1. Electrochemistry Chapter 19 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. 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 Redox = Reduction - Oxidation Redox Reaction - A reaction where one species gains electrons and one species loses electrons. Must have both processes occurring. Remember Bronsted acid-base reactions where protions are transferred. One has to give, the other has to take.

3. 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 - ) 002+2- Reduction Involves Gain of electrons. (oxidation number  ) Oxidation Involves Loss of electrons. (oxidation number  )  OIL RIG Oxidizing agent - reactant of reduction O 0 -2 O 2- Reducing agent - reactant of oxidation Mg 0 +2 Mg 2+

4. Oxidation number The charge the atom would have in a molecule (or an ionic compound) if electrons were completely transferred. 1.Free elements (uncombined state) and purely covalent compounds all have atoms with an oxidation number of zero. Na, Be, K, Pb, H 2, O 2, P 4 = 0 Number assigned to each atom to estimate its excess charge or deficiency. There is ALWAYS a change in oxidation numbers in a redox reaction. Rules for Determination of Oxidation Number

5. 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. 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. 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.

6. HCO 3 - O = -2H = +1 3x(-2) + 1 + ? = -1 C = +4 What are the oxidation numbers of all the atoms in HCO 3 - ? 4.4 From rules we know: Can calculate C:

7. What are the oxidation numbers of Mn in the following compounds and ions ? Mn - element  oxidation # = 0 metal colorless Mn 2+ - ion  oxidation # = +2 colorless MnO 2 - compound  oxidation # = +4 brown +4 + (2x(-2)) = 0 MnO 4 - - ion  oxidation # = +7 purple +7 + (4x(-2)) = -1 MnO 4 -2 - ion  oxidation # = +6 green +6 + (4x(-2)) = -2  Mn can have at least 5 oxidation states.

8. Balancing Redox Reactions Reaction can be balanced in acidic, neutral or basic solutions. MnO 4 - + HSO 3 - Mn 2+ + SO 4 -2 MnO 4 - + HSO 3 - MnO 2 + SO 4 -2 MnO 4 - + HSO 3 - MnO 4 2- + SO 4 -2 acid neutral base Balance mass. Balance charge. Balance number of electrons exchanged (treat electrons as a reactant).

9. Balancing Redox Equations 1.Write the unbalanced equation for the reaction ion 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 Oxidation and reduction must take place simultaneously, but we can consider them separately for simplicity. Can look in tables of standard reduction potentials for half reaction or build the half reaction ourselves.

10. 6e - + 14H + + Cr 2 O 7 2- 2Cr 3+ + 7H 2 O Fe 2+ Fe 3+ + 1e - NOTE: We reversed the direction of the oxidation reaction (table is of reduction potentials.).

11. 19.1 3.Balance the atoms other than O and H in each half-reaction. Cr 2 O 7 2- 2Cr 3+ 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

12. Balancing Redox Equations 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 - 7.Add the two half-reactions together and balance the final equation by inspection. The number of electrons on both sides must cancel. 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 Overall reaction should be balanced if half- reactions are properly balanced.

13. Balancing Redox Equations 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. 14H + + Cr 2 O 7 2- + 6Fe 2+ 6Fe 3+ + 2Cr 3+ + 7H 2 O

14. Balancing Redox Reactions Balance reaction in acidic solution. MnO 4 - + HSO 3 - Mn 2+ + SO 4 -2 MnO 4 - Mn 2+ HSO 3 - SO 4 -2 Reduction: add H 2 O to balance O atoms. MnO 4 - Mn 2+ + 4 H 2 O add H + to balance H atoms. MnO 4 - + 8 H + Mn 2+ + 4 H 2 O add electrons to balance charge. MnO 4 - + 8 H + + 5 e - Mn 2+ + 4 H 2 O -1 + 8  +2 + 0 need 5 e - ’s Oxidation: add H 2 O to balance O atoms. add H + to balance H atoms. add electrons to balance charge. HSO 3 - + H 2 O SO 4 -2 HSO 3 - + H 2 O SO 4 -2 + 3 H + HSO 3 - + H 2 O SO 4 -2 + 3 H + + 2 e - -1 + 0  -2 + 3 need 2 e - ’s

15. equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 2x(MnO 4 - + 8 H + + 5 e - Mn 2+ + 4 H 2 O) 5x(HSO 3 - + H 2 O SO 4 -2 + 3 H + + 2 e - ) 2 MnO 4 - + 16 H + + 10 e - + 5 HSO 3 - + 5 H 2 O 2 Mn 2+ + 8 H 2 O + 5 SO 4 -2 + 15 H + + 10 e - 3 1 2 MnO 4 - + H + + 5 HSO 3 - 2 Mn 2+ + 3 H 2 O + 5 SO 4 -2 Check balance: H 1 + 5 = 3 x 2 Mn 2 = 2 O 8 + 15 = 3 + 20 S 5 = 5 Charge - 2 + 1 - 5 = 4 - 10 -6 = -6

