Course Content
Module 1: Introduction to Electrochemistry
Overview of electrochemistry, its applications, and relevance in daily life and industrial processes.
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Definition and Importance of Electrochemistry
Module 2: Redox Reactions
Understanding oxidation, reduction, oxidation numbers, and balancing redox equations.
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Oxidation and Reduction Concepts
Oxidation Numbers
Balancing Redox Equations
📘 Module 3: Electrolysis
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Introduction to Electrolysis
Faraday’s Laws of Electrolysis
Industrial Applications of Electrolysis
Electrolysis Calculations
📘 Module 4: Redox Titrations
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Introduction and Indicators
KMnO₄ and Fe²⁺ Titration – Theory and Procedure
Oxalic Acid vs KMnO₄ Titration – Theory and Procedure
Iodometric Titrations – Iodine vs Thiosulfate (S₂O₃²⁻)
Electrode Potentials
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Standard Electrode Potentials (E° Values)
Electrochemical Cells and E° Cell Calculations
The Nernst Equation – Effect of Concentration on E°cell
📘 Module 6: Corrosion
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🧪 Lesson 1: Causes, Mechanism, and Prevention of Corrosion
📘 Module 7: Conductivity of Electrolyte Solutions
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🧪 Lesson 1: Electrical Conductivity and Factors Affecting It
📘 Module 8: Electrolytic Cells
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🧪 Lesson 1: Electrolysis Mechanism and Product Prediction
🧪 Lesson 2: Electrolysis of Water and Brine
📘 Module 9: Applications of Electrolysis in Industry
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🧪 Lesson 1: Electrolysis in Metal Extraction, Refining, and Electroplating
📘 Module 10: Redox Titrations and Calculations
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🧪 Lesson 1: Introduction, Procedure, and Redox Calculations
Advanced Chemistry: Electrochemistry

Introduction
Electrolysis calculations allow us to determine the mass, volume, or number of moles of substances deposited or liberated at electrodes based on the current and time passed. These calculations are rooted in Faraday’s Laws and often tested in NECTA exams.

🔹 Key Equations

  1. Q = I × t

    • Q = charge (Coulombs)

    • I = current (Amps)

    • t = time (seconds)

  2. m = (M × I × t) / (n × F)

    • m = mass deposited (grams)

    • M = molar mass (g/mol)

    • n = number of electrons transferred

    • F = Faraday’s constant (96,500 C/mol)

  3. n = Q / (n × F)

    • Used when calculating moles of product

  4. Volume of gas (at STP)

    • 1 mole of gas = 22.4 dm³

    • Volume = (n × 22.4) dm³

🧪 Worked Example 1 – Mass Deposited

Q: A current of 2.5 A is passed through a solution of CuSO₄ for 30 minutes. Calculate the mass of copper deposited.
(M = 63.5, n = 2)

Solution:

Convert time:

ini
t = 30 min × 60 = 1800 s

Use formula:

makefile
m = (M × I × t) / (n × F)
= (63.5 × 2.5 × 1800) / (2 × 96,500)
= 285,750 / 193,000 ≈ 1.48 g

✅ Answer: 1.48 g of copper deposited.

🧪 Worked Example 2 – Volume of Hydrogen Gas

Q: What volume of hydrogen gas (at STP) is produced by passing 4.0 A through acidified water for 40 minutes?

Solution:

  1. Find charge:

ini
t = 40 × 60 = 2400 s
Q = I × t = 4 × 2400 = 9600 C

  1. Reaction at cathode:
    2H⁺ + 2e⁻ → H₂ (n = 2)

  2. Moles of H₂:

ini
n = Q / (n × F) = 9600 / (2 × 96,500) ≈ 0.0497 mol

  1. Volume at STP:

ini
V = 0.0497 × 22.4 ≈ 1.11 dm³

✅ Answer: 1.11 dm³ of hydrogen gas is produced.

🧠 NECTA-Style Question

Q: 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:

makefile
t = 2 × 3600 = 7200 s
Q = 3 × 7200 = 21,600 C
m = (27 × 21,600) / (3 × 96,500)
= 583,200 / 289,500 ≈ 2.02 g

✅ Answer: 2.02 g of aluminum

✅ Summary

  • Use Q = I × t to find charge

  • Apply Faraday’s formula to get mass or moles

  • Always know the value of n from the half-equation

  • Convert gas moles to volume using 22.4 dm³/mol at STP

  • Round answers appropriately for NECTA exams

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