Chemistry part 32, Julia Burdge,2e (2009) pps

786 CHAPTER
19
Electrochemistry
•
Applying
What
You've Learned
The pain caused by having aluminum foil contact a dental filling results from the filling
being made the cathode in an electrochemical cell. Another type
of
discomfort can result
from the filling being made the anode
in
an electrochemical cell. This occurs when the
filling touches a metal with a greater reduction potential than the components
of
the
amalgam, such
as
gold. When an amalgam filling comes into contact with a gold inlay,
the tin in the filling (the most easily oxidized
of
the major amalgam component
s)
is
oxidized creating
an
unpleasant metallic taste in the mouth. A simplified, unbalanced
equation for the redox reaction that takes place is
Problems:
•

Gold inlay
I
",,) +
4H
+(a
q) +
4e
- •
2H20(l)
_.
Sn
LT
filling
Gold inlay touching amalgam dental filling
I
a)
Balance (in acidic medium) the equation for the oxidation
of
tin from an amalgam
filling.
[
~.
Sample
Problem
19.
1]
b) Calculate the standard cell potential for the reaction in part (a).
[
~.
Sample

Problem
19.2]
c)
Which
of
the components
of
dental amalgam (mercury, silver, tin, copper, or zinc)
would
be
oxidized when a filling is brought into contact with lead?
[
~.
Sample
Problem
19.3]
d)
Calculate the standard free-energy change for the reaction in part (a).
[
~.
Sample
Problem
19.4]
e) Calculate the equilibrium constant at
25°C for the reaction in part (a).
[
~.
Sample
Problem
19.5]

I
•
CHAPTER SUMMARY
Section 19.1
o Redox reactions are those in which oxidation numbers change. Half-
reactions are the separated oxidation and reduction reactions that
make up the overall redox reaction.
o Redox equations can
be
balanced via the half-reaction method, which
allows for the addition
of
H
2
0 to balance
0,
H+ to balance H, and
OH- for reactions taking place in basic solution.
Section 19.2
o An electrochemical cell in which a spontaneous chemical reaction
generates a flow
of
electrons through a wire is called a galvanic cell.
o Half-reactions
in
a galvanic cell take place in separate compartments
called
half-cells. Half-cells contain electrodes in solutions and are
connected via an external wire and by a
salt bridge.

o
The
electrode at which reduction occurs is called the cathode; the
electrode at which oxidation occurs is called the
anode.
o The difference in electric potential between the cathode and the anode
is
the cell potential
(E
cell
)'
Section 19.3
o We use standard reduction potentials
(E
O) to calculate the standard
cell voltage or
standard cell potential
(E
~e
ID.
o Half-cell potentials are measured relative to the standard hydrogen
electrode (SHE),
the half-reaction for which has an arbitrarily defined
standard reduction potential
of
zero.
Section 19.4
o
E
~e

ll
is related to the standard free-energy change
(C: G
O) and to the
equilibrium constant,
K.
A positive
E
~e
ll
corresponds to a negative
(C: G
O) value and a large K value.
Section 19.5
o Ec
ell
under other than standard-state conditions is determined from
E
~e
ll
and the reaction quotient,
Q,
using the Nerizst equation.
I(EyWORDS
Anode, 763
Battery, 777
Cathode, 763
Cell potential
(Ec
e

ll)'
764
Concentration cell, 775
Corrosion,
784
Electrode, 763
Electrolysis,
780
Electrolytic cell, 780
Fuel cell, 778
KEY
WORDS
787
o A concentration cell has the same type
of
electrode and the same
ion in solution (at different concentrations) in the anode and
C8
ih
ode
compartments.
Section 19.6
Batteries are portable, self-contained sources
of
electric energy
consisting
of
galvanic cells or series
of
galvanic cells.

o Fuel cells are not really batteries but also supply electric energy via a
spontaneous redox reaction. Reactants must be supplied constantly for
a fuel cell to operate.
Section 19.7
o Electrolysis is the use
of
electric energy to drive a nonspontaneous
redox reaction. An electrochemical cell us
ed
for this purpose is called
an
electrolytic cell.
o
The
voltage that must actually be supplied to drive a nonspontaneous
redox reaction is greater than the calculated amount because
of
overvoltage.
o Electrolysis is used to recharge lead storage batteries, separate
compounds into their constituent elements, and separate and purify
metals.
o We can calculate the amount
of
a substance produced in electrolysis
if
we know the current applied to the cell and the length
of
time for
which it is applied.
Section 19.8

o Corrosion is the undesirable oxidation
of
metals.
o Corrosion can be prevented by coating the metal surface with paint,
a less easily oxidized metal, or a more easily oxidized metal such as
•
ZlllC.
o
The
use
of
a more easily oxidized metal is known
as
cathodic
protection,
wherein the metal being protected is made the cathode in a
galvanic cell.
Galvanization is the cathodic protection
of
iron or steel
. .
USlllg
zmc.
Galvanic cell, 763
Galvanization,
785
Half-cell, 763
Half-reaction,
760
Nemst

equation, 774
•
Overvoltage, 782
Salt bridge, 763
Standard hydrogen electrode
(SHE), 765
Standard reduction potential
(EO), 765
788 CHAPTER 19 Electrochemistry
KEY EQUATIONS
•
19.1
19.2
19.3
19.4
EO - EO EO
cell - cathode - anode
I:: G
= -nFEceli
I:: G
o =
-nFE
~ell
EO II =
RT
In
K
ce
nF
19

5 E
o = 0.0592 V I
ct
K
. ce
ll
n
°b
19.6
19.7
E = EO -
RT
In Q
nF
E = EO _ 0.0592 V log Q
n
QUESTIONS
AND
PROBLEMS
Section 19.
1:
Balancing Redox Equations
Problems
19.1
19.2
Balance the following redox equations by the half-reaction
method:
(a)
H
2

0
2
+
Fe
2+
•
Fe
3+
+ H
2
0 (in acidic solution)
(b) Cu
+ H
N0
3
•
Cu
2+
+
NO
+ H
2
0 (in acidic solution)
(c)
CN-
+
Mn04
•
CNO
- +

Mn0
2 (in
ba
sic solution)
(d)
Br2
•
BrO
} +
Br
- (in
ba
sic solution)
(e)
S20
~-
+ 12 •
1-
+
S40
~
-
(in acidic solution)
Balance the following redox equations
by
the half-reaction
method:
(a)
Mn
2+

+ H
2
0
2


Mn0
2 + H
2
0
(i
n basic solution)
(b)
Bi(OH
)3
+ SnO
i-
•
SnO
~
-
+
Bi
(in
ba
sic solution)
(c)
Cr
2
0

~
-
+
C
2
0
~
-
•
Cr
3+
+ CO
2
(
in
acidic solution)
(d)
CIO} +
Cl
- • Cl
2
+
CI0
2
(in acidic solution)
(e) Mn2+
+
BiO
} • Bi
3+

+
Mn0
4 (in acidic solution)
Section 19.
2:
Galvanic Cells
\
Review Questions
I
19.3 Define the following term
s:
anode, cathode, cell voltage,
electromotive jorce, standard reduction potentia
l.
19.4
19.5
19.6
19.7
Describe the
ba
sic features
of
a gal
va
nic cell.
Wh
y are the two
components
of
the cell separated from each other?

What
is the function
of
a
sa
lt bridge? What kind
of
electrolyte
should
be
used in a salt bridge?
What
is a cell diagram? Write the cell diagram for a galvanic cell
consisting
of
an
Al
electrode placed in a I M
Al
(N0
3
)3
solution
and an Ag electrode placed
in
aIM
AgN0
3
solutio
n.

What
is the difference bet
wee
n the half-reactions discuss
ed
in
redox processes in Chapter 4 and the half-cell reactions discussed
in
Section 19.2?
,
Section 19.3: Standard Reduction Potentials
Review Questions
19.8 Discuss the spontaneity
of
an electrochemical reaction in terms
of
its standard
emf
(E~cll)
'
19.9 After operating a Daniell cell (see Figure 19.1) for a few minutes,
a student notices that the cell
emf
begins to drop.
Why?
Problems
19.10 Calculate the standard
emf
of
a cell that uses the Mg/Mg2+ and

Cu/Cu
2
+ half-cell reactions at 25°C. Write the equation for the
cell reaction that occurs under standard-state conditions.
19.11
19.1?
Calculate the standard
emf
of
a cell that uses Ag/
Ag
+ and
All
AI
3+
half-cell reactions. Write the cell reaction that occurs
.
under standard-state conditions.
Predict whether
Fe
3+
can oxidize
1-
to 12 under standard-state
conditions.
19.13 Which
of
the following reagents can oxidize H
2
0 to 0 2(g) under

standard-state conditions:
H +(aq),
Cl-(aq),
CI
2
(g), Cu
2+
(aq),
Pb
2+
(aq),
Mn0
4 (aq)
(i
n acid)?
19.14
Consider the following half-reactions:
Mn0
4 (aq) + .
8H
+(aq) +
5e
-
-_.
Mn
2+
(aq) +
4H
2
0(I)

NO
} (aq) +
4H
+(aq) +
3e
- • NO(g) +
2H
2
0(l)
Predict whether
NO}
ions will oxidize Mn
2+
to
Mn04
under
standard-state conditions.
19.15 Predict whether the following reactions would occur
spontaneously in aqueous solution at 25°C. Assume that the
initial concentrations
of
dissolved species are all 1.0 M.
(a) Ca(s) +
Cd
2+(
aq) • Ca
2+
(aq) + Cd(s)
(b)
2Br-(aq)

+
Sn
2+
(aq) •
Br
2(l)
+ Sn(s)
(c) 2Ag(s) + Ni2+(aq) •
2Ag
+(aq) + Ni(s)
(d) Cu+(aq) +
Fe
3+
(aq) •
Cu
2+(
aq) +
Fe
2+
(aq)
•
19.
16
Which species in each pair is a better oxidizing agent under
standard-state condition
s:
(a)
Br
2
or

Au
H
, (b) H2
or
Ag +,
(c)
Cd
2+
or
Cr
H
, (d)
O
2
in acidic media
or
O
2
in basic media?
19.17 Which species in each pair is a better reducing agent under
standard-state conditions: (a) Na
or
Li, (b) H2
or
1
2>
(c)
Fe
2
+ or

