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 [Experiment 32!] –We can use the “spontaneity” of the reaction to do electrical work 1

2. (Continued) We can “push” electrons through a cell in order to make a nonspontaneous redox reaction occur. –Electrolysis cell [Experiment 32!] –Doing work to “force” chemical reaction to occur [opposite of voltaic cell] 2

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”) 3

4. Voltaic Cells Recall lab… 4

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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) 6

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8. You used graphite in place of Pt in lab for Fe 2+ /Fe 3+ and I 2 /I - cells. A cheaper “inert” electrode. Used when neither redox species in a half reaction (or electrode) is a neutral metal. Neither is a neutral metal 8

9. Standard Reduction Potentials (E° red ) Recall lab –Make a bunch of different cells, get different E cell values (Eº cell 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.) 9

10. Quick quiz NOTE: Every electrode compartment has one oxidizing agent and one reducing agent (this pair is called the redox “couple”) Ox agent is ___ Red agent is ___ Ni (b/c it’s “more negative”; it has an electron to give) Ni 2+ (b/c it’s “more positive”; it has “room for an electron” If an electrode has Ni(s) and Ni 2+ ions in it, which species is the oxidizing agent and which the reducing agent (of the pair)? 10

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

12. Reducing Cu 2+ is more favorable than reducing Zn 2+ … 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” (E cell, E red, or E ox ), the more favorable the process Where reduction takes place 12

13. Reducing Cu 2+ is more favorable than reducing Zn 2+ … by 1.10 V! (Measure it w/voltmeter!) E° cell = E° red (cathode) - E° red (anode) If E° red (Zn 2+ /Zn) were 0 V, E° red (Cu 2+ /Cu) would be +1.10 V If E° red (Zn 2+ /Zn) were -1.0 V, E° red (Cu 2+ /Cu) would be +0.10 V NOTE: 1.10 V = E° red (Cu 2+ /Cu) - E° red (Zn 2+ /Zn) If E° red (Zn 2+ /Zn) were +1.0 V, E° red (Cu 2+ /Cu) would be +2.10 V  The “zero” is arbitrary, but must be chosen / agreed upon! 13

14. This electrode (SHE) was ultimately the one chosen by the scientific community to be the “zero” of potential. 2 H + + 2 e -  H 2 (g); E° red = 0.0 V Upshot: One can determine any E° red experimentally by just setting up a cell where one of the half cells is SHE! (next slide → ) 14

15. E cell  E cathode - E anode Both as reductions 0.76 V = E° red (SHE) - E° red (Zn 2+ /Zn)  0.76 V = 0 - E° red (Zn 2+ /Zn)  E° red (Zn 2+ /Zn) = - 0.76 V  The reduction of H + is more favorable than the reduction of Zn 2+ …. by 0.76 V! H + (not Zn 2+ ) gets reduced Determining a Standard Reduction Potential using the SHE 15

16. 2 H + (aq) + 2 e - H 2 (g) 0 Zn 2+ (aq) + 2 e - Zn(s) -0.76 Could use this info to predict that this direct reaction would occur: E° cell  E° red (cat) - E° red (an) Use these values to: predict which reactions are spontaneous at standard state and to find any E° cell ! E° cell  0 – (-0.76) = +0.76 V H + gets reduced; Zn 2+ does not (Zn gets oxidized): 2 H + + Zn → H 2 (g) + Zn 2+ is spontaneous: E° cell > 0 16

17. Refers only to species on the left side of the arrow. E.g., F 2, is a better ox agent than H 2 O 2 which is better ox agent than Au 3+ (but all of these species are very good oxidizing agents relative to most!) Refers only to species on the right side of the arrow. E.g., F -, is a poorer red agent than H 2 O which is poorer red agent than Au(s) (but all of these are very poor reducing agents relative to most!) 17

18. Excerpt from Voltaic Cell lab reading 18

19. Table 18.1 (continued) 19

20. Recall earlier slide We define a “standard reduction potential”, E° red, for every reduction half reaction such that: E° cell = E° red (cathode) - E° red (anode) E cell  E red + E ox OR (could also write E cell as… 20

21. 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 “arbitrary- ness” of this. Then it tells you that in reality, Ag + /Ag reduction potential is 0.80 V if the H + /H 2 potential (SHE) is 0.0 V 21

