The quantity of charge required to obtain 2 mol of Mn^2+ from MnO_4^-…

The correct answer is Option (2) → 10 F

The charge needed for an electrode reaction is the number of moles of electrons transferred, multiplied by F (1 F = 96500 C = charge on 1 mol of electrons).

The oxidation state of Mn falls from +7 in \( \ce{MnO4^{-}} \) [x + 4(−2) = −1, x = +7] to +2 in \( \ce{Mn^{2+}} \), so each ion takes up 5 electrons:

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

1 mol of \( \ce{Mn^{2+}} \) needs 5 mol of electrons, so 2 mol needs 10 mol of electrons:

    \( Q = 2 \times 5\ \text{F} = 10\ \text{F} \)

Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. 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⁺?”