Faraday’s Laws of Electrolysis — Equations, Derivation, Examples & Applications

Faraday’s Laws of Electrolysis explain how the amount of electricity passed through an electrolyte is directly related to the mass of substance deposited or liberated at the electrodes. These laws form the foundation of electrochemistry. The key formula is:

m = (Q × M) / (n × F)

Where Q = It, M = molar mass, n = electrons per ion, and F = Faraday constant (96485 C·mol⁻¹).

What Are Faraday’s Laws of Electrolysis?

Faraday’s laws describe how the mass produced at an electrode depends directly on the electric charge passed. These laws are essential for understanding electroplating, refining, metal extraction, and chemical analysis.

Faraday’s First Law (Quantitative Law)

The mass of a substance deposited or liberated is directly proportional to the total electric charge passed.

In simple terms: m ∝ Q

Faraday’s Second Law

If the same amount of electricity passes through different electrolytes, the masses deposited are proportional to their equivalent masses.

Mathematically: m ∝ E = M/n

Key Formula, Units & Explanation

The combined law yields the useful formula:

m = (Q × M) / (n × F)

m = mass deposited (g or kg)
Q = electric charge = I × t (C)
M = molar mass (g/mol)
n = number of electrons involved
F = Faraday constant ≈ 96485 C/mol

Derivation of Faraday’s Law

To derive the relationship, we use the fact that one mole of electrons carries a charge F. If a metal ion requires n electrons to become neutral, then the charge needed to deposit one mole is nF.

Therefore, the number of moles formed = Q / (nF)

Mass = M × (Q / nF) = (Q × M) / (n × F).

Worked Examples

Example 1 — Mass of Copper Deposited

Given: 2 A current for 30 minutes through CuSO₄.

Q = I × t = 2 × (30 × 60) = 3600 C
n = 2, M = 63.55 g/mol

m = (3600 × 63.55) / (2 × 96485) ≈ 1.19 g

Example 2 — Current Required to Deposit Silver

Deposit 2 g Ag in 1 hour.

Q = (m × n × F) / M = (2 × 1 × 96485) / 107.87 ≈ 1788 C
I = Q/t = 1788 / 3600 ≈ 0.50 A

Required current = 0.50 A

Applications of Faraday’s Laws

Electroplating & corrosion protection
Electrowinning and metal purification
Coulometric analysis in laboratories
Battery manufacturing & quality control

Limitations & Practical Notes

Assumes 100% current efficiency (rare in practice)
Side reactions reduce actual yield
Temperature and concentration affect deposition
Electrode geometry and stirring influence results

Frequently Asked Questions (FAQ)

What are Faraday’s laws of electrolysis?

They describe the quantitative relationship between electric charge and mass of substance deposited during electrolysis.

What is the formula of Faraday’s law?

m = (Q × M) / (n × F)

What is Faraday constant value?

Approximately 96485 C mol⁻¹.

What are applications of Faraday’s laws?

Electroplating, metal refining, electro-winning, coulometry, batteries.

What is equivalent mass in electrolysis?

Equivalent mass = Molar mass / Number of electrons transferred (M/n).

What Is the Faraday Constant?

The charge carried by one mole of electrons, approximately 96485 C/mol.

How Do You Apply Current Efficiency?

Correct mass: m(actual) = m(calculated) × η/100

Quiz — Test Your Understanding

If 96500 C passes through Ag⁺ solution, what mass of Ag is deposited?
A: 107.87 g
3 A current for 2 hours deposits how much Cu?
A: ≈ 7.11 g
True/False: Deposited mass depends only on molar mass.
A: False — depends on equivalent mass.
Side reactions like hydrogen evolution require what correction?
A: Use percent current efficiency.

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Faraday’s Laws