1. 1 Chapter 19 Oxidation and Reduction (basic facts) A substance is oxidized if it loses electrons (becomes more positive) A substance is reduced if it gains electrons (becomes more negative) You can not have an oxidation reaction without also having a reduction reaction. The moles of electron lost by the substance being oxidized must equal the moles of electrons gained by the substance being reduced. This fact allows us to balance redox reactions. The substance that causes oxidation is called an oxidizing agent. The substance that causes reduction is called a reducing agent. If we separate the oxidation reaction from the reduction reaction in the correct way, we can use the redox reaction to produce electricity (a battery).

2. 2 The Reaction of Solid Sodium and Gaseous Chlorine to Form Solid Sodium Chloride 2 Na (s) + Cl 2 (g) → 2 Na  + 2 Cl  (ions in the crystal) What happened to Na? What happened to Cl?

3. 3 Na (s) → Na  + e  Each Na is losing an electron to become a cation Cl 2 + 2 e  → 2 Cl  Each Cl is gaining an electron to become an anion Oxidation is losing electrons Reduction is gaining electrons OIL RIG Oxidation Is Loss Reduction Is Gain

4. 4 Since metals have low electronegativities, they tend to lose electrons (be oxidized) while nonmetals which have higher electronegativities tend to gain electrons (be reduced) A substance that is losing electrons is becoming more positive. This means that the oxidation state is increasing. A substance that is gaining electrons is becoming more negative. This means that the oxidation state is decreasing. An agent is a substance that causes something to happen. If it causes an oxidation to occur, it would be an oxidizing agent. If it causes a reduction to occur, it would be a reducing agent.

5. 5 Oxidation Numbers Oxidation numbers are a bookkeeping method that chemists use to compare how many electrons an atom thinks it has at the moment to how many it should have in its elemental state. If an atom has MORE electrons now than it should, it will have an oxidation state that is negative (one negative charge for each extra electron). If an atom has FEWER electrons now than it should, it will have an oxidation state that is positive (one positive charge for each missing electron).

6. 6 General Rules for Oxidation Numbers: (these are like the rules on page 595) An element in its normal state has oxidation number = 0 An element that becomes an ion has an oxidation number equal to the charge of the ion. The sum of the oxidation numbers of all atoms in a compound or ion must equal the charge of the compound or ion. Remember that compounds are neutral (total charge of zero). In most compounds, oxygen has an oxidation number of  2. Oxygen will have an oxidation number of  1 when oxygen is a peroxide. Peroxides must have an oxygen- oxygen bond in the molecule. In compounds containing covalent bonds, the more electronegative element is assigned an oxidation number equal to what that element would have if it were a negative ion. In compounds where hydrogen is bonded to a nonmetal, hydrogen has an oxidation number of +1, when bonded to a metal it will have an oxidation number of  1.

7. 7 Examples/Practice: Assign oxidation numbers to each atom in each formula. PtS8S8 Mg +2 H2OH2ONO 3  Fe(NO 3 ) 3 Cr 2 (SO 3 ) 3 Na 2 O 2 CH 3 F CO 2 NH 4 + NH 4 NO 2 LiH Br 2

8. 8 2 NaBr (aq) + Cl 2 (aq) → 2 NaCl (aq) + Br 2 What substance is being oxidized?What substance is being reduced? What substance is the oxidizing agent?What substance is the reducing agent? How many electrons are transferred in the balanced reaction? We start to answer all of these question by first assigning oxidation numbers to every atom in the equation!!!!! We then use the changes in oxidation numbers to answer the questions. Typical SOL level Redox questions based on the reaction above:

9. 9 2 NaBr (aq) + Cl 2 (aq) → 2 NaCl (aq) + Br 2 What substance is being oxidized?What substance is being reduced? Assign Oxidation Numbers:

10. 10 2 NaBr (aq) + Cl 2 (aq) → 2 NaCl (aq) + Br 2 How many electrons are transferred in the balanced reaction? 2 Br  → Br 2 + 2 e  Cl 2 + 2 e  → 2 Cl  Na + → Na + (spectator ion) Oxidation half reaction Reduction half reaction Since Br  is oxidized, Cl 2 must have caused it to happen, so Cl 2 is the oxidizing agent. Since Cl 2 is reduced, Br  must have caused it to happen, so Br  is the reducing agent. What substance is the oxidizing agent?What substance is the reducing agent?

11. 11 The change in oxidation number can be used to determine if a redox reaction has occurred. If nothing changes oxidation number, a redox reaction has not occurred. The change in oxidation number can also be used to identify the substance being oxidized or reduced. The change in oxidation number can also be used to balance a redox reaction. In order for a redox reaction to be balanced, the number of electrons lost must equal the number of electrons gained. Al → Al +3 + 3 e  2 e  + In +3 → In  If Al and In +3 were reacting together, how many electrons must be transferred in the balanced reaction?

