1. 1
Chapter 18: Electrochemistry
Mid-term Exam 3
F March 27
Redox (Tro Ch. 4.9 & 18.2)
and Electrochemistry
(Tro Ch. 18.1, 18.3-18.9)

2. 2
Electrochemistry
 Electrochemical reactions are redox reactions occurring in
an electrochemical cell.
 The oxidation and reduction half-reactions are physically separated
into half-cells.
 Electrons flow in an external circuit.
 There are two kinds electrochemical cells:
 Non-spontaneous chemical reactions occur in electrolytic cells.
 Spontaneous chemical reactions occur in voltaic or galvanic cells.
 Electrolytic cells require input of electrical energy to force
the reaction to occur.
 Voltaic/galvanic cells release energy from the chemical
reaction as electrical energy.

3. 3
Electrolytic Cells
 Electrical energy is used to force non-
spontaneous chemical reactions to occur.
 The process is called electrolysis.
 Two examples of commercial electrolytic
reactions are:
 The electroplating of jewelry and auto parts.
 The electrolysis of chemical compounds to make
other useful compounds.

4. 4
Electrolysis of Molten Potassium Chloride
l

5. 5
Electrical Conduction and
Electrochemistry
 Metals are good conductors of electrical currents.
 In metallic conduction there is electron flow with no atomic
translational motion of the metal atoms in the electrode.
 Solid metals, can therefore, make good electrodes and
“wires”, (if they don’t participate in the redox reaction!)
 Electrodes are connected to each other by an external
circuit (conducting “wires”), allowing electron flow between
electrodes.
 Inert electrodes do not chemically react with the
liquids or products of the electrochemical
reaction but do conduct electrons.
 Graphite (a form of carbon) and platinum are two
commonly used inert electrodes.

6. 6
Electrolytic Cells
 In all electrolytic cells, electrons are forced
to flow from the positive electrode (anode)
to the negative electrode (cathode) by an
external source of electrical energy.
 In ionic or electrolytic conduction ionic motion
transports oxidizable or reducible species to the
electrodes.
 Positively charged ions, cations, move toward the negative
electrode.
 Negatively charged ions, anions, move toward the positive
electrode.

7. 7
Electrodes
 The following naming convention is used for
either electrolytic or voltaic cells:
 The cathode is the electrode at which
reduction occurs.
 The cathode is negative in electrolytic cells and positive
in voltaic cells.
 The anode is the electrode at which
oxidation occurs.
 The anode is positive in electrolytic cells and negative in
voltaic cells.

8. 8
Electrolysis of Molten Potassium
Chloride

9. 9
Electrolysis of Molten Potassium Chloride
l

10. 10
Electrolysis of Molten Potassium
Chloride
 The non-spontaneous redox reaction that occurs is:

11. 11
Electrolysis of Potassium Chloride
in Aqueous Solution
2 H2O + 2e-
→
H2 (g) + 2 OH-
cathode reaction
Battery, a source
of direct currente-
e-
aqueous KCl
2Cl-
→ Cl2 (g) + 2e-
anode reaction
cathode (-) anode (-)
H2 gas Cl2 gas

12. 12
Electrolysis of Potassium Chloride in
Aqueous Solution
 Electrolysis of an aqueous solution of KCl results in production of
hydrogen gas (H2(g)) at one electrode (the cathode).
 The solution becomes basic near this electrode.
 The reduction half-rxn occurs at this electrode, the cathode.the cathode.
 Gaseous chlorine (Cl2(g)) is produced at the other electrode.
 the oxidation half-rxn occurs at this electrode, the anode.
 These empirical (i.e., experimental) observations lead us to the non-
spontaneous reaction:

13. 13
Electrolysis of Aqueous Potassium Sulfate
e-
e-
cathode (-) anode (+)
H2 gas O2 gas
2H2O + 2e-
→
H2(g) + 2OH-
cathode reaction
2H2O → O2 (g) +
4H+
+ 4e-
anode reaction

14. 14
 In this electrolysis, hydrogen gas is produced at one
electrode (the cathode).
 The solution becomes basic near this electrode.
 Reduction half-rxn occurrs at this electrode.
 Gaseous oxygen is produced at the other electrode
(the anode).
 The solution becomes acidic near this electrode.
 the oxidation half-rxn occurs at this electrode.
 These experimental facts lead us to the following
electrode reactions:
Electrolysis of Aqueous Potassium Sulfate

15. 15
The Electrolysis of
Aqueous Potassium Sulfate

16. 16
Electrolytic Cells
 In all electrolytic cells:
 the most easily reduced species is reduced and,
 the most easily oxidized species is oxidized.

