Chapter 19 Oxidation - Reduction Reactions

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

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

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 Oxidation Half- Reaction Fe → Fe+2 + 2e-

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

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

6. Leo the Lion goes Ger Lose Electrons Oxidation Gain Electrons Reduction Electron Loss Means Oxidation

7. Oxidation - Reduction H2O(l) H2(g) + O2(g) +1 -2 0 0 Charge increased from -2 to 0, so e- were lost Charge reduced from +1 to 0, so e- were gained The hydrogen in water was reduced The oxygen in water was oxidized

8. Example • 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- . OCl- -2 +1

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

10. Example • Balance the following reaction: I- + MnO4- + H+ → MnO2 + I2 + H2O 2I- + MnO4- + 4H+ → MnO2 + I2 + 2H2O 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 • Write the formula equation then ionic equation • Assign oxidation numbers. Exclude anything with an ox. # of zero, or that doesn’t change ox. # • Write the ½ rxn for oxidation • Balance the atoms • Balance the charge (w/ electrons) • Write the ½ rxn for reduction • Balance the atoms • Balance the charge (w/ electrons) • Use coefficients to ensure the # of e- lost in ox. equals the # of e- gained in red. • Combine both ½ rxns and cancel (like Hess’s Law) • Combine ions to form initial compounds.

12. Example Half-Reactions: 2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O 6I- → 3I2 + 6e- Now try to balance the following reaction: I- + MnO4- + H+ → MnO2 + I2 + H2O Overall Balanced Equation: 6KI + 2KMnO4+ 8HCl → 2MnO2 + 3I2 + 4H2O + 8KCl

13. Chapter 19Oxidation - 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. Fe + Cu+2→ Fe+2 + Cu Iron causes Copper to become reduced, so it is the Reducing Agent Fe → Fe+2 + 2e-

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

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