Electrochemistry
Visualizing Concepts
20.1
In a Brensted-Lowry acid-base reaction, H+ is transferred from the acid to the base. In a
redox reaction, the substance being oxidized (the reductant) loses electrons and the
substance being reduced (the oxidant) gains electrons. Furthermore, the number of
electrons gained and lost must be equal. The concept of electron transfer from reductant
to oxidant is clearly applicable to redox reactions. (The path of the transfer mayor may
not be direct, but ultimately electrons are transferred during redox reactions.)
20.2
(a)
If a Zn(s) strip was placed in a CdS0 4(aq) solution, Cd(s) would form on the
strip. Although E r:d for Znv'(aq), -0.763 V, and Cd 2+(aq), -0.403 V, are both
negative, the value for Cd 2+ is larger (less negative), so it will be the reduced
species in the redox reaction.
If a Cu(s) strip was placed in a AgN0 3(aq) solution, Ag(s) would form on the
strip. Although Er:d for Cu 2+(aq), 0.337 V, and Ag+(aq), 0.799 V, are both
(b)
positive, the species with the larger E r:d value, Ag ", will be reduced in the
reaction.
20.3
Ni 2+(aq) + 2e­ ~ Ni(s), E r:d = -0.28 V, cathode
IJ
f;t:1
l
Fe 2+(aq) + 2e­ ~ Fe(s), E r:d = -0.44 V, anode
,'i
Ni 2+(aq) has the larger E r:d, so it will be reduced in the redox reaction. Reduction
occurs at the cathode, so Ni 2+(aq) and Ni(s) will be in the cathode compartment and
Fe 2+(aq) and Fe(s) will be in the anode compartment. The voltmeter will read:
E C:
ll
= Er:d(cathode)-Er:d(anode) = -0.28 V - (-044 V) = 0.16 V
0.16 V
Anode(-)
Fe
Solution contains
Fe 2+(aq), NO) (aq)
~WI,--­
Salt Bridge
K+,NO)
549
Cathode (+)
Ni
Solution contains
Ni 2+(aq), N03(aq)
20q.h6imis±ry @f tite "Envh61LIlie1tt
20.4
20.5
Solutions to Exercises
The species with the largest E r: d is easiest to reduce, while the species with the
smallest, most negative E r: d is easiest to oxidize.
(a)
The species easiest to oxidize is at the bottom of Figure 20.14.
(b)
The species easiest to reduce is at the top of Figure 20.14.
A(aq) + B(aq)
~
A "(aq) + W (aq)
(a)
A gains electrons; it is being reduced. B loses electrons; it is being oxidized.
(b)
Reduction occurs at the cathode; oxidation occurs at the anode.
A(aq) + 1eB(aq)
(c)
~
~
A -(aq) occurs at the cathode.
B+ (aq) + Le" occurs at the anode.
In a voltaic cell, the anode is at higher potential energy than the cathode. The
anode reaction, B(aq) ~ B~ (aq) + 1e-, is higher in potential energy.
fuel _
outlet
20.6
fuel
inlet
__
-
_
product
exhaust
oxidizing
agent inlet
cathode
anode
porous
membrane
The main difference between a fuel cell and a battery is that a fuel cell is not self­
contained. That is, there is a continuous supply of fuel (reductant) and oxidant to the
cell, and continuous exhaust of products. The fuel cell produces electrical current as
long as reactants are supplied. It never goes" dead."
20.7
Zinc, E r: d = -0.763 V, is more easily oxidized than iron, E r: d = -0.440 V. If conditions
are favorable for oxidation, zinc will be preferentially oxidized, preventing iron from
corroding. The protection lasts until all the Zn coating has reacted.
i"
20.8
Unintended oxidation reactions in the body lead to unwanted health effects, just as
unwanted oxidation of metals leads to corrosion. Antioxidants probably have modes of
action similar to anti-corrosion agents. They can preferentially react with oxidizing
agents (cathodic protection), create conditions that are unfavorable to the oxidation­
reduction reaction, or physically coat or surround the molecule being oxidized to
prevent the oxidant from attacking it. The first of these modes of action is likely to be
safest in biological systems. Adjusting reaction conditions in our body can be
dangerous, and physical protection is unlikely to provide lasting protection against
oxidation. Anti-oxidants are likely to be reductants that preferentially react with
oxidizing agents.
Oxidation States
20.9
(a)
Oxidation is the loss of electrons.
550
20Clrerniswy 8£ Lite EiEbir8RIlreHt"
. Solutions to Exercises
(b)
The electrons appear on the products side (right side) of an oxidation half­
reaction.
(c)
The oxidant is the reactant that is reduced; it gains the electrons that are lost by
the substance being oxidized.
