Chapter 20
Electrochemistry
1
Chapter Goals
1.
2.
3.
4.
5.
6.
7.
Electrical Conduction
Electrodes
Electrolytic Cells
The Electrolysis of Molten Potassium Chloride
The Electrolysis of Aqueous Potassium Chloride
The Electrolysis of Aqueous Potassium Sulfate
Faraday’s Law of Electrolysis
Commercial Applications of Electrolytic Cells
2
Chapter Goals
8.
9.
10.
11.
12.
13.
14.
15.
Voltaic or Galvanic Cells
The Construction of Simple Voltaic Cells
The Zinc-Copper Cell
The Copper-Silver Cell
Standard Electrode Potentials
The Standard Hydrogen Electrode
The Zinc-SHE Cell
The Copper-SHE Cell
Standard Electrode Potentials
Uses of Standard Electrode Potentials
3
Chapter Goals
16.
17.
18.
19.
20.
21.
Standard Electrode Potentials for Other HalfReactions
Corrosion
Corrosion Protection
Effect of Concentrations (or Partial Pressures)
on Electrode Potentials
The Nernst Equation
Using Electrohemical Cells to Determine
Concentrations
The Relationship of Eocell to Go and K
4
Chapter Goals
Primary Voltaic Cells
22.
23.
24.
25.
Dry Cells
Secondary Voltaic Cells
The Lead Storage Battery
The Nickel-Cadmium (Nicad) Cell
The Hydrogen-Oxygen Fuel Cell
5
Electrochemistry
Electrochemical reactions are oxidation-reduction
reactions.
The two parts of the reaction are physically
separated.





1.
2.
The oxidation reaction occurs in one cell.
The reduction reaction occurs in the other cell.
There are two kinds electrochemical cells.
Electrochemical cells containing in nonspontaneous
chemical reactions are called electrolytic cells.
Electrochemical cells containing spontaneous
chemical reactions are called voltaic or galvanic
cells.
6
Electrical Conduction



Metals conduct electric currents well in a
process called metallic conduction.
In metallic conduction there is electron flow
with no atomic motion.
In ionic or electrolytic conduction ionic motion
transports the electrons.


Positively charged ions, cations, move toward the
negative electrode.
Negatively charged ions, anions, move toward the
positive electrode.
7
Electrodes


The following convention for electrodes is
correct 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
Electrodes


Inert electrodes do not react with the liquids
or products of the electrochemical reaction.
Two examples of common inert electrodes
are graphite and platinum.
9
Electrolytic Cells
Electrical energy is used to force
nonspontaneous chemical reactions to
occur.
The process is called electrolysis.
Two examples of commercial electrolytic
reactions are:



1.
2.
The electroplating of jewelry and auto parts.
The electrolysis of chemical compounds.
10
Electrolytic Cells
Electrolytic cells consist of:

1.
2.
3.
A container for the reaction mixture.
Two electrodes immersed in the reaction mixture.
A source of direct current.
11
The Electrolysis of Molten Potassium
Chloride

Liquid potassium is produced at one
electrode.



Indicates that the reaction K+() + e-  K() occurs
at this electrode.
Is this electrode the anode or cathode?
Gaseous chlorine is produced at the other
electrode.


Indicates that the reaction 2 Cl-  Cl2(g) + 2 eoccurs at this electrode.
Is this electrode the anode or cathode?
12
The Electrolysis of Molten Potassium
Chloride
Diagram of this electrolytic cell.
eGenerator-source
eof DC
- electrode
+ electrode
molten KCl
K+ + e-  K()
cathode reaction
Porous barrier
2Cl-  Cl2 (g) + 2eanode reaction
13
The Electrolysis of Molten Potassium
Chloride

The nonspontaneous redox reaction that occurs is:
2 Cl  Cl2(g)  2 e
-
Anode reaction

Cathode reaction 2 K   e -  K  
-


Cell reaction 2 Cl  2 K  Cl2g   2 K  
-
14
The Electrolysis of Molten Potassium
Chloride

In all electrolytic cells, electrons are forced
to flow from the positive electrode (anode)
to the negative electrode (cathode).
15
The Electrolysis of Aqueous
Potassium Chloride
In this electrolytic cell, hydrogen
gas is produced
at
Anode reaction
2 Cl  Cl2(g)  2 e
one electrode.

The
aqueous
solution
becomes
basic
near
this
electrode.
Cathode reaction 2 H 2O  2 e  H 2g   2 OH
 What reaction is occurring at this electrode? You do it!

