1. Outline:4/11/07 Today: Finish Chapter 19  Coulomb Calculations  Redox Applications è CAPA 18 due tonight… è CAPA 19 due Friday… è Special seminar Thursday pm è Special seminar Friday 4pm è Pick up Quiz 8 – from me è Pick up grade summary – in front

2. Outline:4/11/07 Today: Finish Chapter 19  Coulomb Calculations  Redox Applications è CAPA 18 due tonight… è CAPA 19 due Friday… è Special seminar Thursday pm è Special seminar Friday 4pm è Pick up Quiz 8 – from me è Pick up grade summary – in front SC 1019 6-8pm Panel @ 8pm

3. Outline:4/11/07 Today: Finish Chapter 19  Coulomb Calculations  Redox Applications è CAPA 18 due tonight… è CAPA 19 due Friday… è Special seminar Thursday pm è Special seminar Friday 4pm è Pick up Quiz 8 – from me è Pick up grade summary – in front

4. Quiz #7 n Average = 8.2 Good!

5. Chapter 19 practice problems: 19.1, 19.3, 19.5, 19.9, 19.11, 19.13, 19.15, 19.17, 19.19, 19.21, 19.23, 19.25, 19.27, 19.33, 19.35, 19.37, 19.39, 19.41, 19.45, 19.49, 19.53, 19.55, 19.63, 19.65, 19.67, 19.69, 19.73, 19.77, 19.81, 19.83, 19.85 19.1, 19.3, 19.5, 19.9, 19.11, 19.13, 19.15, 19.17, 19.19, 19.21, 19.23, 19.25, 19.27, 19.33, 19.35, 19.37, 19.39, 19.41, 19.45, 19.49, 19.53, 19.55, 19.63, 19.65, 19.67, 19.69, 19.73, 19.77, 19.81, 19.83, 19.85

6. Grade Summary Go over grade summary sheet Go over grade summary sheet n 8 quizzes / 70 pts (drop one) n 4 pts/seminar report extra credit n CAPA through 17; w/out a few handins n Exam 1, Exam 2 added at 20 or 15% n Final exam 60% or 75% or 90%...  Please see me if there is an data entry error

7. Finish Chapter 19: 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?

8. n n 250 g PbO 2 × 1 mol/ 239.2 g = 1.045 mol PbO 2 2.090 mol e  × 96485 C/mol = 201654 C (time) × (6 C/sec) = 201654 C  time = 3361 sec = 9.34 hours PbO 2  PbSO 4 (+4  +2) : 2e   1.045 mol PbO 2 × 2 mol e  / mol PbO 2 =  2.090 mol e 

9. Problem #3 - Worksheet #12: n n Current calculation…. n n Which equations? n n What is the limiting reagent? n n How many moles? n n How much time at 0.2 mA?

10. Worksheet #12 HgO  Hg (+2  0) : 2e   Zn  ZnO (0  +2) : 2e  n Moles of everything: 1 g HgO × 1mol/216.6g = 0.00462 mol  1 g Zn × 1mol/65.4g = 0.01530 mol  What is the limiting reagent? HgO n Moles of electrons:  0.00462l mols HgO × 2 mols e  /mol HgO

11. Moles e  to Coulombs conversion:  0.00924 mol e  × 96485 C/mol e   = 891.5 C = 891.5 Amp. sec  How long?  891.5 Amp. sec = (x sec)(0.2×10  3 Amp) x = 4.46×10 6 sec = 1238 hours Practice! Practice! Practice! Practice!

12. Redox applications: n n Batteries n n Fuel Cells n n Electroplating n n Ore Extraction/Mining n n Rust Protection n n Metallurgy/Alloys n n Electrolysis Read pp. 847 - 857 in textbook