1. Electrochemistry Ch. 17

2. Electrochemistry Generate current from a reaction –Spontaneous reaction –Battery Use current to induce reaction –Nonspontaneous reaction –Electroplating

3. Oxidation-Reduction Reaction aka Redox Transfer of electrons Donor + Acceptor (reducing agent)(oxidizing agent) (is oxidized)(is reduced)

4. Mnemonics are cool! O xidation I nvolves L oss of electrons R eduction I nvolves G ain of electrons L oss of E lectrons is O xidation says G ain of E lectrons is R eduction

5. Assigning Oxidation States (1) Covalent bond btw identical atoms => Split electrons evenly (2) Covalent bond btw different atoms => All electrons given to more electronegative atom.

6. (3) For ionic compounds, oxidation states are equal to ionic charge. (4) Oxidation state for an elemental atom is zero.

7. (5) Oxidation state for monatomic ion is the same as the charge. (6) In compounds, fluorine always has an O.S. of -1.

8. (7) Oxygen usually has an O.S. of -2, except when in a peroxide or when in OF 2. H 2 O -2 H 2 O 2 -1 +2 OF 2 (8) With a nonmetal, hydrogen has an O.S. of +1. With a metal, H is assigned an O.S. of -1. NH 3 +1 LiH -1 (9) The sum of the oxidation states must add up to the overall charge.

9. Examples Assign the oxidation states to each atom of the following compounds. CO 2 CH 4 K 2 Cr 2 O 7

10. Redox Reactions CH 4 + O 2  CO 2 + H 2 O Which species is oxidized? Which species is reduced? Which species is the oxidizing agent? Which species is the reducing agent?

11. Balancing Redox Reaction Balance… …# of atoms …# of electrons transferred …overall charge Types of reactions –Acidic conditions –Basic conditions

12. Redox in Acidic Solutions Cr 2 O 7 2- + C 2 H 5 OH  Cr 3+ + CO 2 1.Assign oxidation states 2.Write half reactions Red: Cr 2 O 7 2-  Cr 3+ Ox: C 2 H 5 OH  CO 2

13. 3. Balance elements except H and O Cr 2 O 7 2-  2Cr 3+ C 2 H 5 OH  2CO 2 4. Balance oxygen by adding H 2 O Cr 2 O 7 2-  2Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  2CO 2 5. Balance hydrogen by adding H + 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  2CO 2 + 12H +

14. 6. Balance charge by adding electrons 6e - + 14H + + Cr 2 O 7 2-  2Cr 3+ + 7H 2 O 3H 2 O + C 2 H 5 OH  2CO 2 + 12H + + 12e - 7. Equalize the number of electrons 12e - + 28H + + 2Cr 2 O 7 2-  4Cr 3+ + 14H 2 O 3H 2 O + C 2 H 5 OH  2CO 2 + 12H + + 12e - 8. Cancel like terms and add reactions 16H + + 2Cr 2 O 7 2- + C 2 H 5 OH  4Cr 3+ + 2CO 2 + 11H 2 O 9. Check your answer!

15. Balancing in basic solution Following the same algorithm used for acidic solutions through step #8 then… 9. Add the same # of OH - to both sides of equation as there are H + on one side 10. Combine H + and OH - on same sides of equation to make H 2 O 11. Cancel any like terms and check

16. Galvanic Cells Spontaneous chemistry generating current Some terms –Reducing agent –Oxidizing agent –Half reactions –Anode –Cathode –Cell potential

17. Building a Galvanic Cell Overall Reaction 8H + (aq) + MnO 4 - (aq) + 5Fe 2+ (aq)  Mn 2+ (aq) + 5Fe 3+ (aq) + 4H 2 O(l) Half Reactions Reduction: 8H + + MnO 4 - + 5e - → Mn 2+ + 4H 2 O Oxidation: 5(Fe 2+ → Fe 3+ + e - )

18. Reduction: 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O Oxidation: Fe 2+  Fe 3+ + e - salt bridge KNO 3

19. Calculating Cell Potential Reduction: 8H + + MnO 4 - + 5e -  Mn 2+ + 4H 2 O εº(reduction) = 1.51 V Oxidation: 5(Fe 2+  Fe 3+ + e - ) εº(oxidation) = -0.77 V εº(cell) = εº(red) + εº(ox) = 0.74 V

20. Comments on Cell Potential Potential is an intensive property DO NOT multiply potential by balancing factor The º indicates standard conditions – 1.0 M and 1 atm Potentials references to standard H + red. 2H + + 2e - → H 2 εº = 0.00 V

21. Fe 3+ (aq) + Cu(s) → Cu 2+ (aq) + Fe 2+ (aq) Oxidation:Reduction: salt bridge Cu Fe or Pt Cu 2+ Fe 3+  e - e - e -  ← -← - +→+→ Cu  Cu 2+ + 2e - Fe 3+ + e -  Fe 2+

22. Cu 2+ + Zn  Zn 2+ + Cu Demo

23. Line Notation 2Al 3+ (aq) + 3Mg(s) → 2Al(s) + 3Mg 2+ (aq)

24. Line Notation II MnO 4 - (aq) + H + (aq) + ClO 3 - (aq) → ClO 4 - (aq) + Mn 2+ (aq) + H 2 O(l) Pt(s) │ ClO 3 - (aq), ClO 4 - (aq) ║ MnO 4 - (aq), Mn 2+ (aq) │ Pt(s)

25. To Review Full description of galvanic cell requires: –Composition of solutions –Composition of electrodes –Direction of electron flow –Direction of ion flow –Calculation of cell potential –Labels: “anode” and “cathode”

26. Cell Potential and Free Energy

27. Reconsidering Cell Potential Given: Al 3+ + 3e -  Al ΔG  1 and ε 1  = -1.66V Mg 2+ + 2e -  Mg ΔG  2 and ε 1  = -2.37 V Find ε  (cell) for: 2Al 3+ + 3Mg  2Al + 3Mg 2+

29. Cell Potential and Spontaneity Can bromine oxidize iodide to iodine? Can Cr(II) reduce oxygen gas under acidic conditions to produce water? Can Ag(I) oxidize chloride to chlorine? Can hydrogen reduce Fe(II) to elemental iron?

