The quantity of charge required to obtain one mole of aluminum from \(Al_2O_3\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3 F
In \( \ce{Al2O3} \) aluminium is present as Al3+. At the cathode each ion takes up three electrons:
\( \displaystyle \ce{Al^3+ + 3e^- -> Al} \)
One mole of aluminium therefore needs 3 mol of electrons. The charge on one mole of electrons is one faraday (1 F ≈ 96500 C mol−1), so the charge required is 3 F (about 2.9 × 105 C).
NCERT (2026-27 edition):
Electrolytic Cells and Electrolysis – “… For the electrode reactions: Mg2+(l) + 2e– → Mg(s) and 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.”