Cell
reaction
2 Cl isproduced
2 H 2 O atHthe
 Cl2electrode.
Gaseous
chlorine
2 g  other
g   2 OH
-



What reaction is occurring at this electrode? You do it!
-
K
is
a
spectator
ion.
Note
that
wate
r
is
electrolyz
ed!
 These experimental facts lead us to the following
nonspontaneous electrode reactions:
16
The Electrolysis of Aqueous
Potassium Chloride
Cell diagram
e- flow
- pole of battery
+ pole of battery
Battery, a source
e- flow
of direct current
- electrode
+ electrode
H2 gas
Cl2 gas
aqueous KCl
2 H2O + 2e-  H2 (g) + 2 OHcathode reaction
2Cl-  Cl2 (g) + 2eanode reaction
17
The Electrolysis of Aqueous
Potassium Sulfate

In this electrolysis, hydrogen gas is produced at one
electrode.



Gaseous oxygen is produced at the other electrode



The solution becomes basic near this electrode.
What reaction is occurring at this electrode?
You do it!
The solution becomes acidic near this electrode.
What reaction is occurring at this electrode?
You do it!
These experimental facts lead us to the following
electrode reactions:
18
The Electrolysis of Aqueous
Potassium Sulfate
Anode reaction
2 H 2 O  O 2(g)  4 H   4 e -
Cathode reaction 2(2 H 2 O  2e-  H 2(g)  2OH- )

Cell reaction 6 H 2 O  2 H 2(g)  O 2(g)  4H

4
OH

 

4 H 2O
The overall reaction is 2 H 2 O  2H 2(g)  O 2(g)
19
The Electrolysis of Aqueous
Potassium Sulfate
Cell diagram
e- flow
- pole of battery + pole of battery
Battery, a source
of direct current
e- flow
- electrode
+ electrode
O2 gas
H2 gas
aqueous K2SO4
2 H2O + 2e-  H2 (g) + 2 OHcathode reaction
2H2O  O2 (g) + 4H+ + 4eanode reaction
20
Electrolytic Cells

In all electrolytic cells the most easily reduced
species is reduced and the most easily
oxidized species is oxidized.
21
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 that passes through the
electrolytic cell.
A faraday is the amount of electricity that reduces
one equivalent of a species at the cathode and
oxidizes one equivalent of a species at the anode.
1 faraday of electricit y  6.022 10 e
23
-
22
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis


A coulomb is the amount of charge that passes a
given point when a current of one ampere (A) flows
for one second.
1 amp = 1 coulomb/second
1 faraday  6.022 10 e  96,487 coulombs
23
-
23
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.

This corresponds to the passage of one mole of electrons
through the electrolytic cell.
1 equivalent of oxidizing agent  gain of 6.022 1023 e1 equivalent of reducing agent  loss of 6.022 1023 e-
24
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis

Example 21-1: Calculate the mass of palladium produced by
the reduction of palladium (II) ions during the passage of 3.20
amperes of current through a solution of palladium (II) sulfate
for 30.0 minutes.
2+2 +
-00
Cathode:
Pd
+
2e

Pd
Cathode : Pd + 2e 
11mol
mol 22mol
mol 11mol
mol
106
106gg 2(96,500)
2(96,500)106
106g g
3.20 amp = 3.20 C s
60 s 3.20 C
106 g Pd
? g = 30.0 min 


 316
. g Pd
min
s
296,500 C
25
Counting Electrons: Coulometry and
Faraday’s Law of Electrolysis

Example 21-2: Calculate the volume of oxygen
(measured at STP) produced by the oxidation of
water in example 21-1.
Anode ::22 H
H22O
O
OO22gg ++4H
4H+ +++4e4e- 22 mol
mol
? LSTP
11mol
mol 44mol
mol 44mol
mol
22.4L
4496,500
22.4
LSTP
96,500CC
STP
60 s 3.20 C 22.4 LSTP O 2
O 2  30.0 min 


min
s
496,500 C 
 0.334 LSTP O 2 or 334 mL STP O 2
26
Commercial Applications of
Electrolytic Cells
Electrolytic Refining and Electroplating of Metals
 Impure metallic copper can be purified
electrolytically to  100% pure Cu.



The impurities commonly include some active metals plus
less active metals such as: Ag, Au, and Pt.
The cathode is a thin sheet of copper metal
connected to the negative terminal of a direct
current source.
The anode is large impure bars of copper.
27
Commercial Applications of
Electrolytic Cells



The electrolytic solution is CuSO4 and H2SO4
The impure Cu dissolves to form Cu2+.
The Cu2+ ions are reduced to Cu at the cathode.
0
2

Anode impure Cu s  Cu aq   2e
Cathode very pure
Net rxn.
2+
Cu aq 

 2e 
0
Cu s
No net rxn.
28
Commercial Applications of
Electrolytic Cells

Any active metal impurities are oxidized to cations that are
more difficult to reduce than Cu2+.