Ag, (d)
Br
- or
Co
2+
?
Section 19.4: Spontaneity
of
Redox Reactions
Under
Standard-State Conditions
Review
Questions
19.18 Use the information in Table 2.
1,
and calculate the Faraday
constant.
19.19 Write the equations relating
I1G
o and
Kt
o the standard
emf
of
a
cell. Define all the term
s.
19.
20
Compare the ease

of
measuring the equilibrium constant
electrochemically with that by chemical m
ea
ns [see
Eq
uation
(1
8.15
)]
.
Problems
19.21 What is the equilibrium constant for the follow
in
g reaction at
25°
C?
Mg(s) +
Zn
2+(aq)
:.
=:!:'
Mg
2+
(aq) + Zn(s)
19.22
The
equilibrium constant for the reaction
Sr(s) + Mg
2+

(aq).
' Sr2+(aq) +
Mg
(s)
is 2.69
X 10
12
at 25°C. Calculate EO for a
ce
ll
made up
of
Sr
1
Sr
2
+
and
Mg/Mg
2+
half
-cells.
19.23 Use the standard reduction potentials to find the equilibrium
constant for each
of
the following reactions at
2YC:
(a)
Br
z(l)

+
21
-(
aq)
• ' 2
Br
- (
aq
) + 1
2
(s)
(b) 2Ce
4
+(aq) +
2Cqaq
) • • CI
2
(g) + 2Ce
3
+(aq)
(c) 5Fe
2
+(aq) +
Mn0
4(
aq
) +
8H
+
(aq).

'
Mn
2+(aq) +
4H
2
0 +
5F
e
H
(aq)
19.24 Calculate
I1G
o and Kc
fo
r the following reactions at 25°
C:
(a)
Mg(
s) + Pb2+(
aq)
• •
Mg
2+
(aq) + Pb(s)
(b) 0 2(g) +
4H
+(aq) + 4Fe2+(aq) . ' 2H
2
0
(l)

+ 4Fe
H
(aq)
(c) 2AI(s) + 3lz(s) • '
2AI
H (aq) +
6r-(aq)
19.25 Under standard-state conditions, what spontaneous reaction will
. I . h ' C 4+ C H F
3+
d
occur
1n
aqueous so ulion among t e
IO
ns e , e' , e , an
Fe
2
+?
Calculate
I1G
o and Kc for the reaction.
19.26 Given
th
at EO = 0.52 V for the reduction
Cu
+(aq) + e - •
Cu(s), calculate EO,
I1G
o, and K for the following reaction at

25°C:
Section 19.5: Spontaneity
of
Redox Reactions
Under
Conditions
Other
Than Standard State
Review
Questions
19.
27 Write the Nernst equation, and explain all the term
s.
19.
28 Write the Nernst equation for the following processes at so
me
I temperature T
(
a)
Mg(s) +
Sn
2+(
aq).
•
Mg
2+(aq) + Sn(s)
(b)
2Cr
(s) + 3Pb
2+

(aq) • •
2Cr
H(
aq)
+ 3Pb(s)
QUESTIONS
AND
PROBLEMS
789
Problems
19.29 What is the potential
of
a cell made up
of
Z n
/Z
n
2
+ and C
lLCU
:-
half-cells at 25°C
if
[Zn
2+
] = 0.25 M and [Cu
2
+] = 0.1 5
,\f'?
19.30

Ca
lculate EO,
E,
and
I1G
for the following ce
ll
reaction
s.
(a) Mg(s) +
Sn
2
+(aq) . •
Mg
2+(
aq)
+ Sn(s)
[Mg
2+
] = 0.045 M, [Sn
2+
] = 0.035 M
(b) 3Zn(s) + 2CrH (aq) . •
3Z
n
2+
(aq) +
2C
r(s)
[CrH ]

= 0.010 M, [Zn
2+
] = 0.0085 M
19.31 Calculate the standard potential
of
the cell consist
in
g
of
the
ZnlZn
2
+ half-cell and the
SHE
.
Wh
at will the
emf
of
the cell be
if
[Zn
2
+] = 0.45
M,
PH
= 2.0 atm, and [H+] = 1.8
M?
2
19.32 What is the

emf
of
a cell consisting
of
a Pb2+/Pb half-ce
ll
and
a
PtlH+
1H
2 half-cell if [Pb
2+
] = 0.10
M,
[H] = 0.050
M,
and
PH
= 1.0 atm?
2
19.33 Refe
ning
to the
anangement
in Figure 19.1, calculate the
[Cu
2+
]/[Zn
2
+] ratio at which

th
e following reaction is
spontaneous at 25°C:
Cu(s)
+
Zn
2+
(aq) • Cu
2+
(aq) + Zn(s)
19.
34
Calc
ul
ate the
emf
of
the following concentration cell:
Mg
(s) I
Mg
2+(0.24 M)
II
Mg2 +(0.53 M) I Mg(s)
Section 19.6: Batteries
Review
Questions
19.35 What is a battery? Describe several types
of
batteries.

19.36 Explain the differences between a primary galvanic
cell-one
that is not
rechargeable-and
a stora
ge
ce
ll (for example, the lead
stora
ge
battery), which is rechargeable.
19.37 Discuss the advantages and
di
sadvantages
of
fuel cells ov
er
convention
al
power plants in producing el
ec
tricit
y.
Problems
19.38
The
hydrogen-oxygen fuel
ce
ll
is described in Section 19.6.

(a)
What
volume
of
H
ig),
stored at 25°C at a pressure
of
155 atm, would be ne
eded
to run an electric motor drawing a
cunent
of 8
.5
A for 3.0 h? (b)
What
volume (in liters)
of
air at
25°C and
1.00 atm will have to pass into the
ce
ll per minute to
run the motor? Assume that air is
20
percent O
2
by volume and
that all the
Oz is consumed in the ce

l!.
The other
co
mponents
of
air do
not
affect
th
e fuel-cell reaction
s.
Assume ideal gas
behavior.
19.39 Calculate the standard
emf
of
the propane fuel
ce
ll discussed on
page 779 at
2YC,
given that
I1G
'f
for
pr
opane is - 23.5 kJ/m
o!.
Section 19.7: Electrolysis
Review

Questions
19.40 What is the difference be een a galvanic cell
(s
uch as a Daniell
. cell) and an electrolytic cell?
19.41 What is Faraday's contribution to quantitative electrolysis?
19.42 Define the term
over
volt
age. How does overvoltage affect
electrolytic processes?
790
CHAPTER
19
Electrochemistry
Problems
19.43
The
half-reaction at an electrode is
19.44
. 19.45
19.46
Mg
2+
(molten) + 2e-
~.
Mg(s)
Calculate the number
of
grams

of
magnesium that can be
produced by supplying
1.00 F to the electrode.
Consider the electrolysis
of
molten barium chloride (BaCI
2
) .
(a) Write the half-reactions. (b) How many grams
of
barium
metal can be produced by supplying
0
.5
0 A for
30
min?
•
Considering only the cost
of
electricity, would
it
be
cheaper to
produce a ton
of
sodium or a ton
of
aluminum by electrolysis?

If
the cost
of
electricity to produce magnesium by the electrolysis
of
molten magnesium chloride is $155
per
ton
of
metal, what
is the cost (in dollars)
of
the electricity necessary to produce
(a) 10.0 tons
of
aluminum, (b) 30.0 tons
of
sodium, and
(c)
50.0 tons
of
calcium?
19.47
One
of
the half-reactions for the electrolysis
of
water is
2H
2

0(l)
• 0 2(g) +
4H
+ (aq) +
4e
-
If
0.076 L
of
O
2
is collected at
25
°C and 755
mmHg
, how many
faradays
of
electricity had to
pa
ss through the solution?
19.48 How many faradays
of
electricity are required to produce
(a) 0.84
L
of
O
2
at exactly 1 atm and

25
°C from aqueous H
2
S0
4
solution, (b) 1.50
L
of
Cl
2
at 750 mmHg and 20°C from molten
NaCl, and (c)
6.0 g
of
Sn from molten SnCI
2
?
19.49 Calculate the amounts
of
Cu and
Br
2 produced in 1.0 h at inert
electrodes in a solution
of
CuBr2 by a current
of
4
.5
0 A.
19.50

In
the electrolysis
of
an aqueous
AgN0
3
solution, 0.67 g
of
Ag is deposited after a certain period
of
time. (a) Write the
half-reaction for the reduction
of
Ag +. (b) What is the probable
oxidation half-reaction? (c) Calculate the quantity
of
electricity
us
ed
(i
n coulombs).
19.51 A steady current was passed through molten
CoS0
4
until 2.35 g
of
metallic cobalt was produced. Calculate the number
of
co
ulombs

of
electricity used.
19.52 A constant electric current flows for 3.75 h through two
electrolytic cells connected in series.
One contains a solution
of
AgN0
3
and the second a solution
of
CuCI
2
. During this time,
2.00 g
of
silver is deposited in the first cell. (a) How many grams
of
copper are deposited in the second cell? (b)
What
is the current
flowing (in amperes)?
19.53 What is the hourly production rate
of
chlorine gas (in kg) from
an electrolytic cell using aqueous NaCI electrolyte and carrying a
current
of
1.500 X 10
3
A?

