1. John E. McMurry Robert C. Fay Lecture Notes Alan D. Earhart Southeast Community College Lincoln, NE General Chemistry: Atoms First Chapter 7 Reactions in Aqueous Solution Copyright © 2010 Pearson Prentice Hall, Inc.

2. Chapter 7/2 Reactions in Solution 1.Precipitation Reactions 2.Neutralization Reactions 3.Redox Reactions

3. Chapter 7/3 Some Ways That Chemical Reactions Occur Precipitation Reaction: A process in which soluble ionic reactants yield an insoluble solid product that falls out of solution. 2KNO 3 (aq) + Pb I 2 (s)Pb(NO 3 ) 2 (aq) + 2K I (aq)

4. Chapter 7/4 Some Ways That Chemical Reactions Occur Acid-Base Neutralization Reactions: Processes in which an acid reacts with a base to yield water plus an ionic compound called a salt. H 2 O(l) + NaCl(aq)HCl(aq) + NaOH(aq)

5. Chapter 7/5 Some Ways That Chemical Reactions Occur Oxidation-Reduction (Redox) Reactions: Processes in which one or more electrons are transferred between reaction partners (atoms, molecules, or ions). MgCl 2 (aq) + H 2 (g)Mg(s) + 2HCl(aq)

6. Chapter 7/6 Electrolytes in Aqueous Solution Electrolytes: Substances which dissolve in water to produce conducting solutions of ions. H2OH2O Na 1+ (aq) + Cl 1- (aq)NaCl(s) Nonelectrolytes: Substances which do not produce ions in aqueous solutions. H2OH2O C 12 H 22 O 11 (aq)C 12 H 22 O 11 (s)

8. Chapter 7/8 Electrolytes in Aqueous Solution Weak Electrolytes: Compounds that dissociate to a small extent into ions when dissolved in water. H 1+ (aq) + CH 3 CO 2 1- (aq)CH 3 CO 2 H(aq) Strong Electrolytes: Compounds that dissociate to a large extent into ions when dissolved in water. H 1+ (aq) + Cl 1- (aq)HCl(g) H2OH2O

9. Electrolytes in Aqueous Solution Strong Acids: hydrochloric acid, hydrobromic acid, hydroiodic acid, perchloric acid, nitric acid, sulfuric acid.

10. Chapter 7/10 Electrolytes in Aqueous Solution Ionic compounds

11. Chapter 7/11 Electrolytes in Aqueous Solution Weak acids

12. Chapter 7/12 Electrolytes in Aqueous Solution Molecular compounds (other than any strong or weak electrolytes)

13. Chapter 7/13 Electrolytes in Aqueous Solution Dissociation Equations Fe 3+ (aq) + 3Br 1- (aq)FeBr 3 (s) H2OH2O 2Na 1+ (aq) + SO 4 2- (aq)Na 2 SO 4 (s) H2OH2O

14. Chapter 7/14 Aqueous Reactions and Net Ionic Equations 2KNO 3 (aq) + Pb I 2 (s)Pb(NO 3 ) 2 (aq) + 2K I (aq) Molecular Equation: All substances in the chemical equation are written using their complete formulas as if they were molecules. strong electrolytesprecipitate

15. Chapter 7/15 Aqueous Reactions and Net Ionic Equations Ionic Equation: All of the strong electrolytes are written as ions. 2K 1+ (aq) + 2NO 3 1- (aq) + Pb I 2 (s) Pb 2+ (aq) + 2NO 3 1- (aq) + 2K 1+ (aq) + 2 I 1- (aq) 2KNO 3 (aq) Pb(NO 3 ) 2 (aq) 2K I (aq)

16. Chapter 7/16 Aqueous Reactions and Net Ionic Equations Spectator Ions: Ions that undergo no change during the reaction and appear on both sides of the reaction arrow. 2K 1+ (aq) + 2NO 3 1- (aq) + Pb I 2 (s) Pb 2+ (aq) + 2NO 3 1- (aq) + 2K 1+ (aq) + 2 I 1- (aq)

17. Chapter 7/17 Aqueous Reactions and Net Ionic Equations Pb I 2 (s)Pb 2+ (aq) + 2 I 1- (aq) Net Ionic Equation: Only the ions undergoing change are shown.