16. Balance reaction in neutral solution. MnO 4 - MnO 2 HSO 3 - SO 4 -2 Reduction: Choose half-reaction with neutral product from table. Oxidation: Use same half-reaction as before. 3x(HSO 3 - + H 2 O SO 4 -2 + 3 H + + 2 e - ) MnO 4 - + HSO 3 - MnO 2 + SO 4 -2 MnO 4 - + 4 H + + 3 e - MnO 2 + 2 H 2 O equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 2x(MnO 4 - + 4 H + + 3 e - MnO 2 + 2 H 2 O) 2 MnO 4 - + 8 H + + 6 e - + 3 HSO 3 - + 3 H 2 O 2 MnO 2 + 4 H 2 O + 3 SO 4 -2 + 9 H + + 6 e - 11 HSO 3 - + H 2 O SO 4 -2 + 3 H + + 2 e -

17. Check balance: H 3 = 2 + 1 Mn 2 = 2 O 8 + 9 = 4 + 1 + 12 S 3 = 3 Charge -2 - 3 = 0 + 0 - 6 + 1 -5 = -5 2 MnO 4 - + 3 HSO 3 - 2 MnO 2 + H 2 O + 3 SO 4 -2 + H +

18. Balance reaction in basic solution. MnO 4 - MnO 4 2- HSO 3 - SO 4 -2 Reduction: Oxidation: Use same half-reaction as before. 1x(HSO 3 - + H 2 O SO 4 -2 + 3 H + + 2 e - ) MnO 4 - + e - MnO 4 2- equalize the number of electrons in the two half-reactions by multiplying the half-reactions by appropriate coefficients. 2x(MnO 4 - + e - MnO 4 2- ) 2 MnO 4 - + 2 e - + HSO 3 - + H 2 O 2 MnO 4 2- + SO 4 -2 + 3 H + + + 2 e - HSO 3 - + H 2 O SO 4 -2 + 3 H + + 2 e - MnO 4 - + HSO 3 - MnO 4 2- + SO 4 -2 add electrons to balance charge.

19. Check balance: H 1 + 2 = 3 Mn 2 = 2 O 8 + 3 + 1 = 8 + 4 S 1 = 1 Charge -2 - 1 = - 4 - 2 + 3 -3 = -3 But, are we done ??? Solution needs to be basic and we are generating 3 H + 2 MnO 4 - + HSO 3 - + H 2 O 2 MnO 4 2- + SO 4 -2 + 3 H + 3x(OH - + H + H 2 O)Neutralize H + 2 MnO 4 - + HSO 3 - + 3 OH - 2 MnO 4 2- + SO 4 -2 + 2 H 2 O Check balance: H 1 + 3 = 4 Mn 2 = 2 O 8 + 3 + 3 = 8 + 4 + 2 S 1 = 1 Charge -2 - 1- 3 = -4 - 2 + 0 -6 = -6 2 MnO 4 - + HSO 3 - + H 2 O 2 MnO 4 2- + SO 4 -2 + 3 H +

20. Redox Chemistry and Electrical Energy Electrical Conduction  moving electrical charges In Metals: electric charges that move are electrons in the metal electric potential difference is what causes electrons to move (similar to gravitational forces causing mass to move or chemical potential driving reactions) resistance interferes with the flow of electrons - usually atomic motion circuit - electrons move in circles, must have a place to go and other electrons to replace them electric current is the rate charges move - # charges / time

21. I = Q time 1 A = 1 c / 1 sec c = A. sec 1 Faraday of charge = charge of 1 mole of electrons = 96,500 c Energy = 1 J = (1 V). (1 c) E = I R Ohm’s Law Conduction is good if potential is high and resistance is low.

22. In Solutions (or liquids ionic compounds at temperatures high enough to melt): electric charges that move are ions cations (positively charged) move towards anything negatively charged anions (negatively charged) move towards anything positively charged Electrochemical Cell - A circuit that includes electrolytic and metallic conductors Electrode - connects metal conductor to electrolytic conductor

23. Electrochemical Cells 19.2 spontaneous redox reaction anode oxidation cathode reduction Attracts anionsAttracts cations

25. Electrochemical Cells 19.2 The difference in electrical potential between the anode and cathode is called: cell voltage electromotive force (emf) cell potential Cell Diagram 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 Use || to separate half cells Use | to separate reactants/phases in each half cell

26. Salt Bridge Electrolytic connection between two half cells to complete the circuit Tube containing a solution of inert salt (usually KNO3)

27. Electrochemical Cell - need “direct current source” = an electron pump (ex: battery) Electrons are forced in one direction, independent of sponteneity Electrical energy is used to cause a nonspontaneous reaction to occur - drive the desired chemistry Electrons are added to the cathode by electron pump causing reduction. Two Types of Cells

28. Voltaic Cell or Galvanic Cell - passively electric (no need of “direct current source” = an electron pump) Electrons move because of spontaneous reaction Use chemistry to get desired energy Electrons are taken from the cathode by reduction, causing electrons to flow to it (from outside it appears  ) Can use as a direct current source for electrolytic cell. Two Types of Cells

29. How do we know if reaction is spontaneous? What is the electric potential? How do concentrations effect the process? Electromotive Force (emf) is the electric potential of cell E (emf)  units = volts (V) emf is the potential difference between the anode and the cathode

30. 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)

31. Standard Electrode Potentials 19.3 Standard reduction potential (E 0 ) is the voltage associated with a reduction reaction at an electrode when all solutes are 1 M and all gases are at 1 atm. Defines E 0 = 0 V Standard hydrogen electrode (SHE) 2e - + 2H + (1 M) H 2 (1 atm) Reduction Reaction Use as a reference to measure all other potentials. 

