Chapter19

Contents and Concepts
Half−Reactions
1. Balancing Oxidation–Reduction Reactions in
Acidic and Basic Solutions
Voltaic Cells
2. Construction of Voltaic Cells
3. Notation for Voltaic Cells
4. Cell Potential
5. Standard Cell Potentials and Standard
Electrode Potentials
Copyright © Cengage Learning. All rights reserved.

19 | 2

6.
7.
8.

Equilibrium Constants from Cell Potentials
Dependence of Cell Potentials on Concentration
Some Commercial Voltaic Cells

Electrolytic Cells
9. Electrolysis of Molten Salts
10. Aqueous Electrolysis
11. Stoichiometry of Electrolysis


19 | 3

Learning Objectives

Electrochemistry
1. Balancing Oxidation–Reduction Reactions in
Acidic and Basic Solutions
a. Learn the steps for balancing oxidation–
reduction reactions using the half−reaction
method.
b. Balance equations by the half−reaction
method (acidic solutions).
c. Learn the additional steps for balancing
oxidation–reduction reactions in basic
solution using the half−reaction method.
d. Balance equations using the half−reaction
method (basic solution).

19 | 4

2. Construction of Voltaic Cells
a. Define electrochemical cell, voltaic
(galvanic cell), electrolytic cell, and
half−cell.
b. Describe the function of the salt bridge in a
voltaic cell.
c. State the reaction that occurs at the anode
and the cathode in an electrochemical cell.
d. Define cell reaction.
e. Sketch and label a voltaic cell.

19 | 5

3. Notation for Voltaic Cells
a. Write the cell reaction from the cell notation.
4. Cell Potential
a. Define cell potential and volt.
b. Calculate the quantity of work from a given
amount of cell reactant.


19 | 6

5. Standard Cell Potentials and Standard
Electrode Potentials
a. Explain how the electrode potential of a cell
is an intensive property.
b. Define standard cell potential and standard
electrode potential.
c. Interpret the table of standard reduction
potentials.
d. Determine the relative strengths of oxidizing
and reducing agents.
e. Determine the direction of spontaneity from
electrode potentials.
f. Calculate cell potentials from standard
potentials.

19 | 7

6. Equilibrium Constants from Cell Potentials
a. Calculate the free−energy change from
electrode potentials.
b. Calculate the cell potential from
free−energy change.
c. Calculate the equilibrium constant from cell
potential.
7. Dependence of Cell Potential on Concentration
a. Calculate the cell potential for nonstandard
conditions.
b. Describe how pH can be determined using
a glass electrode.

19 | 8

8. Some Commercial Voltaic Cells
a. Describe the construction and reactions of
a zinc–carbon dry cell, a lithium–iodine
battery, a lead storage cell, and a nickel–
cadmium cell.
b. Explain the operation of a proton−exchange
membrane fuel cell.
c. Explain the electrochemical process of the
rusting of iron.
d. Define cathodic protection.
9. Electrolysis of Molten Salts
a. Define electrolysis.

19 | 9

10. Aqueous Electrolysis
a. Learn the half−reactions for water
undergoing oxidation and reduction.
b. Predict the half−reactions in an aqueous
electrolysis.
11. Stoichiometry of Electrolysis
a. Calculate the amount of charge from the
amount of product in an electrolysis.
b. Calculate the amount of product from the
amount of charge in an electrolysis.

19 | 10

A voltaic cell employs a spontaneous oxidation–
reduction reaction as a source of energy. It separates
the reaction into two half−reactions, physically
separating one half−reaction from the other.


19 | 11

Our first step in studying electrochemical cell is to
balance its oxidation–reduction reaction.
We will use the half−reaction method from Section
4.6 and extend it to acidic or basic solutions.
In this chapter we will focus on electron transfer
rather than proton transfer so the hydronium ion,
H3O+(aq), will be represented by its simpler notation,
H+(aq). Only the notation, not the chemistry, is
different.

19 | 12

We will be studying more complex situations, so our
initial analysis is key.
First, we need to identify what is being oxidized and
what is being reduced.
Then, we determine if the reaction is in acidic or basic
conditions.


19 | 13

For example, consider the skeleton reaction:
+2
+5
+3
+2
Fe2+(aq) + NO3−(aq) → Fe3+(aq) + NO(g)
acidic solution
We determine the oxidation number of Fe and N
in each substance.
Fe2+(aq) is oxidized from +2 to +3 in Fe3+(aq).
N in NO3−(aq) is reduced from +5 to +2 in NO(g).
The reaction is in acidic conditions (HNO3).

19 | 14

Now we split the skeleton reaction into
half−reactions and balance each half.
1. Balance all atoms except H and O.
2. Balance O atoms by adding H2O to the side of the
equation that needs O.
3. Balance H by adding H+ to the side of the equation
that needs H.
4. Balance the electric charge by adding electrons
(e−) to the more positive side of the equation.


