1. 1 Oxidation-Reduction Chapter 17 Hein and Arena Eugene Passer Chemistry Department Bronx Community College © John Wiley and Sons, Inc. Version 1.1

2. 2 Chapter Outline 17.1 Oxidation NumberOxidation Number 17.2 Oxidation-ReductionOxidation-Reduction 17.3 Balancing Oxidation- Reduction EquationsBalancing Oxidation-Reduction Equations 17.6 Electrolytic and Voltaic CellsElectrolytic and VoltaicCells 17.5 Activity Series of MetalsActivity Series of Metals 17.4 Balancing Ionic Redox Redox EquationsBalancing Ionic RedoxRedox Equations

3. 3 Oxidation Number

4. 4 The oxidation number (oxidation state) of an atom represents the number of electrons lost, gained, or unequally shared by an atom.

5. 5 Oxidation numbers can be zero, positive or negative.

6. 6 An oxidation number of zero means the atom has the same number of electrons assigned to it as there are in the free neutral atom.

7. 7 A positive oxidation number means the atom has fewer electrons assigned to it than in the neutral atom.

8. 8 A negative oxidation number means the atom has more electrons assigned to it than in the neutral atom.

9. 9 The oxidation number of an atom that has gained or lost electrons to form an ion is the same as the positive or negative charge of the ion. NaCl Sodium has lost an electron. Chlorine has gained an electron. The charge on sodium is +1. The oxidation number of sodium is +1. The charge on chlorine is –1. The oxidation number of chlorine is -1.

10. 10 In covalently bonded substances, oxidation numbers are assigned by an arbitrary system based on relative electronegativities.

11. 11 For symmetrical covalent molecules each atom is assigned an oxidation number of 0 because the bonding pair of electrons is shared equally between two like atoms of equal electronegativity. Electronegativity 2.1 Oxidation Number 0 Oxidation Number 0

12. 12 For symmetrical covalent molecules each atom is assigned an oxidation number of 0 because the bonding pair of electrons is shared equally between two like atoms of equal electronegativity. Electronegativity 3.0 Oxidation Number 0 Oxidation Number 0

13. 13 When the covalent bond is between two unlike atoms, the bonding electrons are shared unequally because the more electronegative element has a greater attraction for them. Electronegativity 3.0 Electronegativity 2.1 shared pair of electrons unequal electron sharing both shared electrons are assigned to chlorine there is a partial transfer of an electron to chlorine after assignment chlorine has one more electron than neutral chlorine Oxidation Number -1 after assignment hydrogen has one less electron than neutral chlorine Oxidation Number +1

14. 14 N2N2 N2ON2ONON2O3N2O3 NO 2 N2O5N2O5 N oxidation number 0+1+2+3+4+5 Many elements have multiple oxidation numbers

15. 15

16. 16

17. 17 Step 1 Write the oxidation number of each known atom below the atom in the formula. Step 2 Multiply each oxidation number by the number of atoms of that element in the compound. Step 3 Write an expression indicating the sum of all the oxidation numbers in the compound. Remember: The sum of the oxidation numbers in a compound must equal zero. Rules for Determining the Oxidation Number of an Element Within a Compound

18. 18 Determine the oxidation number for sulfur in sulfuric acid. H 2 SO 4 Step 1 -2+1 Write the oxidation number of each known atom below the atom in the formula. Multiply each oxidation number by the number of atoms of that element in the compound. Step 24(-2) = -82(+1) = +2 Step 3+2 + S + (-8) = 0 Write an expression indicating the sum of all the oxidation numbers in the compound. Step 4S = +6 (oxidation number for sulfur)

19. 19 Determine the oxidation number for sulfur in sulfuric acid. Step 1 -2 Write the oxidation number of each known atom below the atom in the formula. Multiply each oxidation number by the number of atoms of that element in the compound. Step 24(-2) = -8 Step 32C + (-8) = -2 (the charge on the ion) Write an expression indicating the sum of all the oxidation numbers in the compound. Step 42C = +6 C = +3 (oxidation number for sulfur)

20. 20Oxidation-Reduction

21. 21 Oxidation-reduction (redox) is a chemical process in which the oxidation number of an element is changed.

22. 22 Redox may involve the complete transfer of electrons to form ionic bonds or a partial transfer of electrons to form covalent bonds.

23. 23 Oxidation occurs when the oxidation number of an element increases as a result of losing electrons. Reduction occurs when the oxidation number of an element decreases as a result of gaining electrons. In a redox reaction oxidation and reduction occur simultaneously, one cannot occur in the absence of the other.

24. 24 Oxidizing agent The substance that causes an increase in the oxidation state of another substance. Reducing agent The substance that causes a decrease in the oxidation state of another substance. –The reducing agent is oxidized in a redox reaction. –The oxidizing agent is reduced in a redox reaction.

