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1. Oxidation and Reduction Reactions

2. Oxidation (Read only) Original definition: When substances combined with oxygen. Ex: All combustion (burning) reactions CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) All “rusting” reactions 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s)

3. Reduction (Read Only) Original Definition: Reaction where a substance “gave up” oxygen. Called “reductions” because they produced products that were “reduced” in mass because gas escaped. Ex: 2Fe 2 O 3 (l) + 3C(s) 4Fe(l) + 3CO 2 (g)

4. Deals with movement of ELECTRONS during a chemical reaction. (Oxygen doesn’t have to be present) Oxidation/Reduction

5. Electron Transfer Reactions Oxidation: LOSS of one or more electrons. Reduction: GAIN of one or more electrons

6. Electron Transfer Reactions Oxidation & reduction always occur together. Electrons travel from what is oxidized towards what is reduced. One atom loses e-, the other gains e-

7. Redox Reactions : ALWAYS involve changes in charge A competition for electrons between atoms!

8. Remember!!

9. Or…Remember

10. Conservation of “Charge” Total electrons lost = Total electrons gained

11. Oxidizing/Reducing Agents Oxidizing Agent : substance reduced Gains electrons Reducing Agent: substance oxidized Loses electrons The “Agent” is the “opposite”

12. Assigning Oxidation Numbers Practice Problems http://www.usca.edu/chemistry/genchem/oxnumb.htm Animation of Oxidation and Reduction http://www.ausetute.com.au/redox.html

13. Identify What is Changing in Charge What is oxidized and reduced? What are the oxidizing and reducing agents? Ex: 3Br 2 + 2AlI 3 2AlBr 3 + 3I 2

14. 0 +3 -1 +3 -1 0 3Br 2 + 2AlI 3 2AlBr 3 + 3I 2 Br 2 is reduced and is the oxidizing agent I -1 is oxidized and is the reducing agent

15. What is oxidized and reduced? What are the oxidizing and reducing agents? Mg + CuSO 4 MgSO 4 + Cu 2K + Br 2 2KBr Cu + 2AgNO 3 Cu(NO 3 ) 2 + 2Ag NOTE: Atoms in a polyatomic ion DO NOT change in charge!

16. 0 +2 +2 0 Mg + CuSO 4 MgSO 4 + Cu Mg oxidized (reducing agent) Cu +2 reduced (oxidizing agent) 0 0 +1 -1 2K + Br 2 2KBr K oxidized (reducing agent) Br 2 reduced (oxidizing agent) 0 +1 +2 0 Cu + 2AgNO 3 Cu(NO 3 ) 2 + 2Ag Cu oxidized (reducing agent) Ag +1 reduced (oxidizing agent)

17. Redox or Not Redox (that is the question…) Redox Reactions: must have atoms changing in charge. Not all reactions are redox. Easy way to spot a redox reaction!!! Look for elements entering and leaving compounds .

18. Is it Redox? Look for Changes in Charge! Are elements entering and leaving compounds? Synthesis: Ex: 2H 2 + O 2 2H 2 O Decomposition: Ex: 2KClO 3 2KCl + 3O 2

19. Is it Redox? Synthesis : YES 0 0 +1 -2 Ex: 2H 2 + O 2 2H 2 O Decomposition : YES +1 +5 -2 +1 -1 0 Ex: 2KClO 3 2KCl + 3O 2

20. Is it Redox? Combustion: CH 4 + 2O 2 CO 2 + 2H 2 Single Replacement: Zn + CuCl 2 ZnCl 2 + Cu

21. Is it Redox? Combustion: YES -4 +1 0 +4 -2 +1 -2 CH 4 + 2O 2 CO 2 + 2H 2 Single Replacement: YES 0 +2 -1 +2 -1 0 Zn + CuCl 2 ZnCl 2 + Cu

22. Is it Redox? Double Replacement: AgNO 3 + LiCl AgCl + LiNO 3

23. Is it Redox? Double Replacement: NO!!!! Ions switch partners, but don’t change in charge +1 +5 -2 +1 -1 +1 -1 +1 +5 -2 AgNO 3 + LiCl AgCl + LiNO 3 Remember charges of atoms inside polyatomic ions do not change!