19 | 15

Oxidation half−reaction
Fe2+(aq) → Fe3+(aq) + e−
Reduction half−reaction
First we balance N, then O:
NO3−(aq) → NO(aq) + 2H2O(l)
Next we balance H:
4H+(aq) + NO3−(aq) → NO(g) + 2H2O(l)
Then we balance e−:
3e− + 4H+(aq) + NO3−(aq) → NO(aq) + 2H2O(l)

19 | 16

Finally, we combine the two half−reactions.
1. Multiply each half−reaction by a factor so that
each has the same number of electrons.
2. Add the two half−reactions (the electrons should
cancel).
3. Simplify the overall reaction by canceling species
that occur on both sides and reducing coefficients
to the smallest whole numbers.
4. Check that the reaction is balanced.


19 | 17

3Fe2+(aq) → 3Fe3+(aq) + 3e−
3e− + 4H+(aq) + NO3−(aq) → NO(g) + 2H2O(l)
4H+(aq) + 3Fe2+(aq) + NO3−(aq) →
3Fe3+(aq) + NO(g) + 2H2O(l)
Check that the reaction is balanced, in terms of both
atoms and charge.
4H; 3Fe; 1N; 3O; charge is +9

19 | 18

?

Dichromate ion in acidic solution is an
oxidizing agent. When it reacts with
zinc, the metal is oxidized to Zn2+, and
nitrate is reduced. Assume that
dichromate ion is reduced to Cr3+.
Write the balanced ionic equation for
this reaction using the half−reaction
method.


19 | 19

First we determine the oxidation numbers of N
and Zn:
+6
+3
0
+2
Cr2O72−(aq) → Cr3+(aq) and Zn(s) → Zn2+(aq)
Cr was reduced from +6 to +3.
Zn was oxidized from 0 to +2.
Now we balance the half−reactions.

19 | 20

Zn(s) → Zn2+(aq) + 2e−
First we balance Cr and O:
Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l)
Next we balance H:
14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l)
6e− + 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l)

19 | 21

Now we combine the two half−reactions by
multiplying the oxidation half reaction by 3.
3Zn(s) → 3Zn2+(aq) + 6e−
6e− + 14H+(aq) + Cr2O72−(aq) → 2Cr3+(aq) + 7H2O(l)
3Zn(s) + 14H+(aq) + Cr2O72−(aq) →
3Zn2+(aq) + 2Cr3+(aq) + 7H2O(l)
Check that atoms and charge are balanced.
3Zn; 14H; 2Cr; 7O; charge +12

19 | 22

To balance a reaction in basic conditions, first
follow the same procedure as for acidic
solution.
Then
1. Add one OH− to both sides for each H+.
2. When H+ and OH− occur on the same side,
combine them to form H2O.
3. Cancel water molecules that occur on both
sides.


19 | 23

?

Lead(II) ion, Pb2+, yields the plumbite
ion, Pb(OH)3−, in basic solution. In turn,
this ion is oxidized in basic hypochlorite
solution, ClO−, to lead(IV) oxide, PbO2.
Balance the equation for this reaction
using the half−reaction method. The
skeleton equation is
Pb(OH)3− + ClO− → PbO2 + Cl−


19 | 24

First we determine the oxidation number of Pb
and Cl in each species.
+2
+1
+4
−1
Pb(OH)3− + ClO− → PbO2 + Cl−
Pb is oxidized from +2 to +4.
Cl is reduced from +1 to −1.
Now we balance the half−reactions.

19 | 25

First we balance Pb and O:
Pb(OH)3−(aq) → PbO2(s) + H2O(l)
Next we balance H:
Pb(OH)3−(aq) → PbO2(s) + H2O(l) + H+(aq)
Pb(OH)3−(aq) → PbO2(s) + H2O(l) + H+(aq) + 2e−


19 | 26

First we balance Cl and O:
ClO−(aq) → Cl−(aq) + H2O(l)
Next we balance H:
2H+(aq) + ClO−(aq) → Cl−(aq) + H2O(l)
Finally we balance e−:
2e− + 2H+(aq) + ClO−(aq) → Cl−(aq) + H2O(l)


19 | 27

Now we combine the half−reactions.
Pb(OH)3−(aq) → PbO2(s) + H2O(l) + H+(aq) + 2e−
2e− + 2H+(aq) + ClO−(aq) → Cl−(aq) + H2O(l)
H+(aq) + Pb(OH)3−(aq) + ClO−(aq) →
PbO2(s) + Cl−(aq) + 2H2O(l)


19 | 28

H+(aq) + Pb(OH)3−(aq) + ClO−(aq) →
PbO2(s) + Cl−(aq) + 2H2O(l)
To convert to basic solution, we add OH− to each
side, converting H+ to H2O.
H2O(l) + Pb(OH)3−(aq) + ClO−(aq) →
PbO2(s) + Cl−(aq) + 2H2O(l) + OH−(aq)
Finally, we cancel the H2O that is on both sides.
Pb(OH)3−(aq) + ClO−(aq) →
PbO2(s) + Cl−(aq) + H2O(l) + OH−(aq)

19 | 29

The next several topics describe battery cells or
voltaic cells (galvanic cells).
An electrochemical cell is a system consisting
of electrodes that dip into an electrolyte and in
which a chemical reaction either uses or
generates an electric current.