25. 25 Zn(s) + H 2 SO 4 (aq) → ZnSO 4 (aq) + H 2 (g) The electron transfer is more clearly expressed as The reaction of zinc with sulfuric acid is a redox reaction. Zinc is oxidized.Zinc is the reducing agent.Zinc transfers electrons to hydrogen.Hydrogen is reduced.Hydrogen is the oxidizing agent.Hydrogen accepts electrons from zinc.

26. 26 17.1

27. 27 Balancing Oxidation- Reduction Equations

28. 28 Change-In-Oxidation Number Method

29. 29 oxidation number of tin increases oxidation number of nitrogen decreases Balance the equation Sn + HNO 3 → SnO 2 + NO 2 +H 2 O Step 1 Assign oxidation numbers to each element to identify the elements being oxidized and those being reduced. Write the oxidation numbers below each element to avoid confusing them with ionic charge. Sn + HNO 3 → SnO 2 + NO 2 + H 2 O 0+1+5-2+4-2+4-2+1-2

30. 30 Balance the equation Sn + HNO 3 → SnO 2 + NO 2 +H 2 O Step 2 Write two new equations, using only the elements that change in oxidation number. Then add electrons to bring the equations into electrical balance. Sn o → Sn 4+ + 4e - 4N 5+ + 1e - → N 4+ + 4e - + 1e -

31. 31 Balance the equation Sn + HNO 3 → SnO 2 + NO 2 +H 2 O Step 3 Multiply the two equations by the smallest whole numbers that will make the electrons lost by oxidation equal to the number of electrons gained by reduction. Sn o → Sn 4+ + 4e - 4N 5+ + 4e - → 4N 4+ 4 4 4

32. 32 Balance the equation Sn + HNO 3 → SnO 2 + NO 2 +H 2 O Step 4 Transfer the coefficient in front of each substance in the balanced oxidation- reduction equations to the corresponding substances in the original equation. Sn + 4HNO 3 → SnO 2 + 4NO 2 + H 2 O 4 4 Sn o → Sn 4+ + 4e - 4N 5+ + 4e - → 4N 4+ 4 4 4

33. 33 Balance the equation Sn + HNO 3 → SnO 2 + NO 2 +H 2 O Step 5 Balance the remaining elements that are not oxidized or reduced to give the final balanced equation. Sn + 4HNO 3 → SnO 2 + 4NO 2 + 2H 2 O 2

34. 34 Balancing Ionic Redox Equations

35. 35 In ionic redox equations both the numbers of atoms and the charges on both sides of the equation must be the same.

36. 36 The Ion-Electron Method

37. 37 Step 2 Balance the elements other than hydrogen and oxygen. Write the two half-reactions that contain the elements being oxidized and reduced. Step 1

38. 38 Step 3 Balance hydrogen and oxygen. Acidic solution: For reactions in acidic solution, use H + and H 2 O to balance oxygen and hydrogen. For each oxygen needed use one H 2 O. Then add H + as needed to balance the hydrogen atoms.

39. 39 Step 3 Balance hydrogen and oxygen. Basic solution: For reactions in alkaline solutions, first balance as thought the reactions were in an acid solution, using Steps 1-3. Then add as many OH - ions to each side of the equation as there H + ions in the equation. Combine OH - ions into water. Rewrite the equation, canceling equal numbers of water molecules that appear on opposite side of the equation. Example: 4H + + 4OH - → 4H 2 O

40. 40 Step 5 Since the loss and gain of electrons must be equal, multiply each half- reaction by the appropriate number to make the number of electrons the same in each half-reaction. Add electrons (e - ) to each half- reaction to bring them into electrical balance. Step 4

41. 41 Step 6 Add the two half-reactions together, canceling electrons and any other identical substances that appear on opposite sides of the equation.

42. 42 Balance the equation Step 1 Write two half-reactions, one containing the element being oxidized and the other the element being reduced (use the entire molecule or ion). Balance elements other than oxygen and hydrogen. (Step 2 is unnecessary, since these elements are already balanced). Step 2

43. 43 Step 3 Balance O and H. The solution is acidic. The oxidation requires neither O nor H, but the reduction equation needs 4H 2 O on the right and 8H + on the left. 4 water molecules are necessary to balance the 4 oxygens in MnO 4. 8H + balance the 8 hydrogens of 4 water molecules. MnO 4. Balance the equation

44. 44 Balance the equation Step 4 Balance each half-reaction electrically with electrons: net charge = +2 on each side net charge = -2 on each side balanced oxidation half-reaction balanced reduction half-reaction

45. 45 Balance the equation Step 5 Equalize loss and gain of electrons. In this case, multiply the oxidation equation by 5 and the reduction equation by 2.

46. 46 Balance the equation Step 6 Add the two half-reactions together, canceling the 10e - from each side, to obtain the balanced equation. The charge on both sides of the balanced equation is +4

47. 47 Balance the equation Step 1 Write two half-reactions, one containing the element being oxidized and the other the element being reduced (use the entire molecule or ion). Step 2 Balance elements other oxygen and hydrogen (Step 2 is unnecessary, since these elements are already balanced).

