1. Chapter 18 Electrochemistry Lesson 2
2007, Prentice Hall

2. Electrochemistry 18.1 Balancing Oxidation–Reduction Reactions
Galvanic Cells
Standard Reduction Potentials
18.4 Cell Potential, Electrical Work, and Free Energy
18.5 Dependence of Cell Potential on Concentration
18.6 Batteries
18.7 Corrosion
18.8 Electrolysis
18.9 Commercial Electrolytic Processes

3. What about non-standard conditions?
Free Energy
Under standard conditions,
G = −nFE
What about non-standard conditions?

4. Nernst Equation Remember that G = G + RT ln Q This means
−nFE = −nFE + RT ln Q

5. Nernst Equation
Dividing both sides by −nF, we get the Nernst equation:
E = E − RT nF
ln Q
or, using base-10 logarithms,
E = E −
2.303 RT nF
log Q

6. Nernst Equation At room temperature (298 K), 2.303 RT F = 0.0592 V
Thus the equation becomes
E = E − V n
log Q

7. Nonstandard Conditions - the Nernst Equation
DG = DG° + RT ln Q
E = E° - (0.0592/n) log Q at 25°C
when Q = K, E = 0
use to calculate E when concentrations not 1 M
Tro, Chemistry: A Molecular Approach

8. Concentration Cells Ecell 
Notice that the Nernst equation implies that a cell could be created that has the same substance at both electrodes.
For such a cell,
would be 0, but Q would not.
Ecell 
Therefore, as long as the concentrations are different, E will not be 0.

9. Example 18.6- Calculate DG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
Given:
Find:
I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)
DG°, (J)
Concept Plan:
Relationships:
E°ox, E°red
E°cell
DG°
Solve:
ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 v
red: I2(l) + 2 e− → 2 I−(aq) E° = v
tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 v
Answer:
since DG° is +, the reaction is not spontaneous in the forward direction under standard conditions
Tro, Chemistry: A Molecular Approach

10. Example 18.7- Calculate K at 25°C for the reaction Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq)
Given:
Find:
Cu(s) + 2 H+(aq) → H2(g) + Cu2+(aq) K
Concept Plan:
Relationships:
E°ox, E°red
E°cell K
Solve:
ox: Cu(s) → Cu2+(aq) + 2 e− E° = −0.34 v
red: 2 H+(aq) + 2 e− → H2(aq) E° = v
tot: Cu(s) + 2H+(aq) → Cu2+(aq) + H2(g) E° = −0.34 v
Answer:
since K < 1, the position of equilibrium lies far to the left under standard conditions
Tro, Chemistry: A Molecular Approach

11. E at Nonstandard Conditions
Tro, Chemistry: A Molecular Approach

12. Example Calculate Ecell at 25°C for the reaction 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
Given:
Find:
3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + Cu2+(aq) + 4 H2O(l)
[Cu2+] = M, [MnO4−] = 2.0 M, [H+] = 1.0 M
Ecell
Concept Plan:
Relationships:
E°ox, E°red
E°cell
Ecell
Solve:
ox: Cu(s) → Cu2+(aq) + 2 e− }x3 E° = −0.34 v
red: MnO4−(aq) + 4 H+(aq) + 3 e− → MnO2(s) + 2 H2O(l) }x2 E° = v
tot: 3 Cu(s) + 2 MnO4−(aq) + 8 H+(aq) → 2 MnO2(s) + 3Cu2+(aq) + 4 H2O(l)) E° = v
Check:
units are correct, Ecell > E°cell as expected because [MnO4−] > 1 M and [Cu2+] < 1 M
Tro, Chemistry: A Molecular Approach

13. Concentration Cells
it is possible to get a spontaneous reaction when the oxidation and reduction reactions are the same, as long as the electrolyte concentrations are different
the difference in energy is due to the entropic difference in the solutions
the more concentrated solution has lower entropy than the less concentrated
electrons will flow from the electrode in the less concentrated solution to the electrode in the more concentrated solution
oxidation of the electrode in the less concentrated solution will increase the ion concentration in the solution – the less concentrated solution has the anode
reduction of the solution ions at the electrode in the more concentrated solution reduces the ion concentration – the more concentrated solution has the cathode
Tro, Chemistry: A Molecular Approach

