Chemistry · JEE Advanced only

Faraday's Laws of Electrolysis (JEE Advanced)

Not in the JEE Main syllabus, but it is in JEE Advanced. Study time: about 3 to 4 hours for concept clarity and formula practice, plus 2 to 3 hours solving varied numerical problems including series cells and gas evolution questions.

Last updated: 26 October 2026

Direct answer

Faraday's Laws of Electrolysis describe the quantitative relationship between the amount of substance deposited or liberated at an electrode and the quantity of electricity passed through the electrolyte.

Faraday's Laws of Electrolysis describe the quantitative relationship between the amount of substance deposited or liberated at an electrode and the quantity of electricity passed through the electrolyte. The first law states that the mass deposited is proportional to the charge passed, while the second law relates the masses of different substances deposited by the same charge to their equivalent weights. These laws form the foundation for all electrochemical calculations involving electrolysis.

Official syllabus

What the JEE Advanced syllabus says

“Faraday's laws of electrolysis”

Source: JEE Advanced 2026 Information Bulletin, Annexure-I. Closest JEE Main topic: Electrochemical cells - Electrolytic and Galvanic cells.

JEE Main covers basic electrochemical cells but does not require you to perform detailed calculations using Faraday's laws. For Advanced, you must be able to calculate masses deposited, volumes of gases evolved, and solve problems involving multiple electrolytic cells connected in series.

Key ideas

What you must understand

Electrolysis involves passing current through an electrolyte to drive a non-spontaneous redox reaction, with reduction at cathode and oxidation at anode
One Faraday (F) equals 96500 C and represents the charge carried by one mole of electrons
Equivalent weight = Molar mass / n-factor, where n-factor is the number of electrons transferred per formula unit
When electrolytic cells are connected in series, the same charge passes through each cell, so you can compare masses using equivalent weights
For gases evolved during electrolysis, use the ideal gas equation after finding moles from Faraday's laws
Current efficiency accounts for side reactions and is defined as (actual mass deposited / theoretical mass) × 100

Formulas

Formulas and key facts

m = ZIt = ZQ, where Z = equivalent weight / 96500 (electrochemical equivalent)
m = (M × I × t) / (n × F), where M = molar mass, n = electrons transferred, F = 96500 C/mol
1 Faraday = 96500 C = charge to deposit 1 gram equivalent of any substance
Equivalent weight = Molar mass / n-factor
For cells in series: m₁/E₁ = m₂/E₂ (E = equivalent weight)
Q = I × t (charge in coulombs = current in amperes × time in seconds)
Moles of electrons = Q / F = It / 96500
Volume of gas at STP = (moles of gas) × 22.4 L

Avoid

Common mistakes

Forgetting to convert time to seconds before calculating charge (minutes or hours must be converted)
Using molar mass instead of equivalent weight, or confusing n-factor for different ions (Cu²⁺ has n=2, Ag⁺ has n=1)
Not recognizing that cells in series receive the same charge, leading to errors in comparative calculations
Ignoring current efficiency in problems where side reactions occur, giving theoretical instead of actual yield

Worked example

A JEE Advanced-level example

Question. Two electrolytic cells are connected in series. The first contains AgNO₃ solution and the second contains CuSO₄ solution. A current is passed until 1.08 g of silver is deposited. Calculate the mass of copper deposited. (Atomic masses: Ag = 108, Cu = 64)

Show solution

Since cells are in series, the same charge passes through both cells
For Ag⁺ + e⁻ → Ag, n-factor = 1, so equivalent weight of Ag = 108/1 = 108 g/eq
For Cu²⁺ + 2e⁻ → Cu, n-factor = 2, so equivalent weight of Cu = 64/2 = 32 g/eq
Using the relation for cells in series: m(Ag)/E(Ag) = m(Cu)/E(Cu)
Substituting: 1.08/108 = m(Cu)/32
Solving: m(Cu) = (1.08 × 32)/108 = 34.56/108 = 0.32 g

Answer: 0.32 g of copper is deposited

Solved twice independently; both solutions agree.

FAQs

Frequently asked questions

Is Faraday's Laws of Electrolysis in the JEE Main syllabus?

No. It is in the JEE Advanced 2026 syllabus only. The closest JEE Main topic is Electrochemical cells - Electrolytic and Galvanic cells.

How long does Faraday's Laws of Electrolysis take to prepare for JEE Advanced?

About 3 to 4 hours for concept clarity and formula practice, plus 2 to 3 hours solving varied numerical problems including series cells and gas evolution questions, if you already know the related JEE Main topic.