(d)
An oxidizing agent is the substance that promotes oxidation. That is, it gains
electrons that are lost by the substance being oxidized. It is the same as the
oxidant.
Reduction is the gain of electrons.
The electrons appear on the reactants side (left side) of a reduction half-reaction.
The reductant is the reactant that is oxidized; it provides the electrons that are
gained by the substance being reduced.
.A reducing agent is the substance that promotes reduction. It donates the electrons
gained by the substance that is reduced. It is the same as the reductant.
True.
False. Fe 3 + is reduced to Fe 2 +, so it is the oxidizing agent, and C 0 2 + is the
reducing agent.
True.
False. If something is reduced, it gains electrons.
True.
True. Oxidation can be thought of as a gain of oxygen atoms. Looking forward,
this view will be useful for organic reactions, Chapter 25.
20.13
20.14
Analyze/plan. Given a chemical equation, we are asked to indicate which elements
undergo a change in oxidation number and the magnitude of the change. Assign
oxidation numbers according to the rules given in Section 4.4. Note the changes and
report the magnitudes.
Solve.
(a)
I is reduced from +5 to 0; C is oxidized from +2 to +4.
(b)
Hg is reduced from +2 to 0; N is oxidized from -2 to O.
(c)
N is reduced from +5 to +2; S is oxidized from -2 to O.
(d)
Cl is reduced from +4 to +3; 0 is oxidized from -1 to O.
(a)
No oxidation-reduction
(b)
I is oxidized from -1 to +5; Cl is reduced from +1 to -1.
(c)
S is oxidized from +4 to +6; N is reduced from +5 to +2.
(d)
5 is reduced from +6 to +4; Br is oxidized from -1 to O.
Balancing Oxidation-Reduction Reactions
20.15
Analyze/plan. Write the balanced chemical equation and assign oxidation numbers. The
Solve.
substance oxidized is the reductant and the substance reduced is the oxidant.
551
~I'I'
Solutions to Exercises
20.16
20.17
20.18
20.19
~
(a)
TiCl 4(g) + 2Mg(l)
Ti(s) + 2MgCI2(1)
(b)
Mg(l) is oxidized; TiCl 4(g) is reduced.
(c)
Mg(l) is the reductant; TiCl 4(g) is the oxidant.
(a)
2N2H4(g)+N204(g)~3N2(g)+4H20(g)
(b)
N 2H4(g) is oxidized; N 204(g) is reduced.
(c)
N 20 4(g) serves as the oxidizing agent; it is itself reduced. N 2H4(g) serves as the
reducing agent; it is itself oxidized.
Analyze/plan. Follow the logic in Sample Exercises 20.2 and 20.3. If the half-reaction
Solve.
occurs in basic solution, balance as in acid, then add OH- to each side.
(a)
Sn 2+(aq)
(b)
Ti0 2(s) + 4H+(aq) + 2e-
(c)
Cl0 3-(aq) + 6H+(aq) + 6e-
(d)
40H-(aq) ~ 02(g) + 2H 20(I) + 4e-, oxidation
(e)
SO/-(aq) + 20H- (aq)
~
S04 2-(aq) + H 20(l) + 2e-, oxidation
(f)
N 2(g) + 8H+(aq) + 6e-
~
2NH 4 +(aq), reduction
(g)
N 2(g) + 6H 20 + 2e-
(a)
Mo3+(aq) + 3e-
(b)
H 2S03(aq) + H 20(l)
(c)
N0 3-(aq) + 4H+(aq) + 3e-
(d)
Mn 2+(aq) + 40H-(aq) ~ Mn02(S) + 2H 20(I) + 2e-, oxidation
(e)
Cr(OHh(s) + 50H-(aq)
(f)
02(g) + 4H+(aq) + 4e-
~
2H 20(l), reduction
(g)
02(g) + 2H 20(l) + 4e-
~
40H-(aq), reduction
~
Sn v'{aq) + 2e-, oxidation
~
~
~
Ti 2+(aq) + 2H 20(I), reduction
~
Cljaq) + 3H 20(l), reduction
2NH 3(g) + Ol-I{aq), reduction
Mo(s), reduction
~
S04 2-(aq) + 4H+(aq) + 2e-, oxidation
~
~
NO(g) + 2H 20(I), reduction
Cr0 42-(aq) + 4H 20(l) + 3e-, oxidation
Analyze/plan. Follow the logic in Sample Exercises 20.2 and 20.3to balance the given
equations. Use the method in Sample Exercise 20.1 to identify oxidizing and reducing
agents.