18. Chapter 7/18 Precipitation Reactions and Solubility Guidelines Solubility: States how much of a compound will dissolve in a given amount of solvent at a given temperature.

19. Chapter 7/19 Precipitation Reactions and Solubility Guidelines 1.A compound is probably soluble if it contains one of the following cations: Group 1A cation: Li 1+, Na 1+, K 1+, Cs 1+ Ammonium ion: NH 4 1+ 2.A compound is probably soluble if it contains one of the following anions: Halide: Cl 1-, Br 1-, I 1- except Ag 1+, Hg 2 2+, and Pb 2+ compounds Nitrate (NO 3 1- ), perchlorate (ClO 4 1- ), acetate (CH 3 CO 2 1- ), sulfate (SO 4 2- ) except Ba 2+, Hg 2 2+, and Pb 2+ sulfates

20. Chapter 7/20 Precipitation Reactions and Solubility Guidelines Write the molecular, ionic, and net ionic equations for the reaction that occurs when aqueous solutions of AgNO 3 and Na 2 CO 3 are mixed.

21. Chapter 7/21 Precipitation Reactions and Solubility Guidelines Write the unbalanced chemical formulas of the products (use proper ionic rules). Ag 2 CO 3 AgNO 3 (aq)+Na 2 CO 3 (aq)+NaNO 3 CB + ADAB + CD double replacement reaction Write the molecular, ionic, and net ionic equations for the reaction that occurs when aqueous solutions of AgNO 3 and Na 2 CO 3 are mixed. 1.

22. Chapter 7/22 Precipitation Reactions and Solubility Guidelines Molecular Equation: Balance the equation and predict the solubility of each possible product. Ag 2 CO 3 (s) + 2NaNO 3 (aq)2AgNO 3 (aq) + Na 2 CO 3 (aq) Contains a group 1A cation. Neither the cation nor the anion is in the solubility list. Write the molecular, ionic, and net ionic equations for the reaction that occurs when aqueous solutions of AgNO 3 and Na 2 CO 3 are mixed. 2.

23. Chapter 7/23 Precipitation Reactions and Solubility Guidelines Ionic Equation: Dissociate the soluble ionic compounds. Write the molecular, ionic, and net ionic equations for the reaction that occurs when aqueous solutions of AgNO 3 and Na 2 CO 3 are mixed. 3. Ag 2 CO 3 (s) + 2Na 1+ (aq) + 2NO 3 1- (aq) 2Ag 1+ (aq) + 2NO 3 1- (aq) + 2Na 1+ (aq) + CO 3 2- (aq) 2NaNO 3 (aq) 2AgNO 3 (aq)Na 2 CO 3 (aq)

24. Chapter 7/24 Precipitation Reactions and Solubility Guidelines Write the molecular, ionic, and net ionic equations for the reaction that occurs when aqueous solutions of AgNO 3 and Na 2 CO 3 are mixed. Net Ionic Equation: Eliminate the spectator ions from the ionic equation. Ag 2 CO 3 (s)2Ag 1+ (aq) + CO 3 2- (aq) 4. Ag 2 CO 3 (s) + 2Na 1+ (aq) + 2NO 3 1- (aq) 2Ag 1+ (aq) + 2NO 3 1- (aq) + 2Na 1+ (aq) + CO 3 2- (aq)

25. Chapter 7/25 Acids, Bases, and Neutralization Reactions H 3 O 1+ (aq) + Cl 1- (aq)HCl(aq) + H 2 O(aq) Arrhenius Acid: A substance that dissociates in water to produce hydrogen ions, H 1+ : H 1+ (aq) + A 1- (aq)HA(aq) In water, acids produce hydronium ions, H 3 O 1+ : H 1+ (aq) + Cl 1- (aq)HCl(aq)

26. Chapter 7/26 Acids, Bases, and Neutralization Reactions Ammonia in water, commonly called “ammonium hydroxide”, is a base: M 1+ (aq) + OH 1- (aq)MOH(aq) Na 1+ (aq) + OH 1- (aq)NaOH(aq) Arrhenius Base: A substance that dissociates in water to produce hydroxide ions, OH 1- : NH 4 1+ (aq) + OH 1- (aq)NH 3 (aq) + H 2 O(aq)

27. Chapter 7/27 Acids, Bases, and Neutralization Reactions Weak acids and weak bases are weak electrolytes. Strong acids and strong bases are strong electrolytes.

