2. Define oxidation and reduction. Determine oxidation numbers for atoms. Identify the oxidizing agent, the reducing agent. Distinguish between redox and non-redox reactions.

3. Burning and corrosion needs oxygen – oxidation. Oxidation-reduction reactions – (redox) Chemical changes when electrons are transferred from one reactant to another. Oxidation - an atom loses one or more electrons. Reduction - an atom gains one or more electrons. "LEO says GER” Losing Electrons is Oxidation, Gaining Electrons is Reduction

4. 2 Mg (s) + O 2 (g) → 2 MgO (s) Mg – neutralMg 2+ ion O – neutral O 2– ion Magnesium is oxidized. Oxygen is reduced. Mg → Mg 2+ + 2e - O + 2e - → O 2-

5. Mg (s) + Cl 2 (g) → MgCl 2 (s) Written difference between ion and oxidation: Chlorine ion – Cl 1- ion charge = 1- oxidation number = -1 Mg → Mg 2+ + 2e - Cl + 1e - → Cl 1- Sometimes these numbers are the same (like above) sometimes they are very different – which is why we write them differently.

6. Oxidation number represents the charge the atom would have if every bond were ionic. 1. Assign known numbers first (below). Then calculate the others. All uncombined elements (& diatomics) – zero. (unless otherwise stated as charge) Monatomic ion in ionic bonds - equals ion charge. **You will assign them to EACH atom**

7. Neutral compound: Sum of ox.numbers from each atom must be zero. Charged compound: Sum of ox.numbers must be charge of compound. Assign oxidation numbers to each atom in SO 2. S = +4 O = –2 In compounds: Alkali metals - always +1. Earth metals - always +2 Al: +3, F: -1, H: +1*, O: –2

8. Assign ox.numbers for each atom in K 2 Cr 2 O 7 Step 1: Start with atoms which are known. O: –2 K: +1 K 2 Cr O 7 -2 +6 +1 -14 +12 +2 = 0 ?? 2 K = +1 Cr = +6 O = –2. Neutral compound: Sum of ox.numbers from each atom must be zero. Step 2: Solve for other atoms.

9. Assign ox. numbers for each atom in Fe(NO 3 ) 3 Step 1: Start with atoms which are known. O: –2 Fe: +3 Fe(NO 3 ) -2 +5 +3 -18 +15 +3 = 0 ?? 3 Fe = +3 N = +5 O = –2. Neutral compound: Sum of ox.numbers from each atom must be zero. Step 2: Solve for other atoms.

10. Use ox.numbers to determine if reaction is a redox reaction. SO 2 + H 2 O → H 2 SO 3 Ox.numbers do not change – no e - transferred – NOT a redox reaction. -2 +4 +1 -2 +1 -2 +4 -6 +4 +2 -2 +2 -4 +4

11. Is the following reaction a redox reaction? Cu – oxidized (loss of electrons). Ag – reduced (gain of electrons). Oxidation cannot occur without reduction. 0 -2 +5 +1 -2 +5 0 -6 +5 +1 -6 +5 +1 0 0 Cu (s) + 2 AgNO 3(aq) → CuNO 3(aq) + 2 Ag (s)

12. Oxidizing agent - causes the oxidation of another substance. AgNO 3 is the oxidizing agent Reducing agent - causes the reduction of another substance. Oxidizing agent becomes reduced and the reducing agent becomes oxidized. Cu is the reducing agent +1 0 0 Cu (s) + 2 AgNO 3(aq) → CuNO 3(aq) + 2 Ag (s)

13. Identify the substance oxidized, the substance reduced, the oxidizing agent and the reducing agent. 2 HNO 3(aq) + 3 H 2 S (g) → 2 NO (g) + 3 S (s) + 4 H 2 O (l) S – oxidized N – reduced H 2 S – reducing agent HNO 3 – oxidizing agent 0 -2 +2 -2 +1 0 -2 +2 -2 +2 -2 +5 -6 +5 +1 -2 +1 -2 +2

14. How many electrons are transferred in the reaction below: Stoichiometry used to determined total electrons transferred. HNO 3(aq) + H 2 S (g) → 2 NO (g) + 3 S (s) + 4 H 2 O (l) 0 -2 +2 -2 +1-2 +5+1 -2 +1 gains 3e - loses 2e - S: (3 atoms) x (2e - lost) = 6 electrons lost N: (2 atoms) x (3e - gained) = 6 electrons gained 2 3

15. Strong oxidizing agents: Are very reactive – will take from anything Oxidizing AgentsReaction Products O 2 O 2–, H 2 O, CO 2 F 2, Cl 2, Br 2, I 2 F –, Cl –, Br –, I – MnO 4 – Mn 2+ Cr 2 O 7 2– Cr 3+ HNO 3 NO, NO 2 H 2 O 2 O 2, H 2 O Strong reducing agents: Are very reactive – will give to anything. Metals, substances that burn easily – H 2, C x H y