1. CHAPTER 16 (pages 776-792) 1.Oxidation and Reduction 2.Galvanic Cells, Half Reactions (E° anode & E° cathode ) 3.Standard Reduction Potential (E°) 4.Nernst Equation, and the dependence of Potential on Concentration 5.Relationship between Equilibrium Constant and Standard Potential 6.Driving Force, ΔG and ε 1

2. REDOX REACTIONS MnO 2 + 4 HBr  MnBr 2 + Br 2 + 2 H 2 O 3 H 2 S + 2 NO 3 – + 2 H +  S + 2 NO + 4 H 2 O 2

3. OBSERVED REDOX PROCESSES 3

4. GALVANIC CELLS 4

6. INERT ELECTRODES 6

7. STANDARD REDUCTION POTENTIALS 7

8. 8

9. MEASURING STANDARD POTENTIALS 9

10. CALCULATING STANDARD CELL POTENTIAL Al (s) + NO 3 − (aq) + 4 H + (aq)  Al 3+ (aq) + NO (g) + 2 H 2 O (l) 10

11. ADDITIONAL EXAMPLE Fe (s) + Mg 2+ (aq)  Fe 2+ (aq) + Mg (s) 11

12. ox: Fe( s )  Fe 2+ ( aq ) + 2 e − E  = +0.45 V red: Pb 2+ ( aq ) + 2 e −  Pb( s ) E  = −0.13 V tot: Pb 2+ ( aq ) + Fe( s )  Fe 2+ ( aq ) + Pb( s ) E  = +0.32 V

13. ELECTROMOTIVE POTENTIAL 13

15. E° CELL, Δ G° AND K Under standard state conditions, a reaction will spontaneously proceeds in the forward direction if: –Δ G° < 1 (negative) – E° > 1 (positive) – K > 1

16. Design a voltaic cell with the following half cells and complete the calculations: Ag + (aq) + 1e -  Ag (s) E o = 0.80 V Pb 2+ (aq) + 2e -  Pb (s) E o = -0.13 V a.Calculate the E o cell (potential at standard conditions) b.Calculate  G o. c.Calculate  d.Calculate the E cell if [Ag + ] = 2.0 M and [Pb 2+ ] = 1.0 x 10 -4 M.

17. Williams, spring 2009 stop here

18. Design a voltaic cell with the following half cells and complete the calculations: Ag + (aq) + 1e -  Ag (s) E o = 0.80 V Pb 2+ (aq) + 2e -  Pb (s) E o = -0.13 V Calculate the E o cell (potential at standard conditions)

19. Design a voltaic cell with the following half cells and complete the calculations: Ag + (aq) + 1e -  Ag (s) E o = 0.80 V Pb 2+ (aq) + 2e -  Pb (s) E o = -0.13 V Calculate  G o.

20. Design a voltaic cell with the following half cells and complete the calculations: Ag + (aq) + 1e -  Ag (s) E o = 0.80 V Pb 2+ (aq) + 2e -  Pb (s) E o = -0.13 V Calculate 

21. Design a voltaic cell with the following half cells and complete the calculations: Ag + (aq) + 1e -  Ag (s) E o = 0.80 V Pb 2+ (aq) + 2e -  Pb (s) E o = -0.13 V Calculate the E cell if [Ag + ] = 2.0 M and Pb 2+ ] = 1.0 x 10 -4 M.

23. OBJECTIVE 11.4: PROVIDE A THOROUGH OVERVIEW OF APPLICATIONS OF ELECTROCHEMICAL CELLS INCLUDING FUEL CELLS, CORROSION, AND OTHER TOPICS AS TIME PERMITS. 23

24. CORROSION corrosion is the spontaneous oxidation of a metal by chemicals in the environment since many materials we use are active metals, corrosion can be a very big problem

25. RUSTING rust is hydrated iron(III) oxide moisture must be present electrolytes promote rusting acids promote rusting – lower pH = lower E° red

26. Dry Cell Batteries

27. Lead – Acid Storage Battery

28. Biological Electrochemistry

29. Lithium Ion Battery