Learning Outcomes

At the end of this section, you should be able to:

  • State Faraday’s first and second law electrolysis.
  • Perform calculations based on Faraday’s laws of electrolysis.

Michael Faraday’s experiments in the 1830s established the fundamental principles that describe the relationship between electricity and chemical reactions in electrolysis. His two laws offer a quantitative framework for understanding this process.

Faraday’s First Law of Electrolysis

Faraday’s first law states that the amount of chemical substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.


m=Z×Qm = Z cdot Q


  • m is the mass of the substance deposited or liberated (in grams),

  • Z is the electrochemical equivalent (in grams per coulomb),

  • Q is the total electric charge passed through the electrolyte (in coulombs).

Relationship with Current

The electric charge Q can be expressed in terms of current I and time t:
Q=I × t
m = Z × I × t

Faraday’s Second Law of Electrolysis

Faraday’s second law states that when the same quantity of electricity is passed through different electrolytes, the masses of the substances deposited or liberated at the electrodes are directly proportional to their equivalent weights.

Equivalent Weight

The equivalent weight of a substance is given by:
text{Equivalent weight} = frac{text{Molar mass}}{text{Number of electrons involved in the reaction}}

Faraday’s Constant

Faraday’s constant (F) is the amount of electric charge per mole of electrons. Its value is approximately 96485 coulombs per mole (C/mol).

Combining the Laws

By combining both laws, the mass of the substance deposited or liberated can also be calculated using:

 text{m} = frac{text{M × I × t}}{text{n × F}}


  • M is the molar mass of the substance,
  • I is the current,
  • t is the time,
  • n is the number of electrons involved in the reaction.

Example Calculations

Example 1: Using Faraday’s First Law

Let’s say you pass a current of 2 A through a copper sulfate solution for 30 minutes. Calculate the mass of copper deposited. The electrochemical equivalent of copper (Cu) is 0.000329 grams per coulomb.

Calculate the total charge: 𝑄 = 𝐼 × 𝑡
= 2   A × ( 30   min × 60   s/min )
= 2 × 1800
= 3600   C

Calculate the mass deposited: 𝑚 = 𝑍 ⋅ 𝑄
= 0.000329   g/C × 3600   C
= 1.1844   g

Example 2: Using Faraday’s Second Law

Suppose you pass the same quantity of electricity through a solution of silver nitrate. If the molar mass of silver (Ag) is 107.87 g/mol and it involves 1 electron (n = 1), calculate the mass of silver deposited using Faraday’s second law.

Calculate the mass deposited:

 text{m} = frac{text{M × I × t}}{text{n × F}}

Using the same current (2 A) and time (30 minutes):

 text{m} = frac{text{107.87g/mol 2A × 1800s}}{text{1 × 96485C/mol}} = {text{4.02g}}