# 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 have been oxidised or reduced.

In this case the sulphur has been reduced from +6 to +4. The sulphur atoms are balance.

2. Balance the oxygen atoms using water

There are 4 on LHS and 2 on RHS so add 2 H20 to the RHS

H2SO4 ----------> SO2 + 2 H2O

3. Balance hydrogen atoms with 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**

The Iodide ions are oxidised from -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 + 2 H20 + I2