Solve.
oxidizing agent, Cr2072-; reducing agent, 1­
(b)
The half-reactions are:
4[Mn0 4-(aq)+8H+(aq)+5e- ~Mn2+(aq)+4HP(I)]
5[CH 30H(aq)+H 20(l)
~
HC0 2H(aq)+4H+(aq)+4e-]
oxidizing agent, Mn0 4 -; reducing agent, CH 30H
552
20 ei Len dstt y of trte EnvIronment
(c)
Solutions..to Exercises
$
12(s)+6H 20(l) ~ 210 3-(aq)+12H+(aq)+10e­
5[OCqaq)+2H+(aq)+2e­ ~ Cl-(aq)+H 20(l)
12(s)+50Cl
(aq)+H 20(1) ~ 2103 (aq)+5Cl (aq)+2H+(aq)]
,
.
oxidizing agent, OCI-; reducing agent, 12.
(d)
AS203(S)+5H 20(l)
~
2H 3As0 4(aq)+4H+(aq)+4e­
2N0 3-(aq)+6H+(aq)+4e­
~
Np3(aq)+3H 2°(l)
oxidizing agent, N0 3 "; reducing agent, As 20 3
(e)
2[Mn0 4-(aq)+2H 20(l)+3e­
~
Mn02(S)+40H-]
Br-(aq)+60H-(aq)
(aqj-s Br (aq)+H 20(l)
~
Br03-(aq)+3H20(l)+6e­
2Mn02(S) + Br03 (aq)+20H (aq)
2Mn0 4
~
oxidizing agent, MnO 4 "; reducing agent, Br'
(f)
Pb(OH)4 2-(aq) + ClO-(aq)
~
Pb0 2(s) + Cl-(aq) + 20H-(aq) + H 20(I)
oxidizing agent, ClO-; reducing agent, Pb(OH)4 2­
20.20
(a)
3[N0 2-(aq)+H 20(1) ~ N0 3-(aq)+2H+(aq)+2e-]
Cr 20/-(aq) + 14H+ (aq) + 6e­ ~ 2Cr 3+(aq) + 7H 20(l)
oxidizing agent, Cr2072-; reducing agent, N0 2­
(b)
4[As(s)+3H 20(l)
~
H 3As03(aq)+3H+(aq)+3e-]
3[CI0 3-(aq) +5H+(aq) +4e­
~
HCIO(aq)+2H 20(1)]
oxidizing agent, C10 3 "; reducing agent, As
(c)
2[Cr20/-(aq)+14H+(aq)+6e- ~. 2Cr 3+(aq)+7H 20(1)]
3[CH 30H(aq)+H20(1)
(d)
~
HC0 2H(aq)+4H+(aq)+4e-]
2[Mn0 4-(aq)+8H+(aq)+5e­ ~ Mn 2+(aq)+4H 20(l)]
5[2Cl- (aq)
~
C1 2(aq) + Ze"]
oxidizing agent, Mn0 4-; reducing agent, Cl
Since the reaction is in base, the H+ can be "neutralized" by adding 2QH- to each
side of the equation to give H 20 2(aq) + 20H-(aq) ~ 02(g) + 2H 20(l) + 2e-. The
other half reaction is 2[CI0 2(aq) + e­ ~ CI0 2-(aq)].
553
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20 Cheri"t"
et tho
.22 oilOIdIll!nt
Solutions to Exercises
Net: H 20 2(aq) + 2Cl0 2(aq) + 20H-(aq) ~ 02(g) + 2Cl0 2-(aq) + 2H 20(l)
oxidizing agent, Cl0 2; reducing agent, H 20 2
(f)
4[H 20 2(aq) + 20H-(aq)
~ 02(g)+
Cl 20 7(aq)+3H 20(l)+8e­
Net:
~
2H 20(l) + 2e-j
2Cl0 2-(aq)+60H-(aq)
V oltaic Cells
20.21
20.22
20.23
20.24
(a)
The reaction Cu 2+(aq) + Zn(s) ~ Cu(s) + Zn 2+(aq) is occurring in both figures. In
Figure 20.3, the reactants are in contact, and the concentrations of the ions in
solution aren't specified. In Figure 20.4 the oxidation half-reaction and reduction
half-reaction are occurring in separate compartments, joined by a porous
connector. The concentrations of the two solutions are initially 1.0 M. In Figure
20.4, electrical current is isolated and flows through the voltmeter. In Figure 20.3,
the flow of electrons cannot be isolated or utilizes}.
(b)
In the cathode compartment of the voltaic cell in Figure 20.5, Cu 2+ cations are
reduced to Cu atoms, decreasing the number of positively charged particles in
the compartment. Na' cations are drawn into the compartment to maintain
charge balance as Cu 2+ ions are removed.
(a)
The porous glass dish in Figure 20.4 provides a mechanism by which ions not
directly involved in the redox reaction can migrate into the anode and cathode
compartments to maintain charge neutrality of the solutions. Ionic conduction
within the cell, through the glass disk, completes the cell circuit.