19 | 30

A voltaic or galvanic cell is an electrochemical
cell in which a spontaneous reaction generates
an electric current.
An electrolytic cell is an electrochemical cell in
which an electric current drives an otherwise
nonspontaneous reaction.


19 | 31

Physically, a voltaic cell consists of two half−cells
that are electrically connected.
Each half−cell is the portion of the electrochemical
cell in which a half−reaction takes place.
The electrical connections allow the flow of
electrons from one electrode to the other.
The cells must also have an internal cell
connection, such as a salt bridge, to allow the flow
of ions.

19 | 32

The Cu2+ ions gain
The zinc metal atom
two electrons,
loses two electrons,
forming solid copper.
forming Zn2+ ions.
The electrons flow through the external circuit from
the zinc electrode to the copper electrode.

Ions flow through the salt bridge to
maintain charge balance.

19 | 33

A salt bridge is a tube of electrolyte in a gel that is
connected to the two half−cells of the voltaic cell. It
allows the flow of ions but prevents the mixing of
the different solutions that would allow direct
reactions of the cell reactants.
In the salt bridge, cations move toward the
cathode and anions move toward the anode.


19 | 34

The electrode at which oxidation takes place is
called the anode.
The electrode at which reduction takes place is
called the cathode.
Electrons flow through the external circuit from
the anode to the cathode.


19 | 35


19 | 36

This figure below illustrates the same reaction,
replacing the light bulb with a voltmeter. The solution
on the right is now blue from the Cu2+ ion formed.


19 | 37

You construct one half−cell of a voltaic
cell by inserting a copper metal strip
into a solution of copper(II) sulfate. You
construct another half−cell by inserting
an aluminum metal strip in a solution of
aluminum nitrate. You now connect the
half−cells by a salt bridge. When connected to
an external circuit, the aluminum is oxidized.
Sketch the resulting voltaic cell. Label the
anode and the cathode, showing the
corresponding half−reactions. Indicate the
direction of electron flow in the external circuit.

?


19 | 38

e−
cathode
Cu

e−

Ca2+
NO3−

anode
Al

Cu2+

Al3+

SO42−

NO3−

Cu2+ + 2e− → Cu

Al → Al3+ + 3e−
19 | 39

Voltaic cell notation is a shorthand method of
describing a voltaic cell.
The oxidation half−cell, the anode, is written on the
left.
The reduction half−cell, the cathode, is written on the
right.


19 | 40

The cell terminal is written on the outside: left for the
anode and right for the cathode.
Phase boundaries are shown with a single vertical
bar, |.
For example, at the anode, the oxidation of Cu(s) to
Cu2+(aq) is shown as Cu(s) | Cu2+(aq).
At the cathode, the reduction of Zn2+(aq) to Zn(s) is
shown as Zn2+(aq) | Zn(s).

19 | 41

Between the anode and cathode the salt bridge is
represented by two vertical bars, ||.
The complete notation for the reaction is
Cu(s) | Cu2+(aq) || Zn2+(aq) | Zn(s)


19 | 42

When the half−reaction involves a gas, the
electrode is an inert material such as platinum, Pt. It
is included as a third substance in the half−cell.
For example, the half−reaction of Cl2 being reduced
to Cl− is written as follows:
Cl2(g) | Cl−(aq) | Pt
Because this is a reduction, the electrode appears
on the far right.
For the oxidation of H2(g) to H+(aq), the notation is
Pt | H2(g) | H+(aq)
In this case, the electrode appears on the far left.

19 | 43

To fully specify a voltaic cell, it is necessary to give
the concentrations of solutions or ions and the
pressures of gases. In the cell notation, these
values are written within parentheses for each
species.
For example, the oxidation of Cu(s) to Cu2+(aq) at
the anode and the reduction of F2(g) to F−(aq) at
the cathode is written as follows:
Cu(s) | Cu2+(1.0 M) || F2(1.0 atm) | F−(1.0 M) | Pt

19 | 44

?

The cell notation for a voltaic cell is
Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s)
Write the cell reaction.

Our strategy is to write and balance each
half−reaction, and then to combine the half−reactions
to give the overall cell reaction.
The skeleton oxidation reaction is Al(s) → Al3+(aq).
The skeleton reduction reaction is Cu2+(aq) → Cu(s).

19 | 45

Balanced oxidation half−reaction
Al(s) → Al3+(aq) + 3e−
Balanced reduction half−reaction
Cu2+(aq) + 2e− → Cu(s)
The common multiple of the electrons is 6. Combine
the half−reactions by multiplying the oxidation
half−reaction by 2 and the reduction half−reaction
by 3.


19 | 46

2Al(s) → 2Al3+(aq) + 6e−
3Cu2+(aq) + 6e− → 3Cu(s)
2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)


19 | 47

Now we shift our focus to the movement of the
electrons in the oxidation–reduction reaction.
To better understand this movement, we can
compare the flow of electrons to the flow of water.


