Faraday’s Laws of Electrolysis

Introduction

Michael Faraday's work on electrolysis led to the formulation of two fundamental laws that state that the amount of chemical change (such as the amount of substance deposited or dissolved) at an electrode during electrolysis is directly proportional to the quantity of electric charge passed through the electrolyte. Faraday's Second Law of Electrolysis (1834): This law states that the amounts of different substances altered at an electrode by the same quantity of electricity are proportional to their equivalent weights. In other words, if the same amount of electric charge is passed through different electrolytes, the masses of substances deposited or dissolved are proportional to their equivalent weights Faraday's laws are foundational in electrochemistry and provide insight into how electrical energy can be used to drive chemical reactions. They are crucial for understanding processes such as electroplating, electrolysis of water, and the functioning of batteries.

JEE Main 2025: Chemistry Formula | Study Materials | High Scoring Topics | Preparation Guide

JEE Main 2025: Syllabus | Sample Papers | Mock Tests | PYQs | Study Plan 100 Days

NEET 2025: Syllabus | High Scoring Topics |  PYQs

This Story also Contains

1. Introduction
2. Faraday’s Laws of Electrolysis
3. Some Solved Examples
4. Summary

This law is very important in understanding and calculating the electrochemical process. It describes the relation between the amount of substance transferred during the process of electrolysis and the amount of electric charge passing through the electrolysis.

Faraday’s Laws of Electrolysis

Faraday's laws of electrolysis are fundamental principles in electrochemistry that relate to the amount of substance transformed during electrolysis to the electric charge passed through the electrolyte. Faraday gives two laws which are the Faraday First Law and the Faraday Second Law

Faraday's First Law of Electrolysis

The amount of chemical change (or substance transformed) at an electrode during electrolysis is directly proportional to the total electric charge passed through the electrolyte.

According to Faraday's first law, "The amount of substance or quantity of chemical reaction at the electrode is directly proportional to the quantity of electricity passed into the cell".

W or m∝qW∝ it W=ZitZ=Mnf= Eq.wt 96500Z= Electrochemical equivalent M= Molar Mass F=96500Cn= Number of electrons transfered q= amount of charge utilized

Faraday's Second Law of Electrolysis

The amounts of different substances transformed or deposited by the same quantity of electricity passing through an electrolyte are proportional to their equivalent weights. In other words, for a given charge, the mass of a substance transformed is inversely proportional to its equivalent weight.

The electrochemical equivalent is the amount of the substance deposited or liberated by one-ampere current passing for one second (that is, one coulomb of charge.)
One gram equivalent of any substance is liberated by one faraday.

Eq. Wt. =Z×96500WE=q96500w=E⋅q96500 W= Edit 96500

As w = a x l x d that is, area x length x density
Here a = area of the object to be electroplated
d = density of metal to be deposited
l = thickness of layer deposited
Hence from here, we can predict charge, current strength, time, thickness of deposited layer etc.

NOTE: One faraday is the quantity of charge carried by one mole of electrons.

1 F=1.6×10−19×6.023×1023≃96500 Coulombs Ideal gas equation

PV=nRT

where;

• P: Pressure of the gas.
• V: Volume of the gas.
• R: Ideal gas constant.
• T: Temperature in Kelvin.

For a better understanding of the topic and to learn more about Faraday’s Laws of Electrolysis with video lesson we provide the link to the

YouTube video:

Some Solved Examples

Example.1

1. A current of 10.0 A flows for 2.00 h through an electrolytic cell containing a molten salt of metal X. This results in the decomposition of 0.250 mol of metal X at the cathode. The oxidation state of X in the molten salt is : (F= 96,500 C)

1)1+

2)2+

3) (correct)3+

4)4+

Solution

i = 10 A, t = 2hr

No. of moles =It96500×(n− factor )

∴ moles of e−=10×2×60×6096500×n− factor ∴0.25=10×2×60×6096500×X∴0.75=0.25×(X)⇒X=3∴ Metal X is present in the form of X3+

Hence, the answer is the option (3).

Example.2

2. How many electrons would be required to deposit 6.35 g of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate? (Atomic mass of copper = 63.5 u, N_A=Avogadro’s constant) :

1)63.5 gNA20

2)46.5 gNA10

3) (correct)6.35gNA5

4) 56.5gNA2

Solution

According to the reaction:

Cu2++2e−→Cu

We require 2 moles of electrons or 2N_A electrons to deposit 1 mol or 63.5 g of Cu.

So, 6.35 g of Cu, requires 2 NA10 electrons.

After simplifying.

2 NA10=NA5

Hence, the answer is the option(3)

Example.3

3. Aluminium oxide may be electrolysed at 1000°C to furnish aluminium metal ( At. Mass = 27 amu,1 Faraday= 96,500 Coulombs ) The cathode reaction is

Al3++3e−→Al0

To prepare 5.12 kg of aluminium metal by this method would require

1) (correct)5.49×107C of electricity

2)1.83×107C of electricity

3)5.49×104C of electricity

4)5.49×101C of electricity

Solution

Moles of Al=5.12×100027≈190∴ Moles of e−=3×190∴ Total charge =3×190×96500≃5.49×107C

Hence, the answer is the option (1).

Example.4

4. Ni(NO3)2 solution is electrolysed between platinum electrodes using 1 faraday electricity. How many moles of Ni will be deposited at the cathode?

1) (correct)0.5

2)0.05

3)1

4)0.25

Solution

The reaction : Ni(NO3)2→Ni2++2NO3− at cathode :Ni2++2e−→Ni
2 mole of e−are required for 1 mole of Ni∴0.5 mole of Ni is deposited by 1 mole of electron
Hence, the answer is the option (1).

EXAMPLE.5

5. The product obtained from the electrolytic oxidation of acidified sulphate solutions, is :

1)Sulphuric acid HSO4−

2) (correct)peroxydisulphuric acid HO3SOOSO3H

3) oxy sulphuric acid )HO2SOSO2H

4) none of these HO3SOSO3H

Solution

As we have learnt,

Electrolytic oxidation of acidified sulphate solution from peroxydisulphuric acid.

Hence, the correct answer is option (2).

Summary

Faraday's laws of electrolysis provide a foundational understanding of how electric currents can drive chemical reactions, and they offer several benefits such as Quantitative Analysis Faraday's laws allow for precise calculations of the amount of substance that will be deposited or dissolved during electrolysis, based on the amount of electric charge passed through the electrolyte. Electroplating and Metallurgy in these laws are essential in industries for electroplating, where they help control the thickness and quality of metal coatings on surfaces. Battery Design The understanding of electrolysis helps in designing more efficient batteries and fuel cells, as it relates to the movement of ions and the overall electrochemical reactions involved. Purification Processes The Faraday law of electrolysis is used in the purification of metals and in various chemical synthesis processes, where precise control over the amount of material processed is crucial. faraday's law of electrolysis also has various benefits in research and development faraday's laws are fundamental in developing new materials and improving existing ones, particularly in fields like electrochemistry and materials science.