2. Electrochemistry Chapter 19

3. Electron Transfer Reactions Electron transfer reactions are oxidation- reduction or redox reactions. Results in the generation of an electric current (electricity) or be caused by imposing an electric current. Therefore, this field of chemistry is often called ELECTROCHEMISTRY.

4. 2Mg (s) + O 2 (g) 2MgO (s) 2Mg 2Mg 2+ + 4e - O 2 + 4e - 2O 2- Oxidation half-reaction (lose e - ) Reduction half-reaction (gain e - ) 19.1 Electrochemical processes are oxidation-reduction reactions in which: the energy released by a spontaneous reaction is converted to electricity or electrical energy is used to cause a nonspontaneous reaction to occur 002+2-

5. Terminology for Redox Reactions OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen.OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen. REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen.REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen. OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel)OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel) REDUCING AGENT—electron donor; species is oxidized.REDUCING AGENT—electron donor; species is oxidized. OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen.OXIDATION—loss of electron(s) by a species; increase in oxidation number; increase in oxygen. REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen.REDUCTION—gain of electron(s); decrease in oxidation number; decrease in oxygen; increase in hydrogen. OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel)OXIDIZING AGENT—electron acceptor; species is reduced. (an agent facilitates something; ex. Travel agents don’t travel, they facilitate travel) REDUCING AGENT—electron donor; species is oxidized.REDUCING AGENT—electron donor; species is oxidized.

6. You can’t have one… without the other! Reduction (gaining electrons) can’t happen without an oxidation to provide the electrons. You can’t have 2 oxidations only or 2 reductions only in the same equation. Reduction has to occur at the cost of oxidation LEO the lion says GER! o s e l e c t r o n s x i d a t i o n a i n l e c t r o n s e d u c t i o n GER!

7. Another way to remember OIL RIG x i d a t i o n s o s e e d u c t i o n s a i n

8. Review of Oxidation numbers The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1.Free elements (uncombined state) have an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 2.In monatomic ions, the oxidation number is equal to the charge on the ion. Li +, Li = +1; Fe 3+, Fe = +3; O 2-, O = -2 3.The oxidation number of oxygen is usually –2. In H 2 O 2 and O 2 2- it is –1. 4.4

9. 4.The oxidation number of hydrogen is +1 except when it is bonded to metals in binary compounds. In these cases, its oxidation number is –1. 6. The sum of the oxidation numbers of all the atoms in a molecule or ion is equal to the charge on the molecule or ion. 5.Group IA metals are +1, IIA metals are +2 and fluorine is always –1. HCO 3 - O = -2H = +1 3x(-2) + 1 + ? = -1 C = +4 Oxidation numbers of all the atoms in HCO 3 - ? 4.4

10. Balancing Redox Equations 19.1 1.Write the unbalanced equation for the reaction in ionic form. The oxidation of Fe 2+ to Fe 3+ by Cr 2 O 7 2- in acid solution? Fe 2+ + Cr 2 O 7 2- Fe 3+ + Cr 3+ 2.Separate the equation into two half-reactions. Oxidation: Cr 2 O 7 2- Cr 3+ +6+3 Reduction: Fe 2+ Fe 3+ +2+3 3.Balance the atoms other than O and H in each half-reaction. Cr 2 O 7 2- 2Cr 3+

11. Balancing Redox Equations 4.For reactions in acid, add H 2 O to balance O atoms and H + to balance H atoms. Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 5.Add electrons to one side of each half-reaction to balance the charges on the half-reaction. Fe 2+ Fe 3+ + 1e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 6.If necessary, equalize the number of electrons in the two half- reactions by multiplying the half-reactions by appropriate coefficients. 6Fe 2+ 6Fe 3+ + 6e - 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 19.1

12. Balancing Redox Equations 7.Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. You should also cancel like species. 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O 6Fe 2+ 6Fe 3+ + 6e - Oxidation: Reduction: 14H + + Cr 2 O 7 2- + 6Fe 2+ 6Fe 3+ + 2Cr 3+ + 7H 2 O 8.Verify that the number of atoms and the charges are balanced. 14x1 – 2 + 6x2 = 24 = 6x3 + 2x3 19.1 9.For reactions in basic solutions, add OH - to both sides of the equation for every H + that appears in the final equation. You should combine H + and OH - to make H 2 O.

