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The End is in Site!

The End is in Site!. Nernst and Electrolysis. Electrochemistry. Galvanic (Electrochemical) Cells. Spontaneous redox processes have:. A positive cell potential, E 0. A negative free energy change, (- G). Zn - Cu Galvanic Cell. Zn 2+ + 2e -  Zn E = -0.76V

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The End is in Site!

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  1. The End is in Site! Nernst and Electrolysis

  2. Electrochemistry

  3. Galvanic (Electrochemical) Cells Spontaneous redox processes have: A positive cell potential, E0 A negative free energy change, (-G)

  4. Zn - Cu Galvanic Cell Zn2+ + 2e- Zn E = -0.76V Cu2+ + 2e-  Cu E = +0.34V From a table of reduction potentials:

  5. Zn - Cu Galvanic Cell Cu2+ + 2e-  Cu E = +0.34V The less positive, or more negative reduction potential becomes the oxidation… Zn  Zn2+ + 2e- E = +0.76V Zn + Cu2+  Zn2+ + Cu E0 = + 1.10 V

  6. Line Notation An abbreviated representation of an electrochemical cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) Anode material Cathode material Anode solution Cathode solution | || |

  7. E0cell E0cell = RT/nF (lnK) DG0 = -nFE0cell DG0 K DG = -RTlnK

  8. Calculating G0 for a Cell G0 = -nFE0 n= moles of electrons in balanced redox equation F = Faraday constant = 96,485 coulombs/mol e- Zn + Cu2+  Zn2+ + Cu E0= + 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 At equilibrium, forward and reverse reactions occur at equal rates, therefore: • The battery is “dead” • The cell potential, E, is zero volts Modifying the Nernst Equation (at 25 C):

  12. Calculating an Equilibrium Constant from a Cell Potential Zn + Cu2+  Zn2+ + Cu E0= + 1.10 V

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

  14. Concentration Cell ??? Both sides have the same components but at different concentrations. Anode Cathode The 1.0 M Zn2+ must decrease in concentration, and the 0.10 M Zn2+ must increase in concentration Zn2+ (1.0M) + 2e- Zn (reduction) Zn  Zn2+ (0.10M) + 2e- (oxidation) Zn2+ (1.0M)  Zn2+ (0.10M)

  15. Concentration Cell Concentration Cell ??? Both sides have the same components but at different concentrations. Anode Cathode Step 2: Calculate cell potential using the Nernst Equation (assuming 25 C). Zn2+ (1.0M)  Zn2+ (0.10M)

  16. Nernst Calculations Zn2+ (1.0M)  Zn2+ (0.10M)

  17. Electrolytic Processes Electrolytic processes are NOT spontaneous. They have: A negative cell potential, (-E0) A positive free energy change, (+G)

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

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

  20. 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.

  21. 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 103 C Hint: C = nF therefore n = C/F n = 1.35 x 103 C / 9.65 x 104 n = 1.40 x 10-2 mole e- therefore 1.40 x 10-2 mole Ag = 1.51g Ag are deposited

  22. Example #2 One ½ reaction occurring in the lead storage battery is: Pb + SO42- PbSO4 + 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 104) C = 4.23 x 105 C Hint: C = amp x sec 4.23 x 105 = 1.50amp x time Time = 2.82 x 105sec = 78.3 hours

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