24. Writing Half Reactions Redox Reactions are composed of two parts or half reactions. Half Reactions Show: Element being oxidized or reduced. Change in charge # of electrons being lost or gained

25. Writing Half Reactions 0 0 +1 -1 2Na + F 2 2NaF Oxidation: Na Na +1 + 1e- or 2Na 2Na +1 + 2e- Note: e- are “lost” (on the right of arrow) Reduction: F + 1e- F -1 or F 2 + 2e- 2F -1 Note: e- are “gained” (on the left of arrow)

26. Ox’s Have Tails!! Oxidation Half reactions always have “tails” of electrons Na Na +1 + 1e-

27. 0 +2 -1 +2 -1 0 Zn + CuCl 2 ZnCl 2 + Cu Ox: Zn Zn +2 + 2e- Red: Cu +2 + 2e- Cu

28. Rules for Assigning Oxidation Numbers All free, uncombined elements have an oxidation number of zero. This includes diatomic elements such as O 2 or others like P 4 and S 8 . Hydrogen, in all its compounds except hydrides, has an oxidation number of +1 (positive one) Oxygen, in all its compounds except peroxides, has an oxidation number of -2 (negative two).

29. Practice Problems What is the oxidation number of . . . 1) N in NO 3 ¯ 2) C in CO 3 2 ¯ 3) Cr in CrO 4 2 ¯ 4) Cr in Cr 2 O 7 2 ¯ 5) Fe in Fe 2 O 3 6) Pb in PbOH + 7) V in VO 2 + 8) V in VO 2+ 9) Mn in MnO 4 ¯ 10) Mn in MnO 4 2 ¯

30. Rules for Assigning Oxidation Numbers All free, uncombined elements have an oxidation number of zero. This includes diatomic elements such as O 2 or others like P 4 and S 8 . Hydrogen, in all its compounds except hydrides, has an oxidation number of +1 (positive one) Oxygen, in all its compounds except peroxides, has an oxidation number of -2 (negative two).

31. What you must be able to do is look at a redox reaction and separate out the two half-reactions in it. To do that, identify the atoms which get reduced and get oxidized. Here are the two half-reactions from the example: Ag + ---> Ag Cu ---> Cu 2+ The silver is being reduced, its oxidation number going from +1 to zero. The copper's oxidation number went from zero to +2, so it was oxidized in the reaction. In order to figure out the half-reactions, you MUST be able to calculate the oxidation number of an atom.

32. When you look at the two half-reactions, you will see they are already balanced for atoms with one Ag on each side and one Cu on each side. So, all we need to do is balance the charge. To do this you add electrons to the more positive side. You add enough to make the total charge on each side become EQUAL. To the silver half-reaction, we add one electron: Ag + + e¯ ---> Ag To the copper half-reaction, we add two electrons: Cu ---> Cu 2+ + 2e¯

33. Half-reactions NEVER occur alone. notice that each half-reaction wound up with a total charge of zero on each side. This is not always the case. You need to strive to get the total charge on each side EQUAL, not zero.

34. Half-Reactions Practice Problems Balance each half-reaction for atoms and charge: 1) Cl 2 ---> Cl ¯ 2) Sn ---> Sn 2+ 3) Fe 2+ ---> Fe 3+ 4) I 3 ¯ ---> I¯ 5) ICl 2 ¯ ---> I¯ 6) Sn + NO 3 ¯ ---> SnO 2 + NO 2 7) HClO + Co ---> Cl 2 + Co 2+ 8) NO 2 ---> NO 3 ¯ + NO

35. Answers 1) Cl 2 + 2e¯ ---> 2Cl¯ 2) Sn ---> Sn 2+ + 2e¯ 3) Fe 2+ ---> Fe 3+ + e¯ 4) I 3 ¯ + 2e¯ ---> 3I¯ 5) ICl 2 ¯ + 2e¯ ---> I¯ + 2Cl¯ 6) Sn ---> SnO 2 and NO 3 ¯ ---> NO 2 7) HClO ---> Cl 2 and Co ---> Co 2+ 8) NO 2 ---> NO 3 ¯ and NO 2 ---> NO

36. Balancing Half-Reactions in Acid Solution MnO 4 ¯ ---> Mn 2+ in an acid solution Before looking at the balancing technique, the fact that it is in acid solution can be signaled to you in several different ways: 1) It is explicitly said in the problem. 2) An acid (usually a strong acid) is included as one of the reactants. 3) An H + is written just above the reaction arrow.

