1. OXIDATION-REDUCTION CHAPTER 19

2. OXIDATION-REDUCTION REACTIONS OXIDATION is the process by which a substance loses one or more electrons. Iron loses three electrons from reactant to product. 0 +3 4 Fe + 3 O 2  2 Fe 2 O 3 REDUCTION is the process by which a substance gains one or more electrons. Iron gains three electrons from reactant to product. +3 0 2 Fe 2 O 3 + 3 C  4 Fe + 3 CO 2

3. OXIDATION-REDUCTION REACTIONS The oxidation number of an atom in a substance is equal to the charge that the atom would have if the electrons in each bond belonged to the most electronegative atom. The more electronegative atom gains electrons from the other atom because it is treated as if it were reduced. The less electronegative atom is oxidized.

4. OXIDATION-REDUCTION REACTIONS You can use one of two mnemonic devices to remember oxidation-reduction: a. OIL RIG (Oxidation Is Loss of electrons and Reduction Is Gain of electrons.) b. LEO GER (Lose Electrons - Oxidation; Gain Electrons - Reduction.)

5. OXIDIZING AND REDUCING AGENTS An OXIDIZING AGENT causes the oxidation of another substance by accepting electrons from that substance. The oxidizing agent contains the atom that shows a decrease in oxidation number. The oxidizing agent is itself reduced. +3 0 0 +4 2 Fe 2 O 3 + 3 C  4 Fe + 3 CO 2 C: 0  +4; C is oxidized, C is the reducing agent

6. OXIDIZING AND REDUCING AGENTS A REDUCING AGENT causes the reduction by providing electrons to another substance. The reducing agent contains the atom that shows an increase in oxidation number. The reducing agent is oxidized. +3 0 0 +4 2 Fe 2 O 3 + 3 C  4 Fe + 3 CO 2 Fe: +3 to 0; Fe is reduced, Fe 2 O 3 is the oxidizing agent

7. ANOTHER EXAMPLE 0 0 +3 -2 4 Fe + 3 O 2  2 Fe 2 O 3 Fe: 0 to +3; Fe is oxidized, Fe is the reducing agent O: 0 to -2; O is reduced, O 2 is the oxidizing agent The substance that is oxidized or reduced is always a single element. The substance that is the oxidizing agent or the reducing agent, is always whatever the reactant is – element or compound.

8. A CLOSER LOOK Okay, we will come back to oxidation and reduction. First we have to learn how to determine oxidation numbers. Get out your periodic table. Watch the video - Copy and paste this into the URL window in Mozilla Firefox: http://www.youtube.com/watch?v=8_CvNPuuhi M&feature=related

9. A CLOSER LOOK In order to identify a redox reaction, you must be able to write the charges for each element. This is known as the oxidation number. I have done these three for you. You write the oxidation number above the compound and do the bookkeeping below. K 2 Cr 2 O 7 SO 4 -2 Al(NO 3 ) 3 +1 +6 -2 +6 -2 +3 +5 -2 K 2 Cr 2 O 7 SO 4 -2 Al(NO 3 ) 3 +2 +12 -14 = 0 +6 -8 = -2 +3 +15 -18 = 0

10. OXIDATION NUMBER RULES 1. The oxidation number for any uncombined (free) element is zero (ex. O 2, Na, Mg) 2. The oxidation number for a monatomic ion equals its ionic charge. (ex. Ca +2, O -2 ) 3. The oxidation number of the more electronegative atom is equal to the charge it would have if it were an ion. Remember that fluorine is the most electronegative element, followed by oxygen and nitrogen.

11. OXIDATION NUMBER 4. Some elements’ oxidation number corresponds to their position on the periodic table:  Elements in group 1A = +1  Elements in group 2A = +2  Aluminum is always +3  Fluorine is always –1 (most electronegative)  Hydrogen has an oxidation number of +1 when combined with nonmetals. If bonded to a metal, it is a –1 because it is more electronegative.  Oxygen has an oxidation number of -2 in most compounds and ions. Peroxides (O 2 -1 ) are the exception and if oxygen is bonded to fluorine, the oxygen’s charge is a +2.