48. 48 Balance the equation Step 3 Balance O and H as though the solution were acidic. Use H 2 O and H +. Since the solution is basic, add 1 OH - to each side. Combine H + and OH - as H 2 O and rewrite, canceling H 2 O on each side. oxidation half-reaction For each unbalanced O add one H 2 O to the other side of the equation. For each unbalanced H add one H + to the other side of the equation.

49. 49 Balance the equation Rewrite, canceling 1 H 2 O from each side: Step 3 Balance O and H as though the solution were acidic. Use H 2 O and H +. Since the solution is basic, add 5 OH - to each side. Combine 5H + + 5OH - → 5H 2 O For each unbalanced H add one H + to the other side of the equation. For each unbalanced O add one H 2 O to the other side of the equation. oxidation half-reaction

50. 50 Balance the equation Step 4 Balance each half-reaction electrically with electrons: net charge = -5 on each side net charge = -1 on each side balanced oxidation half-reaction balanced reduction half-reaction

51. 51 Balance the equation Step 5 Equalize loss and gain of electrons. Multiply the oxidation reaction by 3. It is not necessary to multiply the reduction equation.

52. 52 Balance the equation Step 6 Add the two half-reactions together, canceling 3e - and 3OH - from each side of the equation. The charge on both sides of the balanced equation is -2

53. 53 Activity Series of Metals

54. 54 activity series A listing of metallic elements in descending order of reactivity.

55. 55 Sodium (Na) will displace any element below it from one of its compounds.

56. 56 increasing activity Mg(s) + PbS(s)  MgS(s) + Pb(s) Pb H2H2 K Ba Ca Na Mg Al Zn Cr Fe Ni Sn Cu Magnesium is above lead in the activity series. Magnesium will displace lead from its compounds.

57. 57 increasing activity Silver is below copper in the activity series. Silver will not displace copper from one of its compounds. Ag(s) + CuCl 2 (s)  no reaction Ba Pb H2H2 Na Mg Al Zn Cr Fe Ni Sn Cu Ag Hg

58. 58 Electrolytic and Voltaic Cells

59. 59 electrolysis The process whereby electrical energy is used to bring about a chemical change. electrolytic cell An electrolysis apparatus in which electrical energy from an outside source is used to produce a chemical change.

60. 60 cathode The negative electrode. anode The positive electrode.

61. 61 Electrolysis of Hydrochloric Acid

62. 62 In an electrolytic cell electrical energy from the voltage source is used to bring about nonspontaneous redox reactions.

63. 63 17.3 H 3 O + + 1e - → H o + H 2 O H o + H o → H 2 Cathode Reaction Hydronium ions migrate to the cathode and are reduced.

64. 64 17.3 Cl - → Cl o + e - Cl o + Cl o → Cl 2 Anode Reaction Chloride ions migrate to the anode and are oxidized.

65. 65 The hydrogen and chlorine produced when HCl is electrolyzed have more potential energy than was present in the hydrochloric acid before electrolysis. electrolysis 2HCl(aq) H 2 (g) + Cl 2 (g) electrolysis

66. 66 The Zinc-Copper Voltaic Cell

67. 67 voltaic cell A cell that produces electrical energy from a spontaneous chemical reaction. (Also known as a galvanic cell).

68. 68 When a piece of zinc is put in a copper(II) sulfate solution, the zinc quickly becomes coated with metallic copper. This occurs because zinc is above copper in the activity series.

69. 69 Zn(s) + CuSO 4 (s)  ZnSO 4 (s) + Cu(s) increasing activity Zinc is above copper in the activity series. Zinc will displace copper from one of its compounds. Pb H2H2 K Ba Ca Na Mg Al Zn Cr Fe Ni Sn Cu

70. 70 If this reaction is carried out in a voltaic cell, an electric current is produced.

71. 71

72. 72 Zn o (s) → Zn 2+ (aq) + 2e - anodeoxidationCu 2+ (aq) + 2e - → Cu o (s)reductioncathode loss of electrons gain of electrons Zn o (s) + Cu 2+ (aq) → Zn 2+ (aq) + Cu o (s) Net ionic reaction Zn o (s) + CuSO 4 (aq) → ZnSO 4 (aq) + Cu o (s) Overall equation

73. 73 The reactions of the zinc-copper cell are spontaneous. Spontaneous reactions occur in all voltaic cells. Reactions that occur in electrolytic cells are nonspontaneous. Spontaneity is the crucial difference between all voltaic and electrolytic cells.

74. 74 Key Concepts 17.1 Oxidation NumberOxidation Number 17.2 Oxidation-ReductionOxidation-Reduction 17.3 Balancing Oxidation- Reduction EquationsBalancing Oxidation-Reduction Equations 17.6 Electrolytic and Voltaic CellsElectrolytic and VoltaicCells 17.5 Activity Series of MetalsActivity Series of Metals 17.4 Balancing Ionic Redox Redox EquationsBalancing Ionic RedoxRedox Equations