Electrochemistry

1. Electrochemistry • “Marriage” of redox and thermo • Spontaneous electron-transfer reactions can result in spontaneous electric current if the reactants are separated by a wire • Voltaic (Galvanic) cells • We can use the “spontaneity” of the reaction to do electrical work

2. (Continued) • We can “push” electrons through a cell in order to make a nonspontaneous redox reaction occur. • Electrolysis cell • Doing work to “force” chemical reaction to occur [opposite of voltaic cell]

3. Balancing Redox Equations • Deferred until later • For now just know that: • A half reaction has electrons written as a reactant or a product • Oxidation half reaction: A reactant gets oxidized (loses electrons); electrons appear as a “product” • Reduction half reaction: A reactant gets reduced (gains electrons); electrons appear as a “reactant” • A balanced redox equation does not show electrons explicitly. #e-’s lost = #e-’s gained (called “n”)

4. Voltaic Cells • Recall lab…

6. The spontaneous rxn occurs in the cell • e-’s flow from – to + (“get to go where they want to go”) • Anode = where ox occurs • Cathode = where red occurs • Salt bridge prevents charge buildup (which would stop flow)

8. Used when neither redox species in a half reaction (or electrode) is a neutral metal. You used graphite in place of Pt in lab for Fe2+/Fe3+ and I2/I-cells. A cheaper “inert” electrode. Neither is a neutral metal

9. Standard Reduction Potentials (E°red) • Recall lab • Make a bunch of different cells, get different Ecell values (Eºcell if at standard state). • Clearly some reductions are more favorable than others • How do you know? [Which direction did e-’s flow?] • By how much? • Rank them? (Must pick a zero as reference.)

10. Quick quiz • NOTE: Every electrode compartment has one oxidizing agent and one reducing agent (this pair is called the redox “couple”) • If an electrode has Ni(s) and Ni2+ ions in it, which species is the oxidizing agent and which the reducing agent (of the pair)? • Ni2+ (b/c it’s “more positive”; it has “room for an electron” • Ox agent is ___ • Ni (b/c it’s “more negative”; it has an electron to give) • Red agent is ___

11. Revisit Earlier Cell—Look at this as a “Competition for the electrons”. Which oxidizing agent “wants them more”? Who is the (possible) oxidizing agent on the left? _____ Who is the (possible) oxidizing agent on the right? _____ Zn2+ Cu2+ Hint: The two “players” are Zn and Zn2+ Hint: The two “players” are Cu and Cu2+. Who wins? (Which one “got” the electrons?) ____ Cu2+  Cu2+ “pulled harder” Cu2+ + 2 e- Cu(s) So…which of the half reactions shown at the right is more favorable(greater tendency to happen)? Zn2+ + 2 e- Zn(s) By how much?.....

12. Reducing Cu2+ is more favorable than reducing Zn2+ …by 1.10 V! (Measure it w/voltmeter!) We define a “standard reduction potential”, E°red, for every reduction half reaction such that: E°cell = E°red(cathode) - E°red(anode) The more positive the “E” (Ecell, Ered, or Eox), the more favorable the process

13. Reducing Cu2+ is more favorable than reducing Zn2+ …by 1.10 V! (Measure it w/voltmeter!) E°cell = E°red(cathode) - E°red(anode) 1.10 V = E°red(Cu2+/Cu) - E°red(Zn2+/Zn) NOTE: If E°red(Zn2+/Zn) were 0 V, E°red(Cu2+/Cu) would be +1.10 V If E°red(Zn2+/Zn) were -1.0 V, E°red(Cu2+/Cu) would be +0.10 V If E°red(Zn2+/Zn) were +1.0 V, E°red(Cu2+/Cu) would be +2.10 V The “zero” is arbitrary, but must be chosen / agreed upon!

14. This was chosen to be the “zero” of potential. 2 H+ + 2 e- H2 (g) Ered = 0.0 V

15. Could use this info to predict that this direct reaction would occur: 0.76 V = E°red(SHE) - E°red(Zn2+/Zn)  0.76 V = 0- E°red(Zn2+/Zn)  E°red(Zn2+/Zn) = - 0.76 V Both as reductions Ecell=Ered + Eox OR Ecell=Ecathode - Eanode

18. Table 18.1 (continued)

19. Calculate Ecell (see board)

20. Lab interlude • See overhead / board • The lab manual initially asks you to pretend that the Ag+/Ag reduction potential is 0.0 V just to show you the “arbitraryness” of this. • Then it tells you that in reality, Ag+/Ag reduction potential is 0.80 V if the H+/H2 potential (SHE) is 0.0 V

21. Cu/Cu2+ & Fe3+/Fe2+ +0.45 V -0.13 V Cu/Cu2+ 0.32 V Cu  Cu2+ + 2e- Fe3+ + e- Fe2+ Zn/Zn2+ & Ag/Ag+ 1.50 V Zn/Zn2+ Ag+ + e- Ag 0.0 V 1.50 V Zn  Zn2+ + 2e- Cu/Cu2+ & Zn/Zn2+ 1.05 V Zn/Zn2+ 1.50 V -0.45 V Zn  Zn2+ + 2e- Cu2+ + 2e- Cu **Circle the species that is the better oxidizing agent**

22. From Text Table 0.80 V 0.0 V Ag+ + e- Ag 0.80 V 0.77 V 0.67 V -0.13 V Fe3+ + e- Fe2+ Fe3+ + e- Fe2+ -0.13 V 0.34 V 0.35 V -0.45 V Cu2+ + 2e- Cu Ag+ + e- Ag 0.0 V -0.76 V -0.70 V -1.50 V Zn2+ + 2e-  Zn Cu2+ + 2e- Cu -0.45 V Zn  Zn2+ + 2e- 1.50 V

24. Table 18.1 (continued)

25. Excerpt from Voltaic Cell lab reading

26. Calculate Ecell (see board)

27. Calculate Ecell (see board)

28. Nernst Equation • See handout

29. Standard vs. Nonstandard Cell

30. Explain in detail an in a conceptual way why the cell potential goes up when the NH3 is added. Is the driving force for the cell rxn greater or smaller after the NH3 is added?

31. Relationship between variables (at any conditions) DG=DG +RT ln Q Q

32. Relationship between variables (at standard state conditions)

47. What mass of gold is plated in 25 minutes if the current is 5.5 A? Au3+ (aq) + 3 e- Au(s)