12. Here are some common oxidation numbers. Notice that nonmetals tend to have multiple oxidation numbers.

13. OXIDATION NUMBER 4. The sum of the oxidation numbers of all the atoms in a particle must equal the charge of that ion. (ex. SO 4 -2 ) 5. The sum of all oxidation numbers for all atoms in a neutral compound is zero.

14. PRACTICE Write the oxidation numbers for the following compounds. Write the numbers above each element. 1.HNO 3 8. NO 2.H 3 PO 4 9. CrI 3 3.AgNO 3 10. SO 2 4.Cu(NO 3 ) 2 11. K 2 SO 4 5. 2 Fe12. H 2 S 2 O 7 6.H 2 SO 4 13. P 4 O 6 7.KH 2 PO 4­ 14. AsO 4 -3

15. ANSWERS +1 +5 -2 +2 +5 -2 +1 +1 +5 -2 1.HNO 3 4. Cu(NO 3 ) 2 7. KH 2 PO 4­ +1+5 -6=0 +2 +10 -12=0 +1+2 +5 -8=0 +1 +5 -2 0 +2 -2 2.H 3 PO 4 5. 2 Fe8. NO 3. +3 +5 -8 =0 +2 -2 =0 +1 +5 -2 +1 +6 -2 +3 -1 3. AgNO 3 6. H 2 SO 4 9. CrI 3 +1+5 -6=0 +2 +6 -8 =0 +3 -3 =0

16. ANSWERS +4 -2 10. SO 2 +4 -4 =0 +1 +6 -2 +3 -2 11. K 2 SO 4 13. P 4 O 6 +2+6 -8=0 +12 -12=0 +1 +6 -2 +5 -2 12. H 2 S 2 O 7 14. AsO 4 -3 +2 +12 -14=0 +5 -8 = -3

17. REVIEW Now, put it all back together – see how to determine what is oxidized and what is reduced. Watch the video - Copy and paste this into the URL window in Mozilla Firefox: http://www.youtube.com/watch?v=-vK- OPD3K6g

18. NOW, PRACTICE THIS A.Determine the oxidation number for each atom. B.State what is being oxidized and what is being reduced. C.State the oxidizing agent and the reducing agent. 1. Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag 2. 3 H 2 S + 2 HNO 3  3 S + 2 NO + 4 H 2 O **REMEMBER: you must have something oxidized and something reduced in all redox equations

19. PRACTICE Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag

20. PRACTICE 3 H 2 S + 2 HNO 3  3 S + 2 NO + 4 H 2 O

21. ANSWERS 0 +1 +5 -2 +2 +5 -2 0 A.Cu + 2 AgNO 3  Cu(NO 3 ) 2 + 2 Ag B.Cu 0  +2 lost electrons, oxidized Ag +1  0 gained electrons, reduced C.Cu is oxidized; Cu is the reducing agent Ag is reduced; AgNO 3 is the oxidizing agent

22. ANSWERS +1 -2 +1 +5 -2 0 +2 -2 +1 -2 A. H 2 S + 2 HNO 3  3 S + 2 NO + 4 H 2 O B. S -2  0 lost electrons, oxidized N +5  +2 gained electrons, reduced C.S is oxidized; H 2 S is the reducing agent N is reduced; HNO 3 is the oxidizing agent

23. DO NOW Write the oxidation numbers for the following compounds. Write the charges above the elements: 1.H 2 SO 4 2. Ca 2 (OH)PO 4 3. ClO 4 -1 4. NaHCO 3 State the charge changes, which compound lost or gained electrons, which is oxidized, which is reduced, and name the oxidizing and reducing agents: 1.Cu + AgNO 3  Cu(NO 3 ) 2 + Ag 2. H 2 S + HNO 3  S + NO + H 2 O

24. DO NOW ANSWERS Oxidation Numbers +1 +6 -2 +7 -2 1. H 2 SO 4 3. ClO 4 -1 +2 +6 -8 = 0 +7 -8 = -1 +2 -2 +1 +5 -2 +1 +1+4-2 2.Ca 2 (OH)PO 4 4. NaHCO 3 +4 -2 +1 +5 -8 = 0 +1 +1+4-6 = 0