13. To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire.To obtain a useful current, we separate the oxidizing and reducing agents so that electron transfer occurs thru an external wire. CHEMICAL CHANGE ---> ELECTRIC CURRENT This is accomplished in a GALVANIC, VOLTAIC, or an ELECTROCHEMICAL cell. A group of such cells is called a battery. http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/galvan5.swf

14. Galvanic Cells 19.2 spontaneous redox reaction anode oxidation cathode reduction - +

16. The potential difference between the two electrodes of a voltaic cell provides the driving force that pushes electrons through the external circuit.This potential difference is callled, cell voltage electromotive (causing electron motion) force (emf) cell potential It’s denoted as E cell and is measured in Volts (V).

17. For any cell rxn happening spontaneously, the cell potential will be positive.

18. Galvanic Cells 19.2 Line notation: Zn (s) + Cu 2+ (aq) Cu (s) + Zn 2+ (aq) [Cu 2+ ] = 1 M & [Zn 2+ ] = 1 M Zn (s) | Zn 2+ (1 M) || Cu 2+ (1 M) | Cu (s) anodecathode

19. A common dry cell battery (for watches, calculators etc.)

20. NiO 2, H 2 O, OH - are reacting materials. Anode: Cd, Cathode:Ni

21. Mercury battery, used in calculators

22. Standard Electrode Potentials Standard reduction potential (E 0 red ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M, 25°C and all gases are at 1 atm. E 0 = 0 V Standard hydrogen electrode (SHE) is a reference cell to determine E 0 red of the half cells. 2e - + 2H + (1 M) H 2 (1 atm) Reduction Reaction

23. The superscript ° indicates standard state conditions.

24. Whenever we assign a potential to a half-reaction, we write the reaction as a reduction.

25. Standard Electrode Potentials 19.3 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) 2e - + 2H + (1 M) H 2 (1 atm) Zn (s) Zn 2+ (1 M) + 2e - Anode (oxidation): Cathode (reduction): Zn (s) + 2H + (1 M) Zn 2+ + H 2 (1 atm)

26. 19.3 E 0 is for the reaction as written The more positive E 0 red, the greater the driving force for reduction. The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0

27. Official AP Reduction Table Copyright College Board

28. 19.3 E 0 = 0 + 0.76 V = 0.76 V cell Standard emf (E 0 ) cell Zn 2+ (1 M) + 2e - Zn E 0 = -0.76 V E 0 = E H /H + E Zn /Zn cell 00 + 2 Standard Electrode Potentials E 0 = E cathode + E anode cell 00 Zn (s) | Zn 2+ (1 M) || H + (1 M) | H 2 (1 atm) | Pt (s) If the reaction is backwards, be sure to flip the sign! +2 So E o Zn/Zn = + 0.76 V +2

29. Standard Electrode Potentials 19.3 Pt (s) | H 2 (1 atm) | H + (1 M) || Cu 2+ (1 M) | Cu (s) 2e - + Cu 2+ (1 M) Cu (s) H 2 (1 atm) 2H + (1 M) + 2e - Anode (oxidation): Cathode (reduction): H 2 (1 atm) + Cu 2+ (1 M) Cu (s) + 2H + (1 M) E 0 = E cathode + E anode cell 00 E 0 = 0.34 V cell E cell = E Cu /Cu + E H /H+ 2+ 2 00 0 0.34 = E Cu /Cu + - 0 0 2+ E Cu /Cu = 0.34 V 2+ 0

30. In a voltaic cell,the cathode reaction is always the one that has the more positive value for E 0 red since the greater driving force of the cathode half rxn will force the anode rxn to occur “in reverse,” as an oxidation.