17. 17
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
 Faraday’s Law - The amount of substance undergoing
chemical reaction at each electrode during electrolysis is
directly proportional to the amount of electricity (or
number of electrons) that passes through the electrolytic
cell.
 A Faraday is the amount of electricity that reduces one
equivalent (= 1 mole of electrons) of a species at the
cathode and must, therefore also oxidize one equivalent
of a species at the anode.
1 Faraday of electricity = 6.022 x 1023
e-

18. 18
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
 A coulomb (C) is the amount of charge that passes a
given point when a current of one ampere (A) flows for
one second.
 1 ampere (A) = 1 coulomb (C)/second
 1 Faraday (F) = 96,487 coulombs/mol e-

19. 19
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
 Faraday’s Law states that during electrolysis, one
Faraday of electricity (96,487 coulombs) reduces
and oxidizes, respectively, one equivalent of the
oxidizing agent and the reducing agent.
 A Faraday corresponds to the passage of one mole of
electrons through the electrolytic cell.
 One reducing equivalent ==> gain of 1 mol of e-
.
 One oxidizing equivalent ==> loss of 1 mol of e-
.

20. 20
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
 Example: Calculate the mass of palladium produced by the
reduction of palladium(II) (Pd2+
) ions during the passage of
3.20 amperes of current through an aqueous solution of
palladium (II) sulfate for 30.0 minutes.
Cathode: Pd2+
+ 2e-
 Pd0
1 mol 2 mol 1 mol
2(96,487 C) 106 g
3.20 amp = 3.20 C/s
?g Pd = 30.0 min x (60 s/min) x (3.20 C/s) x (1 mol e-
/96,487 C)
x 1 mol Pd/2 mol e-
x 106 g/mol Pd
= 3.16 g Pd

21. 21
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis
 Example: Calculate the volume of oxygen
(measured at STP) produced by the oxidation of
water in the previous example.
Anode: 2H2O  O2(g) + 4H+
+ 4e-
2 mol 1 mol 4 mole 4(96,487C)
22.4 LSTP
?L O2(g) = 30.0 min x (60 s/min) x 3.20 C/s x (1 mol e-
/96,487 C)
x (1 mol O2(g)/4 mol e-
) x (22.4 L O2(g)/mol)
= 0.334 L O2(g)

22. 25
Voltaic or Galvanic Cells
 Electrochemical cells in which a spontaneous chemical
reaction (∆G < 0) produces electrical energy.
 Half-cells are physically separated so that electrons are
forced to travel through external wires.
 The key concept: the potential difference between the
two half-cells drives the external electron flow.
 Auto batteries
 Flashlight batteries
 Computer, calculator, cell phone batteries
 Batteries are common examples of voltaic cells.
 The electrical energy produced by these spontaneous
reactions can be converted to useful work via the electron
flow in the external circuit.

23. 26
The Standard Zinc-Copper Voltaic Cell
The cell’s initial voltage is 1.10 volts at 25 0
C.

24. 27
The Construction of Simple Voltaic
Cells
 Voltaic cells consist of two half-cells:
 one containing a substance that can be oxidized and the
other containing a substance that can be reduced.
 in electrical contact with each other.
 A simple voltaic cell:
 Two half cells, each containing a metal electrode
immersed in a solution of its ions.
 A wire connecting the two half-cells.
 A salt bridge to complete the circuit, allowing ion diffusion
to maintain charge neutrality.

25. 29
The Standard Zinc-Copper Voltaic Cell
 Cell components for the Zn-Cu cell are:
 A metallic Cu strip immersed in 1.0 M copper (II) sulfate.
 A metallic Zn strip immersed in 1.0 M zinc (II) sulfate.
 A wire and a salt bridge to complete circuit.

26. 30
The Zinc-Copper Voltaic Cell
 In all voltaic cells, electrons flow spontaneously from
the negative electrode where oxidation occurs (the
anode) to the positive electrode where reduction
occurs (the cathode).
Anode half-rxn: Zn(s)  Zn2+
(aq) + 2e-
Cathode half-rxn: Cu2+
(aq) + 2e-
 Cu(s)
Overall rxn: Zn(s) + Cu2+
(aq)  Zn2+
(aq) + Cu(s)
A spontaneous rxn: E0
cell = +1.10 V under standard conditions

27. 31
The Zinc-Copper Cell
 A commonly used short-hand notation for voltaic
cells (using the Zn-Cu cell as an example):
Zn(s) | Zn2+
(aq)(1.0 M) || Cu2+
(aq)(1.0 M) | Cu(s)
Soluble, oxidized species
cathodesalt bridgeanode

28. 32
The Copper - Silver Cell
 Cell components:
 A Cu strip immersed in 1.0 M copper (II) sulfate.
 A Ag strip immersed in 1.0 M silver (I) nitrate.
 A wire and a salt bridge to complete the circuit.
 The initial cell voltage is 0.46 V at 25 0
C.

