1. Outline:4/4/07 Today: Continue Chapter 19  Galvanic cells and  °  Nernst Equation (  and  G) è Pick up CAPA 18 & 19 - outside è Final exam in 4 weeks… è No class/seminar Friday… è Worksheet #11 answers posted

2. Let’s warm up those keypads… How goes balancing redox?

3. n n 2 n n 5 n n 10 n n 20 Let’s try this one: Mn 2+ + NaBiO 3  Bi 3+ + MnO 4  + Na + Mn 2+ + NaBiO 3  Bi 3+ + MnO 4  + Na + (acidic) In the balanced redox equation, how many total electrons are transferred?

4. Let’s try this one: In the balanced redox equation, how many water molecules are there? n n 3 on the left side n n 12 on the right side n n 3 on the right side n n 9 on the right side Se + Cr(OH) 3  Cr + SeO 3 2  (basic)

7. How to find  o ? n For example, what happens when Cu metal is put into a Zn 2+ solution? Look up std reduction potentials:  Cu 2+ + 2e   Cu +0.342 V  Zn 2+ + 2e   Zn  0.762 V

8. How to find  o ? n n Appendix F (p. A-10, A-11) Be careful!

9. How to find  o ? n For example, what happens when Cu metal is put into a Zn 2+ solution? Look up std reduction potentials:  Cu 2+ + 2e   Cu +0.342 V  Zn 2+ + 2e   Zn  0.762 V Reverse one to make it a positive  °

10. How to find  o ? n For example, what happens when Cu metal is put into a Zn 2+ solution? Look up std reduction potentials:  Cu 2+ + 2e   Cu +0.342 V  Zn 2+ + 2e   Zn  0.762 V Cu 2+ + Zn  Zn 2+ + Cu +1.104 V

11. Nomenclature: Cu 2+ + Zn  Cu + Zn 2+ +1.104 V n Galvanic cell: Electrodes:Anode Cathode Oxidation Reduction  Zn  Zn 2+ + 2e   @anode Cu 2+ + 2e   Cu @cathode

12. Galvanic Cells: Nomenclature Half Cell: Anode: Cathode: Salt Bridge

13. Another example:  Cu 2+ + 2e   Cu  V  Ag + + e   Ag  V  Cu + 2Ag +  Cu 2+ + 2Ag  V Practice! Will this be what we measure in lab? Multiply the second equation by two to balance mass/charge, but not  o !

14. Making the link: Spontaneity Reactions with a positive E o will react spontaneously… Will Na or Sn react spontaneously with water? Na  Na + + e   2.71 V Sn  Sn 2+ + 2e  +0.14 V 2H 2 O + 2e   H 2 + 2OH   0.83 V

15. 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

16.  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!

17. 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