How much charge is required for the reduction of \(0.05\) moles of \(MnO_4^-\) to \(MnO_2\)?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(3 × 96500 × 0.05 C\)
The charge on one mole of electrons is one Faraday, 96487 C mol−1 ≈ 96500 C. The charge needed for a reduction is therefore
Q = (moles of substance) × (electrons per formula unit) × F
In \( \ce{MnO4^{-}} \) manganese is +7; in \( \ce{MnO2} \) it is +4. Each ion takes up 3 electrons:
\( \ce{MnO4^- + 2H2O + 3e^- -> MnO2 + 4OH^-} \)
Q = 0.05 mol × 3 × 96500 C mol−1 = 3 × 96500 × 0.05 C (= 14475 C)
NCERT (2026-27 edition):
Quantitative Aspects of Electrolysis – …