The
anode efficiency for the oxidation
of
Cl- is 93.0 percent.
19.54 Chromium plating
is
applied by electrolysis to objects suspended
in a dichromate solution, according to the following (unbalanced)
half-reaction:
How long (in hours) would it take to apply a chromium plating
1.0 X
10-
2
mm
thick to a car bumper with a surface area
of
0.25 m
2
in an electrolytic cell carrying a current
of
25.0 A? (The
density
of
chromium is 7
.1
9 g/cm
3
.)
19.55 The passage
of

a current
of
0.750 A for 25.0 min deposited
0.369 g
of
copper from a CUS04 solution. From this information,
calculate the molar mass
of
copper.
19.56 A quantity
of
0.300 g
of
copper was deposited from a CUS04
solution by passing a current
of
3.00 A through the solution for
304
s.
Calculate the value
of
the Faraday constant.
19.57
19.58
In
a certain electrolysis experiment, 1.44 g
of
Ag were deposited
in one cell (containing an aqueous
AgN0

3
solution), while
0.120 g
of
an unknown metal X was deposited in another cell
(containing an aqueous XCl
3
solution) in series with the
AgN0
3
cell. Calculate the molar mass
of
X.
One
of
the half-reactions for the electrolysis
of
water is
If 0.845 L
of
H2 is collected at 25°C and 782
mmHg
, how many
faradays
of
electricity had to pass through the solution?
Section 19.8: Corrosion
Review
Questions
19.59 Steel hardware, including nuts and bolts, is often coated with a

thin plating
of
cadmium. Explain the function
of
the cadmium
layer.
19.60
"Galvanized iron" is steel sheet that has been coated with zinc;
"tin" cans are made
of
steel sheet coated with tin. Discuss the
functions
of
these coatings and the electrochemistry
of
the
corrosion reactions that occur
if
an electrolyte contacts the
scratched surface
of
a galvanized iron sheet or a tin can.
19.61 Tarnished silver contains Ag
2
S. The tarnish can
be
removed
by placing silverware in an aluminum pan containing an inert
electrolyte solution, such as NaCI. Explain the electrochemical
principle for this procedure. [The standard reduction potential for

the half-cell reaction Ag
2
S(s) +
2e- • 2Ag(s) +
S2
-(a
q) is
-0
.71
V.]
19.62 How does the tendency
of
iron to rust depend on the pH
of
the
solution?
Additional Problems
19.63 For each
of
the following redox reactions, (i) write the half-
reaction
s,
(ii) write a balanced equation for the whole reaction,
(iii) determine in which direction the reaction will proceed
spontaneously under standard-state conditions:
(a) H
2
(g) + Ne+(aq) • H+(aq) + Ni(s)
(b)
Mn0

4 (aq) + Cl - (aq) •
Mn
2+
(aq) + CI
2
(g) (in acid
solution)
(c)
Cr(s) + Zn2+(aq)
+.
CrH(aq)
+ Zn(s)
19.64 The oxidation
of
25.0 mL
of
a solution containing Fe
2
+ requires
26.0 mL
of
0.0250 M K2
Cr
2
07
in
acidic solution. Balance the
following equation, and calculate the molar concentration
of
Fe

2
+:
19.65 The
S0
2 present in air is mainly responsible for the phenomenon
of
acid rain.
The
concentration
of
S0
2 can be determined by
titrating against a standard permanganate solution as follows:
5S0
2
+
2Mn04
+2H
2
0
•
5S0~
-
+
2Mn
2+
+
4H
+
Calculate the number

of
grams
of
S0
2 in a sample
of
air
if
7.37
mL
of
0.00800 M
KMn0
4 solution is required for the
titration.
19.66 A sample
of
iron ore weighing 0.2792 g was dissolved in an
excess
of
a dilute acid solution. All the iron was first converted
to Fe(II) ions.
The
solution then required 23.30 mL
of
0.0194 M
KMn04
for oxidation to Fe(III) ions. Calculate the percent by
mass
of

iron in the ore.
19.67 The concentration
of
a hydrogen peroxide solution can be
conveniently determined by titration against a standardized
potassium permanganate solution in an acidic medium according
to the following unbalanced equation:
(a) Balance this equation. (b)
If
36.44 mL
of
a 0.01652 M
KMn04
solution is required to completely oxidize 25.00
mL
of
an H
2
0
2
solution, calculate the molarity
of
the H
2
0
2
solution.
19.68
Oxalic acid (H
2

C
2
0
4
) is present in many plants and vegetables.
(a) Balance the following equation
in
acid solution:
Mn04
+
C20~-
•
Mn
2+
+
CO
2
(b)
If
a 1.00-g sample
of
plant matter requires 24.0
mL
of
0.0100 M
KMn04
solution to reach the equivalence point, what is
the percent by mass
of
H

2
C
2
0
4
in the sample?
19.69 Calcium oxalate (CaC
2
0
4
) is insoluble in water. This property
has been used to determine the amount
of
Ca
2+
ions in blood.
The calcium oxalate isolated from blood is dissolved in acid and
titrated against a standardized
KMn0
4 solution as described in
Problem 19.68.
In
one test it is found that the calcium oxalate
isolated from a
1O.0-mL sample
of
blood requires 24.2 mL
of
9.56 X 10-
4

M
KMn04
for titration. Calculate the number
of
milligrams
of
calcium per milliliter
of
blood.
19.70 Complete the following table. State whether the cell reaction is
spontaneous, nonspontaneous, or at equilibrium.
19.71
E
D.G
Cell Reaction
> 0
> 0
=0
From the following information, calculate the solubility product
of
AgBr:
Ag
+(aq) + e-
-_.
Ag(s)
AgBr(s)
+ e - • Ag(s) +
Br
- (aq)
eo

=
0.80V
EO =
0.07V
19.72 Consider a galvanic cell composed
of
the
SHE
and a half-cell
using the reaction Ag +
(aq) + e - • Ag(s). (a) Calculate the
standard cell potential. (b)
What
is the spontaneous cell reaction
under standard-state conditions? (c) Calculate the cell potential
when [H+]
in
the hydrogen electrode is changed to (i) 1.0 X
10
-
2
M and (ii) 1.0 X
10-
5
M,
all other reagents being held at
standard-state conditions. (d) Based on this cell arrangement,
suggest a design for a pH meter.
19.73 A galvanic cell consists
of

a silver electrode in contact with 346
mL
of
0.100 M
AgN0
3
solution and a magnesium electrode
in contact with 288
mL
of
0.100 M
Mg(N0
3
h solution. (a)
Calculate
E for the cell at 25°C. (b) A current is drawn from
the cell until
1.20 g
of
silver has been deposited at the silver
electrode. Calculate
E for the cell at this stage
of
operation.
19.74 Explain why chlorine gas can be prepared by electrolyzing an
aqueous solution
of
NaCI but fluorine gas cannot be prepared by
electrolyzing an aqueous solution
of

NaF.
QUESTIONS
AND
PROBLEMS
791
19.75 Calculate the
emf
of
the following concentration cell at
25
°
C:
Cu(s) I Cu
2+
(0.080
M)
II
Cu
2+
(1.2
M)
I Cu(s)
19.76 The cathode reaction in the Leclanche cell is given by
2Mn0
2
(S)
+
Zn
2+
(aq) + 2e- • ZnMn204(S)

If
a Leclanche cell produces a current
of
0.0050 A, calculate how
many hours this current supply will last
if
there is initially 4.0 g
of
Mn0
2 present in the cell. Assume that there is an excess
of
Zn
2
+ ions.
19.77 For a number
of
years, it was not clear whether mercury(I) ions
existed in solution as
Hg
+ or as
Hg~
+
.
To distinguish between
these two possibilities, we could set up the following system:
Hg(l) I soln A II soln B I Hg(l)
where soln A contained 0.263 g mercury(I) nitrate
per
liter
and soln B contained 2.63 g mercury(I) nitrate per liter.

If
the
measured
emf
of
such a cell is 0.0289
Vat
18
°C, what can you
deduce about the nature
of
the mercury(I) ions?
19.78 An aqueous KI solution to which a few drops
of
phenolphthalein
have been added is electrolyzed using an apparatus like the one
shown here:
Describe what you would observe at the anode and the cathode.
(Hint: Molecular iodine is only slightly soluble in water, but in
the presence
of
r ions, it forms the brown color
of
1
:3
ions. See
Problem 13.114.)
19.79 A piece
of
magnesium metal weighing 1.56 g is placed in 100.0

mL
of
0.100 M
AgN0
3
at 25°C. Calculate [Mg2+] and [Ag +] in
solution at equilibrium. What is the mass
of
the magnesium left?
The volume remains constant.
19.80
De
scribe an experiment that would enable you to detennine
which is the cathode and which is the anode in a galvanic cell
using copper and zinc electrodes.
19.81 An acidified solution was electrolyzed using copper electrodes.
A constant current
of
1.18 A caused the anode to lose 0.584 g
after 1.52
X 10
3
s.
(a) What is the gas produced at the cathode,
and what is its volume at
STP? (b) Given that the charge
of
an
electron is
1.6022 X 10-

19
C, calculate Avogadro's number.
Assume that copper
is
oxidized to Cu
2+
ions.
19.82 In a certain electrolysis experiment involving
AI
3+
ions, 60.2 g
of
AI
is recovered when a current
of
0.352 A
is
used. How many
minutes did the electrolysis last?
19.83
Consider the oxidation
of
ammonia:
(a) Calculate the
D.G
o for the reaction. (b)
If
this reaction were
used in a fuel cell, what would the standard cell potential be?
•

•
•
792
CHAPTER
19
Electrochemistry
19.84
When
an aqueous solution contairting gold(III) salt is
electrolyzed, metallic gold is deposited at the cathode and oxygen
gas is generated at the anode. (
a)
If
9.26 g
of
Au is deposited
at the cathode, calculate the volume (in liters)
of
O
2
generated
at 23°C and
747
mmHg. (b)
What
is the current used if the
electroly
ti
c process took 2.00 h?
19.85 In an electrolysis experiment, a student passes the same quantity

of
electricity through two electrolytic cells, one containing a
silver salt and the other a go
ld
sa
lt. Over a certain period
of
time,
the student finds that 2.64
&
of
Ag and 1.61 g
of
Au are deposited
at the cathodes.
What
is the oxidation state
of
gold in the gold
salt?
19.86
People li
vi
ng in cold-climate countries where there is plenty
of
snow are advised not to h
eat
their garages in the winter.
What
is

the electrochemical basis for this recommendation?
19.87 Given that
19.88
19.89
19.90
2Hg
2+(aq) + 2e-

Hg
~
+
(aq)
H
g~+(aq)
+ 2e- • 2Hg(l)
EO = 0.92 V
EO = 0.85 V
calculate GO and K for the following process at 25°
C:
Hg
~
+
(aq)
•
Hg
2+
(aq) +
Hg
(l)
(The preceding reaction is an

example
of
a disproportionation
reaction
in which an element in one oxidation state is both
oxidized and reduced.)
Fluorine (F
2
) is obtained by the electrolysis
of
liquid hydrogen
fluoride (
HF
) containing potassium fluoride (KF). (a) Write
the half-ce
ll
reactions and the overall reaction for the process.
(b) What is the purpose
of
KF? (c) Calculate the volume
of
F2
(in liters) collected at
24.0°C and 1.2 atm after electrolyzing the
solution for
15
h at a current
of
502
A.