25. DO NOW ANSWERS 0 +1+5-2 +2 +5 -2 0 1.Cu + AgNO 3  Cu(NO 3 ) 2 + Ag Cu 0  +2 lose e-, oxidized Cu = reducing agent Ag +1  0 gain e-, reduced AgNO 3 = oxidizing agent

26. DO NOW ANSWERS +1 -2 +1+5-2 0 +2-2 +1 -2 2.H 2 S + HNO 3  S + NO + H 2 O S -2  0 lose e-, oxidized H 2 S = reducing agent N +5  +2 gain e-, reduced HNO 3 = oxidizing agent

27. APPLICATIONS 1. Corrosion This is the oxidation of a metal caused by a reaction between the metal and some substance in the environment. The best example is rusting. Corrosion is a problem for it results in the loss of structural strength of the metal. Can be prevented by coating the metal with paint, plastic or another metal.

28. APPLICATIONS 2. Bleaching A chemical substance that is used to eliminate unwanted color from fabrics and other materials. Bleaches are oxidizing reagents because they remove electrons from the pigments that cause color. Common bleaches are chlorine (Cl 2 ), hydrogen peroxide (H 2 O 2 ), and hypochlorite ion (ClO -1 ).

29. APPLICATIONS 3. Fuels and Explosives Fuels release energy as they are oxidized. Common fuels (gasoline, natural gas) are composed largely of carbon and hydrogen. Once ignited, they are oxidized by oxygen, forming water and carbon dioxide. Nitroglycerin is both an oxidizer (hydrogen and oxygen) and a reducer (nitrogen).

30. APPLICATIONS 4. Photography Photography involves the capturing of a light image on a light- sensitive medium and the processing of the image to make a permanent record. The process is based on the redox of silver halides, such as AgBr.

31. APPLICATIONS 5. Electrochemical cells (page 664) These involve a transfer of electrons from an oxidized substance to a reduced substance. If a zinc strip is in contact with copper(II) sulfate solution, the zinc strip (the anode) loses electrons to the copper(II) ions (the cathode) in solution. The copper(II) ions accept the electrons and drop out of solution as copper atoms (now neutral).

32. APPLICATIONS As the electrons are transferred, energy is released in the form of heat and the temperature rises. If the two metals cannot touch and are separated, a transfer of electrical energy instead of heat accompanies the electron transfer. To complete the circuit, electrons flow in one direction from metal to metal. Then a salt bridge, or some other set-up, allows the ions to pass from one side to another, completing the electrical circuit.

33. APPLICATIONS An electrochemical cell converts chemical energy to electrical energy by a spontaneous redox reaction. It was invented by Allesandro Volta and is also called a voltaic cell.

34. BALANCING REDOX EQUATIONS Many redox reactions are easy to balance by the usual methods. KClO 3  KCl + O 2 Some are harder. Cu + HNO 3  Cu(NO 3 ) 2 + NO 2 + H 2 O

35. BALANCING REDOX EQUATIONS The OXIDATION NUMBER METHOD allows you to apply your knowledge of oxidation numbers in order to balance equations. The fundamental principle in balancing redox equations is that the number of electrons lost in an oxidation process (increase in oxidation number) must equal the number of electrons gained in the reduction process (decrease in oxidation number).

36. BALANCING REDOX EQUATIONS Two methods: 1. Oxidation Number/Bracket Method This will not always work if the reaction is complicated. 2. Half Reaction Method This method always works but takes longer.

37. OXIDATION NUMBER METHOD STEP ONE: Assign oxidation numbers to all atoms in the equations. Write them above the element. S + HNO 3  SO 2 + NO + H 2 O

38. OXIDATION NUMBER METHOD STEP TWO: Identify the element oxidized and the element reduced. Determine the change in oxidation number of each oxidized and reduced element. STEP THREE: Connect the atoms that change oxidation number using a bracket. Write the change in oxidation number at the midpoint.

39. OXIDATION NUMBER METHOD STEP FOUR: Choose coefficients that make the total increase in oxidation number equal the total decrease. STEP FIVE: Balance the remaining elements by inspection using the conventional method and check the final equation.