19 | 48

Just as work is required to pump water from
one point to another, so work is required to
move electrons.
Water flows from areas of high pressure to
areas of low pressure. Similarly, electrons flow
from high electric potential to low electric
potential. Electric potential can be thought of as
electric pressure.


19 | 49

Potential difference is the difference in electric
potential (electrical pressure) between two points.
Potential difference is measured in the SI unit volt
(V).
Electrical work = charge x potential difference
The SI units for this are
J=C×V
J
V=
C

19 | 50

The magnitude of charge on one mole of electrons is
given by the Faraday constant, F. It equals 9.6485 ×
103 C per mole of electrons.
1 F = 9.6485 × 103 C
Substituting this into the equation for work:
w = –F × potential difference
The term is negative because the cell is doing work
in creating the current (that is, electron flow).

19 | 51

Normally the potential across the voltaic cell
electrodes is less than the maximum possible.
One reason for this discrepancy is that it takes
energy (work) to drive the current through the cell:
The greater the current, the less the voltage. The cell
has its maximum voltage only when no current flows.


19 | 52

The situation for electrons is analogous to that seen
with water. The difference between water pressure in
the tap and the pressure of the outside atmosphere
is at its maximum when the tap is off; once the tap is
on, the pressure difference decreases.


19 | 53

The maximum potential difference between the
electrodes of a voltaic cell is called the cell potential
or electromotive force (emf) of the cell, or Ecell.
Ecell can be measured with a digital voltmeter. The
anode of a voltaic cell has negative polarity; the
cathode has positive polarity.


19 | 54

The maximum work is given by the following
equation:
Wmax = –nFEcell


19 | 55

?

The emf of a particular cell is 0.500 V.
The cell reaction is
2Al(s) + 3Cu2+(aq) → 2Al3+(aq) + 3Cu(s)
Calculate the maximum electrical work
of this cell obtained from 1.00 g of
aluminum.


19 | 56

To determine the value of n (that is, the number of
moles of electrons involved in either half−cell
reaction), we need to examine the half−reactions.
2Al(s) → 2Al3+(aq) + 6e−
3Cu2+(aq) + 6e− → 3Cu(s)
n=6
F = 96,485 C/mol
Ecell = 0.500 V = 0.500 J/C
wmax = –nFEcell
wmax = –(6 mol)(96,485 C/mol)(0.500 J/C)
wmax = 2.89 × 105 J

19 | 57

This is the energy corresponding to the balanced
reaction, so we need to convert to one gram of
aluminum.

wmax

2.89 × 105 J 1 mol Al
=
×
= 5.36 × 103 J
2 mol Al
26.98 g Al
wmax = 5.36 kJ


19 | 58

A cell potential is a measure of the driving force of
the cell reaction. It is composed of the oxidation
potential for the oxidation half−reaction at the anode
and the reduction potential, for the reduction
half−reaction at the cathode.
Ecell = oxidation potential + reduction potential
The oxidation potential for a half−reaction =
–(reduction potential for the reverse half−reaction)

19 | 59

Because the oxidation and reduction potentials are
opposites, we need to calculate only one or the
other. By convention, reduction potentials are
calculated and are called electrode potentials, E
(with no subscript).
Electrode potentials are a measure of the oxidizing
ability of the reactant. Table 19.1 shows the
increasing strength of the oxidizing agents.


19 | 60


19 | 61


19 | 62

Ecell = Eoxidation + Ereduction
Because the electrode potentials are for reduction:
Eoxidation = –Eanode
We can rewrite the equation for the cell:
Ecell = − Eanode + Ecathode
Ecell = Ecathode – Eanode


19 | 63

Let’s find Ecell for the following cell:
Al(s) | Al3+(aq) || Cu2+(aq) | Cu(s)
At the cathode, Cu2+(aq) is reduced to Cu(aq).
E = 0.34 V
At the anode, Al is oxidized to Al3+.
E = –[electrode potential for Al3+(aq)]
E = –(–1.66 V) = 1.66 V
Ecell = 0.34 V + 1.66 V
Ecell = 2.00 V

19 | 64

The standard electrode potential, E°, is the
electrode potential when the concentrations of
solutes are 1 M, the gas pressures are 1 atm, and
the temperature has a specified value (usually
25°C). The superscript degree sign (°) signifies
standard−state conditions.


19 | 65

A fuel cell is simply a voltaic cell that uses a
continuous supply of electrode materials to
provide a continuous supply of electrical
energy. A fuel cell employed by NASA on
spacecraft uses hydrogen and oxygen under basic
conditions to produce electricity. The water produced in
this way can be used for drinking. The net reaction is

?

2H2(g) + O2(g) → 2H2O(g)
Calculate the standard emf of the oxygen–hydrogen
fuel cell.
2H2O(l) + 2e− ⇄ H2(g) + 2OH−(aq) E° = –0.83 V
O2(g) + 2H2O(l) + 4e− ⇄ 4OH−(aq) E° = 0.40 V

19 | 66

Anode reaction:
2H2O(l) + 2e− ⇄ H2(g) + 2OH−(aq)

Cathode reaction:
O2(g) + 2H2O(l) + 4e− ⇄ 4OH−(aq)

E°red = –0.83 V

E°red = 0.40 V

Overall reaction:

Ecell = Ecathode – Eanode
Ecell = 0.40 V – (–0.83 V)
Ecell = 1.23 V

19 | 67

Comparing Oxidizing Strengths
The oxidizing agent is itself reduced and is the
species on the left of the reduction half−reaction.
Consequently, the strongest oxidizing agent is the
product of the half−reaction with the largest (most
positive) E° value.