A
300
-mL
solution
of
NaCl was electrolyzed for 6.00 min.
If
t
he
pH
of
the final solution was 12.24, calculate the average current
used.
Industrially,
coppe
r is purified by electrolysis.
The
impure
coppe
r
acts as the anode, and the cat
hod
e is made
of
pure
coppe
r.
The
electrodes are
immer

sed in a CUS04 solution. During electrolysis,
cop
per
at the anode enters the solution as Cu
2+
while Cu
2+
ions
are reduced at the cathode. (a)
Write
half-cell reactions and the
overall rea
ct
ion for the electrolytic process. (b) Suppose the
anode was contaminated with Zn and Ag. Explain what happens
to these impurities during electrolysis. (c)
How
many
hour
s will
it take to obtain
1.00
kg
of
Cu
at a current
of
18.9 A?
19.91 An aqueous solution
of

a platinum salt is electrolyzed at a current
of
2.50 A for 2.00
h.
As a result, 9.09 g
of
metallic
Pt
is formed at
the cathode. Calculate the charge on the
Pt
ions in this solution.
19.92 Consider a galvanic cell consisting
of
a
ma
gnes
ium
electrode
in contact with
1.0 M
Mg(N0
3
)2 and a ca
dmium
electrode in
contact with
1.0 M
Cd
(N0

3
h Calculate EO for the cell, and draw
a diagram showi
ng
th
e cathode, anode, and direction
of
electron
flow.
19.93 A current
of
6.00 A
pa
sses through an electrolytic cell containing
dilute sulfuric acid for
3.40
h.
If
the volume
of
O
2
gas generated
at the anode
is
4.26 L (at STP), calculate the charge (in
coulombs) on an electron.
19.94 Gold will not dissolve in either concentrated nitric acid or
concentrated hydrochloric acid. However, the metal does dissolve
in a mixture

of
the acids (one part
HN0
3
and three parts HCI by
volume), called
aqua regia. (a) Write a balanced equation for
this reactio
n.
(
Hint:
Among
the products are
HAuCl
4
and
NO
z
.)
(
b)
What is the function
of
HCl?
19.95 Explain why most u
sef
ul galvanic cells give voltages
of
no more
than 1.5 to 2.5

V
What
are the prospects for developing practical
galvanic
ce
lls with voltages
of
5 V
or
more
?
19.96 A silver rod and a
SHE
a
re
dipped into a saturated aqueous
solution
of
si
lver oxalate (Ag
2
C
2
0
4
),
at 25°C.
The
mea
sured

potential difference
between
the rod and the
SHE
is 0.589
V,
the
rod being
po
sitive. Calculate the solubility product constant for
s
il
ver oxalate.
19.97 Zinc is an amphoteric metal; that is, it reacts with
both
acids
and bases.
The
standard reduction potential is
-1.36
V for the
reaction
Calculate the formation constant
(Kr) for the reaction
Z n
2+
(aq) +
40W(aq)'
•
Zn(OH)

~
-(aq)
19.98
Use
the data in Table 19.1 to determine whether
or
not hydrogen
peroxide will undergo disproportionation in an acid medium:
19.99
19.100
2H
z
O
z
•
2H
z
O + Oz·
The
magnitudes (
but
not the signs)
of
the standard reduction
potentials
of
two metals X and Y are
y
2+
+

2e
-
__
~.
Y
X
2+
+
2e
-
+.
X
I EO
1=
0.34 V
I EO I = 0.25 V
where the
II
notation denotes that o
nl
y the
ma
gnitude (but not the
sign)
of
the EO val
ue
is show
n.
When the half-cells

of
X and Y
are
con
nected, electrons flow from X to
Y.
When
X is connected
to a SHE, electrons flow from X to SHE. (a) Are the
EO values
of
the half-rea
ct
ions positive
or
ne
gative? (b)
What
is the standard
emf
of
a cell made
up
of
X and Y?
A galvanic ce
ll
is constructed as follows.
One
half-cell consists

of
a platinum wire
immer
sed in a solution containing 1.0 M Sn
2
+
and
1.0 M
Sn
H
;
the other half-cell has a thallium rod
imm
ersed
in a solution
of
1.0 M TI+ (a) Write the
half
-cell reactions and
th
e overall reaction. (b)
What
is the equilibrium constant at 25°
C?
(c)
What
is the cell voltage if the
TI
+ concentration is increased
10-fo1d?

(EO
Ttm
=
-0.34
V)
19.101 Given the standard reduction potential for Au
3
+ in Table 19.1 and
Au+
(aq) +
e-
• Au(s) EO = 1.69 V
answer the following questions. (a)
Why
doe
s gold not
tarnish in air? (b) Will the following disproportionation occur
spontaneously?
3Au+(aq)
•
Au
3+
(aq) + 2Au(s)
(c) Predict the reaction between gold and fluorine gas.
19.102
The
ingestion
of
a very small quantity
of

mercury is not
considered too harmful. Would this sta
tement
still hold
if
the
gastric
juic
e in your stomach were mostly nitric acid instead
of
hydrochloric acid?
19.103
19.104
19.105
19.106
When
25.0
mL
of
a solution containing both
Fe
2
+ and
Fe
3+
ions
is titrated with
23.0
mL
of

0.0200 M
KMn04
(in dilute sulfuric
acid), all the Fe2+ ions are oxidized to Fe
3+
ions. Next, the
solution is treated with Zn metal to convert all the
Fe
3
+ ions to
Fe
2+
ions. Finally, 40.0
mL
of
the same
KMn0
4 solution is added
to the solution in
order
to oxidize the Fe
2+
ions to
Fe
3+
. Calculate
the molar concentrations
of
Fe
2+

and Fe
3+
in the original
solution.
Consider
the Daniell cell in Figure 19.1.
When
viewed externally,
the anode appears negative and the cathode
po
sitive (electrons
are flowing from the anode to the cathode). Yet in solution anions
are moving toward the anode, which means that it must appear
positive to the anions. Because the anode cannot simultaneously
be
negative and positive, give an explanation for this apparently
contradictory situation.
Use
the data in Table 19.1 to show that the decomposition
of
H20
i (a disproportionation reaction) is spontaneous at 25°C:
The
concentration
of
sulfuric acid in the lead-storage battery
of
an automobile over a period
of
time has decreased from 38.0

percent by mass (density = 1.29
g/mL
) to 26.0 percent by
ma
ss
(1.19 g/mL).
Assume
the volume
of
the acid remains constant at
724
mL. (a) Calculate the total charge in coulombs supplied
by
the battery. (b)
How
long (in hours) will it take to recharge the
battery back to the original sulfuric acid concentration using a
current
of
22.4 A?
19.107 Consider a Daniell cell operating under non-standard-state
conditions. Suppose that the
cell's
reaction is mUltiplied by 2.
What
effect does this have on each
of
the following quantities in
the Nernst equation: (a)
E,

(b) E
O,
(c) Q, (d) In Q, (e) n?
19.108 A spoon was silver-plated electrolytically
in
an
AgN0
3
solution.
(a)
Sketch a diagram for the process. (b)
If
0.884 g
of
Ag
was
deposited on the spoon at a constant
cunent
of
18.5 rnA, how
long (in min) did the electrolysis take?
19.109
Comment
on whether
F2
will become a stronger oxidizing agent
with increasing H+ concentration.
19.110 In recent years there has been much interest in electric
car
s. List

some advantages and disadvantages
of
electric cars compared to
automobiles with internal combustion engines.
19.111 Calculate the pressure
of
H2
(in atm) required to maintain
equilibrium with respect to the following reaction at
25°C,
Pb(s)
+
2H
+(
aq).
>
Pb
2+
(aq) + Hi g)
given that [Pb
2+
] = 0.035 M and the solution is buffered at
pH
1.60.
19.112 A piece
of
magnesium ribbon and a copper wire are partially
immer
s
ed

in a 0.1 M HCI solution in a beaker.
The
metals
are
joined
externally by another piece
of
metal wire. Bubbles
are s
een
to
evolve at both the
Mg
and Cu surfaces. (a) Write
equations representing the reactions
occuning
at the metals. (b)
What
visual evidence would you seek
to
show that Cu is not
oxidized to Cu
2+
? (c) At
some
stage,
NaOH
solution is added to
the beaker to neutralize the
HCI acid. Upon further addition

of
NaOH
, a white precipitate forms.
What
is it?
QUESTIONS
AND
PROBLEMS 793
19.113
The
zinc-air battery shows much promise for electric
car
s
because it is lightweight and rechargeable:
·r?
1
Air
cathode
1<
20H
-
H
2
O
I
/
-
Zinc anode
Zn
• Zn(OH)1

•
ZnO
-
The net transformation is Zn(s) +
~
02(g) > ZnO(s). (a)
Write the half-reactions at the zinc-air electrodes, and calculate
the standard
emf
of
the battery at 25°C. (b) Calculate the
emf
under actual operating conditions when the partial pressure
of
oxygen is 0.21 atm. (c)
What
is the energy density (measured
as the energy
in
kilojoules that can be obtained from I
kg
of
the
metal)
of
the zinc electrode? (d)
If
a
CUlTent
of