31. What is the standard emf of an electrochemical cell made of a Cd electrode in a 1.0 M Cd(NO 3 ) 2 solution and a Cr electrode in a 1.0 M Cr(NO 3 ) 3 solution? Cd 2+ (aq) + 2e - Cd (s) E 0 = -0.40 V Cr 3+ (aq) + 3e - Cr (s) E 0 = -0.74 V Cd is the stronger oxidizer Cd will oxidize Cr 2e - + Cd 2+ (1 M) Cd (s) Cr (s) Cr 3+ (1 M) + 3e - Anode (oxidation): Cathode (reduction): 2Cr (s) + 3Cd 2+ (1 M) 3Cd (s) + 2Cr 3+ (1 M) x 2 x 3 E 0 = E cathode + E anode cell 00 E 0 = -0.40 + (+0.74) cell E 0 = 0.34 V cell 19.3

32. 19.4 Spontaneity of Redox Reactions  G = -nFE cell  G 0 = -nFE cell 0 n = number of moles of electrons in reaction F = 96,500 J V mol = 96,500 C/mol  G 0 = -RT ln K = -nFE cell 0 E cell 0 = RT nF ln K (8.314 J/K mol)(298 K) n (96,500 J/V mol) ln K = = 0.0257 V n ln K E cell 0 = 0.0592 V n log K E cell 0

33. Spontaneity of Redox Reactions 19.4

34. 2e - + Fe 2+ Fe 2Ag 2Ag + + 2e - Oxidation: Reduction: What is the equilibrium constant for the following reaction at 25 0 C? Fe 2+ (aq) + 2Ag (s) Fe (s) + 2Ag + (aq) = 0.0257 V n ln K E cell 0 19.4 E 0 = -0.44 + -0.80 E 0 = -1.24 V 0.0257 V x nE0E0 cell exp K = n = 2 0.0257 V x 2x 2-1.24 V = exp K = 1.23 x 10 -42 E 0 = E Fe /Fe + E Ag /Ag 00 2++

35. Factors affecting the cell potential: Specific rxns that occur at the cathode and anode Concentrations of reactants and products temperature

36. Effect of concentration on cell EMF As a voltaic cell discharges, concentrations of reactants and products change. The emf drops until E=0, at which the cell is “dead”, the reactants and products are at equilibrium.

37. How will you calculate the emf of a voltaic cell generated at non- standard conditions? Use the Nernst equation (Walther Nernst, 1864-1941, a German chemist)

38. The Effect of Concentration on Cell Emf  G =  G 0 + RT ln Q  G = -nFE  G 0 = -nFE 0 -nFE = -nFE 0 + RT ln Q E = E 0 - ln Q RT nF Nernst equation At 298 19.5 - 0.0257 V n ln Q E 0 E = - 0.0592 V n log Q E 0 E =

39. n: the number of electrons transferred in the rxn F: Faraday’s constant (96,500 J/V-mol) Q: Rxn quotient, determined in the same w/ Kc except that the concentrations used in Q are those that exist in the rxn mixture at a given moment. E: non-standard voltaic cell potential

40. Will the following reaction occur spontaneously at 25 0 C if [Fe 2+ ] = 0.60 M and [Cd 2+ ] = 0.010 M? Fe 2+ (aq) + Cd (s) Fe (s) + Cd 2+ (aq) 2e - + Fe 2+ 2Fe Cd Cd 2+ + 2e - Oxidation: Reduction: n = 2 E 0 = -0.44 + -(-0.40) E 0 = -0.04 V E 0 = E Fe /Fe + E Cd /Cd 00 2+ - 0.0257 V n ln Q E 0 E = - 0.0257 V 2 ln -0.04 VE = 0.010 0.60 E = 0.013 E > 0Spontaneous 19.5

41. Calculate the emf at 298 K generated by the cell containing K 2 Cr 2 O 7 and H 2 SO 4 solutions in one compartment and a solution of KI in another compartment(a metallic conductor that will not react w/ either solution is suspended in each solution) when [Cr 2 O 7 2- ] = 2.0 M, [H + ] = 1.0 M, [I - ] = 1.0M, and [Cr 3+ ] = 1.0x10 -5 M. Cr 2 O 7 2− + 14H + + 6I −  2Cr 3+ + 7H 2 O + 3I 2

42. Cr 2 O 7 2− + 14H + + 6e −  2Cr 3+ + 7H 2 O E° = 1.33V I 2 + 2e −  2I − E° = 0.54V Answer:.89V

43. CONCENTRATION CELLS

44. The cell operates until the concentrations of reacting ions in two compartments become equal, at which point the cell has reached equilibrium and is “dead.”