19 | 68

Comparing Reducing Strengths
The reducing agent is itself oxidized and is the
species on the right of the reduction half−reaction.
Consequently, the strongest reducing agent is the
reactant in the half−reaction with the smallest (most
negative) E° value.


19 | 69

?

Which is the stronger reducing agent
under standard conditions: Sn2+ (to Sn4+) or
Fe (to Fe2+)?
Which is the stronger oxidizing agent
under standard conditions: Cl2 or MnO4−?

The stronger reducing agent will be oxidized and has
the more negative electrode potential.
The standard (reduction) potentials are
Sn2+ to Sn4+
E = 0.15 V
Fe to Fe2+
E = –0.41 V
The stronger reducing agent is Fe (to Fe 2+).

19 | 70

The stronger oxidizing agent will be reduced.
The standard (reduction) potentials are
Cl2 to Cl−
E = 1.36 V
MnO4− to Mn2+

E = 1.49 V

The stronger oxidizing agent is MnO4−.


19 | 71

Predicting the Direction of Reaction
You can predict the direction of reaction by
comparing the relative oxidizing (or reducing)
strengths.
The stronger oxidizing agent will be reduced. (The
stronger reducing agent will be oxidized.)


19 | 72

?

Will dichromate ion oxidize
manganese(II) ion to permanganate
ion in acid solution under standard
conditions?

Standard potentials
Cr2O72− to Cr3+
MnO4− to Mn2+

E° = 1.33 V
E° = 1.49 V

MnO4− has a larger reduction potential; it will oxidize
Cr3+ to Cr2O72−.
Cr2O72− will not oxidize Mn2+ to MnO4−.

19 | 73

?

Let us define the reduction of I 2 to I−
ions, I2(s) + 2e− → 2I−(aq), as the
standard reduction reaction with E° =
0.00 V. We then construct a new
standard reduction table based on this
definition.

a. What would be the new standard reduction
potential of H+?


19 | 74

b. Would using a new standard reduction table
change the measured value of freshly prepared
voltaic cell made from Cu and Zn? (Assume you
have the appropriate solutions and equipment to
construct the cell.)
c. Would the calculated voltage for the cell in part b
be different if your were using the values given in
Table 19.1? Do the calculations to justify your
answer.


19 | 75

a. When the H+(aq) E° = 0.000 V, the I2 E° = 0.54 V.
So, if the new I2 potential is 0.000 V, then the H+
potential would be –0.54 V.
b. The measured voltaic cell voltage will be
unchanged.
c. No. The calculated voltage is the same using
either standard because it is a difference is
potential.

19 | 76

The free−energy change, ∆G, for a reaction equals
the maximum useful work of the reaction.
∆G° = wmax = –nFE°


19 | 77

?

Calculate the standard free−energy
change for the net reaction used in the
hydrogen–oxygen fuel cell:
2H2(g) + O2(g) → 2H2O(l)
The cell potential is 1.23 V.
How does this compare with ∆Gf° of
H2O(l)?


19 | 78

First we determine the number of electrons involved
in the reaction. Four H are oxidized to H+, and two O
are reduced to O2−. Thus four electrons are involved.
n=4
E° = 1.23 V
∆G° = –nFE°
∆G = –(4 mol)(96,485 C/mol)(1.23 J/C)
∆G° = –4.75 × 105 J
∆G° = –475 kJ
∆Gf° = –285.8 kJ/mol; –550 kJ for 2 mol

19 | 79

?

A voltaic cell consists of one half−cell
with Fe dipping into an aqueous
solution of 1.0 M FeCl2 and the other
half−cell with Ag dipping into an
aqueous solution of 1.0 M AgNO3.

Obtain the standard free−energy change for the cell
reaction using the standard free−energy change for
the reaction (found using standard free energies of
formation). The standard free energies of formation
of the ions are Ag+(aq), 77 kJ/mol, and Fe2+(aq), –85
kJ/mol.
What is the cell potential?

19 | 80

We first determine the reaction by using reduction
potentials.
Fe2+(aq), Fe(s) E° = –0.41 V
Ag+(aq), Ag(s)
E° = 0.80 V
Fe is the anode (oxidation).
Ag is the cathode (reduction).
E° = 0.80 – (–0.41) = 1.21 V
The half−reactions are
Fe(s)  Fe2+(aq) + 2e−
2Ag+(aq) + 2e− → 2Ag(s)
The overall reaction is
Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s)

19 | 81

Fe(s) + 2Ag+(aq) → Fe2+(aq) + 2Ag(s)
∆G° = 1 mol(–85 kJ/mol) – 2 mol(77 kJ/mol)
∆G° = –85 kJ – 154 kJ
∆G° = –239 kJ


19 | 82

Now we can use this value to find E°.
ΔG °
E° = −
nF
J
− 239,000
mol
E° = −
J

(2 mol) 96,485 
V

E ° = 1.24 V

This answer compares favorably to the 1.21 V
determined using standard potentials.