28. Chapter 7/28 “ate”“ic” “ite”“ous”

29. Chapter 7/29 Acids, Bases, and Neutralization Reactions Binary Acids HClhydrochloric acid HBrhydrobromic acid HFhydrofluoric acid

30. Chapter 7/30 Acids, Bases, and Neutralization Reactions MA + H 2 OHA + MOH These acid-base neutralization reactions are double- replacement reactions just like the precipitation reactions: or MA + HOHHA + MOH WaterAcidBaseSalt

31. Chapter 7/31 Write the chemical formulas of the products (use proper ionic rules for the salt). 1. Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous HBr and aqueous Ba(OH) 2. H2OH2OHBr(aq)+Ba(OH) 2 (aq)+BaBr 2 SaltAcidBaseWater

32. Chapter 7/32 Molecular Equation: Balance the equation and predict the solubility of the salt in the products. 2. Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous HBr and aqueous Ba(OH) 2. 2H 2 O(l) + BaBr 2 (aq)2HBr(aq) + Ba(OH) 2 (aq) Use the solubility rules.  Halides are soluble: Cl 1-, Br 1-, I 1-  except compounds of Ag 1+, Hg 2 2+, and Pb 2+

33. Chapter 7/33 Ionic Equation: Dissociate a strong acid and the soluble ionic compounds. 3. Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous HBr and aqueous Ba(OH) 2. 2H 2 O(l) + Ba 2+ (aq) + 2Br 1- (aq) 2H 1+ (aq) + 2Br 1- (aq) + Ba 2+ (aq) + 2OH 1- (aq) BaBr 2 (aq) 2HBr(aq)Ba(OH) 2 (aq)

34. Chapter 7/34 2H 2 O(l) + Ba 2+ (aq) + 2Br 1- (aq) Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous HBr and aqueous Ba(OH) 2. Net Ionic Equation: Eliminate the spectator ions from the ionic equation. 4. 2H 1+ (aq) + 2Br 1- (aq) + Ba 2+ (aq) + 2OH 1- (aq) or H 2 O(l)H 1+ (aq) + OH 1- (aq) 2H 2 O(l)2H 1+ (aq) + 2OH 1- (aq)

35. Chapter 7/35 Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous NaOH and aqueous HF. Write the chemical formulas of the products (use proper ionic rules for the salt). 1. SaltAcidBaseWater H2OH2OHF(aq) + NaOH(aq)+NaF

36. Chapter 7/36 Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous NaOH and aqueous HF. Molecular Equation: Balance the equation and predict the solubility of the salt in the products. 2. H 2 O(l) + NaF(aq)HF(aq) + NaOH(aq) Use the solubility rules. Compounds based on Group 1A cations: Li 1+, Na 1+, K 1+, Cs 1+ are soluble.

37. Chapter 7/37 Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous NaOH and aqueous HF. Ionic Equation: Dissociate a strong acid and the soluble ionic compounds. 3. H 2 O(l) + Na 1+ (aq) + F 1- (aq)HF(aq) + Na 1+ (aq) + OH 1- (aq) NaF(aq) NaOH(aq)

38. Chapter 7/38 Acids, Bases, and Neutralization Reactions Write the molecular, ionic, and net ionic equations for the reaction of aqueous NaOH and aqueous HF. Net Ionic Equation: Eliminate the spectator ions from the ionic equation. 4. H 2 O(l) + Na 1+ (aq) + F 1- (aq)HF(aq) + Na 1+ (aq) + OH 1- (aq) H 2 O(l) + F 1- (aq)HF(aq) + OH 1- (aq)

39. Chapter 7/39 Oxidation-Reduction Reactions (Redox)

40. Chapter 7/40 Redox Reactions - OIL RIG (oxidation is loss / reduction is gain) 2Fe 2 O 3 (s)4Fe(s) + 3O 2 (g) Rusting of iron: an oxidation rxn of Fe (Fe loses e - ) 4Fe(s) + 3CO 2 (g)2Fe 2 O 3 (s) + 3C(s) Manufacture of iron: a reduction rxn of Fe (Fe gains e - )

41. Chapter 7/41 Oxidation-Reduction (Redox) Reactions Oxidation: The loss of one or more electrons by a substance, whether element, compound, or ion. Reduction: The gain of one or more electrons by a substance, whether element, compound, or ion. OIL RIG

