1. Cell EMF Eocell = Eored(cathode) - Eored(anode)
Example: Zn + Cu+2  Zn+2 + Cu
E0cell = V - (-0.76 V) = V

2. A voltaic cell is based on two half-reactions: Cd+2/Cd Sn+2/Sn
Which half-reaction takes place at the cathode?
Which half-reaction takes place at the anode?
What is the standard cell potential?

3. A voltaic cell is based on two half-reactions: Cd+2/Cd Sn+2/Sn
Which half-reaction takes place at the cathode?
Which half-reaction takes place at the anode?
What is the standard cell potential?
Sn e-  Sn
Cd  Cd e-
0.267 V

5. Spontaneity of Redox Reactions
Eo = Eored(reduction) - Eored(oxidation)
E0 = (+) spontaneous
E0 = (-) nonspontaneous

6. Calculate the value of E0.
Cu + 2H+  Cu H2
Calculate the value of E0.

7. Calculate the value of E0.
Cu + 2H+  Cu H2
Calculate the value of E0.
E0 = V
NOT SPONTANEOUS

8. EMF and Free-Energy Change
G = -nFE
n = a positive # (the # of electrons transferred)
F = Faraday’s constant. 1F = 96,500 J/V-mol
E = EMF

9. Use the standard reduction potentials to calculate the standard free-energy change, Go, for the following reaction:
4Ag + O H+  4Ag H2O

10. Go = -(4)(96,500J/V-mol)(+0.43V)
Use the standard reduction potentials to calculate the standard free-energy change, Go, for the following reaction:
4Ag + O H+  4Ag H2O
Go = -(4)(96,500J/V-mol)(+0.43V)
- 170 kJ/mol

11. The Nernst Equation
(at 298 K)

12. Cr2O7-2(aq) + 14H+(aq) + 6I-(aq)  2Cr+3(aq) + 3I2(s) + 7H20(l)
Calculate the emf at 298K generated by the following cell, when (Cr2O7-2) = 2.0 M, (H+) = 1.0 M, (I-) = 1.0 M, and (Cr+3) = 1.0 x 10-5 M:
Cr2O7-2(aq) + 14H+(aq) + 6I-(aq)  2Cr+3(aq) + 3I2(s) + 7H20(l)

13. Calculate the emf at 298K generated by the following cell, when (Cr2O7-2) = 2.0 M, (H+) = 1.0 M, (I-) = 1.0 M, and (Cr+3) = 1.0 x 10-5 M:
Cr2O H+ + 6I-  2Cr+3 + 3I2 + 7H20
E = V V/6 log(5.0 x 10-11)
E = V