2.1 X 10
5
A is to
be drawn from a zinc-air battery system,
what
volume
of
air (in
liters) would need to
be
supplied to the battery every second?
A
ss
ume that the temperature is 25°C and the partial pressure
of
oxygen is 0.21 atm.
19.114 Calculate
EO for the reactions
of
mercury with (a) 1 M HCI
19.115
19.116
19.117
19.118
and (b) I
M
HN0
3
.
Which

acid will oxidize Hg to
Hg~
+
under
standard-state conditions? Can you identify which pictured test
tube contains
HN0
3
and
Hg
and which contains HCI and Hg?
Because all alkali metals react with water, it is not
pos
sible to
mea
sure the standard reduction potentials
of
these metals directly
as in the
ca
se
of
, say, zinc. An indirect method
is
to cons
ider
the
following hypothetical reaction:
Li
+(aq) + m

ig
) > Li(s) + H+(aq)
Using the appropriate equation presented in this chapter and the
thermodynamic data in Appendix 2, calculate
EO for
Li
+(aq)
+ e > Li(s) at 298 K. Use 96,485.338 C/
mol
e- for the
Faraday constant.
Compare
your result with that listed in Table
19.1.
A galvanic cell using
Mg/Mg
2+
and
Cu
/
Cu
2+
half-cells operates
under standard-state conditions at
25°C, and each compartment
ha
s a volume
of
218 mL.
The

cell delivers 0.22 A for 31.6 h. (a)
How many grams
of
Cu
are deposited? (b) What is the [Cu
2+
]
remaining?
Given the followi
ng
standard reduction potentials, calculate the
ion-product,
K
w,
for water at 25°C:
•
2H
+(aq) +
2e
-
+>
H
2
(g)
2H
2
0 (l) +
2e-
• H
2

(g ) +
20W(aq)
EO =
O.OOV
EO =
-0.83
V
Compare the pros and cons
of
a fuel cell, such as the hydrogen-
oxygen fuel cell, and a coal-fired power station for generating
electricity.
794
CHAPTER
19 Electrochemistry
19.119
Lead
storage batteries are rated
by
ampere-hours, that is, the
number
of
ampe
res they
can
deliver in an hour. (a) Show that
1
Ah = 3600 C. (b)
The
lead anodes

of
a certain lead-storage
battery have a total mass
of
406
g. Calculate the
maximum
theoretical capacity
of
the battery in ampere-hours. Explain
why in practice
we
can never extract this much energy from
the batter
y.
(Hint:
Assume
all the lead will
be
used up in the
electrochemical reaction, and refer to t
he
electrode reactions on
page
777.) (c) Calculate
E
~e
ll
and
AG

O for the battery.
19.120
Use
Equations 18.11 and 19.3 to calculate the
emf
values
of
the
Daniell cell at
25°C and
80
°C.
Comment
on
your
results.
What
assumptions are
used
in the derivation? (Hint: You need the
thermodynamic data
in
Appendix 2.)
19.121 A construction company is installing an iron culvert (a long
cylindrical tube) that is
40.0 m long with a radius
of
0.900
m. To
pr

event corrosion, the culvert must be galvanized. This
process is carried out by first
pa
ssing an iron sheet
of
appropriate
dimensions through an electrolytic cell containing Zn
2
+ ions,
using graphite as the anode and the iron sheet as the cathode.
If
the voltage is 3.26
V,
what is the cost
of
electricity for depositing
a layer
0.200
mm
thick if the efficiency
of
the process is 95
percent? The electricity rate is
$0.12
per
kilowatt hour (
kWh
),
where 1 W =
1 lis and the density

of
Zn is 7
.1
4 g/cm
3
19.122 A 9.00 X 10
2
mL
amount
of
0.200 M MgI2 solution was
electrolyzed. As a result, hydrogen gas was generated at the
cathode and i
od
ine was formed at the anode.
The
volume
of
hydrogen collected at 26°C and
779
mmHg
was 1.22 X 10
3
mL.
(a) Calculate the char
ge
in coulombs consumed in the process.
(b)
How
long (in

min)
did the electrolysis last
if
a current
of
7.55 A was used? (c) A white precipitate was formed in t
he
process. What was it, and what was its mass in grams?
Assume
the volume
of
the solution was constant.
19.123
19.124
19.125
19.126
Ba
s
ed
on
the following standard reduction potentials,
Fe
2
+(aq) +
2e
- • Fe(s)
E'l
=
-0.44
V

E
~
= 0.77 V
calculate the standard reduction potential for the half-reaction
Fe
H (aq) +
3e
- • Fe(s) E
'3
=?
To remove the tarnish (Ag
2
S) on a silver spoon, a student
carried out the following steps. First, s
he
pla
ced the spoon in a
large
pan
filled with water so the spoon was totally
immer
se
d.
Next, s
he
added a few tablespoonfuls
of
baking so
da
(sodium

bicarbonate), which readily dissolved. Finally, s
he
placed so
me
aluminum foil at the bottom
of
the pan
in
contact with the spoon
and then heated the solution to about
80°C. After a few minutes,
the spoon was removed a
nd
rinsed with cold water.
The
tarnish
was gone, and the spoon regained its original shiny appearance.
(a) Describe with equations the electrochemical
ba
sis for the
procedure. (b) Adding N aCI instead
of
N
aHC0
3
would also
work because both compounds are strong electrolytes.
What
is
the

ad
ded advantage
of
using
NaHC0
3
?
(Hint: Consider the
pH
of
the solution.) (c)
What
is the purpose
of
he
ating the solution?
(d)
Some
commercial tarnish removers contain a fluid (or paste)
that is a dilute
HCI solution. Rubbing the spoon with the fluid
will also remove the tarnish.
Name
two disadvantages
of
using
this procedure compared to the one described previously.
Calculate the equilibrium constant for the following reaction at
298
K:

Zn(s) + Cu2+(aq)
+.=='
Zn2+(aq) + Cu(s)
Cytochrome-c is a protein involved in biological electron transfer
processes.
The
redox half-reaction is shown by the reduction
of
the
Fe
3
+ ion to the
Fe
2+
ion:
cyt c(Fe
3
+)
+ e -
+.
cyt c(Fe
2
+) EO = 0.254 V
Calculate the number
of
moles
of
cyt c(
Fe
H

)
formed from cyt
c(Fe
2+
) with the Gibbs fr
ee
energy derived from the oxidation
of
I mole
of
glucose.
PRE-PROFESSIONAL PRACTICE
EXAM
PROBLEMS:
PHYSICAL AND BIOLOGICAL SCIENCES
A galvanic cell is constructed by immersing a piece
of
copper wire in 25.0
mL
of
a 0.20 M CUS04 solution and a zinc strip
in
25.0
mL
of
a 0.
20
M
ZnS04
solution. Cu

2+
ions react with aqueous
NH
3 to form the
comp
le
x
ion
Cu(NH
3
)~+:
Cu
2+
(aq) +
4NH
3
(aq)
+.
Cu(NH
3)~+
1.
U si ng the
eq
uati on
E = EO _ 0.0592 V lou Q
n b
calculate the
emf
of
the cell at 25°C.

a) 0.0 V
b) 1.10 V
c)
0.90 V
d) 1.30 V
2.
What
would happen
if
a small
amount
of
concentrated
NH
3 solution
were added to the
CUS0
4 solution?
a) Nothing.
b)
Emf
would increase.
c)
Emf
would decrease.
d)
Not
enough
inf
ormation to determine.

3.
What
would happen
if
a
sma
ll
amount
of
concentrated
NH
3 solution
were added to t
he
ZnS04
solution?
a) Nothing.
b)
Emf
wou
ld
increase.
c)
Emf
would decrease.
d)
Not
enough information to determine.
4. In a separate experiment,
25.0

mL
of
3.00 M
NH
3 is added to the
CUS0
4 solution.
If
the
emf
of
the cell is 0.68 V at equilibrium,
calculate the formation constant
(K
r)
of
Cu(NH
3
)
~+
'
a) 9.4 X 10
22
b) 1.1 X 10-
23
c) 1.5
X 10-
14
d) 1.5 X 10
14

ANSWERS TO IN-CHAPTER MATERIALS 795
ANSWERS TO IN-CHAPTER MATERIALS
Answers
to
Practice Problems
19.1A
Mn04
+
5Fe
2+
+
8H
+ •
Mn
2+
+
5Fe
3+
+
4H
2
0.
19.1B
2Mn0
4 + H
2
0 +
3CN
- •
2Mn0

2 +
20H
- +
3CNO
19.2A
Cd
+ Pb2+ •
Cd
2+
+ Pb,
E
~ell
= 0.27
V.
19.2B 2Al +
3Cu
2+
• 2AI
3+
+ 3Cu,
E
~
ell
=
2.00
V.
19.3A (a)
No
reaction,
(b)

2H
+ +
Pb
•
H2
+
Pb
H
19.3B (a) No reaction, (b)
2H
+ +
Zn • H2 + Zn
2+.
19.4A - 411 kJ/mol. 19.4B 23.2
kJ
/mol.
19.5A 1 X
10-
42
. 19.5B 3 X
10-
45
. 19.6A
Ye
s,
the reaction is
spontaneous.
19.6B 2.2
M.
19.7A

7.44
g Mg. 19.7B 0.96 A.
Answers
to
Checkpoints
19.1.1
a,
c,
e.
19.1.2
c.
19.3.1
d.
19.3.2
c.
19.3.3
b.
19.3.4
e.
19.4.1
b.
19.4.2
b.
19.5.1
b.
19.5.2
b.
19.7.1
a.
19.7.2

e.
Answers
to
Applying
What
You've Learned
a)
2Sn + O
2
+
4H
+ •
2Sn
2+
+
2H
2
0 .
b)
1.37
V.
c) Tin and zinc.
d)
- 5.3 X 10
2
kJ
/mol. e) 4 X 10
92
•
•

lie
ear
emlstr
20.1
Nuclei
and
Nuclear
Reactions
,
20.2
Nuclear Stability
,',
•
•
Patterns
of
Nuclear Stability
•
•
Nuclear Binding Energy
20.3
Natural
Radioactivity
-
•
Kinetics
of
Radioactive
Decay
•

•
Dating Based on
Radioactive Decay
2004
Nuclear
Transmutation
20.5
Nuclear Fission
20.6
Nuclear Fusion
'.
20.7
Uses
of
Isotopes
,
• Chemical Analysis
•
•
Isotopes in Medicine
20.8
Biological Effects
of
•
Radiation
•
•
•
Nuclear
Chemistry

in
the
Treatment
of
Cancer
Brain tumors are some
of
the most difficult cancers to treat because the site
ofthe
malig-
nant growth makes surgical excision difficult or impossible. Likewise, conventional
radiation therapy using X rays or
'Y
rays from outside the skull is usually not effective.
An ingenious approach to this problem is
boron neutron capture therapy (
BNCT
).
This
technique involves first administering a boron-lO compound that is selectively taken
up by tumor cells and then applying a beam
of
low-energy neutrons to the tumor site.
lOB
captures a neutron to produce lIB, which disintegrates via the following nuclear
reaction:
l~
B
+ bn
-_.

