Exhaustive Guide: Electrolysis & Faraday's Laws | Class 12 Chemistry

Electrolytic Cells & Faraday's Laws

Module 4 | CBSE Class 12 Chemistry | Electrochemistry Chapter

Table of Contents

• 1. Electrolysis and Electrolytic Cells
• 2. Faraday’s Laws of Electrolysis
• 3. Factors Affecting Products of Electrolysis
• 4. NCERT Solved Examples
• 5. Previous Year Questions (PYQs)

1. Electrolysis and Electrolytic Cells

In an electrolytic cell, external electrical energy is used to bring about a non-spontaneous chemical change. This process is called electrolysis.

Example: The electrolysis of molten NaCl. In the fused state, Na⁺ and Cl^- ions are free to move. When a potential is applied:
Cathode: Na⁺ + e^- → Na(s) (Reduction)
Anode: Cl^- → ½ Cl₂(g) + e^- (Oxidation)

2. Faraday’s Laws of Electrolysis

Michael Faraday established the quantitative relationship between the amount of electricity passed and the amount of substance deposited/liberated at the electrodes.

Faraday’s First Law:

The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.

w = Z × I × t

Simplified version for students:
Mass (w) = (Molar Mass / n-factor) × (It / F)
Where F (Faraday's Constant) ≈ 96500 C mol^-1.

Faraday’s Second Law:

When the same quantity of electricity is passed through several electrolytes, the masses of the substances deposited at the respective electrodes are directly proportional to their chemical equivalent weights.

3. Factors Affecting Products of Electrolysis

Electrolysis of aqueous solutions is complex because water itself can be oxidized or reduced at the electrodes. The actual products depend on:

• The nature of the material being electrolyzed.
• The concentration of the electrolyte.
• The type of electrodes used:
  - Inert electrodes (Pt, Au, Graphite): Do not participate in the reaction.
  - Active electrodes (Cu, Ag, Zn): Participate in the reaction (e.g., Cu anode dissolves in Cu²⁺ during CuSO₄ electrolysis).
• Overpotential: The extra voltage required to overcome kinetic hindrances (e.g., in the electrolysis of aqueous NaCl, O₂ evolution at the anode is kinetically slower than Cl₂ evolution, so Cl₂ is formed instead).

4. NCERT Solved Examples (Step-by-Step)

NCERT Example 3.10: A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathode?

Solution:
Given: I = 1.5 A; t = 10 min = 600 seconds.
Q = I × t = 1.5 × 600 = 900 C.

The reaction at cathode: Cu²⁺ + 2e^- → Cu(s).
2 moles of electrons (2 × 96500 C) deposit 1 mole of Cu (63.5 g).
Mass = (63.5 × 900) / (2 × 96500) = 0.296 g of Cu.

5. Previous Year Questions (PYQs)

[CBSE 2019]

Q1. What are the products of electrolysis of aqueous NaCl using platinum electrodes?

Answer:
At Cathode: H₂ gas is evolved (Reduction of H₂O is easier than Na⁺).
At Anode: Cl₂ gas is evolved (due to overpotential of oxygen).
Solution: The remaining solution becomes basic (contains NaOH).

[CBSE 2021]

Q2. How much charge is required for the reduction of 1 mole of Al³⁺ to Al?

Answer:
Reaction: Al³⁺ + 3e^- → Al
Since 3 moles of electrons are required for 1 mole of Al, and 1 mole of electrons = 1 Faraday (96500 C):
Total Charge = 3 × 96500 = 289,500 Coulombs.

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