29. 33
The Copper - Silver Cell
 Compare the Zn-Cu cell to the Cu-Ag cell
Anode reaction: Cu(s)  Cu2+
+ 2e-
Cathode reaction: 2(Ag+
+ e-
 Ag(s))
Overall cell reaction: Cu(s) + 2Ag+
 Cu2+
+ 2Ag(s)
 The Cu electrode is the cathode in the Zn-Cu cell.
 The Cu electrode is the anode in the Cu-Ag cell.
 In the Zn-Cu cell, Zn is more active than Cu, i.e., Zn
“wants” to get oxidized more than does Cu.
 In the Cu-Ag cell, Cu “wants” to get oxidized more
than does Ag.

30. 34
The Copper - Silver Cell
 The preceding two experimental observations
demonstrate that Cu2+
is a stronger oxidizing agent than
Zn2+
.
 Cu2+
oxidizes metallic Zn(s) to Zn2+
.
 Similarly, Ag+
is is a stronger oxidizing agent than Cu2+
.
 Ag+
oxidizes metallic Cu(s) to Cu2+
.
 Strength as oxidizing agent: Zn2+
< Cu2+
< Ag+
 Strength as reducing agent: Zn(s) > Cu(s) > Ag(s)
 A measure of the relative “activities” of the metal.
 More active metal is more easily oxidized.
 So order of “activities is: Zn > Cu> Ag
 These species can be, thus, be arranged in order of
increasing oxidizing or reducing strengths:

31.  Combine the SHE with other half-cells to
quantitatively measure relative oxidizing and reducing
strengths.
35
 Establish a reference electrode (half-cell) against
which all other half-reactions can be compared.
 The reference half-cell is the Standard Hydrogen
Electrode (SHE):
 1 M H+
(pH 0) /1 atm H2, 25 0
C.
 The SHE is assigned an arbitrary voltage of 0.000000… V
The Standard Hydrogen
Electrode (SHE)

32. 36
The Zinc-SHE Cell
 For this cell the component are:
1. A Zn strip (electrode) immersed in 1.0 M zinc (II) sulfate.
2. the Standard Hydrogen Electrode.
3. A wire and a salt bridge to complete the circuit.
 Under standard conditions, Zn(s) reduces H+
.

33. 37
The Zinc-SHE Cell
 The cathode is the SHE.
 2H+
is reduced to H2.
 The anode is Zn(s) metal rod.
 Zn(s) metal is oxidized to Zn2+
ions.
 The overall (spontaneous) cell reaction under
standard conditions:
 Zn(s) metal reduces H+
.
Anode rxn: Zn(s)  Zn2+
(aq) + 2e-
E0
= +0.763 V
Cathode rxn: 2H+
(aq) + 2e-
 H2(g) E0
= 0.000 V
Overall rxn: Zn(s) + 2H+
(aq)  Zn2+
(aq) + H2(g) E0
cell = +0.763 V

34. 38
The Copper-SHE Cell
 The two half-cells are:
1. A Cu strip immersed in 1.0 M copper (II) sulfate.
2. The SHE.
Under standard conditions H2 reduces Cu2+
.

35. 39
The Copper-SHE Cell
 In this cell the SHE is the anode
 H2 is oxidized to 2H+
.
 The Cu(s) is the cathode.
 The Cu2+
ions are reduced to Cu metal.
Anode rxn: H2(g)  2H+
(aq) + 2e-
E0
= 0.000 V
Cathode rxn: Cu2+
(aq) + 2e-
 Cu(s) E0
= +0.337 V
Overall rxn: H2(g) + Cu2+
(aq)  2H+
(aq) + Cu(s) E0
cell = +0.337 V

36. 40
Uses of Standard Reduction Potentials
 Half-cells that force the SHE to act as an anode (i.e., H2 gets oxidized)
are assigned positive standard reduction potentials.
 Half-cells that force the SHE to act as the cathode (i.e., H+
gets
reduced) are assigned negative standard reduction potentials.
 Standard reduction potentials, E0
(V) are:
 for the half-cells measured relative to the SHE.
 a quantitative measure of the tendencies of half-reactions to
occur in the direction of reduction.
 Standard potentials are tabulated for the half reactions written in
the direction of reduction, i.e. the Table lists standard reduction
potentials.
 For the previous two examples, the standard reduction potentials are:
Zn2+
(aq) + 2e-
 Zn(s) E0
= -0.763 V
Cu2+
(aq) + 2e-
 Cu(s) E0
= +0.337 V