This effectively removes them from the Cu
0
2

metal.
Zn  Zn  2e
2
Fe  Fe  2e
0

And so forth for other
active metals
29
Commercial Applications of
Electrolytic Cells



The less active metals are not oxidized and
precipitate to the bottom of the cell.
These metal impurities can be isolated and
separated after the cell is disconnected.
Some common metals that precipitate include:
Ag, Au, Pt, Pd
Se, Te
30
Voltaic or Galvanic Cells



Electrochemical cells in which a spontaneous
chemical reaction produces electrical energy.
Cell halves are physically separated so that
electrons (from redox reaction) are forced to travel
through wires and creating a potential difference.
Examples of voltaic cells include:
Auto batteries
Flashlight batteries
Computer and calculator batteries
31
The Construction of Simple Voltaic
Cells


Voltaic cells consist of two half-cells which
contain the oxidized and reduced forms of an
element (or other chemical species) in contact
with each other.
A simple half-cell consists of:



A piece of metal immersed in a solution of its ions.
A wire to connect the two half-cells.
And a salt bridge to complete the circuit, maintain
neutrality, and prevent solution mixing.
32
The Construction of Simple Voltaic
Cells
33
The Zinc-Copper Cell
Cell components for the Zn-Cu cell are:

1.
2.
3.

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
The cell’s initial voltage is 1.10 volts
34
The Zinc-Copper Cell
Anode reaction
Zn 0  Zn 2+  2e 
Cathode reaction
Cu 2+  2e   Cu 0
Overall cell reaction Zn 0  Cu 2+  Zn 2+  Cu 0
This is a spontaneou s reaction w ith E

0
cell
 1.10V
In all voltaic cells, electrons flow spontaneously
from the negative electrode (anode) to the
positive electrode (cathode).
35
The Zinc-Copper Cell

There is a commonly used short hand notation for
voltaic cells.

The Zn-Cu cell provides a good example.
species (and concentrations)
in contact with electrode surfaces
Zn/Zn2+(1.0 M) || Cu2+(1.0 M)/Cu
salt bridge
electrode surfaces
36
The Copper - Silver Cell

Cell components:
1.
2.
3.

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 volts.
37
The Copper - Silver Cell
Anode reaction
Cu 0  Cu 2  2e 
Cathode reaction
2 Ag + + e -  Ag 0
Overall cell reaction


Cu 0 + 2 Ag +  Cu 2+ + 2 Ag 0
This is a spontaneou s reaction w ith E 0cell  0.46V

Compare the Zn-Cu cell to the Cu-Ag cell



The Cu electrode is the cathode in the Zn-Cu cell.
The Cu electrode is the anode in the Cu-Ag cell.
Whether a particular electrode behaves as an anode
or as a cathode depends on what the other
electrode of the cell is.
38
The Copper - Silver Cell

These experimental facts demonstrate that Cu2+ is a stronger
oxidizing agent than Zn2+.


Similarly, Ag+ is is a stronger oxidizing agent than Cu2+.


In other words Cu2+ oxidizes metallic Zn to
Zn2+.
Because Ag+ oxidizes metallic Cu to Cu 2+.
2+
+
If we arrange theseZn
species
increasing strengths,
< Cuin2+order
< Ag of
we see that:
strength as oxidizing agent
Zn 0 > Cu 0 > Ag 0
strength as reducing agent
39
Standard Electrode Potential


To measure relative electrode potentials, we must
establish an arbitrary standard.
That standard is the Standard Hydrogen Electrode
(SHE).

The SHE is assigned an arbitrary voltage of 0.000000… V
40
The Zinc-SHE Cell
For this cell the components are:

1.
2.
3.

A Zn strip immersed in 1.0 M zinc (II) sulfate.
The other electrode is the Standard Hydrogen Electrode.
A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.763 volts.
41
The Zinc-SHE Cell
E0
Anode reaction
Zn 0  Zn 2+ + 2 e -
Cathode reaction 2 H + + 2 e -  H 2g 
0.763 V
0.000 V
Cell reaction Zn 0 + 2H +  Zn 2+ + H 2g  E 0cell  0.763 V

The cathode is the Standard Hydrogen Electrode.


In other words Zn reduces H+ to H2.
The anode is Zn metal.

Zn metal is oxidized to Zn2+ ions.
42
The Copper-SHE Cell
The cell components are:

1.
2.
3.

