2.6b: write ionic half-equations and use them to construct full ionic equations

Eg. HBr + H2SO4 → Br2 + SO2

• oxidation = 2Br- → Br2 + 2e-
• reduction = S6+ + 2e- → S4+

Eg. the reaction of manganate (VII) in acid solution with iron (II) ions

1. write half-equations

• oxidation = Fe2+ → Fe3+ + e-
• reduction = MnO4- + 8H+ + 5e- → Mn2+ + 4H2O

2. multiply oxidation by 5 to get the same number of electrons in both equations

• oxidation = 5Fe2+ → 5Fe3+ + 5e-

3. combine the two half-equations

• 5Fe2+ + MnO4 + 8H+ + 5e- → Mn2+ + 4H2O + Fe3+ + e-

4. cancel out any species that are present on both sides

• 5Fe2+ + MnO4 + 8H+ → Mn2+ + 4H2O + 5Fe3+

BALANCING REDOX EQUATIONS

Eg. the reaction of hydrogen bromide with sulphuric acid to give  bromine and sulphur dioxide

1. write down a simple reaction equation and determine which species has been oxidised and which has been reduced

• HBr + H2SO4 → Br2 + SO2

2. write simple balanced half-equations for each process

• oxidation = 2Br- → Br2
• reduction = H2SO4 → SO2

3. add water to balance for any oxygen needed

• oxidation = 2Br- → Br2

• reduction = H2SO4 → SO2 + 2H2O

4. add hydrogen ions to balance for any hydrogen needed

• oxidation = 2Br- → Br2

• reduction = H2SO4 + 2H+ → SO2 + 2H2O

5. add electrons to each half-equation so that the charges are equal on both sides

• oxidation = 2Br- → Br2 + 2e-

• reduction = H2SO4 + 2H+ + 2e- → SO2 + 2H2O

6. multiply each half-equation so that they contain the same number of electrons
7. combine the half-equations

• 2Br- + H2SO4 + 2H+ + 2e- → SO2 + 2H2O + Br2 + 2e-

8. cancel out any species that appear on both sides

• 2Br- + H2SO4 + 2H+ → SO2 + 2H2O + Br2