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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