19 | 83

Equilibrium Constants from Cell Potentials
nFE° = RT ln K
Rearranging:

RT
E =
ln K
nF
We could now convert from ln to log:
o
cell

2.303 RT
E =
log K
nF
0.0592
o
Ecell =
log K
n
o
cell

At 25°C:


19 | 84

?

Calculate the equilibrium constant K at
25°C for the following reaction for the
standard cell potential:
Pb2+(aq) + Fe(s) ⇄ Pb(s) + Fe2+(aq)

First, we find the cell potential.
E°cathode = –0.13 V (Pb2+)
E°anode = –0.41 V
(Fe2+)
E°cell = E°cathode – E°anode
E°cell = –0.13 V – (–0.41 V)
E°cell = +0.28 V
Now we can find K.

19 | 85

o
cell

nFE
ln K =
RT

C 
J

( 2 mol )  96,485
 0.28 
mol 
C

ln K =
J 

 8.31
( 298 K )
mol K 

ln K = 21.82
K = 3.0 × 109


19 | 86

Summary of
Relationship
sAmong
Variables


19 | 87

Dependence of Cell Potential on Concentration:
Nernst Equation
∆G = ∆G° + RT ln Q
Q is the thermodynamic equilibrium constant.
This equation allows us to relate E to E°.

Ecell = E


o
cell

RT
–
ln Q
nF

19 | 88

At 25°C, this reduces to

Ecell = E


o
cell

0.02568
–
ln Q
n

19 | 89

Determination of pH
Cell measurements can allow us to determine the
pH of a solution by determining [H+]. This is the basis
of the pH meter.


19 | 90

On the left a small, commercial
electrode is pictured.
On the right is a sketch showing the
construction of a glass electrode for
measuring hydrogen concentrations.


19 | 91

Consider a voltaic cell,
Fe(s) | Fe2+(aq) || Cu2+(aq) | Cu(s), being
run under standard conditions.
a. Is ∆G° positive or negative for this process?

?

b. Change the concentrations from their
standard values in such a way that Ecell is
reduced. Write your answer using the
shorthand notation.


19 | 92

a.

∆G° = –nFE°
E° = 0.36 – (–0.41) = 0.75 V
Since E° is positive, ∆G° is negative.

b.

Overall reaction:
Fe(s) + Cu2+(aq) → Fe2+(aq) + Cu(s)
To reduce Ecell, Q must be > 1.

Fe2+ 
Q =  2+  So, [Fe2+] must be > [Cu2+].
Cu 



Larger
Smaller
Fe(s) | Fe2+(aq) (1.1 M) || Cu2+(aq) (0.50 M) | Cu(s)

19 | 93

A pH meter is constructed using
hydrogen gas bubbling over an inert
platinum electrode (the hydrogen
electrode) at a pressure of 1.2 atm.
other electrode is aluminum metal immersed in
The
a 0.20 M Al3+ solution. What is the cell potential
when the hydrogen electrode is immersed in a
sample of acid rain with a pH of 4.0 at 25°C?
If the electrode is placed in a sample of
shampoo solution and the cell potential is 1.17
V, what is the pH of the shampoo solution?
The reaction is
2Al(s) + 6H+(aq) ⇄ 2Al3+(aq) + 3H2(g)

?


19 | 94

E =E
o

Q=

3
H2

[

P Al

]

3+ 2

[H ]

+ 6

o
cell

RT
−
ln Q
nF

(1.2 atm) ( 0.20 M )
=
3

(10

-4

M

)

2

6

= 6.91 × 1022

ln Q = 52.6

0.02568
( 52.6 )
E = 1.66 –
6
E = 1.66 – 0.23
E = 1.43 V

19 | 95

E =E
o

o
cell

RT
−
ln Q
nF

RT
o
ln Q = Ecell − E o
nF

RT
ln Q = 1.66 V − 1.17 V = 0.49 V
nF
nF
ln Q = 0.49 V ×
RT
C 

( 6)  96,485

J
mol 

ln Q = 0.49 ×
J 
C 
 8.3145
( 298 K )
mol K 


19 | 96

ln Q = 114.5
Q = 5.26 × 1049

Q=

[H ]

+ 6

=

3
H2

[

P Al
Q

]

3+ 2

3
H2

[

P Al

]

3+ 2

[H ]

+ 6

(1.2 atm) ( 0.20 M )
=
3

2

(5.26 × 10 )
49

[H+] 6 = 1.31 × 10−51
[H+] = 3.31 × 10−9 M
pH = 8.48

19 | 97

Commercial Voltaic Cells
We next look at several commercial voltaic cells.

Zinc–Carbon Dry Cell: Leclanché
Anode:
Cathode:

Zn(s) → Zn2+(aq) + 2e−
2NH4+(aq) + 2MnO2(s) + 2e−
→ Mn2O3(s) + H2O(l) + 2NH3(aq)

The initial voltage is about 1.5 V, but decreases and
deteriorates rapidly in cold weather.