42. Chapter 7/42 Oxidation-Reduction (Redox) Reactions 1.An atom in its elemental state has an oxidation number of 0. Rules for Assigning Oxidation Numbers Oxidation Number (State): A value which indicates whether an atom is neutral, electron-rich, or electron- poor. NaH2H2 Br 2 SNe Oxidation number 0

43. Chapter 7/43 Oxidation-Reduction (Redox) Reactions 2.An atom in a monatomic ion has an oxidation number identical to its charge. Na 1+ +1 Ca 2+ +2 Al 3+ +3 Cl 1- O 2- -2

44. Chapter 7/44 Oxidation-Reduction (Redox) Reactions b)Oxygen usually has an oxidation number of -2. HO 1- -2+1 HHCa +2 3.An atom in a polyatomic ion or in a molecular compound usually has the same oxidation number it would have if it were a monatomic ion. a)Hydrogen can be either +1 or -1. OHO +1 HHO +1 -2 H +1

45. Chapter 7/45 Oxidation-Reduction (Redox) Reactions HCl +1 c)Halogens usually have an oxidation number of -1.3. Cl O +1 -2

46. Chapter 7/46 START Oxidation-Reduction (Redox) Reactions Cr 2 O 7 2- -2+1x 4.The sum of the oxidation numbers is 0 for a neutral compound and is equal to the net charge for a polyatomic ion. x = +6 2(x) + 7(-2) = -2 (net charge) H2SO3H2SO3 x-2 x = +4 2(+1) + x + 3(-2) = 0 (net charge)

47. Chapter 7/47 Identifying Redox Reactions 2Fe 2 3O2(g)3O2(g)+4Fe(s)O3O3 (s)(s) -20 reduction oxidation +30 Oxidizing Agent (O 2 )Reducing Agent (Fe) Causes reduction Loses one or more e - Undergoes oxidation Atom’s oxidation # increases Causes oxidation Gains one or more e - Undergoes reduction Atom’s oxidation # decreases

48. Chapter 7/48 3 Identifying Redox Reactions 4Fe(s)C(s) +2Fe 2 O 3 +3(g)(g)(s)(s)O2O2 C oxidation +40 0+3 reduction Oxidizing Agent (Fe) Reducing Agent (C)

49. 49 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O

50. 50 Identify the Oxidizing and Reducing Agents in Each of the Following 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O +1 -2 +5 -2 +1 0 +2 -2 +1 -2 ox agentred agent +4 -2 +1 -1 +2 -1 0 +1 -2 oxidation reduction oxidation reduction red agox ag

51. Chapter 7/51 The redox Activity Series of the Elements 2Ag 1+ (aq) + Cu(s)2Ag(s) + Cu 2+ (aq) Cu 2+ (aq) + 2Ag(s)Cu(s) + 2Ag 1+ (aq) Which one of these reactions will occur? The forward rxn or the reverse rxn?

53. Chapter 7/53 The Activity Series of the Elements The elements that are higher up in the table are more likely to be oxidized. Thus, any element higher in the activity series will reduce the ion of any element lower in the activity series.

54. 54

55. 55

56. Chapter 7/56 stop Half Reactions Cu 2+ (aq) + Ag(s)Cu(s) + Ag 1+ (aq) Isolate the reaction of Cu in one equation Cu 2+ (aq)Cu(s) Isolate the reaction of Ag in one equation Ag(s) Ag 1+ (g) Given a total reaction Find the correct half reaction on the following lists

57. Balancing Redox Reactions: The Half-Reaction Method Basic Solution Acidic Solution

58. 58 Balancing Redox Rxns (Basic Solution) 1)Write the unbalanced equation 2)Assign oxidation numbers to the atoms a)determine which element is oxidized and which is reduced 3)Write redox half-reactions, including electrons a)oxidation :electrons on the right of arrow b)reduction: electrons on left of arrow 4)Balance the half-reactions by mass a)first balance elements other than H and O b)next, balance O by adding H 2 O wherever an O is needed c)then add H +1 where a H is needed d)neutralize all H + (acid) with OH - (base) 5)Balance half-reactions by charge a)balance charge by adjusting electrons 6)Balance electrons between half-reactions 7)Add half-reactions together 8)Check the answer