37. There are three other chemical species available in an acidic solution: H 2 O H + e¯ water is present because the reaction is taking place in solution the hydrogen ion is available because it is in acid solution electrons are available because that's what is transferred in redox reactions All three will be used in getting the final answer.

38. Balance the atom being reduced/oxidized. In our example, there is already one Mn on each side of the arrow, so this step is already done. MnO 4 ¯ ---> Mn 2+ Balance the oxygens . Do this by adding water molecules (as many as are needed) to the side needing oxygen. In our case, the left side has 4 oxygens , while the right side has none, so: MnO 4 ¯ ---> Mn 2+ + 4H 2 O

39. Balance the hydrogens. Do this by adding hydrogen ions (as many as are needed) to the side needing hydrogen. In our example, we need 8 (notice the water molecule's formula, then consider 4 x 2 = 8). 8H + + MnO 4 ¯ ---> Mn 2+ + 4H 2 O Balance the total charge. This will be done using electrons. It is ALWAYS the last step. 5e¯ + 8H + + MnO 4 ¯ ---> Mn 2+ + 4H 2 O

40. Here is a second half-reaction, also in acid solution: Cr 2 O 7 2 ¯ ---> Cr 3+ Balance the atom being reduced/oxidized. Cr 2 O 7 2 ¯ ---> 2Cr 3+ 2. Balance the oxygens . Cr 2 O 7 2 ¯ ---> 2Cr 3+ + 7H 2 O 3. Balance the hydrogens . 14H + + Cr 2 O 7 2 ¯ ---> 2Cr 3+ + 7H 2 O 4. Balance the total charge. 6e¯ + 14H + + Cr 2 O 7 2 ¯ ---> 2Cr 3+ + 7H 2 O

41. Another example (in acid solution) SO 2 ---> SO 4 2 ¯ SO 2 ---> SO 4 2 ¯ (the sulfur is already balanced) 2H 2 O + SO 2 ---> SO 4 2 ¯ (now there are 4 oxygens on each side) 2H 2 O + SO 2 ---> SO 4 2 ¯ + 4H + (2 x 2 from the water makes 4 hydrogens) 2H 2 O + SO 2 ---> SO 4 2 ¯ + 4H + + 2e¯ (zero charge on the left; +4 from the hydrogens and -2 from the sulfate, so 2 electrons gives the -2 charge required to make zero on the right)

42. Practice Problems 1) Re ---> ReO 2 2) Cl 2 ---> HClO 3) NO 3 ¯ ---> HNO 2 4) H 2 GeO 3 ---> Ge 5) H 2 SeO 3 ---> SeO 4 2 ¯ 6) Au ---> Au(OH) 3 (this one is a bit odd!) 7) H 3 AsO 4 ---> AsH 3 8) H 2 MoO 4 ---> Mo 9) NO ---> NO 3 ¯ 10) H 2 O 2 ---> H 2 O

43. Answers : 2H 2 O + Re ---> ReO 2 + 4H + + 4e¯ 2H 2 O + Cl 2 ---> 2HClO + 2H + + 2e¯ 2e¯ + 3H + + NO 3 ¯ ---> HNO 2 + H 2 O 4e¯ + 4H + + H 2 GeO 3 ---> Ge + 3H 2 O H 2 O + H 2 SeO 3 ---> SeO 4 2 ¯ + 4H + + 2e¯ 3H 2 O + Au ---> Au(OH) 3 + 3H + + 3e¯ 8e¯ + 8H + + H 3 AsO 4 ---> AsH 3 + 4H 2 O 6e¯ + 6H + + H 2 MoO 4 ---> Mo + 4H 2 O 2H 2 O + NO ---> NO 3 ¯ + 4H + + 3e¯ 2e¯ + 2H + + H 2 O 2 ---> 2 H 2 O

44. Balancing Half-Reactions in Basic Solution Before looking at the balancing technique, the fact that it is in basic solution can be signaled to you in several different ways: It is explicitly said in the problem. A base (usually a strong base) is included as one of the reactants. An OH¯ is written just above the reaction arrow.

