Introduction
Faraday’s Laws of Electrolysis describe the quantitative relationship between electric current passed through an electrolyte and the amount of substance deposited or liberated at the electrodes. These laws form the basis of calculations in NECTA electrolysis questions.

🔹 1. First Law of Electrolysis

“The mass (m) of a substance deposited or liberated at an electrode during electrolysis is directly proportional to the quantity of electricity (Q) passed through the electrolyte.”

Mathematically:

Where:

• m = mass of substance (g)

• Q = total charge passed (Coulombs)

• Z = electrochemical equivalent (g/C)

Since Q = I × t:

Worked Example 1:

A current of 2.0 A is passed through a CuSO₄ solution for 30 minutes. Calculate the mass of copper deposited. (Given: E.C.E of copper = 0.000329 g/C)

Solution:

• I = 2.0 A

• t = 30 × 60 = 1800 s

• Z = 0.000329 g/C

• Q = I × t = 2 × 1800 = 3600 C

✅ Answer: 1.18 g of copper is deposited.

🔹 2. Second Law of Electrolysis

“When the same quantity of electricity is passed through different electrolytes, the masses of substances deposited are proportional to their equivalent masses.”

Mathematically:

Where:

• E = Equivalent mass = Molar mass / n (n = number of electrons exchanged)

Worked Example 2:

If the same current deposits 1.08 g of silver (Ag), how much copper (Cu) will be deposited?
(Given: E.C.E of Ag = 0.001118 g/C, Cu = 0.000329 g/C)

Using:

✅ Answer: 0.318 g of copper will be deposited.

🔹 Faraday Constant

1 Faraday (F) = 96,500 Coulombs
This is the amount of charge needed to deposit or liberate 1 mole of electrons.

🔹 General Formula for Mass of Element Deposited:

m=M×I×tn×Fm = frac{M times I times t}{n times F}m=n×FM×I×t​

Where:

• M = Molar mass of element

• n = Number of electrons involved

• F = Faraday constant (96,500 C/mol)

• I = Current (A), t = time (s)

🧠 NECTA-style Example:

Calculate the mass of aluminum deposited when a current of 3 A is passed through molten AlCl₃ for 2 hours. (M = 27, n = 3)

Solution:

• I = 3 A

• t = 2 × 3600 = 7200 s

• F = 96,500 C/mol

• M = 27, n = 3

✅ Answer: 2.02 g of aluminum

🔹 Applications in Real Life

• Metal purification (e.g. copper refining)

• Electroplating (e.g. silver coating)

• Quantitative analysis (e.g. coulometry)

• Hydrogen fuel generation

✅ Summary

• First Law: Mass ∝ Charge (Q = I × t)

• Second Law: Mass ∝ Equivalent mass

• Use Faraday’s constant in molar-based calculations

• Essential in both practical and theoretical NECTA exams