12. 12 (Al → Al +3 + 3 e  ) (2 e  + In +3 → In  ) 2 3 2 Al → 2 Al +3 + 6 e  6 e  + 3 In +3 → 3 In  2 Al + 3 In +3 → 3 In +1 + 2 Al +3 We see that 6 electrons need to be transferred in order for the Redox reactions to be balanced.

13. 13 ClO 4  + Br  → Cl  + Br 2 in acidic solution 7 -2 -1 -1 0 Cl +7 + 8 e  → Cl  2 Br  → Br 2  + 2 e  4( ) ClO 4  + 8 Br  → Cl  + 4 Br 2 Oxidation Half-Reaction Balancing Redox Reactions ClO 4  + 8 Br  → Cl  + 4 Br 2 Notice that oxygen can not be balanced here! Now we use the fact that the reaction is performed in aqueous acidic solution. This means that we can add water and/or H + ions as needed to balance the equation. + 4 H 2 O 8 H + + Check: The total charge on both sides of the equation must be equal! (+8) + (  9) = (  1) (  1) = (  1) Reduction Half-Reaction

14. 14 If a problem is performed in basic solution, balance it as if it were in acid and then convert to base. See the example below. ClO 4  + 8 Br  → Cl  + 4 Br 2 + 4 H 2 O 8 H + + First: Add enough OH  to both sides to react with all H . ClO 4  + 8 Br  → Cl  + 4 Br 2 + 4 H 2 O ++ 8 H + + 8 OH  Second: H + + OH  makes water. Cancel out any water that appears on both sides. ClO 4  + 8 Br  → Cl  + 4 Br 2 + 4 H 2 O + 8 OH  8 H 2 O + ClO 4  + 8 Br  → Cl  + 4 Br 2 + 8 OH  4 H 2 O + Check for total charge (  9) = (  9)

15. 15 Complete practice problems 1 and 2 on page 605. Cu (s) + H 2 SO 4 (aq) → CuSO 4 (aq) + SO 2 (g) + H 2 O (l) HNO 3 (s) + KI (aq) → KNO 3 (aq) + I 2 (aq) + NO (g) + H 2 O (l)

16. 16 Additional Example (in acid then converted to base): Mn  (aq) + BiO 3  (aq) → Bi  (aq) + MnO 4  (aq) Mn  → Mn  + 5 e  Bi  + 3 e  → Bi  +2 +5 +2 +7 3 Mn  → 3 Mn  + 15 e  5 Bi  + 15 e  → 5 Bi  3 Mn  + 5 Bi  → 5 Bi  + 3 Mn  3 Mn  (aq) + 5 BiO 3  (aq) → 5 Bi  (aq) + 3 MnO 4  (aq)+ 3 H 2 O (l)6 H + (aq) + 3 Mn  (aq) + 5 BiO 3  (aq) → 5 Bi  (aq) + 3 MnO 4  (aq) 3 H 2 O (l) ++ 6 OH  (aq) (+7) = (+7) (+1) = (+1)

17. 17 Example of a disproportionation reaction (a redox reaction where the same substance is both oxidized and reduced): Br 2 (aq) → Br  (aq) + BrO 3  (aq) 0 -1 +5 2 e  + Br 2 → 2 Br  Br 2 → 2 Br  + 10 e  10 e  + 5 Br 2 → 10 Br  Br 2 → 2 Br  + 10 e  6 Br 2 → 10 Br  + 2 Br  3 Br 2 (aq) → 5 Br  (aq) + BrO 3  (aq)3 H 2 O ++ 6 H + (aq) + 6 OH  (aq) 2 Br 2 (aq) → 6 Br  (aq) + 2 BrO 3  (aq)6 OH  1 (aq) ++ 3 H 2 O (aq) (  6) = (  6)

18. 18 Oxidizing and Reducing Agents Metallic elements tend to undergo oxidation (they lose electrons) The more reactive a metal is, the more easily it is oxidized and the better it is as a reducing agent. Non-Metallic elements tend to undergo reduction (they gain electrons) The more reactive a non-metal is, the more easily it is reduced and the better it is as an oxidizing agent. Note that some polyatomic ions that contain metals are good oxidizing agents. Two famous ones are: MnO 4 – and Cr 2 O 7 –2 Look at the relative strengths of Oxidizing and Reducing Agents on page 607. You DO NOT have to memorize the list! In the next chapter, we will use such a list (Standard Reduction Potentials) to solve problems involving bateries.