A Cu strip immersed in 1.0 M copper (II) sulfate.
The other electrode is a Standard Hydrogen Electrode.
A wire and a salt bridge to complete the circuit.
The initial cell voltage is 0.337 volts.
43
The Copper-SHE Cell
E0
Anode reaction
H 2 g   2 H + + 2 e -
Cathode reaction Cu 2+ + 2e -  Cu 0
Cell reaction

0.337 V
H 2g  + Cu 2+  2H + + Cu 0 E 0cell  0.337 V
In this cell the SHE is the anode


0.000 V
The Cu2+ ions oxidize H2 to H+.
The Cu is the cathode.

The Cu2+ ions are reduced to Cu metal.
44
Uses of Standard Electrode Potentials




Electrodes that force the SHE to act as an anode are assigned
positive standard reduction potentials.
Electrodes that force the SHE to act as the cathode are assigned
negative standard reduction potentials.
Standard electrode (reduction) potentials tell us the tendencies of
half-reactions to occur as written.
For example, the half-reaction for the standard potassium electrode
is:


K e K
0
0
E = -2.925 V
The large negative value tells us that this reaction
will occur only under extreme conditions.
45
Uses of Standard Electrode Potentials

Compare the potassium half-reaction to fluorine’s halfreaction:
0
F2



+ 2e  2F
-
-
0
E = +2.87 V
The large positive value denotes that this reaction occurs
readily as written.
Positive E0 values denote that the reaction tends to
occur to the right.
 The larger the value, the greater the tendency to occur
to the right.
It is the opposite for negative values of Eo.
46
Uses of Standard Electrode Potentials



Use standard electrode potentials to predict
whether an electrochemical reaction at
standard state conditions will occur
spontaneously.
Example 21-3: Will silver ions, Ag+, oxidize
metallic zinc to Zn2+ ions, or will Zn2+ ions
oxidize metallic Ag to Ag+ ions?
Steps for obtaining the equation for the
spontaneous reaction.
47
Uses of Standard Electrode Potentials
1.
2.
3.
4
5
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 as an
oxidation with its oxidation potential, i.e. reverse the
tabulated reduction half-reaction and change the sign
of the tabulated E0.
Balance the electron transfer.
Add the reduction and oxidation half-reactions and
their potentials. This produces the equation for the
reaction for which E0cell is positive, which indicates that
the forward reaction is spontaneous.
48
Uses of Standard Electrode Potentials
E0
Reduction
2(Ag + + e -  Ag 0 )
+ 0.7994 V
Oxidation
1(Zn  Zn 2+ + 2 e - )
- (-0.7628 V)
Cell reaction 2Ag + + Zn 0  2Ag 0 + Zn 2+ E 0cell = +1.5662 V
49
Electrode Potentials for
Other Half-Reactions


Example 21-4: Will permanganate ions,
MnO4-, oxidize iron (II) ions to iron (III) ions,
or will iron (III) ions oxidize manganese(II)
ions to permanganate ions in acidic solution?
Follow the steps outlined in the previous
slides.

Note that E0 values are not multiplied by any
stoichiometric relationships in this procedure.
50
Electrode Potentials for
Other Half-Reactions

Example 21-4: Will permanganate ions, MnO4-,
oxidize iron (II) ions to iron (III) ions, or will iron (III)
ions oxidize manganese(II) ions to permanganate
ions in acidic solution?
E0
Reduction
1(MnO -4  8H  + 5e -  Mn 2  4H 2O)
Oxdation
5(Fe 2  Fe3+ + e - )
+ 1.51 V
- (0.771 V)
Cell reaction MnO -4  5Fe2  8H   Mn 2  4H 2 O + 5Fe3+ E 0cell = +0.74 V

Thus permanganate ions will oxidize iron (II) ions to
iron (III) and are reduced to manganese (II) ions in
acidic solution.
51
Electrode Potentials for
Other Half-Reactions

Example 21-5: Will nitric acid, HNO3, oxidize
arsenous acid, H3AsO3, in acidic solution? The
reduction product of HNO3 is NO in this reaction.
You do it!
E0
Reduction
2(NO3-  4H  + 3e -  NO  2H 2 O)
+ 0.96 V
Oxidation
3(H 3AsO 3  H 2 O  H 3AsO 4  2H  + 2e - )
- (0.58 V)
Cell reaction 2NO3-  3H 3 AsO 3  8H   3H 2 O  2NO  4H 2 O + 3H 3 AsO 4  6H 
2NO3-  3H 3 AsO 3  2H   2NO  H 2 O + 3H 3AsO 4  6H  E 0cell = +0.38 V
52
Corrosion

Metallic corrosion is the oxidation-reduction
reactions of a metal with atmospheric components
such as CO2, O2, and H2O.
4 Fe0 + 3 O02  2 Fe2O3 overall reaction 
The reaction occurs rapidly at exposed points.
53
Corrosion Protection

