1. Outline:4/9/07 Today: Continue Chapter 19  Nernst Equation (  and  G)  Coulomb Calculations  Redox Applications è CAPA 18 due Wednesday… è Worksheet #11 answers posted… è Final exam in 3 weeks… è Special seminar Tuesday/Thursday

2. Spontaneous? Relation to  G  G =  nF  o n n Definitions:    o = Electromotive force (at STP) Coulomb = unit of charge  e  = 1.602  10  19 Coulomb  1 mol of e  = 6.022  10 23. 1.602  10  19 = 9.6485  10 4 Coulomb F =Faraday Const. = 96,485 C/mol

3.  G,  o, and equilibrium.... Since  G =  G o + RTlnQ.....   Then:  =  o  (RT/nF) lnQ n Which is often written as:  Then:  =  o  (0.05916/n) log Q (@298K) (Nernst Equation) n What good is this?  Can calculate  at non-standard conditions!

4. Example   What is the voltage produced by the following cell ?  o   MnO 4  +8H + +5e   Mn 2+ +4H 2 O 1.51   O 2(g) + 4H + + 4e   2 H 2 O 1.23 at pH=7.00, p O 2 =0.20 atm,   [MnO 4  ] = [Mn 2+ ] = 0.10 M ? n First balance redox equation for cell.... n Set up expression for Q n Solve in Nernst equation

5. Set up redox equation for cell: n n Which equation gets reversed? n What is n? What is  o ?  MnO 4  +8H + +5e   Mn 2+ +4H 2 O 1.51  2 H 2 O  O 2 + 4H + + 4e   1.23 n=20  MnO 4  +8H + +5e   Mn 2+ + 4H 2 O  4  2 H 2 O  O 2 + 4H + + 4e   5  4MnO 4  +12H +  4Mn 2+ +5O 2 + 6H 2 O +0.28V

6. What is Q? n n Products/reactants : What is  ?  =  o  (0.05916/n) log Q  = 0.28  0.24 = 0.04 V [Mn 2+ ] 4 p O 2 5 = 3.2  10 80  [H + ] 12 [MnO 4  ] 4 PRACTICE!

7. Using the Nernst equation….  =  o  (0.05916/n) log Q

8. Example : What is the voltage produced by the concentration cell shown with two copper electrodes. One is immersed in 1M CuSO 4 and the other in 0.001M? What is  o for this cell? = 0.00 V Only the Q part of the Nernst equation is non-zero…..  =  o  (0.05916/n) log Q

9. What is Q? n n Products/reactants : What is  ?  =  o  (0.05916/n) log Q  = 0.00  (0.05916/2)  3 = 0.089 V [Cu 2+ ] (less concentrated)  = 1  10  3 [Cu 2+ ] (more concentrated)

10. Worksheet #12 n n A galvanic cell is constructed from a silver-silver iodide electrode and a Zn wire in 0.005 M solution of ZnCl 2. What is the cell potential? n First, what is the galvanic cell? Balance redox equation for cell....  o n Set up expression for Q….prods/rcts n Solve with Nernst equation   =  o  (0.05916/n)logQ

11. Set up redox equation for cell: n n Which equations? n n Which equation get reversed? n What is n? What is  o ?  AgI (s) + e   Ag (s) + I   0.152  Zn (s)  Zn 2+ + 2e   0.762 n=2  2AgI (s) + Zn (s)  2Ag (s) + Zn 2+ + 2I    0.610 V

12. What is Q? n n Products/reactants :   [Zn 2+ ] [I  ] 2 = (0.005)(1) 2 1   = 5 × 10  3 What is  ?  =  o  (0.05916/n)logQ = 0.610 + 0.068 = 0.678 V

13. Problem 2 n n You need to know:  o  0.050 V +0.695 V N 2 O (g) + 6H + + H 2 O + 4e -  2NH 3 OH + O 2(g) + 2H + + 2e -  H 2 O 2 Answer: + 0.745 V

14. More definitions: n n Current = charge/time 1 Ampere = 1 Coulomb / second Calculational Example: n A lead-acid battery:  PbO 2 +HSO 4  +2H + +Pb  2PbSO 4 +2H 2 O Assume 250 g of PbO 2 and that this battery supplies 6 Amps until it dies; how long will it last?

15. Quiz #7 n Put away your books & papers n When you are finished, turn your quiz into me