14. Concentration Cell
when the cell concentrations are different, electrons flow from the side with the less concentrated solution (anode) to the side with the more concentrated solution (cathode)
when the cell concentrations are equal there is no difference in energy between the half-cells and no electrons flow
Cu(s) Cu2+(aq) (0.010 M)  Cu2+(aq) (2.0 M) Cu(s)
Tro, Chemistry: A Molecular Approach

15. LeClanche’ Acidic Dry Cell
electrolyte in paste form
ZnCl2 + NH4Cl
or MgBr2
anode = Zn (or Mg)
Zn(s) ® Zn2+(aq) + 2 e-
cathode = graphite rod
MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-
® 2 NH4OH(aq) + 2 Mn(O)OH(s)
cell voltage = 1.5 v
expensive, nonrechargeable, heavy, easily corroded
Tro, Chemistry: A Molecular Approach

16. Alkaline Dry Cell
same basic cell as acidic dry cell, except electrolyte is alkaline KOH paste
anode = Zn (or Mg)
Zn(s) ® Zn2+(aq) + 2 e-
cathode = brass rod
MnO2 is reduced
2 MnO2(s) + 2 NH4+(aq) + 2 H2O(l) + 2 e-
® 2 NH4OH(aq) + 2 Mn(O)OH(s)
cell voltage = 1.54 v
longer shelf life than acidic dry cells and rechargeable, little corrosion of zinc
Tro, Chemistry: A Molecular Approach

17. PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e-
Lead Storage Battery
6 cells in series
electrolyte = 30% H2SO4
anode = Pb
Pb(s) + SO42-(aq) ® PbSO4(s) + 2 e-
cathode = Pb coated with PbO2
PbO2 is reduced
PbO2(s) + 4 H+(aq) + SO42-(aq) + 2 e-
® PbSO4(s) + 2 H2O(l)
cell voltage = 2.09 v
rechargeable, heavy
Tro, Chemistry: A Molecular Approach

18. NiCad Battery electrolyte is concentrated KOH solution anode = Cd
Cd(s) + 2 OH-1(aq) ® Cd(OH)2(s) + 2 e-1 E0 = 0.81 v
cathode = Ni coated with NiO2
NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0 = 0.49 v
cell voltage = 1.30 v
rechargeable, long life, light – however recharging incorrectly can lead to battery breakdown
Tro, Chemistry: A Molecular Approach

19. Ni-MH Battery electrolyte is concentrated KOH solution
anode = metal alloy with dissolved hydrogen
oxidation of H from H0 to H+1
M∙H(s) + OH-1(aq) ® M(s) + H2O(l) + e-1 E° = 0.89 v
cathode = Ni coated with NiO2
NiO2 is reduced
NiO2(s) + 2 H2O(l) + 2 e-1 ® Ni(OH)2(s) + 2OH-1 E0 = 0.49 v
cell voltage = 1.30 v
rechargeable, long life, light, more environmentally friendly than NiCad, greater energy density than NiCad
Tro, Chemistry: A Molecular Approach

20. Lithium Ion Battery electrolyte is concentrated KOH solution
anode = graphite impregnated with Li ions
cathode = Li - transition metal oxide
reduction of transition metal
work on Li ion migration from anode to cathode causing a corresponding migration of electrons from anode to cathode
rechargeable, long life, very light, more environmentally friendly, greater energy density
Tro, Chemistry: A Molecular Approach

21. Fuel Cells
like batteries in which reactants are constantly being added
so it never runs down!
Anode and Cathode both Pt coated metal
Electrolyte is OH– solution
Anode Reaction:
2 H2 + 4 OH–
→ 4 H2O(l) + 4 e-
Cathode Reaction:
O2 + 4 H2O + 4 e-
→ 4 OH–
Tro, Chemistry: A Molecular Approach

22. Electrolytic Cell
uses electrical energy to overcome the energy barrier and cause a non-spontaneous reaction
must be DC source
the + terminal of the battery = anode
the - terminal of the battery = cathode
cations attracted to the cathode, anions to the anode
cations pick up electrons from the cathode and are reduced, anions release electrons to the anode and are oxidized
some electrolysis reactions require more voltage than Etot, called the overvoltage
Tro, Chemistry: A Molecular Approach