19 | 98


19 | 99

Zinc–Carbon Dry Cell: Alkaline
Anode:
Zn(s) + 2OH−(aq) → Zn(OH)2(s) +
2e−Cathode:
2MnO2(s) + H2O(l) + 2e−
→ Mn2O3(s) + 2OH− (aq)
This cell performs better under current drain and in
cold weather. It isn’t truly “dry” but rather uses an
aqueous paste.


19 | 100


19 | 101

Lithium–Iodine Battery
In this solid−state
battery, the electrodes
are separated by a thin
crystalline layer of
lithium iodide. Diffusion
of the Li+ ion carries the
current. The cell has
high resistance and,
therefore, low voltage. It
is very reliable and is
used for pacemakers.

19 | 102

Lead Storage Cell
The electrodes are lead alloy grids: one is packed
with a spongy lead to form the anode, and the other
is packed with lead dioxide to form the cathode. Both
electrodes are in an aqueous solution of H 2SO4.
Anode:
Cathode:

Pb(s) + HSO4−(aq)
→ PbSO4(s) + H+(aq) + 2e−
PbO2(s) + 3H+(aq) + HSO4−(aq) + 2e−
→ PbSO4(s) + 2H2O(l)

Unlike dry cells, after discharge, lead storage cells
can be recharged.

19 | 103

Each of the six cells in the lead storage cell
generates 2 V, yielding 12 V. During discharge, white
PbSO4(s) coats each electrode.
To recharge the cell, an external current is used,
reversing the previous reactions. Some water
decomposes into hydrogen and oxygen gas, so
more water may need to be added.
Newer batteries use electrodes with calcium in the
lead, which resists decomposition by water. These
versions are maintenance free.

19 | 104


19 | 105

Nickel–Cadmium Cell
Anode:

Cd(s) + 2OH−(aq) → Cd(OH)2(s) + 2e−

Cathode:

NiOOH(s) + H2O(l) + e−
→ Ni(OH)2(s) + OH−(aq)

These cells are used in calculators,
portable power tools, shavers, and
toothbrushes. During recharge, the
reactions are reversed, which can
be done many times.
When cadmium is replaced with a metal hydride
(MH), nickel metal hydride and lithium hydride cells
result. They are less toxic.

19 | 106

Fuel Cell
Fuel cells require a continuous supply of reactants
(fuel).
Anode:

H2(g) → 2H+(aq) + 2e−

Cathode:

O2(g) + 4H+(aq) + 4e− → 2H2O(l)

Fuel cells were originally used in space
applications, but are now being explored for more
uses.

19 | 107


19 | 108

Corrosion Control: Cathodic Protection
Voltaic cells can be used to control corrosion of
underground pipelines and tanks. Rusting occurs
when water comes in contact with iron. The edge
of the water drop, when exposed to air, becomes
one pole of a voltaic cell where oxygen is reduced
to hydroxide.
Anode:
Cathode:

Fe(s) → Fe2+(aq) + 2e−
O2(g) + 2H2O(l) + 4e− → 4OH−(aq)


19 | 109


19 | 110

When the buried metal is connected to a more
active metal such as magnesium, the magnesium
becomes the anode and the iron becomes the
cathode. The iron is, therefore, protected from
oxidation.
This phenomenon is called cathodic protection.


19 | 111


19 | 112

Cathodic protection is illustrated below. The nails
are in a gel containing phenolphthalein indicator
and potassium ferricyanide.
Iron corrosion yields Fe2+, which reacts with
ferricyanide ion to give a dark blue precipitate.
Where OH− forms, phenolphthalein appears pink.

Unprotected

Protected
19 | 113

?

Keeping in mind that seawater contains
a number of ions, explain why
seawater corrodes iron much faster
than freshwater.

Many of the ions in seawater have very high
reduction potentials—higher than the reduction
potential of Fe(s). This means spontaneous
electrochemical reactions will occur with Fe(s),
causing the iron to form ions and go into solution. At
the same time, the ions in the sea will be reduced
and plate out on the surface of the iron.

19 | 114

Electrolytic Cell
An electrolytic cell is an electrochemical cell in
which an electric current drives an otherwise
nonspontaneous reaction.
The process of producing a chemical change in an
electrolytic cell is called electrolysis. Many
important substances are produced commercially
by electrolysis—for example, aluminum and
chlorine.


19 | 115

Downs Cell
A Downs cell is an electrolytic cell used to obtain
sodium metal by electrolysis of sodium chloride.
The products must be kept separated or they
would react.
Anode:

Cl−(l) → ½Cl2(g) + e−

Cathode:

Na+(l) + e− → Na(l)


19 | 116


19 | 117


19 | 118

Aqueous Electrolysis
For a molten salt, the possible reactions are limited
to those involving the ions from the salt.
In aqueous situations, however, the possible
reactions of water must also be included.
2H2O(l) + 2e− → H2(g) + 2OH−(aq)
2H2O(l) → O2(g) + 4H+(aq) + 4e−


19 | 119

Unlike voltaic cell reactions, which maximize the
cell potential, the reaction requiring the smallest
voltage will be the one that occurs. To determine
the reaction, it is necessary to consider all possible
half−reactions that might occur.
First, examine the possible oxidation reactions.
The one with the least negative E° value will occur.
Next examine the possible reduction reactions.
The one with the more positive E° value will occur.