1
Some examples of corrosion protection.
Plate a metal with a thin layer of a less active (less
easily oxidized) metal.
"Tin plate" for steel.
54
Corrosion Protection
2.
Connect the metal to a sacrificial anode, a piece of
a more active metal.
Soil pipes and ship hulls have
Mg and Zn on the exterior
as sacrificia l anodes.
55
Corrosion Protection
3.
Allow a protective film to form naturally.
0
0
2
2 3
4 Al + 3 O  2 Al O
Al 2 O 3 forms a hard, transpare nt
film on exterior of Al foil.
56
Corrosion Protection
4
Galvanizing, the coating of steel with zinc, provides
a more active metal on the exterior.
The thin coat of Zn must be
oxidized before Fe begins to rust.
Zinc
Steel
57
Corrosion Protection
5.
Paint or coat with a polymeric material such as
plastic or ceramic.
Steel bathtubs are coated with ceramic.
58
Effect of Concentrations (or Partial
Pressures) on Electrode Potentials


The Nernst Equation
Standard electrode potentials, those compiled
in appendices, are determined at
thermodynamic standard conditions.
Reminder of standard conditions.
1.00 M solution concentrations
1.00 atm of pressure for gases
All liquids and solids in their standard thermodynamic states.
Temperature of 250 C.
59
The Nernst Equation



The value of the cell potentials change if conditions
are nonstandard.
The Nernst equation describes the electrode
potentials at nonstandard conditions.
The Nernst equation is:
60
The Nernst Equation
2.303RT
log Q
E=E nF
E = potential under condition of interest
0
E 0  potential under standard conditions
R = universal gas constant = 8.314 J
mol K
T = temperatur e in K
n = number of electrons transferr ed
F = the faraday = (96,487 C/mol e - )  1 J


CV
.
 96,487 J/V mol e Q = reaction quotient
61
The Nernst Equation

Substitution of the values of the constants into the
Nernst equation at 25o C gives:
2.303 RT 2.303  8.314 J mol K  298 K

 0.0592
F
96,487 J/V mol e
0 0.0592
Thus E = E log Q
n
62
The Nernst Equation

For this half-reaction:


Cu  e  Cu
2+

-
0
E = +0.153 V
The corresponding Nernst equation is:
 
 
+
Cu
0.0592
E=E log
2
1
Cu
0
63
The Nernst Equation

Substituting E0 into the above expression gives:
 
 
+
Cu
0.0592
E = 0.153 V log
2
1
Cu

If [Cu2+] and [Cu+] are both 1.0 M, i.e. at standard
conditions, then E = E0 because the concentration
term equals zero.
0.0592
1
E = 0.153 V log
1
1
64
The Nernst Equation
65
The Nernst Equation

Example 21-6: Calculate the potential for the
Cu2+/ Cu+ electrode at 250C when the
concentration of Cu+ ions is three times that
of Cu2+ ions.
Cu
2+


+ e  Cu
-

Cu 
3 Cu 
Q=

3
Cu  Cu 
+
2+
2+
2+
66
The Nernst Equation

Example 21-6: Calculate the potential for the
Cu2+/ Cu+ electrode at 250C when the
concentration of Cu+ ions is three times that
of Cu2+ ions.
0.0592
log Q
1
0.0592
E = 0.153 V log 3
1
E = E0 -
67
The Nernst Equation

Example 21-6: Calculate the potential for the
Cu2+/ Cu+ electrode at 250C when the
concentration of Cu+ ions is three times that
of Cu2+ ions.
0.0592
log Q
1
0.0592
E = 0.153 V log 3
1
E = 0.153V - 0.05920.477  V
E = 0.153 - 0.0282  V
E = 0.125 V
E = E0 -
68
The Nernst Equation

Example 21-7: Calculate the potential for the Cu2+/Cu+
electrode at 250C when the Cu+ ion concentration is 1/3
of the Cu2+ ion concentration.
Cu
2+


+ e  Cu
-

Cu 

Cu 
Q=
 3
 0.333
Cu  Cu 
+
2+
1
2+
2+
You do it!
69
The Nernst Equation

Example 21-7: Calculate the potential for the
Cu2+/Cu+ electrode at 250C when the concentration
of Cu+ ions is 1/3 that of Cu2+ ions.
0.0592
E=E log Q
1
0.0592
E = 0.153 V log 1
3
1
0
70
The Nernst Equation
0 0.0592
E=E log Q
1
0.0592
E = 0.153 V log 1
3
1
E = 0.153V - 0.0592- 0.477  V
E = 0.153 + 0.0282 V
E = 0.181 V
71
The Nernst Equation

Example 21-8: Calculate the electrode potential for a
hydrogen electrode in which the [H+] is 1.0 x 10-3 M and
the H2 pressure is 0.50 atmosphere.
H
2H +2e 
2
+
Q=
-
PH 2

0.50
H  1.0 10 
+ 2
Q  5.0 10
3 2
5
72
The Nernst Equation

Example 21-8: Calculate the electrode potential for a
hydrogen electrode in which the [H+] is 1.0 x 10-3 M and
the H2 pressure is 0.50 atmosphere.

