3 \)

The sulphur balance gives x:

    \( \displaystyle x=2b=6 \)

Now balance the charge:

    \( \displaystyle \text{left}=8(-1)+3(-2)=-14 \)

    \( \displaystyle \text{right}=6(-2)+y(-1)=-12-y \)

    \( \displaystyle -14=-12-y\quad\Rightarrow\quad y=2 \)

Oxygen then gives c = 1, and the hydrogen count (2c = y = 2) checks out:

    \( \displaystyle \ce{8MnO4^{-} + 3S2O3^{2-} + H2O -> 8MnO2 + 6SO4^{2-} + 2OH^{-}} \)

Substituting a = 8, b = 3 and y = 2:

    \( \displaystyle 2a+\frac{2}{9}b-\frac{3}{2}y=2(8)+\frac{2}{9}(3)-\frac{3}{2}(2) \)

    \( \displaystyle =16+\frac{2}{3}-3 \)

    \( \displaystyle =\frac{41}{3} \)

Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
In Neutral or Faintly Alkaline Solutions