3. Electrochemistry Terminology  Oxidation  Oxidation – A process in which an element attains a more positive oxidation state Na(s)  Na + + e -  Reduction  Reduction – A process in which an element attains a more negative oxidation state Cl 2 + 2e -  2Cl -  Oxidizing agent  Oxidizing agent -The substance that is reduced is the oxidizing agent  Reducing agent  Reducing agent - The substance that is oxidized is the reducing agent LEO says GER

4. Electrochemistry Terminology Anode -  Anode -The electrode where oxidation occurs Cathode -  Cathode - The electrode where reduction occurs Reduction at the Cathode Leo is a

7. Step 4:Multiply each half-reaction by a factor so that the reducing agent supplies as many electrons as the oxidizing agent requires. Reducing agent Cu ---> Cu 2+ + 2e- Oxidizing agent 2 Ag + + 2 e- ---> 2 Ag Step 5:Add half-reactions to give the overall equation. Cu + 2 Ag + ---> Cu 2+ + 2Ag The equation is now balanced for both charge and mass.

8. Balance the following in acid solution— VO 2 + + Zn ---> VO 2+ + Zn 2+ VO 2 + + Zn ---> VO 2+ + Zn 2+ Step 1:Write the half-reactions OxZn ---> Zn 2+ RedVO 2 + ---> VO 2+ Step 2:Balance each half-reaction for mass. OxZn ---> Zn 2+ Red VO 2 + ---> VO 2+ + H 2 O 2 H + + Add H 2 O on O-deficient side and add H + on other side for H-balance.

9. Step 3:Balance half-reactions for charge. OxZn ---> Zn 2+ + 2e- Rede- + 2 H + + VO 2 + --> VO 2+ + H 2 O Step 4:Multiply by an appropriate factor. OxZn ---> Zn 2+ + 2e- Red 2e- + 4 H + + 2 VO 2 + ---> 2 VO 2+ + 2 H 2 O Step 5:Add balanced half-reactions Zn + 4 H + + 2 VO 2 + ---> Zn 2+ + 2 VO 2+ + 2 H 2 O Draw it!

10. A great example of a thermodynamically spontaneous reaction is the thermite reaction. Here, iron oxide (Fe 2 O 3 = rust) and aluminum metal powder undergo a redox (reduction-oxidation) reaction to form iron metal and aluminum oxide (Al 2 O 3 = alumina):reduction-oxidation Fe 2 O 3 (s) + Al(s) ↔ Al 2 O 3 (s) + Fe(l) Al = 0 Fe = +3 Fe = 0 Al = +3

12. 1. 12 2. 5 3. 2 4. 30 5. 10 15 Response Counter

13. 1. Zn → Zn 2+ + 2e – 2. Zn 2+ + 2e – → Zn 3. Zn 2+ + Cu → Zn + Cu 2+ 4. Zn + Cu 2+ → Zn 2+ + Cu 5. two of these 30 Response Counter

14. 1. S(s) 2. NO 3 – (ag) 3. Cr 2 O 7 2– (aq) 4. I – (aq) 5. MnO 4 – (aq) 15 Response Counter

15.  An apparatus that allows a redox reaction to occur by transferring electrons through an external connector.  Product favored reaction --- > voltaic or galvanic cell ---- > electric current  Reactant favored reaction --- > electrolytic cell ---> electric current used to cause chemical change. Batteries are voltaic cells

16. AnodeCathode Basic Concepts of Electrochemical Cells

19. Zn(s) ---> Zn 2+ (aq) + 2e- Cu 2+ (aq) + 2e- ---> Cu(s) -------------------------------------------- Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) Zn(s) ---> Zn 2+ (aq) + 2e- Cu 2+ (aq) + 2e- ---> Cu(s) -------------------------------------------- Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s)

20. Measuring Standard Electrode Potential Potentials are measured against a hydrogen ion reduction reaction, which is arbitrarily assigned a potential of zero volts.

21. Table of Reduction Potentials Measured against the Standard Hydrogen Electrode

22. 2 E o (V) Cu 2+ + 2e- Cu+0.34 2 H + + 2e- H0.00 Zn 2+ + 2e- Zn-0.76 oxidizing ability of ion reducing ability of element To determine an oxidation from a reduction table, just take the opposite sign of the reduction!

23. Zn(s) ---> Zn 2+ (aq) + 2e-E o = +0.76 V Cu 2+ (aq) + 2e- ---> Cu(s)E o = +0.34 V --------------------------------------------------------------- Cu 2+ (aq) + Zn(s) ---> Zn 2+ (aq) + Cu(s) E o = +1.10 V Cathode, positive, sink for electrons Anode, negative, source of electrons +

25. All ingredients are present. Which way does reaction proceed?

27. e-

28. Assume I - ion can reduce water. 2 H 2 O + 2e- ---> H 2 + 2 OH - 2 I - ---> I 2 + 2e- ------------------------------------------------- 2 I - + 2 H 2 O --> I 2 + 2 OH - + H 2 2 H 2 O + 2e- ---> H 2 + 2 OH - 2 I - ---> I 2 + 2e- ------------------------------------------------- 2 I - + 2 H 2 O --> I 2 + 2 OH - + H 2 Cathode Anode

