1. Chapter 19 Oxidation - Reduction Reactions 19.1 Oxidation and Reduction

2. Oxidation – Reduction Rules for assigning oxidation numbers 1.Most substances have the same oxidation number as their individual charge (the more electronegative element 1 st ) 2.All oxidation numbers in a compound must add up to equal the total charge on the compound 3.All single elements have a oxidation number of zero 4.All single ions have the same oxidation number as their charge 5.Oxygen has a charge of -2, except with F (+2), or peroxides (-1) 6.Hydrogen is usually +1, unless with a metal (-1)

3. Oxidation Reactions in which the atoms or ions of an element experience an increase in oxidation state A species whose oxidation number increases is oxidized Reactions in which the atoms or ions of an element experience an increase in oxidation state A species whose oxidation number increases is oxidized Fe → Fe +2 + 2e- Oxidation Half- Reaction

4. Reduction Reactions in which the oxidation state of an element decreases. A species that undergoes a decrease in oxidation state is reduced. Reactions in which the oxidation state of an element decreases. A species that undergoes a decrease in oxidation state is reduced. Cu +2 + 2e- → Cu Fe → Fe +2 + 2e- + Fe + Cu +2 → Fe +2 + Cu Overall Rxn Reduction Half- Reaction

5. Oxidation - Reduction AgCl (aq) + Na (s)  Ag (s) + NaCl (aq) +100 +1 OIL RIG Oxidation is loss, reduction is gain Charge reduced from +1 to 0, so e- were gained The silver in silver chloride was reduced Charge increased from 0 to +1, so e- were lost The sodium was oxidized

6. Leo the Lion goes Ger Lose Electrons Oxidation Gain Electrons Reduction E lectron L oss M eans O xidation

7. Oxidation - Reduction H 2 O (l)  H 2(g) + O 2(g) -2+100 Charge reduced from +1 to 0, so e- were gained The hydrogen in water was reduced The oxygen in water was oxidized Charge increased from -2 to 0, so e- were lost

8. Example Household Bleach removes stains through a redox reaction:Household Bleach removes stains through a redox reaction: Stain molecules (s) + OCl - (aq) → colorless molecules (s) + Cl - (aq) Determine the oxidation numbers of oxygen & chlorine in OCl -.Determine the oxidation numbers of oxygen & chlorine in OCl -. Household Bleach removes stains through a redox reaction:Household Bleach removes stains through a redox reaction: Stain molecules (s) + OCl - (aq) → colorless molecules (s) + Cl - (aq) Determine the oxidation numbers of oxygen & chlorine in OCl -.Determine the oxidation numbers of oxygen & chlorine in OCl -. OCl - -2 +1

9. Chapter 19 Oxidation - Reduction Reactions 19.2 Balancing Redox Equations

10. Example Balance the following reaction:Balance the following reaction: I - + MnO 4 - + H + → MnO 2 + I 2 + H 2 O Balance the following reaction:Balance the following reaction: I - + MnO 4 - + H + → MnO 2 + I 2 + H 2 O 2I - + MnO 4 - + 4H + → MnO 2 + I 2 + 2H 2 O But this reaction is balanced for mass not charge! A half-reaction system has to be used to balance for charge. But this reaction is balanced for mass not charge! A half-reaction system has to be used to balance for charge.

11. Half-Reaction Method 1.Write the formula equation then ionic equation 2.Assign oxidation numbers. Exclude anything with an ox. # of zero, or that doesn’t change ox. # 3.Write the ½ rxn for oxidation 4.Balance the atoms 5.Balance the charge (w/ electrons) 6.Write the ½ rxn for reduction 7.Balance the atoms 8.Balance the charge (w/ electrons) 9.Use coefficients to ensure the # of e- lost in ox. equals the # of e- gained in red. 10.Combine both ½ rxns and cancel (like Hess’s Law) 11.Combine ions to form initial compounds. 1.Write the formula equation then ionic equation 2.Assign oxidation numbers. Exclude anything with an ox. # of zero, or that doesn’t change ox. # 3.Write the ½ rxn for oxidation 4.Balance the atoms 5.Balance the charge (w/ electrons) 6.Write the ½ rxn for reduction 7.Balance the atoms 8.Balance the charge (w/ electrons) 9.Use coefficients to ensure the # of e- lost in ox. equals the # of e- gained in red. 10.Combine both ½ rxns and cancel (like Hess’s Law) 11.Combine ions to form initial compounds.

12. Example Now try to balance the following reaction: I - + MnO 4 - + H + → MnO 2 + I 2 + H 2 O Now try to balance the following reaction: I - + MnO 4 - + H + → MnO 2 + I 2 + H 2 O Half-Reactions: 2MnO 4 - + 8H + + 6e - → 2MnO 2 + 4H 2 O 6I - → 3I 2 + 6e - 6I - → 3I 2 + 6e -Half-Reactions: 2MnO 4 - + 8H + + 6e - → 2MnO 2 + 4H 2 O 6I - → 3I 2 + 6e - 6I - → 3I 2 + 6e - Overall Balanced Equation: 6KI + 2KMnO 4 + 8HCl → 2MnO 2 + 3I 2 + 4H 2 O + 8KCl Overall Balanced Equation: 6KI + 2KMnO 4 + 8HCl → 2MnO 2 + 3I 2 + 4H 2 O + 8KCl

13. Chapter 19 Oxidation - Reduction Reactions 19.3 Oxidizing and Reducing Agents

14. Reducing Agent Substance that has the potential to cause another substance to be reduced. They lose electrons; are oxidized. Substance that has the potential to cause another substance to be reduced. They lose electrons; are oxidized. Fe + Cu +2 → Fe +2 + Cu Fe → Fe +2 + 2e- Iron causes Copper to become reduced, so it is the Reducing Agent

15. Oxidizing Agent Substance that has the potential to cause another substance to be oxidized. They gain electrons; are reduced. Substance that has the potential to cause another substance to be oxidized. They gain electrons; are reduced. Fe + Cu +2 → Fe +2 + Cu Copper causes Iron to become oxidized, so it is the Oxidizing Agent Cu +2 + 2e- → Cu

16. Disproportionation/ Autooxidation A process by which a substance acts as both an oxidizing and reducing agent