jLi +
ia
The highly energetic particles produced by this reaction destroy the tumor cells in which
the
lOB
is
concentrated. Because the particles are confined to
ju
st a few micrometers,
they preferentially destroy tumor cells without damaging neighboring normal cells.
BNCT
is a highly promising treatment and is
an
active area
of
researc
h.
One
of
the major
goals
of
the research is to develop suitable compounds to deliver
lOB
to the desired site.
For such a compound to be effective, it must meet several criteria. It must have a high
affinity for tumor cells, be able to pass through membrane barriers to reach the tumor
site, and have minimal toxic effects on the human body.
This is one example
of

how nuclear chemistry is impOitant in the treatment
of
cancer.
In
Th
is
Chapter, You Will Learn some
of
the fundamentals
of
nuclear chemistry and how nuclear reactions are
important to living systems and to society.
Before you begin, you should review
• Radioactivity
[~~
Sect
ion 2.2]
• Atomic number, mass number, and isotopes
[~~
Section
2.3]
•
First-orderkinetics
[~
Section
14.3]
The
CT
scan shows a brain
tum

or that might be difficult or
impossible to tr
ea
t by conventional surgical methods. Nuclear
medicine, including boron neutron
ca
ptu
re
therapy (BNCT
),
enables doctors to tr
ea
t cancers
of
this type.
Media
Playerl
MPEG
Content
Chapter
in
Rev
iew
797
798 CHAPTER
20
Nuclear Chemistry
•
An
CI.

pa
rt
i
cl
e
is
identical
to a helium-4
nucleus
and
ca
n be
repres
e
nted
eithe
r as
ju
or
j
He
.
Nuclei
and
Nuclear Reactions
With the exception
of
hydrogen
cl
B), all nuclei contain protons and neutrons.

Some
nuclei are
unstable a
nd
undergo radioactive decay, emitting particles and/or electromagnetic radiation
[
~~
Section 2.2]. Spontaneous emission of particles or electromagnetic radiation is known as radio-
activity. All elements having an atomic number greater than 83 are unstable and are therefore
radioactive. Polonium-210
e~~
p
o),
for example, decays spontaneously to Pb by emitting an a
particle.
Another type
of
nuclear process, known as nuclear transmutation, results from the bom-
bardment of nuclei by neutrons, protons, or other nuclei. An example
of
a nuclear transmutation is
the conversion of atmospheric
IjN to
I
~
C
and
IB
, which results when the nitrogen isotope is bom-
barded by neutrons (from the sun). In some

ca
se
s,
heavier elements are synthesized from lighter
elements. This type
of
transmutation occurs naturally in outer space, but it can also be achieved
artificiall
y,
as we will see in Section 20.4.
Radioactive decay and nuclear transmutation are nucle
ar
rea
ctions, which differ signifi-
cantly from ordinary chemical reactions. Table
20.1 summarizes the differences.
To discuss nuclear reactions in
an
y depth, we must understand how to write and balance
nuclear equations. Writing a nuclear equation differs somewhat from writing equations for chemi-
cal reactions. In addition to writing the symbols for the various chemical elements, we must also
explicitly indicate the number
of
subatomic particles in every species involved in the reaction.
Th
e symbols for subatomic particles are as follows:
lB
or lp bn
_
~e

or
-
~
f3
proton neutron electron
° 0
f3
+ I e or + I
positron
ia or i
Be
CI.
particle
In
accordance with the notation introduced in Section 2.3, the superscript in each case denotes the
mass number (the total number
of
neutrons and protons present) and the subscript is the atomic
number (the number
of
protons). Thus, the "atomic number"
of
a proton is 1, because there is one
proton present, and the
"
ma
ss number" is also
1,
because there is one proton but no neutrons pres-
ent.

On the other hand, the mass number
of
a neutron is 1, but its atomic number is zero, because
there are no protons present. For the electron, the
ma
ss
number is zero (there are neither protons
nor neutrons present), but the atomic number is
-
1,
because the electron possesses a unit negative
charge.
The symbol
_?e represents an electron in or from an atomic orbital. The symbol _
?f3,
on the
other hand, represents an electron that, although physically identical to any other electron, comes
from a nucleus (in a decay process in which a neutron is converted to a proton and an electron)
and not from an atomic orbital.
The
positron has the same mass as the electron, but bears a charge
of
+
1.
The a particle has two protons and two neutrons, so its atomic number is 2 and its mass
.


.
number is 4.

In balancing any nuclear equation, we must balance the total
of
all atomic numbers and the
total
of
all ma
ss
numbers for the products and reactants.
If
we know the atomic numbers and mass
numbers
of
all but one
of
the species in a nuclear equation, we can identify the unknown species
by applying these rule
s,
as shown in Sample Problem 20.1.
Chemical
Reactions
1.
Atoms are rearranged by the breaking and
fOIlning
of
chemical bonds.
2.
Only electrons in atomic or molecular
orbitals are involved in the reaction.
3.
Reactions are accompanied by the

absorption or release
of
relatively small
amounts
of
energy.
4.
Rates
of
reaction are influenced by
temperature, pressure, concentration, and
catalysts.
Nuclear
Reactions
1.
Elements are converted to other elements
(or
isotopes).
2. Protons, neutrons, electrons, and other
subatomic particles such as
a particles may
be involved.
3.
Reactions are accompanied by the
absorption or release
of
tremendous
amounts
of
energy.

4. Rates
of
reaction normally are not affected
by temperature, pressure, or catalysts.
SECTION
20.2
Nuclear
Stability
799
Sample
Problem 20.1 .
Identify the missing species X in each
of
the following nuclear equations:
(a)
2~~
PO
•
2~~
Pb
+ X
(b)
~~
Sr
" X + -
?f3
(c) X •
I~
O
+ +

?f3
Strategy
Determine the mass number for the unknown species, X, by summing the mass numbers
on both sides
of
the equation:
~
reactant mass numbers =
~
product
ma
ss numbers
Similarly, determine the atomic number for the unknown species:
~
reactant atomic numbers =
~
product atomic numbers
Use the mass number and atomic number to determine the identity
of
the unknown species.
Setup
(a) 212 = (208 + mass number
of
X); mass number
of
X =
4.
84 = (82 + atomic number
of
X); atomic number

of
X =
2.
(b) 90 = (mass number
of
X + 0);
ma
ss number
of
X = 90.
38
= [atomic number
of
X + (
-1
)
];
atomic number
of
X = 39.
(c) Mass number
of
X = (18 + 0);
ma
ss number
of
X = 18. Atomic number
of
X = (8 +
1)

; atomic
number
of
X =
9.
Solution
(a) X = i
a:
2~~
PO
-_
I
2~~
Pb
+
ia
(b) X =
~~
Y:
~~
Sr
"
~~
Y
+ -?
f3
(c) X =
I
~
F:

I~
F
•
I~
O
+
+?
f3
Practice
Problem
A Identify X in each
of
the following nuclear equations:
(a)
~~
As
• X +
-?
f3
(
b)jH
+
iHe
·X
(c)
T
6~
Fm

" T

6Z
Fm
+ X
Practice
Problem
B Identify X
in
each
of
the following nuclear equations:
(a) X +
-?f3
•
2~PU
(b)
2~~
U
" X + iHe
(c) X " liN +
_?
e
Checkpoint 20.1
Nuclei and Nuclear Reactions
20
.1.1 Identify the species X in the following
nuclear equation:
2~~
Rn
• X +
ia

a)
2~~
PO
b)
2~~
Ra
c)
2
~~
PO
d)
2~~
PO
e)
2~~
Ra
Nuclear Stability
20
.1.2 Identify the species X in the following
nuclear equation:
I
~
O
• X + -?
f3
a)
I
~
F
b)

I~
O
c)
I
~F
d) liN
e) li N
The nucleus occupies a very small portion
of
the total volume
of
an
atom, but it contains most
of
the atom's mass because both the protons and the neutrons reside there. In studying the stability
of
the atomic nucleus, it is helpful to know something about its density, because it tells us how
tightly the particles are packed together.
As
a sample calculation, let
us
as
sume that a nucleus has
Think
About
It
The
rules
of
summation we apply to balance

nuclear equations can
be
thought
of
as the conservation
of
mass number
and the conservation
of
atomic
numbe
r.
•
800 .
CHAPTER
20 Nuclear Chemistry
•
The
significance
of
these
numbers
for
nuclear
stab
ili
ty
is
similar
to

the
numbers
of
electrons
associated
with the
very
stable
noble
gases
(i.
e.,
2,
10,
18,
36,
54,
and
86
electrons).
Of
the
two
st
able
isotopes
of antim
ony
mentioned
in

ru
le
1,
both
have
even
numbers
of
neutrons
:
'~lSb
and
'~~Sb.


a radius
of
5 X 10-
3
pm
and a
mass
of
1 X
10
-
22
g.
The
se figures

conespond
roughly to a nucleus
containing
30
protons and 30 neutrons, Density is mass/volume, and
we
can calculate the volume
from
the
known
radius (the volume
of
a sphere is
~'ITr3,
where
r is the radius
of
the sphere). First
we
convert the
picometer
units to centimeters.
Then,
we
calculate the density in
g/cm
3
:
r =
(5

X 10-
3
pm
) 1
\1~:2
m
(1~0
~m
)
= 5 X 10-
13
cm
10
-~~
.
ma
ss 1 X

g
density = =

-
volume
~
'ITr
3
1 X 10-
22
g
~

'IT(5
X 10-
13
cm)3
= 2 X 10
14
g/
cm
3
This is an exceedingly high density.
The
highest
density
known
for an
element
is
22
.6 g/cm3, for
osmium
(Os).
Thu
s, the average atomic nucleus is roughly 9 X
10
12
(or
9 trillion) times as dense
as the densest
element
known!

