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