45. There are three other chemical species available in a basic solution besides the ones shown above. They are: H 2 O OH¯ e¯ water is present because the reaction is taking place in solution the hydroxide ion is available because it is in basic solution electrons are available because that's what is transferred in redox reactions. All three will be used in getting the final answer.

46. PbO 2 ---> PbO It is to be balanced in basic solution. Step One to Four: Balance the half-reaction AS IF it were in acid solution. Balance the atom being reduced/oxidized. Balance the oxygens (using H 2 O). Balance the hydrogens (using H + ). Balance the charge. When you do that to the above half-reaction, you get: 2e¯ + 2H + + PbO 2 ---> PbO + H 2 O

47. Step Five: Convert all H + to H 2 O. Do this by adding OH¯ ions to both sides. The side with the H + will determine how many hydroxide to add. In our case, the left side has 2 hydrogen ions, while the right side has none, so: 2e¯ + 2H 2 O + PbO 2 ---> PbO + H 2 O + 2OH¯

48. Notice that, when the two hydroxide ions on the left were added, they immediately reacted with the hydrogen ion present. The reaction is: H + + OH¯ ---> H 2 O

49. Step Six: Remove any duplicate molecules or ions. In our example, there are two water molecules on the left and one on the right. This means one water molecule may be removed from each side, giving: 2e¯ + H 2 O + PbO 2 ---> PbO + 2OH¯ The half-reaction is now correctly balanced.

50. Here is a second half-reaction, also in basic solution: MnO 4 ¯ ---> MnO 2 Step One: Balance the half-reaction AS IF it were in acid solution. 3e¯ + 4H + + MnO 4 ¯ ---> MnO 2 + 2H 2 O Step Two: Convert all H + to H 2 O. 3e¯ + 4H 2 O + MnO 4 ¯ ---> MnO 2 + 2H 2 O + 4OH¯ Step Three: Remove any duplicate molecules or ions. 3e¯ + 2H 2 O + MnO 4 ¯ ---> MnO 2 + 4OH¯

51. Balancing Simple Redox Rxns Must be: Balanced for Mass ATOMS balance Balanced for Charge Total e- Lost = Total e- Gained

52. Balancing Harder Redox Reactions (Honors)

53. Oxidation Number Method (Balancing in Acid Solution) Find ox #’s and use brackets to connect elements changing in charge. Balance atoms changing in charge Find total e- involved in each change If necessary balance e- by multiplication Balance all other atoms except H and O Balance oxygen by adding H 2 O to side deficient Balance hydrogen by adding H +1 to side deficient Check for balance with respect to atoms and charge.

54. Half Reaction Method (Ion/Electron Method) (In acid solution) Separate equation into two “basic” half reactions Balance all atoms except H and O Balance oxygen by adding H 2 O Balance hydrogen by adding H +1 Balance charge by adding electrons to more positive side If necessary balance e- by multiplication Add together half reactions and simplify Check for balance of atoms and charge

55. Applications of Redox Reactions Corrosion of Metals the metal gets oxidized forming metal oxides on the surface Prevention: Use paint, oil, plating or attach to negative terminal of a battery. Gold doesn’t rust…Why?

56. Photograph Development involves oxidation and reduction of silver atoms and ions

57. Bleach acts on stains by oxidizing them, getting reduced in the process Explosives form neutral gases like N 2 from compounds!

58. Reactivity of Metals Reference Table J Metals Higher on Table J are more ‘active” It is easier for more “active” metals to be oxidized or lose electrons.

59. Copper replaces silver! Cu (s) + AgNO 3 (aq) Ag (s) + CuNO 3 (aq) Ag (s) + CuNO 3 (aq) wouldn’t happen!!!

60. Reactivity of Nonmetals Reference Table J Nonmetals higher on Table J are more “active” It is easier for more “active” nonmetals to “gain” electrons and be reduced.