19 | 120

Electrolysis of Sulfuric Acid Solutions
Possible oxidation reactions:
2SO42−(aq) → S2O82−(aq) + 2e− E°red = 2.01 V
2H2O(l) → O2(g) + 4H+(aq) + 4e−

E°red = 1.23 V

Smaller reduction potential
Possible reduction reactions:
H+(aq) + e− → ½ H2(g)
E°red = 0.00 V
2H2O(l) + 2e− → H2(g) + 2OH−(aq) E°red = –0.83 V
Larger (more positive) reduction potential

19 | 121

2H2O(l) → O2(g) + 4H+(aq) + 4e−
4H+(aq) + 4e− → 2H2(g)
2H2O(l) → 2H2(g) + O2(g)
Ecell°= E°cathode – E°anode
Ecell°= 0.00 – 1.23
Ecell°= –1.23 V


19 | 122

Electrolysis of Sodium Chloride Solution
Smaller
Possible oxidation reactions: reduction potential
2Cl−(aq) → Cl2(g) + 2e−
E°red = 1.36 V
2H2O(l) → O2(g) + 4H+(aq) + 4e−

E°red = 1.23 V

Na+(aq) + e− → Na(s)

E°red = –2.71 V

H+(aq) + e− → ½H2(g)

E°red = 0.00 V
Larger (more positive) value


19 | 123

Cl2(g) + 2e− → 2Cl−(aq)
2H+(aq) + 2e− → H2(g)
2H+(aq) + Cl2 (g) → H2(g) + 2Cl−(aq)
Ecell° = 0.00 – 1.36
Ecell° = –1.36 V


19 | 124

?

Describe what you expect to happen at
the electrodes when an aqueous
solution of sodium iodide is
electrolyzed.


19 | 125

Electrolysis of Sodium Iodide Solution
Smaller
reduction potential
Possible oxidation reactions:
2I−(aq) → I2(g) + 2e−
E°red = 0.54 V
2H2O(l) → O2(g) + 4H+(aq) + 4e−

E°red = 1.23 V

Na+(aq) + e− → Na(s)

E°red = –2.71 V

H+(aq) + e− → ½H2(g)

E°red = 0.00 V
Larger (more positive) value


19 | 126

I2(g) + 2e−

→ 2I−(aq)

2H+(aq) + 2e− → H2(g)
2H+(aq) + I2 (g) → H2(g) + 2I−(aq)
Ecell°= 0.00 – 0.54
Ecell°= –0.54 V


19 | 127

The electrolysis of sodium chloride is the basis of
the chlor−alkali industry, which produces chlorine
and sodium hydroxide.


19 | 128

The chlor−alkali mercury cell uses mercury
metal as the cathode. Sodium is reduced, rather
than water, forming a sodium mercury amalgam.


19 | 129

Metals can also be protected from corrosion by
plating them with other metals.
For example, zinc can be plated on steel by
electrogalvanizing to prevent rusting.


19 | 130

Metals can also be purified by electrolysis. Here
we see copper being purified.


19 | 131

Stoichiometry of Electrolysis
In the 1830s, Michael Faraday showed that the
total charge that flows in a circuit is related to the
amount of substance released at the electrodes.
One faraday of charge is the charge on one mole
of electrons and is equal to 96,485 coulombs.


19 | 132

For quantitative considerations, we also need to
know the magnitude of the current and the time it
has flowed.
Electric charge = current × time
Coulombs = amperes × seconds
The ampere, A, is the base unit of current.
The coulomb, C, is equal to an ampere−second.


19 | 133

?

What electric charge is required to
plate a piece of automobile molding
with 1.00 g of chromium metal using a
chromium(III) ion solution?
If the electrolysis current is 2.00 A, how
long does the plating take?


19 | 134

First, we convert mass to moles Cr; then to moles
e−; then, using current, to seconds.

1 mol Cr
3 mol e - 96,485 C
1s
1.00 g Cr ×
×
×
×
52.00 g Cr 1 mol Cr 1 mol e
2.00 C
= 2.78 × 103 s = 46.4 min


19 | 135

?

A solution of nickel salt is electrolyzed
to nickel metal by a current of 2.43 A. If
this current flows for 10.0 min, how
many coulombs is this?
How much nickel metal is deposited in
the electrolysis?


19 | 136

We first find the charge:

C
60 s
2.43 × 10.0 min ×
s
1 min
= 1458 C
Given the half−reaction: Ni2+(aq) + 2e− → Ni(s)

1 mol e−
1 mol Ni 58.69 g
1458 C ×
×
×
−
96,485 C 2 mol e
1 mol Ni
= 0.443 g Ni

19 | 137

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