The
enormously
high density
of
the nucleus means that
some
very strong force is needed
to
hold
the particles together so tightly.
From
Coulomb's law
we
know that like charges repel
and
unlike charges attract
one
another. We would thus expect the protons to repel one another strongly,
particularly when
we
consider how close they
mu
st
be
to
each
other.
This
indeed is so. However,
in

addition to the repulsion, there are also short-range attractions between proton and proton, proton
and neutron, and neutron and neutron.
The
stability
of
any nucleus is determined by the difference
between
coulombic
repulsion
and
the shOlt-range attraction.
If
repulsion outweighs attraction, the
nucleus disintegrates, emitting
particles
and/or
radiation, If attractive forces prevail, the nucleus
is stable.
Patterns
of
Nuclear Stability
The
principal factor
that
determines whether a nucleus is stable is the neutron-to-proton ratio
(
nip
).
For
stable

atoms
of
elements having low atomic
number
«20),
the nip value is close to
1.
As the atomic
number
increases, the neutron-to-proton ratios
of
the stable nuclei also increase.
This deviation at higher atomic numbers arises because
more
neutrons are needed to counteract
the strong repulsion
among
the protons and stabilize the nucleus,
The
following rules are useful
in
gauging whether or
not
a particular nucleus is expected to
be
stable:

1.
There
are

more
stable nuclei containing 2, 8, 20, 50, 82, or 126 protons
or
neutrons than there
are containing
other
numbers
of
protons
or
neutrons.
For
example, there are
10
stable iso-
topes
of
tin (Sn) with the atomic number
50
and
only 2 stable isotopes
of
antimony (Sb) with
. . . . . . .

. . . . .

. . . . . .

. .

.
the atomic
number
51.
The
numbers
2,8,20
,
50
, 82, and 126 are called magic numbers.
2.
There
are
many
more
stable nuclei with even
number
s
of
both protons and neutrons than
with odd numbers
of
these particles (Table
20
.2),
3. All isotopes
of
the elements with atomic numbers higher than 83 are radioactive.
4. All isotopes

of
technetium (Tc, Z = 43) and
promethium
(Pm
, Z = 61) are radioactive,
Figure
20.1 shows a
plot
of
the
number
of
neutrons versus the
number
of
protons
in
various iso-
topes.
The
stable nuclei are located in an area
of
the graph known as the belt
of
stability.
Most
radioactive nuclei lie outside this belt. Above the belt
of
stability, the nuclei have
higher

neutron-
to-proton ratios than those within the belt (for the
same
number
of
protons). To lower this ratio
(and
hence
move
down
toward the belt
of
stability), th
ese
nuclei undergo the following process,
called
{3-particle emission:
Protons
Neutrons
Number
of
Stable
Isotopes
Odd
Odd
4
Odd
Even
50
Even

Odd
53
Even
Even
164
SECTION 20.2 Nuclear
Stability
801
120
100
'" 80
g
"
1l
'-<
II
60
~
(nIp ratio 1.3: 1)
~~
Kr
Z
40
20
(nIp ratio 1.5: 1)
2~~
Pb ,.
c :
Belt
of

stability
·

•
• •
• •
••
•



• •
•
••

'" .
•
• • •
•

.

•



•
•
04
•

•
• • •
• •
• • •
•


1:
1 neutron-t
o-
proton ratio
(nIp ratio
1.1:1 )
~~
N
a
o
o
20
40
60 80
Number
of
protons
\
,
Beta-particle emission leads to
an
increase in the number
of

protons in the nucleus and a simulta-
neous decrease in the number
of
neutrons. Some examples are
I~C
+.
lj N + -7
/3
ig
K
+.
ig
ca
+
-7
/3
~lZr
·
nNb
+ -
7/3
Below the belt
of
stability, the nuclei have lower neutron-to-proton ratios than those in the belt (for
the same number
of
protons).
To
increase this ratio (and hence move up toward the belt
of

stabil-
ity), these nuclei may eIllit a positron,
An example
of
positron emission is
Alternatively, a nucleus may undergo electron capture,
1 0 I
IP
+ - le ' on
Electron capture is the capture
of
an electron usually a Is electron by the nucleus.
The
cap-
tured electron combines with a proton in the nucleus to form a neutron so that the atoIllic number
Figure
20.1 Plot
of
neutron \'
erslb
protons for various stable isotopes.
represented by dots. The straight line
represents the points at which the
neutron-to-proton ratio is
1.
The haded
area represents the belt
of
stabili
ty.

decreases by 1 while the mass number remains the same. Electron capture has the same net effect •
on the nucleus as positron emission. Examples
of
electron capture are
Nuclear Binding Energy
A quantitative measure
of
nuclear stability is the nuclear binding energy, which is the energy
required to break up a nucleus into its component protons and neutrons. This quantity represents
the conversion
of
mass to energy that occurs during an exothermic nuclear reaction.
802
CHAPTER
20
Nuclear
Chemistry
•
When
you
apply
Einstein's
equation, E = m
el,
it
is
important to
remember
that
mass

de
f
ect
must
be
expressed
in
kilograms
in
order for the
units
to
cancel
properly.
1
kg
= 6.022 x
10
26
amu
Remember
that joule is a
derived
unit:
1 J = 1
kg
. m
2
/s
2

[
~~
Section
5.1)
.
Note that when
we
report a
nuclear
binding
energy
per
mole,
we
give
just the
magn
itu
de
without the negati
ve
si
gn
.
The concept
of
nuclear binding energy evolved from studies
of
nuclear properties showing
that the masses

of
nuclei are always less than the sum
of
the masses
of
the nucleons, which is a
general term for the protons and neutrons in a nucleu
s.
For example, the I§F isotope has an atomic
mass
of
18.9984
am
u.
The
nucleus has 9 protons and 10 neutrons and therefore a total
of
19 nucle-
ons. Using the known
ma
sses
of
the
lH
atom
(l.00782
5 amu) and the neutron (
l.008665
amu),
we can carry out the following analysis. The

ma
ss
of
9 1 H atoms (i.e., the mass
of
9 protons and
9 electrons) is
9
X l.007825 amu = 9.070425 amu
and the mass
of
10 neutrons is
10
X l.008665 amu = 10.08665 amu
Therefore, the atomic mass
of
an I§F atom calculated from the known numbers
of
electrons, pro-
tons, and neutrons is
9.070425 amu
+ 10.08665 amu = 19.15708 amu
This value is larger than 18.9984 amu (the
mea
sured mass
of
I§F)
by 0.1587 amu.
The
difference between the mass

of
an atom and the
sum
of
the
ma
sses
of
its protons, neu-
trons, and electrons is called the
mass defect. According to relativity theory, the loss in
ma
ss
shows up as energy (heat) given
off
to the surroundings. Thus, the formation
of
I§F
is exothermic.
According to Einstein's mass-energy equivalence relationship (E = me
2
,
where E is energy, m is
ma
ss, and e is the velocity
of
light), we can calculate the amount
of
energy release
d.

We start by
writing
Equation 20.1
AE
= (Am)c
2
where
AE
and
Am
are defined as follows:
AE
= energy
of
product - energy
of
reactants
Am
= mass
of
product -
ma
ss
of
reactants
Thus, the change in mass is
or
Am
= 18.9984 amu - 19.15708 amu
=

-0.1587
amu
l.00
kg
Am
= (
-0.1587
amu)
6.022
X 10
26
amu
. . . . . . . . . . . .
. .


.

. . . . .

'28

=
-2.635
X 10- kg
Because
Ig
F
has a
ma

ss that is less than the
ma
ss calculated from the number
of
electrons and
nucleons present,
Am
is a negative quantity, Consequently,
AE
is also a negative quantity; that i
s,
energy is released to the surroundings as a result
of
the formation
of
the fluorine-19 nucleus. We
. . . . . . . . .
calculate
AE
as
follows:
AE
= (
-2,635
X 10-
28
kg)(3,00 X 10
8
m/S)
2

=
-2.37
X
10-
11
kg . m
2
/s2
=
-2.37
X 10-
11
J
This is the amount
of
energy released when one fluorine-19 nucleus is formed from 9 protons and
10 neutrons. The nuclear binding energy
of
the nucleus is 2,38 X 10-
11
J, which is the amount
of
energy needed to decompose the nucleus into separate protons and neutrons. In the formation
of
1
mole
of
fluorine nuclei, for instance, the energy released is
AE
= (

-2.3
7 X 10-
11
J)(6.022 X
10
23
/mol)
=
-1.43
X
1013
J/mol
=
-1.43
X
1O
Io
kJ/mol
The
nucie~
binding energy,
the~~fore
:
is·
l.43
X
1010
kJ
for 1 mole
of

fluorine-19 nuclei, which
is a tremendously large quantity when we consider that the enthalpies
of
ordinary chemical reac-
SECTION
20.2
Nuclear
Stability
803
4He
815XlO-
12
-
0-
~
9 X
10-
13
-
"
on
"
'g
6 X
lO-
I L
.~
.D
til
!l

3 X
10-
13
_
(j
Z
I:
1
I",
" , . , , , . ,

,.
. ,
'fW'
~
.
;-
o 20 40 60
80
100 120 140 160 180 200 220 240 260
Mass number
tions are on the order
of
only 200
kJ.
The procedure we have followed can be used to calculate the
nuclear binding energy
of
any nucleus.
As we have noted, nuclear binding energy is an indication

of
the stability
of
a nucleus. When
comparing the stability
of
any two nuclei, however, we must account for the fact that they have dif-
ferent numbers
of
nucleons. It makes more sense, therefore, to compare nuclei using the nuclear

.

.

