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

2. (Continued)
We can “push” electrons through a cell in order to make a nonspontaneous redox reaction occur.
Electrolysis cell [Experiment 31!]
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…

5. Chapter 18, Figure 18.1
Spontaneous Oxidation–Reduction Reaction

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)
Chapter 18, Figure 18.2
A Voltaic Cell

7. Chapter 18, Figure 18.3
An Analogy for Electrical Current

8. Neither is a neutral metal
Used when neither redox species in a half reaction (or electrode) is a neutral metal. (i.e.both are solution species)
You used graphite in place of Pt in lab for Fe3+/Fe2+ and I2/I-cells. A cheaper “inert” electrode.
Chapter 18, Figure 18.4
Inert Platinum 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)?
Ox agent is ___
Ni2+ (b/c it’s “more positive”; it has “room for an electron”
Ni (b/c it’s “more negative”; it has an electron to give)
Red agent is ___

11. Board Work / Intro

12. Revisit Earlier Cell—Look at this as a “Competition for the electrons”
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”
Cu e-  Cu(s)
So…which of the half reactions shown at the right is more favorable (greater tendency to happen)?
Zn e-  Zn(s)
By how much?.....

13. Where reduction takes place
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)
Where reduction takes place
The more positive the “E” (Ecell, Ered, or Eox), the more favorable the process

14. Reducing Cu2+ is more favorable than reducing Zn2+ …by 1. 10 V
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 V
If E°red(Zn2+/Zn) were -1.0 V, E°red(Cu2+/Cu) would be V
If E°red(Zn2+/Zn) were +1.0 V, E°red(Cu2+/Cu) would be V
 The “zero” is arbitrary, but must be chosen / agreed upon!

15. This electrode (SHE) was ultimately the one chosen by the scientific community to be the “zero” of potential.
2 H+ + 2 e-  H2 (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 → )
Chapter 18, Figure 18.6
The Standard Hydrogen Electrode

16. 0.76 V = E°red(SHE) - E°red(Zn2+/Zn)
Determining a Standard Reduction Potential using the SHE
0.76 V = E°red(SHE) - E°red(Zn2+/Zn)
 0.76 V = 0 - E°red(Zn2+/Zn)
 E°red(Zn2+/Zn) = V
Chapter 18, Figure 18.7
Measuring Electrode Potential
Both as reductions
Ecell = Ecathode - Eanode
The reduction of H+ is more favorable than the reduction of Zn2+….
by 0.76 V!
H+ (not Zn2+) gets reduced

17. Could use this info to predict that this direct reaction would occur:
Use these values to:
predict which reactions are spontaneous at standard state
and to
find any E°cell!
E°cell = E°red(cat) - E°red(an)
2 H+(aq) + 2 e H2(g)
Click to add notes
E°cell = 0 – (-0.76) = V
H+ gets reduced; Zn2+ does not (Zn gets oxidized):
Zn2+(aq) + 2 e Zn(s)
2 H+ + Zn → H2(g) + Zn2+
is spontaneous: E°cell > 0

18. Refers only to species on the left side of the arrow. E. g
Refers only to species on the left side of the arrow. E.g., F2, is a better ox agent than H2O2 which is better ox agent than Au3+ (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 H2O which is poorer red agent than Au(s) (but all of these are very poor reducing agents relative to most!)
Click to add notes

19. Excerpt from Voltaic Cell lab reading

20. Table 18.1 (continued)
Click to add notes

21. 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)
OR (could also write Ecell as…
Ecell = Ered + Eox

22. Lab interlude
**Example on next slide is not with the same cells that you used! It illustrates the idea and “procedure” only!**
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+/H2 potential (SHE) is 0.0 V

23. Cu2+/Cu & Fe3+/Fe2+ 0.32 V Cu2+/Cu +0.45 V -0.13 V Zn2+/Zn & Ag+/Ag
Determined by the color of the leads when voltmeter gave positive reading.
Electrons flow away from (-) electrode  Oxidation occurs there. RA loses e-s
Cu2+/Cu & Fe3+/Fe2+
0.32 V
Cu2+/Cu
Cu  Cu2+ + 2e-
+0.45 V
Fe3+ + e- Fe2+
-0.13 V
Zn2+/Zn & Ag+/Ag
1.50 V
Zn2+/Zn
Zn  Zn2+ + 2e-
1.50 V
Ag+ + e- Ag
0.0 V
Cu2+/Cu & Zn2+/Zn
1.05 V
Zn2+/Zn
Zn  Zn2+ + 2e-
1.50 V
Cu2+ + 2e- Cu
-0.45 V
**Circle the species (not the “electrode”!) that is the better oxidizing agent**

24. From Text Table 0.0 V 1.50 V -0.45 V -0.13 V 0.0 V 0.80 V 0.80 V
(See next slide)
Cu2+ + 2e- Cu
Zn  Zn2+ + 2e-
Fe3+ + e- Fe2+
Ag+ + e- Ag
0.0 V
1.50 V
-0.45 V
-0.13 V
From Table A.1:
0.0 V
Ag+ + e- Ag
0.80 V
0.80 V
-0.13 V
Fe3+ + e- Fe2+
0.67 V
0.77 V
-0.45 V
Cu2+ + 2e- Cu
0.35 V
0.34 V
-1.50 V
Zn2+ + 2e-  Zn
-0.70 V
-0.76 V
Flip
Zn2+ + 2e-  Zn
-1.50 V

25. Click to add notes

26. Table 18.1 (continued)
Click to add notes

27. State the better oxidizing agent, determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow
Ni2+(aq) + 2 e Ni(s) V
Mn2+(aq) + 2 e Mn(s) V
The better oxidizing agent is:___
Ni2+
(Because its Ered is more positive)
So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.
Ni2+
right
Chapter 18, Figure 18.8
Mn/Ni2+ Electrochemical Cell
cath
So the Ni electrode must be ______ive
posit
E°cell = _____ - _____
-0.23 V
-1.18 V
OR E°cell = _____ + _____
-0.23 V
-(-1.18 V)
= _______
+0.95 V

28. State the better oxidizing agent, determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow
Fe2+(aq) + 2 e Fe(s) V
Pb2+(aq) + 2 e Pb(s) V e
The better oxidizing agent is:___
Pb(s)
Fe(s)
Pb2+
So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.
Pb2+
Salt
bridge
Chapter 18, Figure 18.10
Mn/Ni2+ Electrochemical Cell
left
cath
So the Pb electrode must be ______ive
1 M Fe2+
1 M Pb2+
posit
E°cell = _____ - _____
-0.13 V
-0.45 V
= _______
+0.32 V

29. Determine the cell reaction, calculate Ecell , identify the cathode and anode, label the (+) and (-) electrodes, and show electron flow (see board)
Fe2+(aq) + 2 e Fe(s) V
Mg2+(aq) + 2 e Mg(s) V
The better oxidizing agent is:___
Fe2+
So ___ actually gets reduced, and thus electrons flow to the _______ side, which must be the ____ode.
Fe2+
Chapter 18, Figure 18.8
Mn/Ni2+ Electrochemical Cell
right
cath
So the Fe electrode must be ______ive
posit
E°cell = _____ - _____
-0.45 V
-2.37 V
= _______
+1.92 V

30. Electrical Quantities

31. Relation of Ecell to DG Done on board. Look at units!
Think about meaning of voltage!
Explains why we don’t multiply through the Ered by a coefficient in a half reaction!

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

33. 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?

34. Standard vs. Nonstandard Cell
Zn + Cu2+  Zn2+ + Cu; Q = ??
Recall lab—adding NH3 to Cu2+ side!
Chapter 18, Figure 18.11
Cell Potential and Concentration
(T = 25C)
**Always write out the balanced redox equation before using the Nernst Equation or predicting whether Ecell should increase or decrease!!**

35. Relationship between variables (at any conditions; Mines Fig)
DG = DG + RT ln Q Q
Chapter 18, Unnumbered Figure, Page 835
(T = 25C)

36. Relationship between variables (at standard state conditions; From Tro)
Chapter 18, Unnumbered Figure, Page 835
(T = 25C)

37. E°cell = E°red(cat) - E°red(an) OR E°cell = E°red(cat) + E°ox(an)
EXAMPLE Calculating Ecell 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)
Chapter 18, Unnumbered Figure, Page 837
**Need to write balanced equation before using Nernst! What will “n” be here?**

38. Chapter 18, Figure 18.12
Cu/Cu2+ Concentration Cell

39. Chapter 18, Figure 18.13
Concentration Changes in Nerve Cells

40. Chapter 18, Figure 18.14
Potential Changes across the Nerve Cell Membrane

41. Chapter 18, Figure 18.15A
Dry-Cell Batteries

42. Chapter 18, Figure 18.15B
Dry-Cell Batteries

43. Chapter 18, Figure 18.16
Lead–Acid Storage Battery

44. Chapter 18, Figure 18.17
Lithium Ion Battery

45. Chapter 18, Figure 18.18
Hydrogen-Oxygen Fuel Cell

46. Chapter 18, Figure 18.19
Fuel-Cell Breathalyzer

47. Fig
Chapter 18, Figure 18.22
Voltaic versus Electrolytic Cells

48. Chapter 18, Figure 18.20
Electrolysis of Water

49. Chapter 18, Figure 18.21
Silver Plating

50. Chapter 18, Figure 18.26
Electrolytic Cell for Copper Plating

51. What mass of gold is plated in 25 minutes if the current is 5.5 A?
Au3+ (aq) e-  Au(s)
Chapter 18, Unnumbered Figure, Page 852

52. Chapter 18, Figure 18.23
Electrolysis of Molten NaCl

53. Chapter 18, Figure 18.25
Electrolysis of Aqueous NaCl: The Effect of Overvoltage