23. Tro, Chemistry: A Molecular Approach

24. electroplating In electroplating, the work piece is the cathode.
Cations are reduced at cathode and plate to the surface of the work piece.
The anode is made of the plate metal. The anode oxidizes and replaces the metal cations in the solution
Tro, Chemistry: A Molecular Approach

25. Electrochemical Cells
in all electrochemical cells, oxidation occurs at the anode, reduction occurs at the cathode
in voltaic cells,
anode is the source of electrons and has a (−) charge
cathode draws electrons and has a (+) charge
in electrolytic cells
electrons are drawn off the anode, so it must have a place to release the electrons, the + terminal of the battery
electrons are forced toward the anode, so it must have a source of electrons, the − terminal of the battery
Tro, Chemistry: A Molecular Approach

26. Electrolysis
electrolysis is the process of using electricity to break a compound apart
electrolysis is done in an electrolytic cell
electrolytic cells can be used to separate elements from their compounds
generate H2 from water for fuel cells
recover metals from their ores
Tro, Chemistry: A Molecular Approach

27. Electrolysis of Water
Tro, Chemistry: A Molecular Approach

28. Electrolysis of Pure Compounds
must be in molten (liquid) state
electrodes normally graphite
cations are reduced at the cathode to metal element
anions oxidized at anode to nonmetal element
Tro, Chemistry: A Molecular Approach

29. Electrolysis of NaCl(l)
Tro, Chemistry: A Molecular Approach

30. Mixtures of Ions
when more than one cation is present, the cation that is easiest to reduce will be reduced first at the cathode
least negative or most positive E°red
when more than one anion is present, the anion that is easiest to oxidize will be oxidized first at the anode
least negative or most positive E°ox
Tro, Chemistry: A Molecular Approach

31. Electrolysis of NaI(aq) with Inert Electrodes
possible oxidations
2 I-1 ® I2 + 2 e-1 E° = −0.54 v
2 H2O ® O2 + 4e-1 + 4H+1 E° = −0.82 v
possible oxidations
2 I-1 ® I2 + 2 e-1 E° = −0.54 v
2 H2O ® O2 + 4e-1 + 4H+1 E° = −0.82 v
possible reductions
Na+1 + 1e-1 ® Na0 E° = −2.71 v
2 H2O + 2 e-1 ® H2 + 2 OH-1 E° = −0.41 v
possible reductions
Na+1 + 1e-1 ® Na0 E° = −2.71 v
2 H2O + 2 e-1 ® H2 + 2 OH-1 E° = −0.41 v
overall reaction
2 I−(aq) + 2 H2O(l) ® I2(aq) + H2(g) + 2 OH-1(aq)
Tro, Chemistry: A Molecular Approach

32. Faraday’s Law
the amount of metal deposited during electrolysis is directly proportional to the charge on the cation, the current, and the length of time the cell runs
charge that flows through the cell = current x time
Tro, Chemistry: A Molecular Approach

33. Example Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction Au3+(aq) + 3 e− → Au(s)
Given:
Find:
3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min
mass Au, g
Concept Plan:
Relationships:
t(s), amp
charge (C)
mol e−
mol Au
g Au
Solve:
Check:
units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−
Tro, Chemistry: A Molecular Approach

34. Corrosion
corrosion is the spontaneous oxidation of a metal by chemicals in the environment
since many materials we use are active metals, corrosion can be a very big problem
Tro, Chemistry: A Molecular Approach

35. Rusting rust is hydrated iron(III) oxide moisture must be present
water is a reactant
required for flow between cathode and anode
electrolytes promote rusting
enhances current flow
acids promote rusting
lower pH = lower E°red
Tro, Chemistry: A Molecular Approach

36. Tro, Chemistry: A Molecular Approach

37. Preventing Corrosion
one way to reduce or slow corrosion is to coat the metal surface to keep it from contacting corrosive chemicals in the environment
paint
some metals, like Al, form an oxide that strongly attaches to the metal surface, preventing the rest from corroding
another method to protect one metal is to attach it to a more reactive metal that is cheap
sacrificial electrode
galvanized nails
Tro, Chemistry: A Molecular Approach

38. Sacrificial Anode
Tro, Chemistry: A Molecular Approach