1. The End is in Site! Nernst and Electrolysis

2. Electrochemistry

3. Galvanic (Electrochemical) Cells Spontaneous redox processes have: A positive cell potential, E 0 A negative free energy change, (-  G)

4. Zn - Cu Galvanic Cell Zn 2+ + 2e -  Zn E = -0.76V Cu 2+ + 2e -  Cu E = +0.34V From a table of reduction potentials:

5. Zn - Cu Galvanic Cell Cu 2+ + 2e -  Cu E = +0.34V The less positive, or more negative reduction potential becomes the oxidation… Zn  Zn 2+ + 2e - E = +0.76V Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V

6. Line Notation Zn(s) | Zn 2+ (aq) || Cu 2+ (aq) | Cu(s) An abbreviated representation of an electrochemical cell Anodesolution Anodematerial Cathodesolution Cathodematerial ||||

7. E 0 cell G0G0 K E 0 cell = RT/nF (lnK)  G 0 = -nFE 0 cell  G = -RTlnK

8. Calculating  G 0 for a Cell  G 0 = -nFE 0 n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e - E 0 Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V

9. The Nernst Equation Standard potentials assume a concentration of 1 M. The Nernst equation allows us to calculate potential when the two cells are not 1.0 M. R = 8.31 J/(mol  K) T = Temperature in K n = moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e -

10. Nernst Equation Simplified At 25  C (298 K) the Nernst Equation is simplified this way:

11. Equilibrium Constants and Cell Potential equilibrium At equilibrium, forward and reverse reactions occur at equal rates, therefore: 1. 1. The battery is “dead” 2. The cell potential, E, is zero volts Modifying the Nernst Equation (at 25  C):

12. E 0 Zn + Cu 2+  Zn 2+ + Cu E 0 = + 1.10 V Calculating an Equilibrium Constant from a Cell Potential

13. Concentration Cell Step 1: Determine which side undergoes oxidation, and which side undergoes reduction. Both sides have the same components but at different concentrations. ???

14. Concentration Cell Both sides have the same components but at different concentrations. The 1.0 M Zn 2+ must decrease in concentration, and the 0.10 M Zn 2+ must increase in concentration Zn 2+ (1.0M) + 2e -  Zn (reduction) Zn  Zn 2+ (0.10M) + 2e - (oxidation) ??? Cathode Anode Zn 2+ (1.0M)  Zn 2+ (0.10M)

15. Concentration Cell Both sides have the same components but at different concentrations. Spontaneous: e- move from anode to cathode (+Voltage) Use Le Chatelier: Increasing [Zn 2+ (1.0M)] forces the reverse to take place Therefore, Decreasing the E cell ??? Cathode Anode

16. Concentration Cell Step 2: Calculate cell potential using the Nernst Equation (assuming 25  C). Both sides have the same components but at different concentrations. ??? Cathode Anode Zn 2+ (1.0M)  Zn 2+ (0.10M) Concentration Cell

17. Nernst Calculations Zn 2+ (1.0M)  Zn 2+ (0.10M)

18. Electrolytic Processes A negative -E 0 cell potential, ( -E 0 ) +  G A positive free energy change, ( +  G ) NOT Electrolytic processes are NOT spontaneous. They have:

19. Electrolysis of Water In acidic solution Anode rxn: Cathode rxn: -1.23 V -0.83 V -2.06 V

20. Electroplating of Silver Anode reaction: Ag  Ag + + e - Electroplating requirements: 1. Solution of the plating metal 3. Cathode with the object to be plated 2. Anode made of the plating metal 4. Source of current Cathode reaction: Ag + + e -  Ag

21. A couple of important relationships to remember: C = coulomb = charge transported by a steady current of one ampere in one second C = amp x sec F = C/n then C = nF If you know the moles of e- generated, then you use stoichiometry to determine moles of the metal that are deposited.

22. Example #1 Assume that 1.50amps of current flow through a solution containing silver ions for 15.0 minutes. The voltage is such that silver is deposited at the cathode. How many grams of silver metal are deposited? Ag + + e - Ag Hint: C = amp x sec C = 1.50 amp x (15min x 60sec/min) C = 1.35 x 10 3 C Hint: C = nF therefore n = C/F n = 1.35 x 10 3 C / 9.65 x 10 4 n = 1.40 x 10 -2 mole e- therefore 1.40 x 10 -2 mole Ag = 1.51g Ag are deposited

23. Example #2 One ½ reaction occurring in the lead storage battery is: Pb + SO 4 2- PbSO 4 + 2e - If the battery delivers 1.50 amps and if its lead electrode contains 454g of Pb, how long can current flow? Hint: convert to mole Pb 454g = 2.19mol Pb which means 4.38 mole e - are produced C = nF = 4.38 (9.65 x 10 4 ) C = 4.23 x 10 5 C Hint: C = amp x sec 4.23 x 10 5 = 1.50amp x time Time = 2.82 x 10 5 sec = 78.3 hours