Number of coulombs required for the reduction of 1 mol of MnO_4^- to…

The correct answer is Option (2) → \(4.825 × 10^5 C\)

By Faraday’s laws, the charge needed for a reduction is the number of moles of electrons taken up multiplied by the Faraday constant, F = 96500 C mol⁻¹ (the charge on one mole of electrons).

In \( \ce{MnO4^{-}} \), x + 4(−2) = −1, so Mn is +7. Reduction to \( \ce{Mn^{2+}} \) therefore takes 7 − 2 = 5 electrons:

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

    \( Q = 5 \times 96500\ \mathrm{C} = 482500\ \mathrm{C} = 4.825 \times 10^5\ \mathrm{C} \)

So 4.825 × 10⁵ C (5 F) are needed for 1 mol of \( \ce{MnO4^{-}} \).

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