How much charge is required for the reduction of 1 mol of MnO4− to Mn…

The correct answer is Option (2) → 5F

By Faraday’s laws, the charge needed to reduce one mole of a species equals the number of electrons (n) it gains per formula unit multiplied by the Faraday constant, the charge on one mole of electrons:

    \( \displaystyle Q=n\,F \quad \quad 1\ F=96487\ \mathrm{C\,mol^{-1}} \)

Find n from the change in oxidation state of manganese. In \( \ce{MnO4^-} \), Mn is +7 (x + 4(−2) = −1); in \( \ce{Mn^{2+}} \), it is +2:

    \( \displaystyle n=(+7)-(+2)=5 \)

    \( \displaystyle \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \)

    \( \displaystyle Q=5\times F=5F \)

So the quantity of charge required is 5F (5 × 96487 C = 482435 C).

Note 1: 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):
Exercises – “2.12 How much charge is required for the following reductions: … (iii) 1 mol of MnO₄^– to Mn²⁺?”

Note 2: The same/similar question has been asked in different years by NTA as per the details given below: