1. Zn + Cu2+  Zn2+ + Cu

2. Applied Electrochemistry
Voltaic (or galvanic) cells: spontaneous
redox reaction  electricity (or electrical work)
Electrolytic cells: non-spontaneous
electricity  redox
To produce electricity, we must direct the electron flow through an external circuit. We cannot have direct redox.
Daniell cell
Zn – Cu rxn:

3. Lemon Battery

4. 17.1 Galvanic Cells
3:25

5. Galvanic Cells To produce electricity, we need:
Isolated half-reactions, using half-cells
Conductive solids (electrodes) connected by external circuits
May consist of a reactant/product or be an inert substance such as platinum or graphite
Anode: oxidation half-reaction
Cathode: reduction half-reaction

6. Galvanic Cells

7. Galvanic Cells
Externally, the anode has the negative charge; internally, it has a positive charge
Anions flow towards the anode; cations move away from it and towards the cathode.
Two half-cells must be connected to pass ions
Salt bridge or porous glass

8. Molecular View of Electrode Processes

9. 17.2 Cell Potentials
If the half-reactions are carried out separately (but coupled), we find they generate an electrical current characterized by a voltage

10. Cell Potentials
The voltage produced by a voltaic cell is called the cell potential, Eocell (also the reaction potential, Eorxn, when the half-reactions are not separated)
Under standard conditions, the voltage is also called the standard electromotive force (emf), Eo.
Unit = Volt (V)
Volt = 1 Joule of energy / coulomb of charge transferred = 1 J/C
J = Joule C = coulomb
1 mol e- = 96,500 coulombs

11. Cell Potentials
Reference for Eo is the standard hydrogen electrode, using the reaction:
2H+(aq) + 2e-  H2(g) Eo = 0 V = 0 J/C
1 M atm (std conditions)

12. Cell Potentials
Then get other half-reaction potentials from measured Eocell values.
Zn + 2H+  Zn2+ + H2 Eocell = 0.76 V
Zn  Zn2+ + 2e- Eoox = 0.76 V
2H+ + 2e-  H2 Eored = 0.00 V
How do these reactions
relate?

13. Cell Potentials
Cell potential values can thus be determined relative to the standard hydrogen electrode, with Eo = 0 V
Eored = cell potential for the reduction half-rxn
Eoox = cell potential for the oxidation half-rxn
Thus, Eocell = Eored + Eoox
We can measure Eocell, using the standard reference of 0 V, we can measure Eoox and Eored for half reactions paired with the H2 half reaction.

14. Determining Cell Potentials Values
If we reverse a half-reaction, what happens to the sign of Eo.
Zn  Zn2+ + 2e- Eoox = 0.76 V
Zn2+ + 2e-  Zn Eored = ?
Now consider data for Eocell = 0.63 V for the following reaction.
Zn + Pb2+  Zn2+ + Pb
What is the Eored of
Pb2+ + 2e-  Pb ?

15. Cell Potential Values Zn  Zn2+ + 2e- Eoox = 0.76 V
Pb2+ + 2e-  Pb Eored = ?
Zn + Pb2+  Zn2+ + Pb Eocell = 0.63 V
Eocell = Eored + Eoox
0.63 V = Eored V
Eored = 0.63 V V = V
Values determined in this way are listed in Table 17.1 and Appendix 5.5

16. Reduction Potentials

17. Relative Strengths of Oxidizing and Reducing Agents

18. Cell Potential
Reduction or oxidation values can also be measured from Eocell with other known half-cells.
AgCl + e-  Ag + Cl- Eredo = 0.22 V
Hg2Cl2 + 2e-  2Hg + 2Cl- Eredo = V
Half-cells such as these are used as reference electrodes. The Ag/AgCl electrode, along with a glass electrode, is used in a pH meter.

19. 17.2 Describing Galvanic Cells
What will happen if we place a piece of Zn and a piece of Cu in a solution that contains a mixture of Zn2+ and Cu2+?
Two possibilities:
Zn + Cu2+  Zn2+ + Cu
Cu + Zn2+  Cu2+ + Zn

20. Spontaneous Redox Two possible reduction half-reactions:
Zn2+ + 2e-  Zn Eored = V
Cu2+ + 2e-  Cu Eored = V
Two possible oxidation half-reactions:
Zn  Zn2+ + 2e- Eoox = V
Cu  Cu2+ + 2e- Eoox = V
One way to combine them:
Cu  Cu2+ + 2e- Eoox = V
———————— ————————
Cu + Zn2+  Cu2+ + Zn Eorxn = V

21. Spontaneous Redox Other way to combine them:
Cu2+ + 2e-  Cu Eored = 0.34 V
Zn  Zn2+ + 2e- Eoox = 0.76 V
———————— ———————
Zn + Cu2+  Zn2+ + Cu Eorxn = 1.10 V
Which combination is observed to be spontaneous.
skip

22. Group Quiz
(a) If we mix chromium(II) and vanadium(III) ions, will a reaction occur?
(b) If we mix chromium(III) and vanadium(II) ions, will a reaction occur?
(c) Write a balanced equation for any reaction that occurs.
(d) What is its E° ?
Cr3+(aq) + e-  Cr2+(aq) Eored = V
V3+(aq) + e-  V2+(aq) Eored = V

23. Predicting Reactions Vanadium Reduction Potentials:
VO2+ + 2H+ + e-  VO2+ + H2O E°red = V
VO2+ + 2H+ + e-  V3+ + H2O E°red = V
V3+ + e-  V E°red = V
V2+ + 2e-  V E°red = V

24. Predicting Reactions Chromium Reduction Potentials:
Cr2O H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l) E°red = V
Cr3+(aq) + e-  Cr2+(aq) E°red = V
Cr2+(aq) + 2e-  Cr(s) E°red = V

25. Predicting Reactions Manganese Reduction Potentials:
MnO4-(aq) + e-  MnO42-(aq) E°red = V
MnO42-(aq) + 4H+(aq) + 2e-  MnO2(s) + 2H2O(l) E°red = V
MnO2(s) + 4H+(aq) + e-  Mn3+(aq) + 2H2O(l) E°red = V
Mn3+(aq) + e-  Mn2+(aq) E°red = V
Mn2+(aq) + 2e-  Mn(s) E°red = V

26. Predicting Reactions Add excess Cr2+ to VO2+. What is the product?
Cr3+(aq) + e-  Cr2+(aq) E°red = V
Cr2+(aq)  Cr3+(aq) + e E°ox = V
VO2+ + 2H+ + e-  VO2+ + H2O E°red = V
VO2+ + 2H+ + e-  V3+ + H2O E°red = V
V3+ + e-  V E°red = V
V2+ + 2e-  V E°red = V
Product is V2+, because successive Eorxn = 1.41, 0.77, 0.15, V

27. Group Work: Predicting Reactions
Add excess Cr to Cr2O72-. What is the product?
Cr2O H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l) E°red = V
Cr3+(aq) + e-  Cr2+(aq) E°red = V
Cr2+(aq) + 2e-  Cr(s) E°red = V
Product is Cr2+; the successive values of Eorxn are 2.24 and 0.50 V.

28. Group Work: Predicting Reactions
Add excess Cr2+ to Cr2O72-. What is the product?
Cr2O H+(aq) + 6e-  2Cr3+(aq) + 7H2O(l) E°red = V
Cr3+(aq) + e-  Cr2+(aq) E°red = V
Cr2+(aq) + 2e-  Cr(s) E°red = V
Product is Cr3+; the value of Eorxn is 1.74 V.

29. Group Work: Predicting Reactions
Add excess V2+ to MnO4-. What is the product?
V2+  V3+ + e E°ox = 0.26 V
MnO4-(aq) + e-  MnO42-(aq) E°red = V
MnO42-(aq) + 4H+(aq) + 2e-  MnO2(s) + 2H2O(l) E°red = V
MnO2(s) + 4H+(aq) + e-  Mn3+(aq) + 2H2O(l) E°red = V
Mn3+(aq) + e-  Mn2+(aq) E°red = V
Mn2+(aq) + 2e-  Mn(s) E°red = V
Product is Mn2+

30. Group Work: Predicting Reactions
Add Mn2+ to MnO4-. What is the product?
MnO4-(aq) + e-  MnO42-(aq) E°red = V
MnO42-(aq) + 4H+(aq) + 2e-  MnO2(s) + 2H2O(l) E°red = V
MnO2(s) + 4H+(aq) + e-  Mn3+(aq) + 2H2O(l) E°red = V
Mn3+(aq) + e-  Mn2+(aq) E°red = V
Mn2+(aq) + 2e-  Mn(s) E°red = V

31. Group Work: Predicting Reactions
Predict no reaction, but reaction actually occurs to form Mn3+ or MnO2. We will deal with this discrepancy in the next section. It arises from the fact that there are more half-reactions that could be considered, which arise from combinations of these half-reactions.

32. Stability in Aqueous Systems
Disproportionation Reactions
A substance reacts with itself to form new substances with higher and lower oxidation numbers.
No examples with V or Cr.
MnO42- will disproportionate:
MnO4-(aq) + e-  MnO42-(aq) E°red = V
MnO42-(aq) + 4H+(aq) + 2e-  MnO2(s) + 2H2O(l) E°red = V
—————————————————————
3MnO H+  2MnO4- + MnO2 + 2H2O Eorxn = 1.70 V

33. Stability in Aqueous Systems
Mn3+ will disproportionate:
MnO2(s) + 4H+(aq) + e-  Mn3+(aq) + 2H2O(l) E°red = V
Mn3+(aq) + e-  Mn2+(aq) E°red = V
—————————————————————
2Mn3+ + 2H2O  Mn2+ + MnO2 + 4H Eorxn = 0.56 V

34. Stability in Aqueous Systems
Reaction with Water
Reduce hydronium ion to release hydrogen gas:
2H+(aq) + 2e-  H2(g) E°red = V
Any substance with Eoox > 0 will reduce H+ to H2
Examples are V, V2+, Cr, Cr2+, Mn
The ions will react, but tend to react only very slowly. There seems to be a kinetic factor that results in a fast reaction only if Eorxn > V (called an overvoltage).

35. Stability in Aqueous Systems
Reaction with Water
Oxidize water to release oxygen gas:
2H2O(l)  O2(g) + 4H+(aq) + 4e- E°ox = V
Any Eored > 1.23 V will result in production of O2. Generally need Eorxn > V for fast reaction.
Examples are Cr2O72- (very slow),
MnO42- (disproportionates faster),
Mn3+ (disproportionates faster)

36. Stability in Aqueous Systems
Oxidation by O2 in Air
O2(g) + 4H+(aq) + 4e-  2H2O(l) E°red = 1.23 V
Any Eoox > V will result in oxidation by air. Many substances fall into this category (Eorxn > V for fast reaction).
V Cr Mn
V2+ Cr2+ not Mn2+
V3+ not Cr3+ Mn3+
VO2+ (very slow) not MnO2
MnO (disproportionates faster)

37. Eocell and Spontaneity
We have seen three criteria for spontaneity:
Eo > 0
DGo < 0 1 V = 1 J/C, so 1 J = 1 C x 1 V
These criteria are related:
K >> 1
DGo = - RT lnK
DGo = - n F Eo, DG = - nFE
where n = number of e- transferred and
F = Faraday constant (charge on 1 mole e-)
1 F = 96,500 coul/mol e- = 96,500 J/V mol e-

38. Thermodynamics
These relationships work for half-reactions or complete redox reactions.
Zn + Cu2+  Zn2+ + Cu Eo = 1.10
n = 2
DGo = -2mol e- x J/V mol e- x 1.10 V
DGo = -212,300 J = kJ
DGo depends on the number of moles, but Eo does not

39. Voltage and Moles
Note that different size alkaline cells all deliver the same voltage, in spite of different number of moles of reactants.

40. Thermodynamics DGoox = - 2 x 96500 x 0.44 = -84900 J
We can add Eoox to Eored to give Eocell or Eorxn in the same way that we can add half-reactions to give an overall reaction.
Fe  Fe2+ + 2e- Eoox = V
DGoox = - 2 x x 0.44 = J
Cl2 + 2e-  2Cl- Eored = 1.36 V
DGored = - 2 x x 1.36 = J
Fe + Cl2  Fe2+ + 2Cl- Eorxn = 1.80 V
DGo = - 2 x x 1.80 = J
DGorxn = DGoox + DGored
= = J

41. Thermodynamics
From Chapter 16, we know that DGo values are additive when we add reactions.
Eos are additive when we add half-reactions to give a complete reaction because the value of n is the same for the half-reactions and the complete reaction.
Eos are not additive when adding two half-reactions to give a third half-reaction because the value of n is not constant.

42. Thermodynamics We can add DGo under all circumstances:
DGo3 = DGo1 + DGo2
-n3FEo3 = -n1FEo1 - n2FEo2
n3Eo3 = n1Eo1 + n2Eo2
Eo3 = (n1Eo1 + n2Eo2)/n3
V  V2+ + 2e- Eo1 = 1.20 V
V2+  V3+ + e- Eo2 = 0.26 V
V  V3+ + 3e- Eo3 <
Eo3 = (2 x x 0.26)/3 = V

43. Group Work
Given the first two half-reactions, what is the value of E°red for the third?
MnO4-(aq) + e-  MnO42-(aq) E°1,red = V
MnO42-(aq) + 4H+(aq) + 2e-  MnO2(s) + 2H2O(l) E°2.red = V
MnO4-(aq) + 4H+(aq) + 3e-  MnO2(s) + 2H2O(l) E°3,red = ?
E°3,red = 1.69 V
This is why we could not make correct predictions for this system.

44. Thermodynamics We can calculate Keq from Eo: DGo = - nFEo = - RT ln K
Eo = (RT/nF) ln K = (RT/nF) log K
At 25oC, RT/F =
Eo = ( /n) log K at 25oC
Thus, we can measure Eo for a redox reaction and then calculate the equilibrium constant for that reaction.

45. 17.4 Effect of Concentration on Cell EMF
So far, we have been using standard state conditions, but we don’t always have 1 M solutions. We can correct Eo to E by using the Nernst equation.
DG = DGo + RT ln Q
But, DG = - nFE and DGo = -nFEo, so
- nFE = -nFEo RT log Q
E = Eo - (2.303 RT/nF) log Q
At 25oC, E = Eo - ( /n) log Q

46. Nernst Equation
At 25oC, E = Eo - ( /n) log Q
E = Eo if Q = 1
When the system reaches equilibrium, Q = K, and E = 0, because Eo = ( /n) log K, and the cell has “run down”.
Consider the Zn/Cu2+ reaction if more Cu2+ is added to the cell. The voltage becomes greater than 1.10 V.
21m07an1

47. Nernst Equation
What is E of the Zn/Cu2+ reaction if [Cu2+] = M and [Zn2+] =1.99 M? Note that this corresponds to starting with standard conditions and changing to 99% completion of reaction. Eo = 1.10 V (with [Cu2+] = [Zn2+] = 1.00 M)
Zn + Cu2+  Zn2+ + Cu
For this reaction, n = 2.
Q = [Zn2+]/[Cu2+]

48. Nernst Equation E = 1.10 V - (0.05916/2) log (1.99/0.010)
E = Eo - ( /n) log Q
E = 1.10 V - ( /2) log (1.99/0.010)
E = 1.10 V - ( /2) log 199
E = = 1.03 V
For 99.9% reaction (1.999 M Zn2+, M Cu2+), E = = 1.00 V
For 99.99% reaction, M Zn2+, M Cu2+), E = = 0.97 V

49. Concentration Cells
We can generate a voltage with a cell that contains the same materials in the cathode and anode compartments, but at different concentrations.