59. 59 Balancing Redox Rxns (Acid Solution) 1)Write the unbalanced equation 2)Assign oxidation numbers to the atoms a)determine which element is oxidized and which is reduced 3)Write redox half-reactions, including electrons a)oxidation :electrons on the right of arrow b)reduction: electrons on left of arrow 4)Balance the half-reactions by mass a)first balance elements other than H and O b)next, balance O by adding H 2 O wherever an O is needed c)then add H +1 where a H is needed 5)Balance half-reactions by charge a)balance charge by adjusting electrons 6)Balance electrons between half-reactions 7)Add half-reactions together 8)Check the answer

60. Chapter 7/60 Balancing Redox Reactions: The Half-Reaction Method Balance the following net ionic equation in acidic solution: STEP 1: WRITE THE UNBALANCED EQUATION Cr 3+ (aq) + I O 3 1- (aq) I 1- (aq) + Cr 2 O 7 2- (aq) STEP 2: ASSIGN OXIDATION STATES Cr 3+ (aq) + I O 3 1- (aq) I 1- (aq) + Cr 2 O 7 2- (aq) 1- (6+)(2-) 3+ (5+)(2-) I goes from 1- to 5+ (LOSES e-) = OXIDIZED Cr goes from 6+ to 3+ (GAINS e-) = REDUCED

61. Chapter 7/61 Balancing Redox Reactions: The Half-Reaction Method STEP 3: Write the two unbalanced half-reactions Cr 3+ (aq) Cr 2 O 7 2- (aq) + e - I 1- (aq) IO 3 1- (aq) + e - OX: RED: STEP 4: Balance the half-reactions by mass (excluding O & H) 2Cr 3+ (aq)Cr 2 O 7 2- (aq) I 1- (aq) IO 3 1- (aq) OX: RED: one Iodine two Chromium

62. Chapter 7/62 Balancing Redox Reactions: The Half-Reaction Method STEP 4: Balance the half-reactions by mass – add water wherever O is needed. 2Cr 3+ (aq) + 7H 2 OCr 2 O 7 2- (aq) I 1- (aq) + 3H 2 O IO 3 1- (aq) OX: RED: STEP 4: Balance the half-reactions by mass – add H + wherever H is needed. 2Cr 3+ (aq) + 7H 2 OCr 2 O 7 2- (aq) + 14H + I 1- (aq) + 3H 2 O IO 3 1- (aq) + 6H + OX: RED:

63. Chapter 7/63 Balancing Redox Reactions: The Half-Reaction Method STEP 5: Balance the half-reactions by charge I 1- (aq) + 3H 2 O IO 3 1- (aq) + 6H + + e - Cr 2 O 7 2- (aq) + 14H + + e - 2Cr 3+ (aq)+ 7H 2 O (1-)=1-(1-)+(6+)+(x)=1- (2-)+(14+)+(x)=6+(6+)=6+ X=6- 6x x 6

64. Chapter 7/64 Balancing Redox Reactions: The Half-Reaction Method I 1- (aq) + 3H 2 O IO 3 1- (aq) + 6H + + 6e - Cr 2 O 7 2- (aq) + 14H + + 6e - 2Cr 3+ (aq) + 7H 2 O STEP 6 : Balance electrons between half-reactions STEP 7 : Add the half-reactions I 1- (aq) + Cr 2 O 7 2- (aq) + 8H + IO 3 1- (aq) +2Cr 3+ (aq) +4H 2 O Electrons are balanced as is I 1- (aq) + 3H 2 O IO 3 1- (aq) + 6H + + 6e - Cr 2 O 7 2- (aq) + 14H + + 6e - 2Cr 3+ (aq) + 7H 2 O 4 8

65. Chapter 7/65 MnO 2 (s) + BrO 3 1- (aq)MnO 4 1- (aq) + Br 1- (aq) Balancing Redox Reactions: The Half-Reaction Method Balance the following net ionic equation in basic solution:

66. Chapter 7/66 Balancing Redox Reactions: The Half-Reaction Method Write the two unbalanced half-reactions. Br O 3 1- (aq) Br 1- (aq) MnO 2 (s)MnO 4 1- (aq) Ox: Red: Write oxidation states Br- = -1BrO3- MnO 4 1- X+ 3(-2)=1- X=+5 X+ 4(-2)=1- X=+7 X+ 2(-2)=0 X=+4 MnO 2 -6 e- +3 e-