22. Cu/Cu 2+ & Zn/Zn 2+ Cu/Cu 2+ & Fe 3+ /Fe 2+ Zn/Zn 2+ & Ag/Ag + **Circle the species that is the better oxidizing agent** 1.50 VZn/Zn 2+ 1.05 VZn/Zn 2+ 0.32 V Cu/Cu 2+ Cu  Cu 2+ + 2e - Cu 2+ + 2e -  Cu Zn  Zn 2+ + 2e - Fe 3+ + e -  Fe 2+ Ag + + e -  Ag 0.0 V 1.50 V -0.45 V +0.45 V-0.13 V 22

23. Ag + + e -  Ag 0.0 V Fe 3+ + e -  Fe 2+ -0.13 V Cu 2+ + 2e -  Cu -0.45 V Zn 2+ + 2e -  Zn -1.50 V 0.80 V 0.67 V 0.35 V -0.70 V 0.80 V 0.77 V 0.34 V -0.76 V From Text Table (See next slide) Zn 2+ + 2e -  Zn -1.50 V Flip Cu 2+ + 2e -  Cu Zn  Zn 2+ + 2e - Fe 3+ + e -  Fe 2+ Ag + + e -  Ag 0.0 V 1.50 V -0.45 V -0.13 V From Table A.1: 23

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25. Table 18.1 (continued) 25

26. Determine the cell reaction, calculate E  cell, identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board) Ni 2+ (aq) + 2 e - Ni(s) -0.23 V Mn 2+ (aq) + 2 e - Mn(s) -1.18 V The better oxidizing agent is:___ So ___ actually gets reduced, and thus electrons flow to the _____ __ side, which must be the ____ode. So the Ni electrode must be ______ive E° cell = _____ - _____ = _______ Ni 2+ right cath posit -0.23 V -1.18 V +0.95 V 26 (Because its E red is more positive)

27. Determine the cell reaction, calculate E  cell, identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board) Fe 2+ (aq) + 2 e - Fe(s) -0.45 V Mg 2+ (aq) + 2 e - Mg(s) -2.37 V The better oxidizing agent is:___ So ___ actually gets reduced, and thus electrons flow to the _____ __ side, which must be the ____ode. So the Fe electrode must be ______ive E° cell = _____ - _____ = _______ Fe 2+ right cath posit -0.45 V -2.37 V +1.92 V 27

28. Salt bridge 1 M Fe 2+ 1 M Pb 2+ ee Fe 2+ (aq) + 2 e - Fe(s) -0.45 V Pb 2+ (aq) + 2 e - Pb(s) -0.13 V The better oxidizing agent is:___ So ___ actually gets reduced, and thus electrons flow to the _____ __ side, which must be the ____ode. So the Pb electrode must be ______ive E° cell = _____ - _____ = _______ Pb 2+ left cath posit -0.13 V -0.45 V +0.32 V Pb(s) Fe(s) Determine the cell reaction, calculate E  cell, identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board) 28

29. See last week’s pink lab handout (Voltaic Cells), board, and below –Start with  G =  G  + RTlnQ and substitute in  G = -nFE cell and  G  = -nFE  cell After some algebra (and substituting in values for R, assuming T = 298 K, and converting to a base 10 log): Nernst Equation (T = 25  C) 29

30. Standard vs. Nonstandard Cell 30 **Always write out the balanced redox equation before using the Nernst Equation or predicting whether E cell should increase or decrease.** Zn + Cu 2+  Zn 2+ + Cu; Q = ?? Recall lab—adding NH 3 to Cu 2+ side! (T = 25  C)

31. Explain in detail an in a conceptual way why the cell potential goes up when the NH 3 is added. Is the driving force for the cell rxn greater or smaller after the NH 3 is added? 31

32. Relationship between variables (at any conditions; Mines Fig)  G   G   RT ln Q Q 32 (T = 25  C)

33. Relationship between variables (at standard state conditions; From Tro) 33 (T = 25  C)

34. 34 EXAMPLE 18.8 Calculating E cell Under Nonstandard Conditions Determine the cell potential for an electrochemical cell based on the following two half-reactions: Oxidation: Reduction: E° cell = E° red (cat) - E° red (an) OR E° cell = E° red (cat) + E° ox (an) **Need to write balanced equation before using Nernst! What will “n” be here?**

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44. 44 Fig. 18.22

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48. What mass of gold is plated in 25 minutes if the current is 5.5 A? Au 3+ (aq) + 3 e -  Au (s) 48

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