37. Midterm Exam 3 information
 Covers material through today’s lecture.
 Special office hours:
 W March 25 11 AM (right after class period).
 Th March 26 3 PM.
 Both office hours in my office (BSE 4.324).
42

38. 43

39. 44
Uses of Standard Reduction Potentials
 Can use standard reduction potentials to predict
whether an electrochemical (redox) reaction will
occur spontaneously under standard conditions.
 Example: Will silver ions, Ag+
, oxidize metallic
Zn to Zn2+
ions, or will Zn2+
ions oxidize metallic
Ag to Ag+
ions (under standard conditions)?
 Use the following sequence to determine which
direction is the spontaneous electrochemical
(redox) reaction:

40. 45
Uses of Standard Electrode Potentials
 Choose the appropriate half-reactions from a table of
standard reduction potentials.
 Write the equation for the half-reaction with the more
positive E0
value first, along with its E0
value.
 Write the equation for the other half-reaction in the reverse
direction (i.e., as an oxidation), and reverse the sign of the
tabulated E0
.
 Multiply the half rxns by appropriate coefficients to balance
the electrons. DON’T MULTIPLY THE E0
values!
 Add the half-rxns and corresponding E0
values to obtain
the overall rxn and its potential, E0
cell.
 This procedure always produces an equation for the
reaction with positive E0
cell, which indicates that the forward
reaction is spontaneous.

41. 46
Uses of Standard Reduction Potentials
Ag+
+ e-
 Ag(s) E0
=+0.799 V
Zn(s)  Zn2+
+ 2e-
E0
=-(-0.763 V)
2( )
2Ag+
+ Zn(s)  2Ag(s) + Zn2+
Eo
cell = +1.562 V
 The positive E0
cell tells us:
 the spontaneous (forward) reaction is oxidation of
Zn(s) by Ag+
.
 Zn(s) is a better reducing agent than Ag(s).
 Conversely, Ag+
is a better oxidizing agent than Zn2+
.

42. 48
Uses of Standard Reduction Potentials
 Example: Will MnO4
-
oxidize Fe2+
to Fe3+
, or will Fe3+
oxidize Mn2+
to MnO4
-
in acidic solution?
 Thus, the answer is, yes, MnO4
-
will oxidize Fe2+
ions to Fe3+
,
and MnO4
-
will be reduced to Mn2+
in acidic solution.
Reduction: MnO4
-
+ 8H+
5e-
 Mn2+
+ 4H2O E0
= +1.51 V
Oxidation: 5(Fe2+
 Fe3+
+ e-
) -E0
= -(+0.77 V)
Cell rxn: MnO4
-
+ 8H+
+ 5Fe2+
 Mn2+
+ 4H2O + 5Fe3+
E0
cell = +0.74 V

43. 50
Uses of Standard Reduction Potentials
 Example: Will nitric acid, HNO3, oxidize arsenous
acid, H3AsO3,to arsenic acid, H3AsO4, in acidic
solution? The reduction product of HNO3 is NO in
this reaction.
Reduction: 2(NO3
-
+ 4H+
+ 3e-
 NO + 2H2O) E0
= +0.96 V
Oxidation: 3(H3AsO3 + H2O  H3AsO4 + 2H+
+ 2e-
) -E0
= -(+0.58 V)
Cell rxn: 2NO3
-
+ 2H+
+ 3H3AsO3  2NO + H2O + 3H3AsO4 E0
cell = +0.38 V

44. 52
Corrosion: Oxidation of Metal by O2.
 Metallic corrosion consists of oxidation of the metal by O2,
aided by diffusion through surface moisture and acidification
by dissolved CO2 or acidic pollutants.
pit
Overall rust reaction: 4Fe + 3O2 + xH2O(l) → 2Fe2O3 · xH2O

45. 53
Corrosion Protection
 Examples of corrosion protection:
1 Coat the corrosion-prone metal surface with with a thin
layer of a less active (less easily oxidized) metal.
“Tin cans” are actually made of steel coated with
a thin layer of tin.
Tin is harder to oxidize than is iron (i.e., Sn(s) is
less active than Fe(s).

