Michael Faraday was the first scientist who described the quantitative aspects of electrolysis. Now Faraday’s laws also flow from what has been discussed earlier. Faraday’s Laws of Electrolysis After his extensive investigations on electrolysis of solutions and melts of electrolytes, Faraday published his results during 1833-34 in the form of the following well known Faraday’s two 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).
(ii) Second Law: The amounts of different substances liberated by the same quantity of electricity passing through the electrolytic solution are proportional to their chemical equivalent weights (Atomic Mass of Metal ÷ Number of electrons required to reduce the cation).
When electricity is passed through molten AlCl3, 13.5 g of Al is deposited. What is the number of faradays involved? (Al = 27 g mol−1)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.5
By Faraday’s first law, the amount of substance deposited is proportional to the charge passed. One faraday (1 F) is the charge on one mole of electrons, so the number of faradays equals the number of moles of electrons the electrode reaction needs.
In molten \( \ce{AlCl3} \), each \( \ce{Al^3+} \) ion takes three electrons at the cathode:
\( \displaystyle \ce{Al^3+(l) + 3e^- -> Al(s)} \)
So 1 mol of Al needs 3 F. With Al = 27 g mol−1:
\( \displaystyle n(\ce{Al})=\frac{13.5\,\mathrm{g}}{27\,\mathrm{g\,mol^{-1}}}=0.5\,\mathrm{mol} \)
\( \displaystyle \text{Faradays}=0.5\times 3=1.5\,\mathrm{F} \)
The same result comes from the second law in the passage: the equivalent weight of Al is \( \frac{27}{3}=9 \) g, 1 F deposits 9 g, and \( \frac{13.5}{9}=1.5 \), so 1.5 F is needed.
NCERT (2026-27 edition):
Faraday’s Laws of Electrolysis – “For the electrode reactions:
\( \displaystyle \ce{Mg^2+(l) + 2e^- -> Mg(s)} \qquad (2.31) \)
\( \displaystyle \ce{Al^3+(l) + 3e^- -> Al(s)} \qquad (2.32) \)
It is obvious that one mole of Mg2+ and Al3+ require 2 mol of electrons (2F) and 3 mol of electrons (3F) respectively.”