30. Non-Standard Cell Potential

31. Practice Problems Determine ε for the following reaction and conditions: 2Al(s) + 3Mn 2+ (aq)  2Al 3+ (aq) + 3Mn (a) [Al 3+ ] = 2.0 M; [Mn 2+ ] = 1.0 M @ 25°C (b) [Al 3+ ] = 1.0 M; [Mn 2+ ] = 3.0 M @ 25°C

32. Do Worksheet

33. Potential and Equilibrium Cell potential at equilibrium = 0.0 V Q = K at equilibrium

34. Types of Batteries Lead storage (car battery) Dry cell battery –Acidic –Alkaline –Rechargeable Lithium ion battery Fuel cell

35. Lead Storage Battery Anode: Pb + HSO 4 - → PbSO 4 + H + + 2e - Cathode: PbO 2 + HSO 4 - + 3H + + 2e - → PbSO 4 + 2H 2 O Overall Potential: εº = 2.04 V

37. Dry Cell Battery (acidic) Anode: Zn → Zn 2+ + 2e - Cathode: 2NH 4 + + 2MnO 2 + 2e - → Mn 2 O 3 + 2NH 3 + H 2 O Overall Potential: εº = 1.5 V

38. Dry Cell Battery (alkaline) Anode: Zn → Zn 2+ + 2e - Cathode: 2MnO 2 + H 2 O + 2e - → Mn 2 O 3 + 2OH - Overall Potential: εº = 1.5 V

40. Lithium Ion Battery Anode: Li  Li + + e - Cathode: MnO 2 + Li + + e -  LiMnO 2 Cell Potential: εº = 3.6 V

42. Fuel Cell Anode: 2H 2 (g) + 4OH - (aq)  4H 2 O + 4e - Cathode: O 2 + 2H 2 O + 4e -  4OH - Cell Potential: εº = 1.23 V

44. Corrosion A significant portion of construction is done to replace corroded materials. Cathode: O 2 + 2H 2 O + 4e -  4OH - ε  = 0.40 V Anode: Fe  Fe 2+ + 2e - ε  = 0.44 V ε  (cell) = 0.84 V

46. Corrosion and Acidic Conditions Cathode: O 2 + 4H + + 4e -  2H 2 Oε  = 1.23 V Anode: Fe  Fe 2+ + 2e - ε  = 0.44 V ε  (cell) = 1.67 V

47. Electrolysis Supply current to perform chemistry Performed in an electrolytic cell Stoichiometric relationship btw. charge and chemical amount Factor-label fun! Current measured in Ampere = 1 coulomb per second

48. Example Problem I How long will it take to plate out 1.00 kg of aluminum from an aqueous solution of Al 3+ using a current of 100.0 A?

49. Al 3+ + 3e -  Al

50. Example Problem II What volume of F 2 gas, at 25  C and 1.00 atm, is produced when molten KF is electrolyzed by a current of 10.0 A for 2.00 hours? What mass of K metal is produced? At which electrode does each reaction occur?

51. Solution II Molten KF contains K + and F - Cathode: K + + e -  K Anode: F -  1/2F 2 + e -

52. Volume of F 2

53. Mass of K K + + e -  K F -  1/2F 2 + e - 0.373 mol F 2  0.746 mol K (0.746 mol K)(39.10 g/mol) = 29.2 g K

54. Electrolysis in Water Anode: 2H 2 O  O 2 + 4H + + 4e - ε  = -1.23V Cathode: 4H 2 O + 4e -  2H 2 + 4OH - ε  = -0.83V 2H 2 O  O 2 + 2H 2 ε  (cell) = -2.06 V Assuming [H + ] = [OH - ] = 1.0 M

55. Electrolysis in Pure Water In pure water: [H + ] = [OH - ] = 1.0 x 10 -7 M Use Nernst equation to determine ε Anode: Cathode:

56. Electrolysis in Pure Water The overall potential for the electrolysis of pure water is -1.24 V. Need to consider several possible oxidations and reductions when performing electrolysis of aqueous salt solutions. Consider the electrolysis of 1.0 M NaCl(aq)

57. 1.0 M NaCl(aq) contains: 1.0 M Na + 1.0 M Cl - H 2 O10 -7 M H + 10 -7 M OH - Reducible species: Na + H + H 2 O Oxidizable species: Cl - OH - H 2 O

58. Possible Reductions Using pH = 7.00 Na + + e -  Naε = -2.71 V H + + e -  1/2H 2 ε = -0.414 V** H 2 O + e -  1/2H 2 + OH - ε = -0.416 V** So, H 2 produced at the cathode **Potentials found using Nernst equation

59. Possible Oxidations Using pH = 7.00 Cl -  1/2Cl 2 + e - ε = -1.36 V 2OH -  1/2O 2 + H 2 O + 2e - ε = -0.814 V** H 2 O  1/2O 2 + 2H + + 2e - ε = -0.816 V** So, O 2 expected to form at the anode. But…