46. 54
Corrosion Protection
2. Connect the metal to a “sacrificial anode”, a piece of a
more active (i.e., more easily oxidized) metal.
• Ships hulls often have Zn exterior “sacrificial anodes”.
Fe2+
+ 2e-
 Fe(s) E0
= -0.41V
Zn2+
+ 2e-
 Zn(s) E0
= -0.76V
Mg2+
+ 2e-
 Mg(s) E0
= -2.39V
Ship’s hull showing
lighter colored Zn
blocks, which
function as
sacrificial anodes.
• The Mg or Zn are more active than iron and will, therefore,
be preferentially oxidized by O2.
• Mg and Zn are “sacrificed” to protect iron.

47. 55
Corrosion Protection

48. 56
Mg bar
Corrosion Protection

49. 57
Corrosion Protection
3. Coating or mixing with a more active (more
easily oxidized) metal.
 Stainless steel is an alloy of Cr (~10%) and Fe. The Cr
is more easily oxidized than is Fe, and the Cr2O3 forms
a protective coating on the grains of Fe.
 Galvanizing, either dipping steel in molten zinc or
electrochemical reduction of Zn2+
on the steel
surface), provides a more active metal on the
exterior.
 The thin coat of Zn functions as a sacrificial anode
inhibiting the “rusting” of iron underneath. The
oxidized Zn also forms an additional hard protective
layer of zinc oxide (ZnO).

50. 58
Corrosion Protection
4. Paint or coat with a polymeric material such as
plastic or ceramic.
Steel bathtubs coated with porcelain, a
ceramic material akin to glass.

51. 59
Corrosion Protection
5. Allow a protective film to form naturally.
4Al(s) + 3O2  2Al2O3
Al2O3 forms a hard, transparent film
on aluminum surfaces.

52. 60

53. 61
Effect of Non-Standard Conditions on
Electrode Potentials
 Standard electrode potentials are determined at:
 1.00 M solution concentrations.
 1.00 atm for gases.
 Pure liquids or solids (activities = 1).
 25o
C (298 K).
 Electrode and cell potentials under non-standard
conditions can be calculated using the Nernst
equation.
 Non-standard potentials are often more useful for
“real-world” conditions.

54. 62
The Nernst Equation
E = E0
-
2.303RT
nF
logQ
• E is the electrode (half-cell) potential under the non-standard conditions.
• E0
is the standard electrode (half-cell) potential (from the table).
• R is the universal gas constant, 8.314 J/mol-K (not the same units as for gas
law).
• T is absolute temperature (Kelvins).
• n is number of electrons transferred in the half-reaction listed in the table.
• 1 V = 1J/C
• F is the Faraday (96,487 C/mol e-
)
• Q is the reaction quotient.
The Nernst equation for electrochemical half-reactions under
non-standard conditions:
x (1J/C·V) = 96,487 J/V · mol e-
(J V-1
mol -1
).

55. 63
The Nernst Equation
 Substitution of the values of the constants into the
Nernst equation for half-reactions at 25o
C gives:
MEMORIZE this form of the Nernst equation for non-
standard half-cells at 25 o
C!
2.303RT
F
2.303 x (8.314J/mol-K) x 298K
96,487 J/V-mol e-
= = 0.0592
E = E0
-
0.0592
n
logQ

56. 64
The Nernst Equation
 Can write Nernst equations for half-reactions, for
example:
 The corresponding Nernst equation is:
E = E0
- 0.0592
1
log
[Cu+
]
[Cu2+
]
Cu2+
+ e-
 Cu+
E0
= +0.153 V

57. 65
The Nernst Equation
 Substituting E0
into the above expression gives:
 If [Cu2+
] and [Cu+
] are both 1.0 M, then E = E0
(= 0.153
V) because the concentration ratio, Q = 1 and,
therefore, logQ equals zero.
E = 0.153 - 0.0592
1
log
[Cu+
]
[Cu2+
]
E = 0.153 - 0.0592
1
log
1
1
= 0.153

58. 66
The Nernst Equation
 Example: Calculate the potential for the Cu2+
/Cu+
electrode at 250
C when the concentration of Cu+
ions
is three times that of Cu2+
ions.
Cu2+
+ e-
 Cu+
[Cu+
]
[Cu2+
]
Q = = 3
E = +0.153 -0.0592
1
log 3
E = +0.153 - 0.0282 = +0.125 V

59. 67
The Nernst Equation
 Example: Calculate the potential for the Cu2+
/Cu+
electrode at 25 0
C when the Cu+
ion concentration
is 1/3 of the Cu2+
ion concentration.
Cu2+
+ e-
 Cu+
[Cu+
]
[Cu2+
]
Q = = 1/3
E = 0.153 - 0.0592
1
log 0.333
E = 0.153 + 0.0282 = +0.181 V

