How many faraday of charge is required to deposit 2 moles of Al^3+ fr…

The correct answer is Option (3) → 6F

By Faraday’s laws, the charge needed depends on the number of electrons in the electrode reaction. One faraday (F ≈ 96500 C) is the charge on one mole of electrons. In the electrolysis of molten alumina (\( \ce{Al2O3} \)), aluminium is deposited at the cathode:

    \( \displaystyle \ce{Al^3+ + 3e- -> Al} \)

One mole of \( \ce{Al^{3+}} \) takes 3 mol of electrons (3F). For 2 moles:

    \( \displaystyle 2 \times 3\,\text{F} = 6\,\text{F} \)

So 6 faraday are required.

Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.

NCERT (2026-27 edition):
Exercises – “2.12 How much charge is required for the following reductions:
(i) 1 mol of Al³⁺ to Al?
(ii) 1 mol of Cu²⁺ to Cu?
(iii) 1 mol of MnO₄^– to Mn²⁺?”

Electrolytic Cells and Electrolysis – “… For the electrode reactions:
Mg²⁺(l) + 2e^– → Mg(s)
Al³⁺(l) + 3e^– → Al(s)
It is obvious that one mole of Mg and Al require 2 mol of electrons (2F) and 3 mol of electrons (3F) respectively.”