40. PRACTICE 1. Cu + HNO 3  Cu(NO 3 ) 2 + NO 2 + H 2 O

41. PRACTICE 2. Cl 2 + KBr  KCl + Br 2

42. BALANCING REDOX REACTIONS HALF REACTIONS These reactions are used to balance redox equations. The reaction is broken into two parts – the oxidation part and the reduction part. The half –reaction method allows you to apply your knowledge of oxidation numbers in order to balance equations.

43. HALF REACTION METHOD STEPS TO FOLLOW: 1. Write out the equation. Then change it to a net ionic equation if it is not one already, omit the spectator ions, and assign oxidation numbers to all atoms in the equation. Write them above the element. HS -1 + IO 3 -1  I -1 + S + H 2 O

44. HALF REACTION METHOD 2. Write the separate half-reactions. HS -1  SIO 3 -1  I -1 3. Balance all the elements except O and H (already balanced in this one). HS -1  SIO 3 -1  I -1

45. HALF REACTION METHOD 4. If the oxygen is unbalanced, add enough water (H 2 O) to the side deficient in oxygen. HS -1  SIO 3 -1  I -1 + 3H 2 O 5. Add sufficient hydrogen ions (H + ) to the side deficient in hydrogen to balance the hydrogen. HS -1  S + H +1 6H +1 + IO 3 -1  I -1 + 3H 2 O

46. HALF REACTION METHOD 6. Write the electrons in each half reaction. -2 0 HS -1  S + H +1 + 2e- +5 -1 6e- + 6H +1 + IO 3 -1  I -1 + 3H 2 O

47. HALF REACTION METHOD 7. Determine the least common multiple and multiply each to get it so that the number of electrons gained equals the number of electrons lost. (x3)3HS -1  3S + 3H +1 + 6e- (x1) 6e- + 6H +1 + IO 3 -1  I -1 + 3H 2 O

48. HALF REACTION METHOD 8.Add the two half reactions together and return the spectator atoms. Delete anything that exactly occurs on both sides. (Notice how the water showed back up!) 3HS -1 + 6e- + 6H +1 + IO 3 -1  3S + 3H +1 + 6e- + I -1 + 3H 2 O becomes 3H +1 3HS -1 + 3H +1 + IO 3 -1  3S + I -1 + 3H 2 O

49. HALF REACTION METHOD PRACTICE (ACIDIC): MnO 4 -1 + H 2 SO 3  Mn +2 + HSO 4 -1 + H 2 O

50. BALANCING REDOX REACTIONS BASIC: Basic solutions add OH -1 to get rid of the H +1. Balance them the same way that you balance acidic solutions (H 2 O and then H +1 ) only add one more step. Anywhere that you added an H +1, you need to add an OH -1 to both sides. Combine any OH -1 with any H +1 on the same side to made water.

51. HALF REACTION METHOD PRACTICE (Basic): I -1 + OCl -1  I 2 + Cl -1 + H 2 O

52. PRACTICE (BASIC) -1 -2 +1 0 -1 +1 -2 I -1 + OCl -1  I 2 + Cl -1 + H 2 O +1 0 -1 0 2e- + 2H + + OCl -1  Cl -1 + H 2 O2I -1  I 2 + 2e- 2e- + 2H + + OCl -1 + 2I -1  Cl -1 + H 2 O +  I 2 + 2e- 2H + + OCl -1 + 2I -1  Cl -1 + H 2 O + I 2

53. HALF REACTION METHOD -1 -2 +1 0 -1 +1 -2 I -1 + OCl -1  I 2 + Cl -1 + H 2 O + 1 0 -1 0 2e- + 2H + + OCl -1  Cl -1 + H 2 O 2I -1  I 2 + 2e- 2e- + 2H + + OCl -1 + 2I -1  Cl -1 + H 2 O +  I 2 + 2e- 2H + + OCl -1 + 2I -1  Cl -1 + H 2 O + I 2 BASIC2OH - + 2H + + OCl -1 + 2I -1  Cl -1 + H 2 O + I 2 + 2OH - 2H 2 O H 2 O + OCl -1 + 2I -1  Cl -1 + I 2 + 2OH -