60. 68
The Nernst Equation
 Example: Calculate the reduction potential for a
hydrogen electrode in which the [H+
] is 1.0 x 10-3
M and the H2 pressure is 0.50 atmosphere.
0.50
(1.0 x 10-3
)2
Q = = 5.0 x 105
2H+
+ 2e-
 H2
E = 0 - 0.0592
2
log 5.0 x 105
E = 0 - 0.168 = -0.168 V

61. 69
The Nernst Equation
 The Nernst equation also applies to overall redox
reactions occurring in a voltaic cell.
 For an overall cell reaction, substitute Ecell and E0
cell in
place of E and E0
, respectively:
 The Nernst equation can, therefore, be used to calculate
the cell potential, Ecell, and, thus, spontaneity of a reaction
in a voltaic cell that consists of two non-standard
electrodes.
Ecell = E0
cell -
0.0592
n
logQ

62. 70
The Nernst Equation
Zn(s) + Cu2+
(aq)  Zn2+
(aq) + Cu(s)
A spontaneous rxn: E0
cell = +1.10 V under
standard conditions

63. 71
The Nernst Equation
Cu2+
+ 2e-
 Cu(s) Eo
= +0.340 V
Zn2+
+ 2e-
 Zn(s) Eo
= -0.763 V
Cu2+
+ Zn(s)  Cu(s) + Zn2+
Eo
cell = + 1.10 V
Reverse the zinc half-reaction and add:
Example: Starting from the standard Zn|Zn2+
||
Cu2+
|Cu voltaic cell, calculate the cell potential,
Ecell, when [Cu2+
] is changed to 3 M.

64. 72
The Nernst Equation
0.0592 log([Zn2+
]/[Cu2+
])
0.0592 log 0.333
n
2
Ecell = 1.10 - (-0.477)
Ecell = +1.58 V
Calculate the cell potential, Ecell, when [Cu2+
] is
changed to 3 M.
Ecell = E0
cell -
Ecell = 1.10 -

65. 73
The Nernst Equation
Can even construct a voltaic cell consisting of the same
half reaction at different starting concentrations!

66. If both half-cells contained either 1 M
Cu2+
or 0.1 M Cu2+ :
74
The Nernst Equation
0.0592 log (0.10/1.00)
0.0592 log 0.10
n
2
Ecell = 0 - (-0.030)
Ecell = +0.030 V
Cu|Cu2+
(0.10M)||Cu2+
(1.00M)|Cu
E0
cell = 0, no net rxn
Ecell = E0
cell -
Ecell = 0 -
For unequal [Cu2+
]:

67. 75
The Nernst Equation
 Example: Calculate the initial potential, Ecell, of a voltaic
cell consisting of an Fe3+
/Fe2+
half-cell connected to a
Sn4+
/Sn2+
half-cell with initial concentrations:
 [Fe3+
] = 1.0 x 10-2
M and [Fe2+
] = 0.1 M
 [Sn4+
] = 1.0 M and [Sn2+
] = 0.10 M.
Ecell = E0
cell -
0.0592
n
logQ

68. 76
The Nernst Equation
 Calculate the E0
cell by the usual procedure.
Fe3+
+ e-
 Fe2+
E0
= +0.771 V
Sn2+
 Sn4+
+ 2e-
-E0
= -(0.15 V)
2Fe3+
+ Sn2+
 2Fe2+
+ Sn4+
Eo
cell = +0.62 V
2( )
 Make sure the overall redox equation is balanced!

69. 77
The Nernst Equation
 Substitute the ion concentrations into Q to calculate
Ecell.

70. 78
The Nernst Equation

71.  When the redox reaction is at equilibrium, Ecell = 0, (no net
reaction or electron flow) so from the Nernst equation:
expressing constants explicitly:
E0
cell = RT
or rearrange to: nFE0
cell = RTlnK
 From previous chapters the relationship between ∆G0
and
the equilibrium constant, K, is:
∆G0
rxn = -RTlnK
 Equating the two expressions for RTlnK yields:
0 0
0.0592
logK
79
Relationship of E0
cell to ∆G0
and K
MEMORIZE!
MEMORIZE!E0
cell =
n
nF
lnK

72. Tro, Chemistry: A Molecular
Approach 80
Relationship of E0
cell to ∆G0
and K
 For a spontaneous redox reaction with the reactants
and products in their standard states:
 ∆G° < 0 (negative)
 E° > 0 (positive)
 K > 1
 ∆G° = −RTlnK = −nFE°cell
 n is the number of electrons transferred in the balanced
reaction.
 F = 96,485 C/mol e-
or J/V-mol e-