67. Chapter 7/67 Balancing Redox Reactions: The Half-Reaction Method Balance both half-reactions for all atoms except O and H. Br O 3 1- (aq) + e- Br 1- (aq) MnO 2 (s)e- + MnO 4 1- (aq)  Both Br and Mn are balanced Balance O by adding H2O Br O 3 1- (aq) 3H2O + Br 1- (aq) MnO 2 (s) + 2H2OMnO 4 1- (aq)

68. Chapter 7/68 Balancing Redox Reactions: The Half-Reaction Method Br O 3 1- (aq) + 6H 1+ (aq)3H 2 O(l) + Br 1- (aq) MnO 2 (s) + 2H 2 O(l)+4H 1+ (aq) + MnO 4 1- (aq) Balance each half-reaction for H by adding H 1+. Since in basic solution, add OH - for each H 1+. Br O 3 1- (aq) + 6H 1+ + 6OH - 3H 2 O(l) + Br 1- (aq) + 6OH - MnO 2 (s) + 2H 2 O(l) + 4OH - 4OH - + 4H 1+ + MnO 4 1- (aq) Br O 3 1- (aq) + 6H 2 O 3H 2 O(l) + Br 1- (aq) + 6OH - MnO 2 (s) + 2H 2 O(l) + 4OH - 4H 2 O(l) + MnO 4 1- (aq) 3 2

69. Chapter 7/69 Balancing Redox Reactions: The Half-Reaction Method Br O 3 1- (aq) + 3H 2 O(l) + e - Br 1- (aq) + 6OH - MnO 2 (s) + 4OH - e - + 2H 2 O(l) + MnO 4 1- (aq) Balance each half-reaction for charge by adding electrons to the side with greater positive charge. (1-)+(6-)=7- (1-)+(x)=7-x=6- x6 x 4- x+(1-)=4- X=3- 3

70. Chapter 7/70 Balancing Redox Reactions: The Half-Reaction Method Multiply each half-reaction by a factor to make the electron count the same in both half-reactions. 2 Br O 3 1- (aq) + 3H 2 O(l) + 6e - Br 1- (aq) + 6OH - MnO 2 (s) + 4OH - 3e - + 2H 2 O(l) + MnO 4 1- (aq)

71. Chapter 7/71 stop 2MnO 2 (s) + 2OH-(aq) + BrO 3 1- (aq) H 2 O(l) + 2MnO 4 1- (aq) + Br 1- (aq) Balancing Redox Reactions: The Half-Reaction Method Br O 3 1- (aq) + 3H2O(l) + 6e - 6OH-(aq) + Br 1- (aq) 2MnO 2 (s) + 8OH-(aq)6e - + 4H2O(l) + 2MnO 4 1- (aq) Add the two balanced half-reactions together and cancel species that appear on both sides of the equation. 2

72. 72 Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Assign Oxidation States I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) Separate into half- reactions ox: red: Assign Oxidation States Separate into half- reactions ox: I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s)

73. 73 Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Balance half- reactions by mass ox: I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s) Balance half- reactions by mass ox: 2 I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s) Balance half- reactions by mass, then O by adding H 2 O ox: 2 I  (aq)  I 2(aq) red: MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l)

74. 74 Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Balance half-rxns by mass then H by adding H + ox: 2 I  (aq)  I 2(aq) red: 4 H + (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) Balance half-rxns by mass in base, neutralize the H + with OH - ox: 2 I  (aq)  I 2(aq) red: 4 H + (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) 4 H + (aq) + 4 OH  (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) + 4 OH  (aq) 4 H 2 O (aq) + MnO 4  (aq)  MnO 2(s) + 2 H 2 O (l) + 4 OH  (aq) MnO 4  (aq) + 2 H 2 O (l)  MnO 2(s) + 4 OH  (aq) 2 H 2 O

75. 75 Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Balance Half- rxns by charge ox: 2 I  (aq)  I 2(aq) + 2 e  red: MnO 4  (aq) + 2 H 2 O (l) + 3 e   MnO 2(s) + 4 OH  (aq) Balance electrons between half-rxns ox: 2 I  (aq)  I 2(aq) + 2 e  } x 3 red: MnO 4  (aq) + 2 H 2 O (l) + 3 e   MnO 2(s) + 4 OH  (aq) } x 2 ox: 6 I  (aq)  3 I 2(aq) + 6 e  red: 2 MnO 4  (aq) + 4 H 2 O (l) + 6 e   2 MnO 2(s) + 8 OH  (aq)

