2. 10.3 The Half-Reaction Method for Balancing Equations SCH4U1 Dec 8 th, 2009

3. Balancing Half-Reactions Half-Reaction: describes either the oxidation or reduction of a REDOX reaction. Oxidation : K  K + + e - Reduction : Cl 2 + 2e -  2Cl - Net Reaction :K + Cl 2  2KCl

4. Balanced Half Reactions and Net Reactions Net Reaction :K + Cl 2  2KCl Balance :2K + Cl 2  2KCl » 2 electrons lost 2 electrons gained

5. Balancing Half- Reactions in Acidic Media Write unbalanced half-reactions Balance all atoms EXCEPT for O and H Balance O by adding H 2 O Balance H by adding H + ions Balance the charges by adding electrons

6. Balancing Half- Reactions in Acidic Media ClO 3 -  Cl - ClO 3 -  Cl - + 3H 2 O (add water) 6H + + ClO 3 -  Cl - + 3H 2 O (add H ions) 6e - + 6H + + ClO 3 -  Cl - + 3H 2 O (add electrons)

7. Balancing Half- Reactions in Basic Media Write unbalanced half-reactions Balance all atoms EXCEPT for O and H Balance O by adding H 2 O Balance H by adding H + ions To BOTH sides, add as many OH - as H + were added Simplify OH - + H + to H 2 O if possible Remove excess H 2 O from both sides Balance the charges by adding electrons

8. Balancing Half- Reactions in Basic Media MnO 4 -  MnO 2 MnO 4 -  MnO 2 + 2H 2 O 4H + + MnO 4 -  MnO 2 + 2H 2 O 4OH - + 4H + + MnO 4 -  MnO 2 + 2H 2 O + 4OH - 4OH - + 3e - + 4H + + MnO 4 -  MnO 2 + 2H 2 O + 4OH - 4H 2 O + 3e - +MnO 4 -  MnO 2 + 2H 2 O + 4OH - 2H 2 O + 3e - + MnO 4 -  MnO 2 + 4OH -

9. Balancing the Net Ionic Reaction Electrons are neither created nor destroyed Electrons lost = Electrons gained  Break the reaction into 2 half-reactions  Balance the half-reactions separately  Multiply both by the lowest common multiple  Recombine the two half-reactions  Simplify by removing electrons and excess species

10. Balancing the Net Ionic Reaction MnO 4 - + Ag  Mn 2+ + Ag + (acidic conditions) Oxidation Ag  Ag + + e - (balanced) For a 5e - transfer: 5Ag  5Ag+ + 5e - Reduction MnO 4 -  Mn 2+ MnO 4 -  Mn 2+ + 4H 2 O 8H + + MnO 4 -  Mn 2+ + 4H 2 O 8H + + 5e - + MnO 4 -  Mn 2+ + 4H 2 O

11. Balancing the Net Ionic Reaction 5Ag  5Ag + + 5e - 8H + + 5e - + MnO 4 -  Mn 2+ + 4H 2 O 5Ag + 8H + + 5e - + MnO 4 -  5Ag + + 5e - + Mn 2+ + 4H 2 O 5Ag + 8H + + MnO 4 -  5Ag + + Mn 2+ + 4H 2 O

12. Homework p. 662 # 18 p. 668 # 2a, 3a p. 673 # 6a, 7b