Faraday explained that the decomposition of electrolytes by an electric current is governed by two laws.
First law: The amount of the substance liberated or deposited or dissolved at an electrode during electrolysis of an electrolyte is directly proportional to the quantity of electricity passing through the solution of electrolyte or melt. Mathematically Faraday’s first law is \(m \propto q\) or \(m = eq\) or \(m = ect\), where m is the mass of the substance liberated or deposited or dissolved, q is the quantity of electricity in coulomb, t is the time in seconds, c is the strength of current in amperes and e is the electrochemical equivalent of the ion or metal or molecule deposited or liberated or dissolved at the electrode. Electrochemical equivalent of a substance is the amount deposited or liberated or dissolved or underwent electrode reaction at an electrode by passing one ampere current for 1 s, i.e., 1 C. Chemical equivalent of a substance is the amount of substance deposited or liberated or dissolved or had undergone electrode reaction at an electrode during the passage of one Faraday of electricity during the electrolysis of electrolyte solution or melt.
\(\text{Chemical equivalent of an element or ion = }\frac{\text{Atomic weight}}{\text{Valency or charge of the ion}}\)
The electrochemical equivalent of an element is directly proportional to its chemical equivalent
\( \displaystyle e \propto E \text{ or }E = F.e \)
Unit of electrochemical equivalent is gram C–1. One Faraday, i.e., 96,500 C is equal to the charge present on one mole (6.023×1023) [Avogadro’s number] of electrons or protons
\( \displaystyle m= \frac{ECt}{96500} \)
Second law of Faraday states that if same quantity of electricity is passed through different electrolyte solutions or melts, the amount of the different substances liberated or deposited or dissolved or had undergone reaction at electrode is directly proportional to their chemical equivalents.
\( \displaystyle \frac{W_1}{E_1} = \frac{W_2}{E_2} = \frac{W_3}{E_3} \)
The chemical equivalents depend on the number of electrons participated at the electrode reaction. The chemical equivalents or equivalent weights of NaCl, KCl, KBr, NaOH, etc., are equal to their molecular weights since only one electron takes part in electrode reaction. The equivalent weights of other electrolytes depend on the number of electrons.
\( \displaystyle \text{Equivalent weight} =\frac{\text{Molecular weight}}{\text{Number of electrons involved in electrode reaction}} \)
According to Faraday’s laws of electrolysis, which of the following statements is false?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → One Faraday of electricity liberates one mole of atoms of any metal from its solution
One Faraday (1 F ≈ 96500 C) is the charge on one mole of electrons. To liberate one mole of a metal from its ion Mn+, n moles of electrons are needed, so the number of Faradays depends on the charge on the ion:
\( \displaystyle \ce{M^{n+} + ne^- -> M} \)
Option (1) The number of Faradays required to liberate 1 mol of an element from its ions is equal to the charge on the ion — A true statement. One mole of \( \ce{Mg^2+} \) needs 2 F and one mole of \( \ce{Al^3+} \) needs 3 F.
Option (2) Faraday’s first law of electrolysis can be expressed as \(m \propto q\) — A true statement. By the first law, the amount of chemical reaction at an electrode is proportional to the quantity of electricity passed.
Option (3) One Faraday is the charge on one mole of electrons — A true statement. This is the definition of the Faraday: 96487 C mol−1.
Option (4) One Faraday of electricity liberates one mole of atoms of any metal from its solution — A false statement. 1 F liberates one mole only of a metal whose ion carries a single charge, such as \( \ce{Ag^+} \). From \( \ce{Mg^2+} \) it gives only half a mole of Mg, and from \( \ce{Al^3+} \) one third of a mole of Al.
NCERT (2026-27 edition):
2.5 Electrolytic Cells and 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.”