0.0592
5
E=E log 5.0 10
2
0.0592
5.699
E  0
2
E  0.168 V
0

73
The Nernst Equation


The Nernst equation can also be used to calculate
the potential for a cell that consists of two
nonstandard electrodes.
Example 21-9: Calculate the initial potential of a cell
that consists of an Fe3+/Fe2+ electrode in which
[Fe3+]=1.0 x 10-2 M and [Fe2+]=0.1 M connected to a
Sn4+/Sn2+ electrode in which [Sn4+]=1.0 M and
[Sn2+]=0.10 M . A wire and salt bridge complete the
circuit.
74
The Nernst Equation

Calculate the E0 cell by the usual procedure.
Reduction
Oxidation

1Sn
2 Fe3+ + e  Fe2+
2+

 Sn 4+ + 2 e -
E

0
0.771 V
- 0.15 V
Cell reaction 2 Fe3+  Sn 2+  2 Fe2+  Sn 4+ E 0cell  0.62 V
75
The Nernst Equation

Substitute the ion concentrations into Q to calculate
Ecell.
E cell = E 0cell -
0.0592
log Q
2
= 0.62 V -


Fe
 
 Sn 
2 2
Sn4
2
2
0.0592
log
2
Fe3
0.0592
010
.  10
. 

= 0.62 V log
2 2
2
10
.  10 010
. 
2
E cell


76
The Nernst Equation

0.10 1.0 
0.0592
log
E cell = 0.62 V 2
2

2
1.0 10 0.10 
0.0592
3.00
E cell  0.62 V 2
E cell  0.62  0.09 V
2


E cell = 0.53 V
77
Relationship of E0cell to G0 and K

From previous chapters we know the relationship of
G0 and K for a reaction.
G  -RT lnK
or
0
G  2.303 RT log K
0
78
Relationship of E0cell to G0 and K

The relationship between G0 and E0cell is also a
simple one.
G  -n F E
0
0
cell
where F  96,487 J/V mol e
n  number of e
-
-
79
Relationship of E0cell to G0 and K

Combine these two relationships into a single
relationship to relate E0cell to K.
-n FE
0
cell
 - RT lnK
or
n FE
ln K 
RT
0
cell
80
Relationship of E0cell to G0 and K

Example 21-10: Calculate the standard Gibbs
free energy change, G0 , at 250C for the
following reaction.
2
2
Pb  Cu  Pb  Cu
81
Relationship of E0cell to G0 and K
1.
Calculate E0cell using the appropriate halfreactions.
E
2

Cu  2 e  Cu
2
Pb  Pb  2 e
2
0
0.337 V
- - 0.126 V 

2
0
Cu  Pb  Pb  Cu E
0
0
cell
 0.463 V
82
Relationship of E0cell to G0 and K
2.
Now that we know E0cell , we can calculate G0 .
GG nnFFEE
00
00
cell
cell
 

 96,500
mole e 96
,500J J
0.463
GG 22mol
V V 
- .463
- 0
mol
VV
mol
ee
0
4
G  8.94 10 J (per " mol of rxn" )
00


G  -89.4 kJ
0
83
Relationship of E0cell to G0 and K

Example 21-11: Calculate the thermodynamic
equilibrium constant for the reaction in
example 21-10 at 250C.
G   RT ln K 
0
 8.94 10 J
G

ln K 
 RT - 8.314 J
0

ln K  36.1 K  5 1015
4
mol
298 K 

mol K

Pb 

Cu 
2
2
84
Relationship of E0cell to G0 and K

Example 21-12: Calculate the Gibbs Free
Energy change, G and the equilibrium
constant at 250C for the following reaction
with the indicated concentrations.
Zn  2 Ag 0.30 M   2Ag  Zn

2
0.50 M 
85
Relationship of E0cell to G0 and K
1.
Calculate the standard cell potential E0cell.