61. Electrochemical Cells (Batteries) Chemical reaction that produces electricity. Called “voltaic cells” as they produce voltage This happens SPONTANEOUSLY.

62. Moving Electrons = Electricity Electrons given off by oxidized substance travel towards substance being reduced. Traveling electrons move through “external circuit” where they do work.

63. How do the Electrons Move? Batteries often contain 2 metals. Start with Table J Electrons travel from the more “Active metal” toward the less active metal. Metal above = oxidized Ion on Metal below = reduced

64. Electrons flow “ Down Table J ” From metal above to ion of metal below e-

65. Parts of a Simple Battery (Voltaic Cell) Made of Two “Half Cells” containing: 2 Metal Electrodes 2 Solutions of Ions External Wire Salt Bridge

66. Electrons need to flow in a “circuit” that is connected. External Wire : allows electrons to flow between metal electrodes Salt Bridge: allows ions to flow between solutions

67. Zn/Zn +2 //Cu +2 /Cu What is Ox/Red? See Table J Metal above is oxidized Zn Ion of metal below reduced Cu +2

68. Which way do electrons flow in the external wire? See Table J Electrons flow “Down” the table from what is oxidized towards what is reduced. from Zn to Cu e-

69. Which electrode is negative? Which electrode is positive? Electrons flow from negative to positive electrode. Negative electrode: Zn Positive electrode: Cu e-

70. Which electrode is the anode and cathode? Anode : metal electrode where oxidation occurs Zn Cathode : metal electrode where reduction occurs Cu

71. Remember AN OX RED CAT Anode is where oxidation happens Cathode is where reduction happens

72. What are the Half Reactions? What is the Net Equation? Ox: Zn Zn +2 + 2e- Red: Cu +2 + 2e- Cu Net: (add ½ reactions) Zn + Cu +2 Zn +2 + Cu Make sure final net equation is balanced for electrons and atoms! e-

73. Which electrode gains/loses weight? Look at half reactions!! Which one forms solid metal? Which one forms dissolved ions? Ox: Zn Zn +2 + 2e- Red: Cu +2 + 2e- Cu Zinc electrode loses mass Copper electrode gains mass

74. Which way to do the ions in the salt bridge “migrate” or move? Remember: “The negative ions complete the circuit” (The ions actually end up moving towards the solution of opposite charge that forms.)

75. Follow the ions

76. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

77. Dead Battery Voltage = 0 Means the reaction in the battery has reached EQUILIBRIUM.

78. You try it… Mg/Mg +2 //Al +3 /Al Draw and label Battery What is oxidized/reduced? What are the half reactions and net(balanced)? What is the neg/pos electrode? What is the anode/cathode? Which way do e- flow in wire? Which way do -/+ ions flow in salt bridge? Which electrode gains/loses mass?

79. Finding Voltage of a Battery (Honors) Use Voltage Table Find your half reactions and record voltage Note: All ½ reactions shown are reductions. For oxidation, reverse the sign of the voltage

80. Nerntz Equation (Honors) Find voltage of a battery when the conc. of dissolved ions is not 1 Molar (as on “standard voltage” table) E cell = E – 0.0592 log [product ion] x n [reactant ion] y n = total # of moles electrons being transferred The concentration of dissolved ions can affect voltage. Greater concentration of reactant ions (see net) increases the overall voltage.

81. Electrolytic Cells & Electrolysis Reactions Uses electricity to “split” or “lyse” a compound into it’s neutral elements An outside electrical source provides electrons to force a non-spontaneous redox reaction to occur.

82. Electrolysis Set Up Single Cell filled with electrolyte with +/- ions Attach battery to two electrodes. Electrodes are made of an inert substance (like platinum or graphite) that conducts. Electrodes don’t chemically change like in a battery, they just provide current

83. Pulls electrons off one electrode Making it POSITIVE Adds electrons onto one electrode Making it NEGATIVE Role of the Battery

84. Which Way do the Ions Move? To electrode of opposite charge

85. At neg. electrode electrons are gained by ion (reduction at CATHODE) At positive electrode electrons are lost by ion (oxidation at ANODE) What is Oxidized/Reduced?

