How To Balance Redox Equations (Step-By-Step A-Level Chemistry Guide)
- Dec 9, 2017
- 2 min read
Updated: 6 days ago
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This video walks through the same redox balancing method using a different A-Level Chemistry example.
First in acid and then in alkaline solution.
Balancing redox equations is a common topic in OCR, AQA and Edexcel A-Level Chemistry.
Many students understand oxidation and reduction but struggle when they need to balance half equations and combine them into a full redox equation.
In this example, sulphuric acid is reduced by iodide ions to form sulphur dioxide, while iodide ions are oxidised to iodine.
By following the method below step by step, you can apply the same approach to any redox equation you encounter in exams.
Sulphuric Acid is reduced by Iodide ions to form Sulphur Dioxide.
Iodide ions are oxidised to Iodine.
Reduction Half Equation
H₂SO₄ -----> SO₂
1. Balance the species that has been oxidised or reduced.
In this case the sulphur has been reduced from +6 to +4. The sulphur atoms are balanced.
2. Balance oxygen using H₂O
There are 4 on LHS and 2 on RHS so add 2 H2O to the RHS
H₂SO₄ ----> SO₂ + 2 H₂O
3. Balance hydrogen atoms (H+)
There are 2 on LHS and 4 on RHS so add 2H+ to LHS
H₂SO₄ + 2 H⁺ ----> SO₂ + 2 H₂O
4. Balance any charges with electrons
There is 2+ on LHS and 0 on RHS so add 2e to the LHS
H₂SO₄ + 2 H⁺ + 2e⁻ --> SO₂ + 2 H₂O
The reduction equation is balanced.
Oxidation Half Equation
Iodide ions are oxidised -1 to 0...
I⁻ ----------> I₂
1. Balance the I atoms
2 I⁻ -------> I₂
2. Balance the charges with electrons
2 I⁻ -------> I₂ + 2e⁻ This is now balanced!
Full Redox Equation
Combine the 2 half equations
H₂SO₄ + 2H⁺ + 2e⁻ -> SO₂+ 2 H₂O
2 I- -----> I₂ + 2e⁻
As both equations have 2e⁻ then there is no need to multiply them and the electrons cancel and we have the full balanced redox equation...
H₂SO₄ + 2H⁺ + 2 I⁻ > SO₂ + 2H₂O + I₂
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