1 / 16

Cell Potential and Nernst Equation

Cell Potential and Nernst Equation. Cell Potential (E cell ). Because of the moving electrons in a redox reaction, electricity is generated. We can figure out how much electricity is generated in each redox reaction. Electricity is measured in Volts (V). Example #1.

george-wiley
Download Presentation

Cell Potential and Nernst Equation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Cell Potential and Nernst Equation

  2. Cell Potential (Ecell) • Because of the moving electrons in a redox reaction, electricity is generated. • We can figure out how much electricity is generated in each redox reaction. • Electricity is measured in Volts (V).

  3. Example #1 Calculate the cell potential (Ecell) for the following reaction. Zn(s) + 2H+(aq) Zn+2(aq) + H2(g) Oxidation: Zn  Zn+2 + 2e- Reduction: 2e- + 2H+(aq) H2(g) Look at the Reduction Reaction Table

  4. Standard Conditions • Reactions occur at 25oC • 1M concentrations for reactants and products in solutions • 1 atm or 101.3 kPa for gases. • However, this table is of REDUCTION reactions only! What do we do with the oxidation reaction???????

  5. Example #1 Zn(s) + 2H+(aq) Zn+2(aq) + H2(g) Oxidation: Zn(s) Zn+2 + 2e- Reduction: 2e- + 2H+(aq) H2(g)

  6. Example #1 continued Reduction: 2e- + 2H+(aq) H2(g) Ered=0 Reduction: Zn(s) Zn+2 + 2e- Ered=-0.76 But Zn is not being reduced. It is being oxidized. Thus, we need to take the equation and potential value from the chart and flip them around. Oxidation: Zn+2 + 2e- Zn(s)Eox=+0.76 Ecell = Ered + Eox Ecell = 0V + 0.76V = 0.76V

  7. Example #2 Zn + Cu+2 Cu + Zn+2 Reduction: 2e- + Cu+2 Cu Oxidation: Zn  Zn+2 + 2e-

  8. Example #2 continued Reduction: 2e- + Cu+2 Cu +0.34V Reduction according to table: Zn  Zn+2 + 2e- -0.76V Oxidation made into Reduction: Just flip the sign. Zn+2 + 2e-  Zn +0.76 Ecell = Ered + Eox Ecell = 0.34V + 0.76V = 1.10 V

  9. Spontaneous or Not? • If the total cell potential (Ecell) is positive—a spontaneous reaction occurs. • If the total cell potential (Ecell)is negative—a non-spontaneous reaction occurs.

  10. Let’s practice • On pink sheet, do #1-5 1. +0.780, spontaneous 2. -1.646, not spontaneous 3. -2.9149, not spontaneous 4. +1.1105, spontaneous 5. -3.716, not spontaneous

  11. What if we don’t have standard conditions???? Use the Nernst Equation n= number of electrons transferred K = equilibrium constant

  12. Example #1 Cu+2 + Fe  Fe+2 + Cu We need to calculate E0 for reaction n= number of electrons moving K=equilibrium constant

  13. Example #1 continued Cu+2 + 2e-  Cu Reduction +0.34 Fe  Fe+2 + 2e- Oxidation -0.44 But flip around to reduction. So +0.44 Thus, Eo=Ered+Eox = +0.78

  14. Example #1 continued n = number of electrons transferred In this reaction, 2 electrons are transferred so n=2 Cu+2 + 2e-  Cu Fe  Fe+2 + 2e-

  15. Example #1 continued Cu+2 + Fe  Fe+2 + Cu • Let’s assume [Cu+2]=0.200 M and [Fe+2]=0.800 M

  16. Example #1 continued

More Related