29. Galvanic (Electrochemical) Cells Spontaneous redox processes have: A positive cell potential, E 0 A negative free energy change, (-  G)

30. Zn - Cu Galvanic Cell Zn 2+ + 2e -  Zn E = -0.76V Cu 2+ + 2e -  Cu E = +0.34V From a table of reduction potentials:

31. Zn - Cu Galvanic Cell Cu 2+ + 2e -  Cu E = +0.34V The less positive, or more negative reduction potential becomes the oxidation… Zn  Zn 2+ + 2e - E = +0.76V Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V

32. Line Notation Zn(s) | Zn 2+ (aq) (1.0M) || Cu 2+ (aq) (1.0M) | Cu(s) An abbreviated representation of an electrochemical cell Anodesolution Anodematerial Cathodesolution Cathodematerial |||| Line notation is cool, just like AC

33. Zn (s) | Zn 2+ (aq) (1.0M) || H + (aq) (1.0M) | H 2(g) (1.00 atm) | Pt (s)

35. Calculating  G 0 for a Cell  G 0 = -nFE 0 n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e - Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V

39. Day 3 (dahditdadahditahh…Charge!)

40. The Nernst Equation Standard potentials assume a concentration of 1.0 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M. R = 8.31 J/(mol  K) T = Temperature in K n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e -

41. Nernst Equation Simplified At 25  C (298 K) the Nernst Equation is simplified this way:

42. Try Problem 62

43. Equilibrium Constants and Cell Potential At equilibrium, forward and reverse reactions occur at equal rates, therefore: 1. 1. The battery is “dead” 2. The cell potential, E, is zero volts Modifying the Nernst Equation (at 25  C):

44. Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V Calculating an Equilibrium Constant from a Cell Potential

45. Let’s Do Problems: 26a, 28a, 32a, 38a 26a, 28a, 32a, 38a

46. Apply to Problem 70

49. Concentration Cell Step 1: Determine which side undergoes oxidation, and which side undergoes reduction. Both sides have the same components but at different concentrations. ???

50. Concentration Cell Both sides have the same components but at different concentrations. The 1.0 M Zn 2+ must decrease in concentration, and the 0.10 M Zn 2+ must increase in concentration Zn 2+ (1.0M) + 2e -  Zn (reduction) Zn  Zn 2+ (0.10M) + 2e - (oxidation) ??? Cathode Anode Zn 2+ (1.0M)  Zn 2+ (0.10M)

51. Concentration Cell Step 2: Calculate cell potential using the Nernst Equation (assuming 25  C). Both sides have the same components but at different concentrations. ??? Cathode Anode Zn 2+ (1.0M)  Zn 2+ (0.10M) Concentration Cell

52. Nernst Calculations Zn 2+ (1.0M)  Zn 2+ (0.10M)

56. Nearly same reactions as in common dry cell, but under basic conditions. Anode (-): Zn + 2 OH - ---> ZnO + H 2 O + 2e- Cathode (+): 2 MnO 2 + H 2 O + 2e- ---> Mn 2 O 3 + 2 OH -

59. Anode (-) Cd + 2 OH - ---> Cd(OH) 2 + 2e- Cathode (+) NiO(OH) + H 2 O + e- ---> Ni(OH) 2 + OH -

60. The positive electrode is made of Lithium cobalt oxide, or LiCoO 2 The negative electrode is made of carbon.

62. The Electrochemical Corrosion of Iron

63. Electrolytic Processes A negative cell potential, (-E 0 ) A positive free energy change, (+  G) Electrolytic processes are NOT spontaneous. They have:

64. Electrolysis of Water In acidic solution Anode rxn: Cathode rxn: -1.23 V -0.83 V -2.06 V

65. Electroplating of Silver Anode reaction: Ag  Ag + + e - Electroplating requirements: 1. Solution of the plating metal 3. Cathode with the object to be plated 2. Anode made of the plating metal 4. Source of current Cathode reaction: Ag + + e -  Ag

66. Step 1 – convert current and time to quantity of charge in coulombs a. Amps x time = total charge transferred in coulombs (Coulomb/sec) x sec = coulombs Step 2 – convert quantity of charge in coulombs to moles of electrons coulombs /(96,485 coulombs/mol e-) = mol e- Step 3 – Convert moles of electrons to moles of substance mol e- x (mole substance/mol e-) = mol substance Step 4 – Convert moles of substance to grams of substance mol substance x formula mass of substance = mass of substance Calculating plating

67. 1 mol Pb Suppose that in starting a car on a cold morning a current of 125 amperes is drawn for 15.0 seconds from a cell of the type described above. How many grams of Pb would be consumed? (The atomic weight of Pb is 207.19.) 125 C 15 sec1 mol e - 1 mol Pb 96 485 C 2 mol e - 207.19 g 1 sec Pb 2+ + 2e -  Pb 2.01 g Pb

68. Solving an Electroplating Problem Q: How many seconds will it take to plate out 5.0 grams of silver from a solution of AgNO 3 using a 20.0 Ampere current? 5.0 g Ag + + e -  Ag 1 mol Ag 107.87 g 1 mol e - 1 mol Ag 96 485 C 1 mol e - 1 s 20.0 C = 2.2 x 10 2 s

69. Do 79!!!

70. Do 81!!!