How much charge is required for the reduction of 1 mol of MnO4¯ to Mn2+?
Answer & explanation
Correct answer: option 2
\(MnO_4^− + 8H^+ + 5e^− ————→ Mn^{2+} + 4H_2O\)
Number of electrons = 5
we know,
1 mole of electron = 1 Faraday
5 moles of electron = 5 Faraday