Course Content
Module 1: Introduction to Electrochemistry
Overview of electrochemistry, its applications, and relevance in daily life and industrial processes.
0/1
Module 2: Redox Reactions
Understanding oxidation, reduction, oxidation numbers, and balancing redox equations.
0/3
📘 Module 7: Conductivity of Electrolyte Solutions
0/1
📘 Module 9: Applications of Electrolysis in Industry
0/1
📘 Module 10: Redox Titrations and Calculations
0/1
Advanced Chemistry: Electrochemistry

Introduction
Redox equations must be balanced not only for mass (atoms) but also for charge, since electrons are transferred in the process. A properly balanced redox reaction is essential in both theory and NECTA practical exams, especially in titrations and electrolysis.


Key Concepts

There are two major methods of balancing redox reactions:

  1. Oxidation Number Method

  2. Ion-Electron (Half-Reaction) Method

This lesson focuses on the Ion-Electron method, which is most used in aqueous (acidic or basic) solutions.


🔹 Steps for Balancing Redox Equations Using the Half-Reaction Method


🧪 Worked Example 1:

Balance the redox reaction below in acidic solution:

 
MnO₄⁻ + Fe²⁺ → Mn²⁺ + Fe³⁺

Step 1: Write two half-reactions

Oxidation (Fe²⁺ is oxidized):

 
Fe²⁺ → Fe³⁺

Reduction (MnO₄⁻ is reduced):

 
MnO₄⁻ → Mn²⁺

Step 2: Balance all elements except H and O

Already balanced.


Step 3: Balance oxygen by adding H₂O

MnO₄⁻ has 4 oxygen atoms, so we add 4 H₂O to the right:

mathematica
MnO₄⁻ → Mn²⁺ + 4HO

Step 4: Balance hydrogen by adding H⁺

There are 8 H atoms in 4H₂O, so add 8 H⁺ to the left:

mathematica
8H+ MnO₄⁻ → Mn²⁺ + 4HO

Step 5: Balance charge by adding electrons

Left side charge: +8 (H⁺) + (-1) = +7
Right side charge: +2 (Mn²⁺)
Add 5 electrons to the left to equalize charge:

mathematica
8H+ MnO₄⁻ + 5e⁻ → Mn²⁺ + 4HO

Fe²⁺ → Fe³⁺ is already balanced for atoms; add 1 electron to the right:

 
Fe²⁺ → Fe³⁺ + e⁻

Step 6: Equalize electrons transferred

Multiply Fe reaction by 5:

 
5Fe²⁺ → 5Fe³⁺ + 5e⁻

Now, both half-reactions involve 5 electrons.


Step 7: Add the half-reactions

mathematica
8H+ MnO₄⁻ + 5Fe²⁺ → Mn²⁺ + 4HO + 5Fe³⁺

✅ Final balanced redox equation.


🔹 Worked Example 2:

Balance this redox reaction in acidic solution:

mathematica
CrO₇²⁻ + I⁻ → Cr³⁺ + I

Step 1: Half-reactions
Oxidation:

mathematica
2I⁻ → I+ 2e

Reduction:

mathematica
CrO₇²⁻ → 2Cr³⁺

Balance O:

mathematica
CrO₇²⁻ → 2Cr³⁺ + 7HO

Balance H:

mathematica
14H+ CrO₇²⁻ → 2Cr³⁺ + 7HO

Balance charge:
Total charge left = +14 + (-2) = +12
Right = +6 → Add 6e⁻ to left:

mathematica
14H+ CrO₇²⁻ + 6e⁻ → 2Cr³⁺ + 7HO

Now multiply I⁻ reaction by 3:

mathematica
6I⁻ → 3I+ 6e

Final redox equation:

mathematica
14H+ CrO₇²⁻ + 6I⁻ → 2Cr³⁺ + 3I+ 7HO

🔹 Oxidation Number Method (Summary)

Steps:

  1. Assign oxidation numbers to all elements

  2. Identify which ones increase (oxidation) or decrease (reduction)

  3. Calculate total electrons lost and gained

  4. Balance them

  5. Balance all atoms and charges


💡 NECTA Tips:

  • Always show oxidation numbers above each element when asked

  • State clearly which element is oxidized or reduced

  • Indicate oxidizing and reducing agents


🧪 Real-Life Applications

  • Redox titrations (e.g. KMnO₄ vs Fe²⁺)

  • Industrial electrolysis

  • Battery reactions


✅ Summary

  • Redox equations must be balanced for both mass and charge

  • Use half-reactions to organize the process

  • Always balance O using H₂O, H using H⁺ (in acidic medium)

  • Equalize electrons and combine the two halves