' .
""'"

'
.,

.
binding energy
per
nucleon:
nuclear binding energy
nuclear binding energy per nucleon
=

-

number
of
nucleons
For the fluorine-l 9 nucleus,
I
b' d' I 2.38 X
10-
11
J
nuc ear m mg energy per nuc eon
=
19
nucleons
= 1.25 X
10-
12
J/nucleon
The nuclear binding energy per nucleon makes it possible to compare the stability
of
all nuclei on
a common basis. Figure
20.2 shows the variation
of
nuclear binding energy per nucleon plotted
against mass number. As you can see, the curve rises rather steeply. The highest binding energies
per nucleon belong to elements with intermediate mass numbers between
40 and
100-and
are
greatest for elements in the iron, cobalt, and nickel region (the Group

8B
elements)
of
the periodic
table. This means that the net attractive forces among the particles (protons and neutrons) are
greatest for the nuclei
of
these elements.
Nuclear binding energy and nuclear binding energy per nucleon are calculated for an iodine
nucleus in Sample
Problem 20.2.
The atomic mass
of
l~jI
is 126.9004 amu. Calculate the nuclear binding energy
of
this nucleus and
the corresponding nuclear binding energy per nucleon.
Strategy
To calculate the nuclear binding energy,
we
first determine the difference
be
t
ween
the
mass
of
the nucleus and the mass
of

all the protons and neutrons, which yields the mass defect. Next,
we
must
apply Einstein's mass-energy relationship
[ilE
=
Cilm)c
2
].
Solution
There are 53 protons and
74
neutrons in the iodine nucleus.
The
mass
of
53 IH atoms is
53 X
1.007825 amu = 53.41473 amu
and the mass
of
74
neutrons is
74 X
1.008665 amu = 74.64121 amu
(Continued)
Figure 20.2 Plot
of
nuclear
binding energy per nucleon versus mass

number.
In
general,
the greater the
nuclear
binding
energy
per
nu
cleon,
the
m
ore
stable
the
nucleus.
,
804
CHAPTER
20
Nuclear Chemistry
•
Therefore
, the predicted mass for
In
r is 53.41473 +
74
.64121 = 128.05594 amu, and the mass
defect is
I:lm = 126.9004 amu - 128.05594

amu
= - 1.1555
amu
= (
-1.1555
amu
)
6.022 x 10
26
amu
1.00
kg
= - 1.919 X
lO-
27
kg
The
energy released is
I:lE = (I:lm )c
2
=
(-
1.919 X
lO
-27
kg)
(3
.00 X
lO
8

mJsi
= 1.73 X lO-IO
kg·
m
2
/
s2
= 1.73 X lO-10 J
Thus
the
nuclear
binding
energy is 1.73 X
lO-
10
J.
The
nuclear binding
energy
per
nucleon is
obtained as follows:
1.73 X
lO-
IO
I
136
X 10-
12
II

I
' '
- = .
nuc
eon
127 nucleons
Practice Problem A Calculate the nuclear binding energy (in
joules)
and
the
nuclear binding
energy
per
nucleon of
2~§
Bi
(2
08.9804
amu).
Practice Problem B Calculate the nuclear binding energy (in
joules)
and the nuclear binding energy
per
nucleon
of
1~~
Au
(196.9665 amu).
Checkpoint
20.2

Nuclear Stability
20.2.1
Determine
the binding energy 20.2.2
What
is the energy associated with a
per
nucleon
in
a
238
U nucleus
mass
defect
of
2.000
amu?
(238.0507847 amu).
a)
3.321 X
lO-
27
J
a)
2.891 X
lO
-
IO
I/nucleon
b)

2.989 X
lO
- 7 I
b)
1.215 X
lO-
12
J/nucleon
c)
1.800 X
lO17
I
c)
1.186 X
lO-
12
I/nucleon
d)
2.989 X
lO
-
10
I
d)
2.823 X
lO-1O
J/nucleon
e)
3.309 X lO-45 J
e)

3.212 X
lO-
24
I/nucleon
Natural
Radioactivity
Nuclei that do not lie within the belt
of
stability, as well as nuclei with more than
83
protons,
tend to be unstable. The spontaneous emission by unstable nuclei
of
particles or electromagnetic
radiation, or both, is known as radioactivity. The main types
of
radioactivity are the emission
of
il'
particles (doubly charged helium nuclei, He
?+
); the emission
of
f3
particles (electrons
of
nuclear
origin); the emission
of
'Y

rays, which are very-short-wavelength (0.1 nm to 104 nm) electromag-
netic waves; the emission
of
positrons; and electron capture.
The disintegration
of
a radioactive nucleus often
is
the beginning
of
a radioactive decay
series,
which
is
a sequence
of
nuclear reactions that ultimately result in the formation
of
a stable
isotope. Figure
20.3 shows the decay series
of
naturally
OCCUlTing
uranium-238, which involves
14
steps. This decay scheme, known as the uranium decay series, also shows the half-lives
of
all
the nuclei involved.

It
is important
to
be able to balance the nuclear reaction for each
of
the steps
in
a radioactive
decay series. For example, the first step
in
the uranium decay series is the decay
of
uranium-238 to
thorium-234, with the emission
of
an
il'
particle. Hence, the reaction is represented by
SECTION
20.3 Natu
ra
l Radioact iv
ity
805
•
238
U

+' a
92

4.51 X 10
9
yr
2§
6
Th

+.
f3
24
.1 days
2§i
Pa

+'
f3
1.17 min
234
U
__
.'
a
92
2.47 X 10
5
yr
2§8
Th

••

a
7.5 X 10
4
yr
2§~
Ra
__
••
a
1.60 X 103
yr
2§g
Rn • a
3.
82
da
ys
f3
~

2I8
p

-+
-f 8
40

a
0.
04

% 3.05 min
a
~.
_ _
2~~A
t
2~i
pb
•
f3
26.8 min
2s
\
f3
•
2§j
Bi

.'
a
99
.
96
%
19.
7
min
a
~.
__

2~
P
O
2~?
Tl
•
f3
1.6 X
lO
-
4
s \
2
~g
Pb

+.
f3
20.4 yr
f3
•
2IO
B'
__
-+
83
1 • a
1.32 min
-
100

% / \ 5.
01
days
a
••
_ _ 2IDp
2D6
T1
__
••
f3
138
da
ys
84
\
81
4.20 min
2~~
Pb
The next step is represented by
2~6
Th
-
_.
2~
iPa
+ -?
f3
and so on.

In
a discussion of radioactive decay steps, the beginning radioactive isotope is called
the
par
e
nt
and the product isotope is called the daughter. Thus,
2~~
U
is the parent in the first step
of
the uranium decay series, and
2§6
Th is the daughter.
Kinetics
of
Radioactive Decay
All radioactive decays obey first-order kinetics. Therefore, the rate
of
radioactive decay at any
time
t is given by
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
rate
of
decay at time t =
kN
where k is the first-order rate constant and N is the number
of
radioactive nuclei present at time

{,
According to Equation 14.3, the number of radioactive nuclei at time zero (No) and time t (Nt) is





,
.
Nt
In
No =
-kt
and the corresponding hal
f-
life
of
the reaction is given by Equation 14.
5:
•
••
Figure 20.3
uranium-238.
Decay series for
' Most n
uclea
r
sci
e
ntists

(an
d
so
me g
enera
l
chem
i
stry
boo
ks
) u
se
the
sy
mb
ol A
ins
te
ad
of k
for
the
rate
co
n
st
ant of nu
clear
rea

ctions.
Althou
gh
Equation
14
.3
u
ses
co
n
centrati
ons
of
a reactant
at
t
im
es
t and 0, it is the
ratio
of the
two t
ha
t
is
impo
rta
nt,
so
we c

an
also
use
the
number of
radi
o
active
nuclei
in
t
hi
s
equa
ti
on
.
806
CHAPTER
20 Nuclear Chemistry
r"
"""
ij
.

. -
Multimedia
Nuclear Chemistry- radioactive half-life.
The half-lives and, therefore, the rate constants
of

radioactive isotopes vary greatly from nucleus to
nucleus. Looking at Figure
20.3, for example, we find that
2~~U
and
2~:PO
are two extreme cases:
2~~U
•
2~riT
h
+
ia
t1/2 = 4.51 X 10
9
yr
These two rate constants, after conversion to the same time unit, differ by many orders
of
magni-
tude. Furthermore, the rate constants are unaffected by changes in environmental conditions such
as temperature and pressure. These highly unusual features are not seen in ordinary chemical reac-
I tions (see Table 20.1).
Think
About
It
Carbon dating
cannot be used for objects older
than about
60,000 years (about 10
half-lives). After that much time

has passed, the activity
of
carbon-
14 has fallen to a level too low to
be
measured reliably.
Dating
Based
on Radioactive Decay
The half-lives
of
radioactive isotopes have been used as "atomic clocks" to determine the ages
of
cer-
tain objects. Some examples
of
dating by radioactive decay measurements will be described here.
The
carbon-14 isotope is produced when atmospheric nitrogen is bombarded by cosmic
rays:
The radioactive carbon-14 isotope decays according to the equation
I~C
• I
jN
+
-
~
f3
This reaction is the basis
of

the radiocarbon
or
"carbon-14" dating technique described
on
page
563.
In
order to determine the age
of
an object,
we
measure the activity (disintegrations per sec-
ond)
of
14C
and compare
it
to the activity
of
14C
in living matter.
Sample
Problem 20.3 shows how to use radiocarbon dating to determine the age
of
an
artifact.
Sample Problem 20.3 ·
.•
A wooden artifact is found to have a 1
4C

activity
of
9.1 disintegrations per second. Given that the
14C
activity
of
an equal mass
of
fresh-cut wood has a constant value
of
15.2 disintegrations
per
second,
determine the age
of
the artifact.
The
half-life
of
carbon-14 is 5715 years.
Strategy
The
activity
of
a radioactive sample is proportional to the number
of
radioactive nuclei.
Thu
s, we can use Equation 14.3 with activity
in

place
of
concentration:
14C
activity in artifact
In
4 =
-kt
1 C activity in fresh-cut wood
To determine
k,
though, we must solve Equation 14.5, using the value
of
t1
/, for carbon-14 (5715
years) given in the problem statement.
Setup Solving Equation 14.5 for k gives
k = 0.693 =
121
X 10-
4
- I
5715
yr'
yr
Solution
9.1 disintegrations per second 4 1
In

= - 1.21 X

10-
yr- (t)
15.2 dlsmtegratlOns per second
-0.513
t =
=
4~Oyr
- 1.21 X
1O
-
4
yr-
1
Therefore, the age
of
the artifact is 4.2 X 10
3
years.
Practice Problem A A piece
of
linen cloth found at an ancient burial site is found to have a
14C
activity
of
4.8 disintegrations
per
minute. Determine the age
of
the cloth. Assume that the carbon-14
activity

of
an equal mass
of
living flax (the plant from which linen is made) is 14.8 disintegrations
per minute.
Practice Problem B
What
would
be
the
14
C activity in a 2500-year-old wooden object? Assume that
the 14C activity
of
an equal mass
of
fresh-cut wood is 13.9 disintegrations per second.