76. 76 Balance the equation: I  (aq) + MnO 4  (aq)  I 2(aq) + MnO 2(s) in basic solution Add the Half- reactions ox: 6 I  (aq)  3 I 2(aq) + 6 e  red: 2 MnO 4  (aq) + 4 H 2 O (l) + 6 e   2 MnO 2(s) + 8 OH  (aq) 6 I  (aq) + 2 MnO 4  (aq) + 4 H 2 O (l)  3 I 2(aq) + 2 MnO 2(s) + 8 OH  (aq) Check Reactant CountElement Product Count 6I6 2Mn2 12O 8H8 88 charge 88

77. 77 Practice Balancing Equations H 2 O 2 + KI + H 2 SO 4  K 2 SO 4 + I 2 + H 2 O +1 -1 +1 -1 +1 +6 -2 +1 +6 -2 0 +1 -2 oxidation reduction ox:2 I -1  I 2 + 2e -1 red:H 2 O 2 + 2e -1 + 2 H +  2 H 2 O tot2 I -1 + H 2 O 2 + 2 H +  I 2 + 2 H 2 O 1 H 2 O 2 + 2 KI + H 2 SO 4  K 2 SO 4 + 1 I 2 + 2 H 2 O

78. 78 Practice - Balance the Equation ClO 3 -1 + Cl -1  Cl 2 (in acid) +5 -2 -1 0 oxidation reduction ox:2 Cl -1  Cl 2 + 2 e -1 } x5 red:2 ClO 3 -1 + 10 e -1 + 12 H +  Cl 2 + 6 H 2 O} x1 tot10 Cl -1 + 2 ClO 3 -1 + 12 H +  6 Cl 2 + 6 H 2 O 1 ClO 3 -1 + 5 Cl -1 + 6 H +1  3 Cl 2 + 3 H 2 O

79. Chapter 7/79 Redox Stoichiometry Titration: A procedure for determining the concentration of a solution by allowing a carefully measured volume to react with a solution of another substance (the standard solution) whose concentration is known. 5H 2 C 2 O 4 (aq) + 2MnO 4 1- (aq) + 6H 1+ (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) (Oxalic acid) + Potassium permanganate K + is a spectator ion and it is not shown in the net ionic equation

80. Chapter 7/80

81. Chapter 7/81 Redox Stoichiometry 5H 2 C 2 O 4 (aq) + 2MnO 4 1- (aq) + 6H 1+ (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) A solution is prepared with 0.2585 g of oxalic acid (H 2 C 2 O 4 ). 22.35 mL of an unknown potassium permanganate solution are needed to titrate the oxalic acid solution. What is the concentration of the potassium permanganate solution? Moles of H 2 C 2 O 4 Mass of H 2 C 2 O 4 Moles of KMnO 4 Molarity of KMnO 4 Mole Ratio 2 mol KMnO 4 5 mol H 2 C 2 O 4 Molarity of KMnO 4 158 g KMnO 4 1 mole KMnO 4 Molar Mass of H 2 C 2 O 4 1 mol 90.04 g oxalic acid

82. Chapter 7/82 Redox Stoichiometry 5H 2 C 2 O 4 (aq) + 2MnO 4 1- (aq) + 6H 1+ (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) Moles of H 2 C 2 O 4 available: 90.04 g 1 mol = 0.002871 mol H 2 C 2 O 4 0.2585 g H 2 C 2 O 4 Moles of KMnO 4 reacted: 5 mol H 2 C 2 O 4 2 mol KMnO 4 = 0.001148 mol KMnO 4 0.002871 mol H 2 C 2 O 4 x x

83. Chapter 7/83 1 L 1000 mL Redox Stoichiometry 5H 2 C 2 O 4 (aq) + 2MnO 4 1- (aq) + 6H 1+ (aq) 10CO 2 (g) + 2Mn 2+ (aq) + 8H 2 O(l) Concentration of KMnO 4 solution: = 0.05136 M KMnO 4 22.35 mL x 0.001148 mol KMnO 4