How much calcium will be obtained when 4 Faradays of electricity is passed through molten \(CaCl_2\)?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 80 g
By Faraday’s first law, the amount deposited is proportional to the charge passed. At the cathode in molten \( \ce{CaCl2} \):
\( \ce{Ca^2+ + 2e- -> Ca} \)
So 2 F (2 mol of electrons) deposit 1 mol of Ca (40 g).
\( \displaystyle n(\ce{Ca}) = \frac{4\ \text{F}}{2\ \text{F mol}^{-1}} = 2\ \text{mol} \)
Mass of Ca = 2 × 40 g = 80 g
NCERT (2026-27 edition):
Faraday’s Laws of Electrolysis – “… (i) First Law: The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte (solution or melt).”