1. V. Everything’s Related Positive E° cell means spontaneous. Negative ΔG° means spontaneous. K > 0 means spontaneous. Thus, all of these must be related somehow.

2. V. Potential and Work Potential difference can be expressed as a function of work. We will use this new view to derive the relationship between E° cell and ΔG°.

3. V. Eqn. Relating E° cell and ΔG°

4. V. Sample Problem Using tabulated half-cell potentials, calculate ΔG° for the reaction 2Na (s) + 2H 2 O (l)  H 2(g) + 2OH - (aq) + 2Na + (aq).

5. V. Potential and Equilibrium Constants Using our new equation that relates standard cell potential to standard free energy, we can derive an equation between E° cell and K. We start with the equation that relates ΔG° to K.

6. V. Eqn. Relating E° cell and K

7. V. Sample Problem Use tabulated half-cell potentials to calculate the equilibrium constant for the two electron oxidation of copper metal by H +.

8. V. Relating ΔG°, K, and E° cell With the last equation derived, we summarize how we can convert between ΔG°, K, and E° cell.

9. V. Nonstandard Potentials Just like any other system, electrochemical cells may not be under standard conditions. Why is the potential higher?

10. V. Calculating Nonstandard Cell Potentials Although we can qualitatively predict whether E cell is higher or lower than E° cell, we’d like to calculate an exact value. We can derive an equation starting with the nonstandard ΔG equation.

11. V. Deriving the Nernst Equation

12. V. Nernst Eqn. Generalizations Under standard conditions, log Q = log 1 = 0, so E cell will equal E° cell. If Q than E° cell. If Q > 1, there are more products than reactants, so redox reaction shifts left; E cell will be < than E° cell. If Q = K, then E cell = 0.

13. V. Sample Problem Calculate the cell potential for the electrochemical cell represented by Ni (s) |Ni 2+ (aq, 2.0 M)||VO 2 + (aq, 0.010 M), H + (aq, 1.0 M), VO 2+ (aq, 2.0 M)|Pt (s).

14. V. Concentration Cells

15. If the two half-cells are the same, there is no reason to reduce something on one side and oxidize the same thing on the other side. However, if there are [ ] differences, there is a push to get to equilibrium. Electrons flow in order to increase the [ ] of the dilute cell and decrease the [ ] of the concentrated cell.

16. VI. Using Electrochemistry If constructed correctly, electrochemical cells can be used to store and deliver electricity. We briefly look at dry-cell batteries, lithium ion batteries, and fuel cells.

17. VI. Dry-cell Batteries Called dry-cell because there’s very little water. Voltage derived from the oxidation of Zn (s) and the reduction of MnO 2(s). In acidic battery, MnO 2(s) reduced to Mn 2 O 3(s). In alkaline battery, MnO 2(s) reduced to MnO(OH) (s).

18. VI. Lithium Ion Batteries This rechargeable battery works a bit differently. Motion of Li + from anode to cathode causes e-’s to flow externally and reduce the transition metal. Battery is recharged by oxidizing the transition metal in the cathode.

19. VI. Fuel Cells

20. VII. Nonspontaneous Redox Reactions Nonspontaneous reactions can be forced to occur by inputting energy. Nonspontaneous redox reactions can be forced to occur by inputting energy in the form of electrical current. This type of electrochemical cell is called an electrolytic cell.

21. VII. Voltaic vs. Electrolytic Cells

22. VII. Predicting Products In an aqueous electrolytic cell, it’s possible that H 2 O can be reduced or oxidized. To predict products, must consider potentials of all processes that could occur. Processes that are easiest (least negative or most positive half-cell potential) will occur.

23. VII. Overvoltage There is one problem in predicting electrolysis products. Some half-cell reactions don’t occur at their expected voltage potentials! e.g. 2H 2 O (l)  O 2(g) + 4H + (aq) + 4e - has E ox = -0.82 V when [H + ] = 1 x 10 -7 M.  However, kinetic factors require a voltage of 1.4 V for this to occur.

24. VII. Example Electrolysis What happens when a solution of NaI undergoes electrolysis? Oxidation  2I - (aq)  I 2(aq) + 2e - E° ox = -0.54 V  2H 2 O (l)  O 2(g) + 4H + (aq) + 4e - E° ox = -1.4 V Reduction  Na + (aq) + e -  Na (s) E° red = -2.71 V  2H 2 O (l) + 2e -  H 2(g) + 2OH - (aq) E° red = -0.41 V

25. VII. Electrolysis of NaI Solution

26. VII. Electrolysis Stoichiometry The # of electrons in any redox reaction can be used as a stoichiometric ratio. If we know how long an electrolysis takes place, and the magnitude of current that flowed, we can do stoichiometric calculations. Important relationships:  1 A = 1 C/s  F = 96,485 C/mole e-

27. VII. Sample Problem Copper can be plated out of a solution containing Cu 2+ according to the half- reaction: Cu 2+ (aq) + 2e-  Cu (s). How long (in minutes) will it take to plate 10.0 g of copper using a current of 2.0 A?