An electrochemical cell consists of two metallic electrodes (anode and cathode) dipping in electrolytic solutions. Oxidation takes place at the anode and reduction takes place at the cathode. These cells are of two types:
1) Galvanic cells, in which the chemical energy of a spontaneous redox reaction is converted into electrical energy. Eocell = Eocathode - Eoanode, where Eocell is the standard potential of the cell, and ΔrGo = -nFEocell, where ΔrGo is the standard Gibbs energy change of the cell reaction.
2) Electrolytic cells, in which electrical energy is used to carry out a non-spontaneous redox reaction. The amount of substance produced at an electrode depends on the quantity of electricity passed, Q = I × t (where I is the current in ampere and t is the time in seconds). One faraday is the charge carried by 1 mole of electrons, that is 96500 C.
The conductivity (κ) of an electrolytic solution depends on the nature and the concentration of the electrolyte. According to Kohlrausch’s law of independent migration of ions, the limiting molar conductivity of an electrolyte is the sum of the limiting molar conductivities of its cation and anion, for example Λom(NaCl) = λoNa+ + λoCl–, where λo is the limiting molar conductivity of an ion. The law can be used to calculate the limiting molar conductivity of weak electrolytes.
How much charge is required for the reduction of one mole of MnO4- to Mn2+?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5F
By Faraday’s laws, the charge needed to reduce one mole of a species equals the number of electrons (n) it gains per formula unit multiplied by the Faraday constant, the charge on one mole of electrons:
\( \displaystyle Q=n\,F \quad \quad 1\ F=96487\ \mathrm{C\,mol^{-1}} \)
Find n from the change in oxidation state of manganese. In \( \ce{MnO4^-} \), Mn is +7 (x + 4(−2) = −1); in \( \ce{Mn^{2+}} \), it is +2:
\( \displaystyle n=(+7)-(+2)=5 \)
\( \displaystyle \ce{MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O} \)
\( \displaystyle Q=5\times F=5F \)
So the quantity of charge required is 5F (5 × 96487 C = 482435 C).
Note 1: 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+?”
Note 2: The same/similar question has been asked in different years by NTA as per the details given below:
| 2026 | 2025 | 2024 | 2023 | 2022 |
|---|---|---|---|---|
| 07-06 S2 25-05 S2 | – | – | – | 20-08 S1 10-08 S1 |