How many C mol–1 are required to convert 1 mol of MnO4– to Mn+2?
MnO4– → Mn2+
[1F = 96500 C]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 482500 \(C\, \ mol^{-1}\)
The oxidation state of manganese falls from +7 in \( \ce{MnO4^{-}} \) to +2 in \( \ce{Mn^{2+}} \), so each ion gains 5 electrons:
\( \ce{MnO4- + 8H+ + 5e- -> Mn^2+ + 4H2O} \)
One mole of electrons carries 1 F of charge, so 1 mol of \( \ce{MnO4^{-}} \) needs 5 F:
\( Q = 5 \times 96500\ \text{C mol}^{-1} = 482500\ \text{C mol}^{-1} \)
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+?”