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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.

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Cell Potential and Nernst Equation

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  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

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