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V. Everything’s Related

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. V. Potential and Work. Potential difference can be expressed as a function of work.

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V. Everything’s Related

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  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) + 2H2O(l) H2(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 Ecell 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 Ecell will equal E°cell. • If Q < 1, there are more reactants than products, so redox reaction shifts right; Ecell will be > than E°cell. • If Q > 1, there are more products than reactants, so redox reaction shifts left; Ecell will be < than E°cell. • If Q = K, then Ecell = 0.

  13. V. Sample Problem • Calculate the cell potential for the electrochemical cell represented by Ni(s)|Ni2+(aq, 2.0 M)||VO2+(aq, 0.010 M), H+(aq, 1.0 M), VO2+(aq, 2.0 M)|Pt(s).

  14. V. Concentration Cells

  15. V. Concentration Cells • 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 MnO2(s). • In acidic battery, MnO2(s) reduced to Mn2O3(s). • In alkaline battery, MnO2(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 H2O 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. 2H2O(l) O2(g) + 4H+(aq) + 4e- has Eox = -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) I2(aq) + 2e- E°ox = -0.54 V • 2H2O(l)  O2(g) + 4H+(aq) + 4e- E°ox = -1.4 V • Reduction • Na+(aq) + e- Na(s) E°red = -2.71 V • 2H2O(l) + 2e-  H2(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 Cu2+ according to the half-reaction: Cu2+(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?

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