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 H20 to the RHS H₂SO₄ ----> SO₂ + 2 H₂O 3. Balance hydrogen atoms (H+) There are 2 on LHS and 4 on RHS