86. Remember AN OX RED CAT Anode is where oxidation happens Cathode is where reduction happens

87. Half Reactions & Net Equation Rxn at Anode: (Ox) Cl - Cl + 1e- Or more correctly DIATOMIC!!!!!! 2Cl - Cl 2 + 2e- Rxn at Cathode: (Red) Na + + 1e- Na (Multiply by 2 to balance electrons) NET: 2Na + + 2Cl - 2Na + Cl 2

88. Determining Voltage Needed Use the Voltage Table to determine the total voltage needed to run the Electrolytic cell. Total voltage should be a NEGATIVE number

89. Electrolysis of Molten NaCl (l)

90. Electrolysis of PbCl 2 (l) What is oxidized? What is reduced? What are the ox/red half reactions? What is the net equation? Negative Electrode Positive Electrode

91. Electrolysis of PbCl 2 (l) Oxidized: Cl - Reduced: Pb +2 Half Reactions Ox: Cl -1 Cl + 1e- 2Cl -1 Cl 2 + 2e- Red: Pb +2 + 2e- Pb Net: Pb +2 + 2Cl -1 Pb + Cl 2

92. Electrolysis of NaCl(aq)

93. Electrolysis of NaCl (aq)

94. Electrolysis of Water At Positive Electrode: Ox: O -2 O + 2e- but there is a diatomic! 2O -2 O 2 + 4e- At Negative Electrode Red: H +1 + 1e- H but there is a diatomic! 2H +1 + 2e- H 2 Net: 2H 2 O 2H 2 + O 2

95. Lemon Battery Demo http://youtu.be/AY9qcDCFeVI Electrolysis of Copper Sulfate http://youtu.be/xBz9HJ32Ouw Electrolysis of Water/ Silver Nitrate and Cu reaction http://youtu.be/Bcfp8VtcrSA Electrolysis of Water (Animation) http://youtu.be/2t13S-KpGeE Electrolysis of Water (Simple) http://youtu.be/HQ9Fhd7P_HA

96. Electroplating Electrolysis reaction used to coat a substance with a thin layer of metal. Often coating is a less reactive metal that is not easily oxidized or corroded.

97. Electroplating Negative Electrode Is the OBJECT TO BE PLATED so the positive metal ions would go towards it and be REDUCED. It is the CATHODE Red: Ag + + 1e- Ag

98. Electroplating Positive Electrode Made of plating metal It dissolves into solution as metal strip gets OXIDIZED. It is the ANODE This replenishes the ions for plating. Ox: Ag Ag+ + 1e-

99. Electroplating Problems (Honors) Coulomb = measure of electrical charge 1 mole e- = 96,500 coulombs # coulombs = # amps x seconds

100. Electroplating Problems (Honors) Reduction: Happens on object to be plated Look at Reduction half reaction Look at mole relationships between electrons and metal atoms. Ex: Ag + + 1e- Ag

101. Electroplating Problems (Honors) You can now answer questions regarding the amount of a substance in moles or grams that can be electroplated over a certain amount of time.

102. Electroplating Problems (Honors) If 10 amps are run through a CuSO 4 solution for 5 minutes, calculate the grams of Cu that will plate onto the spoon. We Know: 1 mole e- = 96,500 coulombs # coulombs = # amps x seconds Red: Cu +2 + 2e- Cu 1 mole Cu = 63.5 grams

103. So….Let’s start here # coulombs = 10 amps x 300 seconds = 3000 coulombs 3000 coul. x 1 mole e- x 1 mole Cu x 63.5g Cu = .987 grams 96,500 coul 2 mole e- 1 mole Cu Mole ratio from Reduction half reaction

104. You Try One How long will it take to deposit 20 grams of silver from a solution of AgCl onto a copper tray if a current of 5 amps is used? Answer = 3, 574 sec or 59.5 minutes or about 1 hour

105. You Try One How many amps are needed to deposit .504g. of Iron in 40 minutes by passing a current through a solution of Iron II Sulfate? Answer: .72 amps