73.  Example: Calculate K at 25 0
C for oxidation of Zn
by Ag+
:
85
Relationship of E0
cell to ∆G0
and K
First calculate the standard cell potential E0
cell:
Ag+
+ e-
--> Ag0
E0
=+0.799 V
Zn --> Zn2+
+ 2e-
E0
=-(-0.763 V)
2( )
2Ag+
+ Zn0
--> 2Ag0
+ Zn2+
Eo
cell = +1.562 V

74. [Zn2+
]
[Ag+
]2
86
Relationship of E0
cell to ∆G0
and K
 Use the value of E0
cell (1.562 V) and n (= 2) to
obtain K:
K = 5.9 x 1052
=
∆G0
rxn = -nFE0
cell
= (2 mol e-
)(96,487J/V-mol e-
)(1.562V)
= -301.4 kJ/mol rxn
0.0592 logKE0
cell =
n 0.0592
logK =
E0
celln
0.0592
logK = 2(1.562)
= 52.7

75. 87
Relationship of E0
cell to ∆G0
and K
Now calculate Ecell and ∆G starting with [Ag+
] = 0.30
M and [Zn2+
] = 0.50 M.
[Zn2+
]
[Ag+
]2
Q =
(0.50)
(0.30)2
= = 5.6
Ecell = Eo
cell - 0.0592 log Q
n
Ecell = 1.562 - 0.0592 log 5.6
2
Ecell = 1.562 - 0.022 = 1.540 V

76. 88
Relationship of E0
cell to ∆G0
and K
• So the forward reaction is still spontaneous under these
non-standard conditions, but somewhat less so than under
standard conditions.
∆Grxn = -nFEcell
= -(2 mol e-
)(96,487 J/V-mol e-
)(1.540 V)
= 2.97 x 105
J/mol of rxn
∆Grxn = -297 kJ/mol
Now calculate ∆Grxn under these non-standard conditions:

77. 89
Primary Voltaic Cells (Non-Rechargeable
Batteries)
 As a voltaic cell discharges, its reagents are consumed.
 Once the reagents are consumed (or the concentrations
reach equilibrium), further chemical reaction is
impossible, and therefore, current can no longer flow.
 In primary voltaic cells, the electrodes and electrolytes
cannot be regenerated by reversing current flow through
the cell.
 Primary cells are, thus, not rechargable.

78. 96
The Alkaline Dry Cell
(Non-Rechargeable)

79. 97
The Alkaline Dry Cell (Non-Rechargeable)
 Half reactions for an alkaline dry cell are:
Anode half-rxn: Zn0
(s) + 2OH-
--> Zn(OH)2(s) + 2e-
Cathode half-rxn: 2MnO2(s) + 2H2O + 2e-
--> 2OH-
+ 2MnO(OH)(s)
Cell rxn: Zn0
(s) + 2MnO2(s) + 2H2O --> Zn(OH)2(s) + 2MnO(OH)(s)
E0
cell = 1.5V
 All the reactants and products are solids or water (activity = 1),
so the cell voltage remains nearly constant until most of the
reactants are consumed.
 Alkaline dry cells contain KOH, which precipitates the redox
products, i.e., keeps them solids, so activity = 1.

80. 99
Secondary Voltaic Cells
 Secondary cells are reversible and, therefore,
rechargeable.
 The electrodes in a secondary cell can be
regenerated (recharged) by application of an
external voltage.
 Application of an external voltage switches the cell from
voltaic to electrolytic.
 One example of a secondary voltaic cell is the
lead acid storage or car battery.

81. 100
The Lead Acid Storage (Car) Battery
 In the lead storage battery the electrodes are two
sets of lead alloy grids (plates).
 Holes in one of the grids (cathodes) are filled with
lead(IV) oxide, PbO2.
 The holes in the other grids (anodes) are filled with
“spongy” lead (Pb0
).
 The electrolyte is dilute sulfuric acid.

