# How to Balance Redox Equations

Sulphuric Acid is reduced by Iodide ions to form Sulphur Dioxide and 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 H20 to the RHS

H₂SO₄ ----> SO₂ + 2H₂O

3. Balance hydrogen atoms (H+)

There are 2 on LHS and 4 on RHS so add 2H+ to LHS

H₂SO₄ + 2H⁺ ----> 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 + 2H+ + 2e --> SO₂ + 2 H₂

The reduction equation is balanced.

**Oxidation Half Equation**

Iodide ions are oxidised -1 to 0...

I⁻ ----------> I₂

1. Balance the atoms

2 I⁻ -------> I₂

2. Balance the charges with electrons

2 I⁻ -------> I₂ + 2e This is balanced.

**Full Redox Equation**

Combine the 2 half equations

H₂SO₄ + 2H⁺ + 2e -> SO₂+ 2 H₂0

2 I- -----> I₂ + 2e

As both equations have 2e then there is no need to multiply them and the elections cancel and we have the full balanced redox equation...

H₂SO₄ + 2H⁺ + 2 I⁻ > SO₂ + 2H₂0 + I₂

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