Zn

2 Ag  e  Ag
0

-
2
0

 Zn  2e
-
E
0.799 V

- - 0.763 V 
2 Ag  Zn  2 Ag  Zn
0
0
0
2
E
0
cell
 1.562 V
86
Relationship of E0cell to G0 and K
2.
Use the Nernst equation to calculate Ecell for
the given concentrations.


Zn 
0.50 
Q=

Ag  0.30
2+
+ 2
E cell
2
 5.6
log 5.6 = 0.748
0.0592
0
 E cell 
log Q
2
87
Relationship of E0cell to G0 and K

Zn  0.50
Q=

 5.6
Ag  0.30
2+
+ 2
2
log 5.6 = 0.748
0.0592
0
E cell  E cell 
log Q
2
0.0592
E cell  1.562 V log 5.6 V
2
88
Relationship of E0cell to G0 and K

Zn  0.50 
Q=

 5. 6
Ag  0.30
2+
+ 2
2
log 5.6 = 0.748
0.0592
E cell  E 
log Q
2
0.0592
E cell  1.562 V log 5.6  V
2
E cell = 1.562 - 0.022  V
0
cell
E cell = 1.540 V
89
Relationship of E0cell to G0 and K


Ecell = +1.540 V, compared to E0cell = +1.562 V.
We can use this information to calculate G.
G = - n F E cell


G   2 mol e - 96,500 J


1.540 V 
V mol e
-
G  2.97 10 J per mol of rxn.
G = -297 kJ/mol
5

The negative G tells us that the reaction is
spontaneous.
90
Relationship of E0cell to G0 and K


Equilibrium constants do not change with reactant
concentration.
We can use the value of E0cell at 250C to get K.
- n F E 0cell   2.303 RT log K
- n F E 0cell
log K =
 2.303 RT
 2 96,500 1.562 
log K 
 2.3038.314 298
log K  52.8
K = 6  10

Zn 

Ag 
2+
52
+ 2
91
Primary Voltaic Cells



As a voltaic cell discharges, its chemicals are
consumed.
Once the chemicals are consumed, further
chemical action is impossible.
The electrodes and electrolytes cannot be
regenerated by reversing current flow through
cell.

These cells are not rechargable.
92
The Dry Cell
One example of a dry cell is flashlight, and radio,
batteries.
The cell’s container is made of zinc which acts as
an electrode.
A graphite rod is in the center of the cell which acts
as the other electrode.
The space between the electrodes is filled with a
mixture of:




1.
2.
3.
4.
ammonium chloride, NH4Cl
manganese (IV) oxide, MnO2
zinc chloride, ZnCl2
and a porous inactive solid.
93
The Dry Cell


As electric current is produced, Zn dissolves and
goes into solution as Zn2+ ions.
The Zn electrode is negative and acts as the anode.
94
The Dry Cell

The anode reaction is:
Zn  Zn + 2 e
0


2
2+
-
The graphite rod is the positive electrode (cathode).
Ammonium ions from the NH4Cl are reduced at the cathode.
+
NH 4
 2 e  2 NH 3 g   H 2 g 
-
95
The Dry Cell

The cell reaction is:
Zn 0  Zn 2+ + 2 e -
Anode reaction
Cathode reaction 2 NH +4 + 2 e -  2 NH3g   H 2g 
Cell reaction
Zn + 2 NH  Zn
0
+
4
2+
 2 NH3g   H 2g 
96
The Dry Cell


The other components in the cell are included to remove the
byproducts of the reaction.
MnO2 prevents H2 from collecting on graphite rod.
H 2 g   2 MnO2 s  2 MnO(OH) s

At the anode, NH3 combines with Zn2+ to form a soluble
complex and removing the Zn2+ ions from the reaction.
Zn
2+
 4 NH 3  Zn

2
NH 3 4
97
The Dry Cell
98
The Dry Cell


Alkaline dry cells are similar to ordinary dry cells
except that KOH, an alkaline substance, is added to
the mixture.
Half reactions for an alkaline cell are:
Anode reaction
Zn 0s  + 2 OH-  Zn OH2s   2 e 
Cathode reaction 2 MnO 2s   2 H 2 O + 2 e -  2 MnO(OH) s  + 2 OHCell reaction
Zn 0s  + 2 MnO 2s   2 H 2 O  Zn OH2s   2 MnO(OH) s 
E 0cell  1.5 V
99
The Dry Cell


Alkaline dry cells are similar to ordinary dry cells
except that KOH, an alkaline substance, is added to
the mixture.
Half reactions for an alkaline cell are:
100
Secondary Voltaic Cells


Secondary cells are reversible, rechargeable.
The electrodes in a secondary cell can be
regenerated by the addition of electricity.