32. 19.3 E 0 = 0.76 V cell Standard emf (E 0 ) cell 0.76 V = 0 - E Zn /Zn 0 2+ E Zn /Zn = -0.76 V 0 2+ Zn 2+ (1 M) + 2e - Zn E 0 = -0.76 V E 0 = E H /H - E Zn /Zn cell 00 + 2+ 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) Convention: E° > 0 spontaneous reaction reductionoxidation

33. 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++ 2 000 0.34 = E Cu /Cu - 0 0 2+ E Cu /Cu = 0.34 V 2+ 0

34. 19.3 E 0 is for the reaction as written The half-cell reactions are reversible The sign of E 0 changes when the reaction is reversed (E° red = -E° oxid ) Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E 0

35. The more positive E 0 the greater the tendency for the substance to be reduced Strongest oxidizing agent Strongest reducing agent Zero reference point

36. E 0 = 0.76 V cell E 0 = 0.34 V cell Combine Zn (s) Zn 2+ (1 M) + 2e - 2e - + Cu 2+ (1 M) Cu (s) Zn (s) + Cu 2+ (1 M) Cu (s) + Zn 2+ (1 M) E 0 = 0.76 V + 0.34 V = 1.10 V cell

37. 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  spontaneous

38. 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 Sponteneity  G < 0 Energy = Q E = -nFE cell Total charge

39. Spontaneity of Redox Reactions 19.4  G 0 = -RT ln K  G = -nFE cell

40. 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+ +

41. 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 K 19.5 - 0.0257 V n ln Q E 0 E = - 0.0592 V n log Q E 0 E = Can look at concentration effects / non- standard conditions

42. 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 V E > 0Spontaneous 19.5

43. Will the following reaction occur spontaneously at 25 0 C if [Ag + ] = 0.10 M and [Ag + ] = 0.010 M? Ag + (aq) + Ag (s) Ag (s) + Ag + (aq) e - + Ag + Ag Ag Ag + + e - Oxidation: Reduction: n = 1 E 0 = -0.7991 V – (-0.7991 V) E 0 = -0.000 V E 0 = E Ag /Ag – E Ag /Ag 00 + + - 0.0257 V n ln Q E 0 E = 0.0257 Vln = - 0.10 0.010 = - 0.0592 V E < 0 Non-spontaneous Concentration Cells have the same half-reaction in each cell, but at different concentrations. Ag (s) | Ag + (0.10 M) || Ag + (0.010 M) | Ag (s) Ag oxid + Ag + red Ag red + Ag + oxid - 0.0257 V 1 ln -0.000 V E = [Ag + oxid ] [Ag + red ]

44. 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)

45. 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

46. 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

47. Batteries 19.6 Solid State Lithium Battery

48. 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)

49. Corrosion 19.7 Dissolved oxygen in water causes oxidation E° red = -0.44 V Rust Fe 2 O 3 E° red = 1.23 V Since E° red (Fe 3+ ) < E° red (O 2 ) Fe can be oxidized by oxygen

50. Cathodic Protection of an Iron Storage Tank 19.7 E° red = -2.37 V E° red = 1.23 V Mg reduction potential is too negative to be overcome by oxygen.

51. 19.8 Electrolysis is the process in which electrical energy is used to cause a nonspontaneous chemical reaction to occur.

52. Electrolysis of Water 19.8

53. Electrolysis and Mass Changes charge (C) = current (A) x time (s) 1 mole e - = 96,500 C 19.8 Quantitative Analysis How much current? time ? product?

54. 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

55. e - + Ag + AgCathode: 2 H 2 O O 2 (g) + 4 H + + 4 e - Anode: Consider the stoichiometryof the electrolytic cell: What current (in amps) is required to convert 0.100 mole Ag + to Ag in 10.0 min? 1 mole electrons = 1 F A = c/sec Q = nF I = Q / t Find Q Q = 0.10 mol Ag 1 mole electrons 1 F 96,500 c = mole Ag mole electrons F Q = 9,650 c t = 10 min 60 seconds = 600 seconds 1 minute I = 9,6500 c / 600 s = 16 c/s = 16 A

56. e - + Ag + AgCathode: 2 H 2 O O 2 (g) + 4 H + + 4 e - Anode: Stoichiometry of prodcuts at different electrodes What is the pH of the anode half cell (assume volume = 0.100 L) after 6.00 g of Ag plates out at the cathode? Find [H + ] 6.00 g Ag 1 mol Ag 1 mole electrons 4 mol H + = 107.9 g Ag 1 mole Ag 1 mol electrons = 0.05567 mol H + [H + ] = 0.0556 mol = 0.56 M 0.10 L pH = 0.25