45. Anode: Cu  Cu 2+ (.01M) + 2e - Cathode: Cu 2+ (1M) + 2e -  Cu Overall: Cu 2+ (1M)  Cu 2+ (.01M) At 298K; E= 0-.0592/2 log.01/1 =.0592V

46. How can you increase the cell potential of a concentration cell? By making the difference between concentrations of the ions higher.

47. 2) Temperature Temp ↑, voltaic cell potential ↓

48. Charging a Battery When you charge a battery, you are forcing the electrons backwards (from the + to the -). To do this, you will need a higher voltage backwards than forwards. This is why the ammeter in your car often goes slightly higher while your battery is charging, and then returns to normal. In your car, the battery charger is called an alternator. If you have a dead battery, it could be the battery needs to be replaced OR the alternator is not charging the battery properly.

49. Batteries 19.6 Leclanché cell Dry cell Zn (s) Zn 2+ (aq) + 2e - Anode: Cathode: 2NH 4 (aq) + 2MnO 2 (s) + 2e - Mn 2 O 3 (s) + 2NH 3 (aq) + H 2 O (l) + Zn (s) + 2NH 4 (aq) + 2MnO 2 (s) Zn 2+ (aq) + 2NH 3 (aq) + H 2 O (l) + Mn 2 O 3 (s)

50. Batteries Zn(Hg) + 2OH - (aq) ZnO (s) + H 2 O (l) + 2e - Anode: Cathode: HgO (s) + H 2 O (l) + 2e - Hg (l) + 2OH - (aq) Zn(Hg) + HgO (s) ZnO (s) + Hg (l) Mercury Battery 19.6

51. Batteries 19.6 Anode: Cathode: Lead storage battery PbO 2 (s) + 4H + (aq) + SO 2- (aq) + 2e - PbSO 4 (s) + 2H 2 O (l) 4 Pb (s) + SO 2- (aq) PbSO 4 (s) + 2e - 4 Pb (s) + PbO 2 (s) + 4H + (aq) + 2SO 2- (aq) 2PbSO 4 (s) + 2H 2 O (l) 4

52. Batteries 19.6 Solid State Lithium Battery

53. Batteries 19.6 A fuel cell is an electrochemical cell that requires a continuous supply of reactants to keep functioning Anode: Cathode: O 2 (g) + 2H 2 O (l) + 4e - 4OH - (aq) 2H 2 (g) + 4OH - (aq) 4H 2 O (l) + 4e - 2H 2 (g) + O 2 (g) 2H 2 O (l)

54. Corrosion 19.7

55. Cathodic Protection of an Iron Storage Tank 19.7

56. 19.8 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

57. Electrolysis of Water 19.8

58. Chemistry In Action: Dental Filling Discomfort Hg 2 /Ag 2 Hg 3 0.85 V 2+ Sn /Ag 3 Sn -0.05 V 2+ Sn /Ag 3 Sn -0.05 V 2+

59. Electrolysis and Mass Changes charge (Coulombs) = current (Amperes) x time (sec) 1 mole e - = 96,500 C = 1 Faraday 19.8 1 amp = 1 Coulomb / sec

60. How much Ca will be produced in an electrolytic cell of molten CaCl 2 if a current of 0.452 A is passed through the cell for 1.5 hours? Anode: Cathode: Ca 2+ (l) + 2e - Ca (s) 2Cl - (l) Cl 2 (g) + 2e - Ca 2+ (l) + 2Cl - (l) Ca (s) + Cl 2 (g) 2 mole e - = 1 mole Ca mol Ca = 0.452 C s x 1.5 hr x 3600 s hr96,500 C 1 mol e - x 2 mol e - 1 mol Ca x = 0.0126 mol Ca = 0.50 g Ca 19.8