These cells can be switched from voltaic to electrolytic cells.
One example of a secondary voltaic cell is the lead
storage or car battery.
101
The Lead Storage Battery




In the lead storage battery the electrodes are two
sets of lead alloy grids (plates).
Holes in one of the grids are filled with lead (IV)
oxide, PbO2.
The other holes are filled with spongy lead.
The electrolyte is dilute sulfuric acid.
102
The Lead Storage Battery

Diagram of the lead storage battery.
103
The Lead Storage Battery

As the battery discharges, spongy lead is oxidized
to lead ions and the plate becomes negatively
charged.
Pb 0s   Pb 2+  2 e This is the anode reaction.
The anode is the negative battery post during discharge.

The Pb2+ ions that are formed combine with SO42- from
sulfuric acid to form solid lead sulfate on the Pb
electrode.
2+
2Pb  SO4  PbSO4 s
104
The Lead Storage Battery

The net reaction at the anode during discharge is:
0
Pb s






2SO4
 PbSO4 s + 2 e
-
Electrons are produced at the Pb electrode.
These electrons flow through an external circuit
(the wire and starter) to the PbO2 electrode.
PbO2 is reduced to Pb2+ ions, in the acidic solution.
The Pb2+ ions combine with SO42- to form PbSO4
and coat the PbO2 electrode.
PbO2 electrode is the positive electrode (cathode).
105
The Lead Storage Battery

As the cell discharges, the cathode reaction is:
PbO2 s + 4 H + + SO2+
2
e
 PbSO4 s + 2 H 2O
4

The cell reaction for a discharging lead storage battery is:
Anode reaction
Pb s   SO 24-  PbSO 4s   2 e -
Cathode reaction PbO 2s  + 4 H + + SO 24- + 2 e -  PbSO 4s  + 2 H 2 O
Cell reaction
Pb s   PbO 2s  + 4 H + + SO 24-  2 PbSO 4s  + 2 H 2 O
106
The Lead Storage Battery

As the cell discharges, the cathode reaction is:
PbO2 s + 4 H + + SO2+
2
e
 PbSO4 s + 2 H 2O
4

The cell reaction for a discharging lead storage battery is:
107
The Lead Storage Battery


What happens at each electrode during recharging?
At the lead (IV) oxide, PbO2, electrode, lead ions are oxidized
to lead (IV) oxide.
PbSO4s   2 H 2O  PbO2 s   4 H  SO  2 e
+


24
-
The concentration of the H2SO4 decreases as the cell discharges.
Recharging the cell regenerates the H2SO4.
discharge



 PbSO  2H O
Pbs   PbO 2 s  + 4H + 2SO 


4 s 
2
charge
+
24
108
The Lead Storage Battery


What happens at each electrode during recharging?
At the lead (IV) oxide, PbO2, electrode, lead ions are oxidized
to lead (IV) oxide.
PbSO4s   2 H2O  PbO2s   4 H+  SO24-  2 e

The concentration of the H2SO4 decreases as the cell discharges.
Recharging the cell regenerates the H2SO4.
109
The Nickel-Cadmium (Nicad) Cell


Nicad batteries are the rechargeable cells used
in calculators, cameras, watches, etc.
As the battery discharges, the half-reactions are:
Anode Cd s   2 OH-  Cd(OH) 2  2 e Cathode NiO 2s  + 2 H 2O + 2 e  Ni(OH) 2s  + 2 OH
-
-
Cell rxn Cd s   NiO 2s  + 2 H 2O  Cd(OH) 2s  + Ni(OH) 2s 
E 0  1.4V
110
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 used in the
space shuttle

The fuel cell is what exploded in Apollo 13.


Hydrogen is oxidized at the anode.
Oxygen is reduced at the cathode.
111
The Hydrogen-Oxygen Fuel Cell
Anode reaction
Cathode reaction
Cell reaction



2g
2

4
e

4
OH
()
(aq)
2 H 2 g   O 2 g   2 H 2 O  
Notice that the overall reaction is the combination of
hydrogen and oxygen to form water.



O    2 H O
2 H 2g   2 OH-(aq)  2 H 2O (  )  2 e -
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!
112
Synthesis Question

What are the explosive chemicals in the fuel
cell that exploded aboard Apollo 13?
113
Synthesis Question

The Apollo 13 fuel cells contained hydrogen
and oxygen. Both are explosive, especially
when mixed. The oxygen tank aboard Apollo
13 exploded.
114
Group Question

Some of the deadliest snakes in the world, for
example the cobra, have venoms that are
neurotoxins. Neurotoxins have an
electrochemical basis. How do neurotoxins
disrupt normal chemistry and eventually kill their
prey?
115
End of Chapter 21

Electrochemistry is an important part of the
electronics industry.
116