82. 101
The Lead Acid Storage (Car) Battery
Pb0
(s) + SO4
2-
-->
PbSO4(s) + 2e-
PbO2(s) + 4H+
+ SO4
2-
+ 2e-
-->
PbSO4(s) + 4H+
+ 2H2O

83. 102
Lead Acid Storage
(Car) Battery
 electrolyte = 30% H2SO4
 Anode half-rxn: Pb(s) + SO4
2-
(aq) → PbSO4(s) + 2 e-
 Cathode half-rxn: PbO2(s) + 4 H+
(aq) + SO4
2-
(aq) + 2 e-
→ PbSO4(s) + 2 H2O(l)
 Cell rxn: Pb0
(s) + PbO2(s) + 4H+
+ 2SO4
2-
--> 2PbSO4(s) + 2H2O Ecell ~ 2V
 A comproportionation reaction: two Pb oxidation states converting to one.
 PbSO4(s) coats both anodes and cathodes.
 Battery voltage = ~2 V/cell x 6 cells connected in series = ~12 V

84. 107
The Lead Acid Storage (Car) Battery
 During charging (or recharging), Pb2+
in PbSO4(s) (coating
both electrodes) is either reoxidized to PbO2(s) or re-
reduced to Pb0
.
 The concentration of the H2SO4 decreases as the cell discharges.
 Recharging the cell regenerates the H2SO4.

85. , e.g., LiCoO2 or LMnO4

86. 111
The Hydrogen-Oxygen Fuel Cell
 The overall reaction is the combination of hydrogen
and oxygen to form water.
 The cell provides a drinking water supply for the astronauts
as well as the electricity for the lights, computers, etc. on
board.
 Fuel cells are very efficient.
 Energy conversion rates of 60-70% are common.
Anode rxn: 2(H2(g) + 2OH-
(aq) --> 2H2O + 2e-
)
Cathode rxn: O2(g) + 2H2O + 4e-
--> 4OH-
(aq)
Cell rxn: 2H2(g) + O2(g) --> 2H2O(l)

87. 113
The Hydrogen-Oxygen Fuel Cell
 Fuel cells are batteries that must have their reactants
continuously supplied in the presence of appropriate
catalysts.
 A hydrogen-oxygen fuel cell is the most common.
 H2 is oxidized at the anode.
 O2 is reduced at the cathode.
 Reagents are very light, so saves weight.
 A potentially “green” alternative the internal
combustion engine for transportation vehicles.
 However, a source of energy is needed to generate
the H2.

88. 114
The Hydrogen-Oxygen Fuel Cell

89. 115
Biological Redox Chemistry and
Reduction Potentials
 Living organisms derive their energy from redox reactions.
 Respiration is essentially combustion of the the food we
eat, for example:
∆G0
= -2878 kJ/mol
Glucose “combustion”:
 In this biological “combustion” of organic molecules, the
electrons are passed through several intermediary protein
molecules called redox enzymes rather than directly from
organic molecules to dioxygen.

90. 116
Biological Redox Chemistry and
Reduction Potentials
• E0
’ (pH 7) of some
biologically accessible
molecules and ions.
 Biological redox reactions are often analyzed in
terms of their half-reaction reduction potentials rather
than free energies.
• Energy can be derived
from reduction of
molecules with higher
E0
’ by any molecule
with a lower E0
’.
glucose
reductants
glucoseCO2
O2 H2O

91. 117
After studying Tro Chapter 18 text and lecture slides and
working the problems, you should be able to:
1. Explain the difference between electrolytic and voltaic cells.
2. Diagram electrolytic and galvanic cells, identify cathodes and anodes and
their half-reactions.
3. For a given amount of current passed, calculate grams and moles of
substances oxidized or reduced in an electrochemical cell.
4. Not responsible for commercial applications of electrolysis.
5. Determine relative strengths of oxidizing and reducing agents from
electrochemical reactions in voltaic cells or given half-reaction E0
values.
6. Calculate E0
cell given half-reaction E0
values.
7. Describe and diagram the process of iron corrosion (rusting).
8. Explain various methods for preventing corrosion of metals based on metal
“activities”, i.e. relative ease of oxidation or reduction.
9. Use the Nernst equation (memorize) to calculate Ecell under non-standard
conditions given half-reaction E0
values, initial concentrations and the Nernst
equation.
10. Calculate values of K and ∆Grxn from E0
cell using the Nernst equation and ∆G =
-nFEcell (memorize).
11. Describe and calculate Eo
for a concentration cell.

92. 118
After studying Tro Chapter 18 text and lecture slides and
working the problems, you should be able to:
12. Identify components in an alkaline dry cell given the half reactions and a
diagram of the cell. Show direction of current flow, and explain why voltage
stays fairly constant during discharge.
13. Draw a diagram of a lead acid car battery given the half reactions, and
describe the discharge and recharge cycles.
14. Describe and draw a diagram of a H2/O2 fuel cell.
15. Given E0
’ values, describe how energy can be derived from biological redox
reactions.