(b)
In the anode compartment of Figure 20.5, Zn atoms are oxidized to 2n 2+ cations,
increasing the number of positively charged particles in the compartment. N0 3 ­
anions migrate into the compartment to maintain charge balance as Zn 2+ ions are
produced.
AnalyzejPlan. Follow the logic in Sample Exercise 20.4.
Solve.
(a)
Fe(s) is oxidized, Ag+(aq) is reduced.
(b)
Ag ' (aq) + le-
(c)
Fe(s) is the anode, Ag(s) is the cathode.
(d)
Fe(s) is negative; Ag(s) is positive.
(e)
Electrons flow from the Fe(-) electrode toward the Ag(+) electrode.
(f)
Cations migrate toward the Ag(s) cathode; anions migrate toward the Fe(s)
anode.
(a)
AI(s) is oxidized, Ni 2+(aq) is reduced.
(b)
AI(s) ~ AP+(aq) + 3e-; Ni 2+(aq) + 2e­
(c)
AI(s) is the anode; Ni(s) is the cathode.
~
Ag(s); Fe(s) ~ Fe 2+(aq) + 2e­
554
~
Ni(s)
20 {J~en dstI, of the Ell innUil'ii'iQt
0
Solutions to Exercises
(d)
AI(s) is negative (-); Ni(s) is positive (+).
(e)
Electrons flow from the AI(-) electrode toward the Ni(+) electrode.
(e)
Cations migrate toward the Ni(s) cathode; anions migrate toward the AI(s)
anode.
Cell EMF under Standard Conditions
20.25
20.26
(a)
Electromotive force, emf, is the driving force that causes electrons to flow through
the external circuit of a voltaic cell. It is the potential energy difference between
an electron at the anode and an electron at the cathode.
(b)
One volt is the potential energy difference required to impart 1 J of energy to a
charge of 1 coulomb. 1 V = 1 J/C.
(c)
Cell potential, E c ell, is the emf of an electrochemical cell.
(a)
In a voltaic cell, the anode has the higher potential energy for electrons. To
achieve a lower potential energy, electrons flow from the anode to the cathode.
20.27
20.28
(b)
The units of electrical potential are volts. A potential of one volt imparts one joule
of energy to one coulomb of charge.
(c)
A standard cell potential describes the potential of an electrochemical cell where
all components are present at standard conditions: elements in their standard
states, gases at 1 atm pressure and 1 M aqueous solutions.
(a)
2H+ (aq) + 2e­
(b)
A standard hydrogen electrode is a hydrogen electrode where the components are
at standard conditions, 1 M H+ (aq) and H 2(g) at 1 atm.
(c)
The platinum foil in an SHE serves as an inert electron carrier and a solid reaction
surface.
(a)
H 2(g)
(b)
The platinum electrode serves as a reaction surface; the greater the surface area,
the more H 2 or H+ that can be adsorbed onto the surface to facilitate the flow of
electrons.
~
~
H 2(g)
2H+ (aq) + 2e­
(c)
o
o
0
0
o
0°
0
o
o
°
o
o
0
0
0
Pt
0
0
H+(aq, 1 M)
555
20~i
20.29
20.30
5'~* aLZnt
Solutions to Exercises
(a)
A standard reduction potential is the relative potential of a reduction half-reaction
measured at standard conditions, 1 M aqueous solution and 1 atm gas pressure.
(b)
E r:d = 0 V for a standard hydrogen electrode.
(c)
The reduction of Ag ' (aq) to Ag(s) is much more energetically favorable, because
it has a substantially more positive E r:d (0.799 V) than the reduction of Sn 2+(aq)
to Sn(s) (-0.136 V).
(a)
It is not possible to measure the standard reduction potential of a single half­
reaction because each voltaic cell consists of two half-reactions and only the
potential of a complete cell can be measured.
(b)
The standard reduction potential of a half-reaction is determined by combining it
with a reference half-reaction of known potential and measuring the cell
potential. Assuming the half-reaction of interest is the reduction half-reaction:
EC: lI
=E r:d (cathode) - Er~ (anode) =Er~d (unknown) - Er~d (reference);
Er~ (unknown)
(c)
=
E C: lI + E r:d (reference).
Cd 2+(aq) + 2e­
~
Cd(s) P = -0.403 V
Ca 2+(aq) + 2e­
~
Ca(s) P = -2.87 V
The reduction of Ca 2+(aq) to Ca(s) is the more energetically unfavorable
reduction because it has a more negative P value.
20.31
Analyze/plan. Follow the logic in Sample Exercise 20.5.
(a)
The two half-reactions are:
TI 3+(aq) + 2e- ~ TI+(aq)
'~'
cathode Er~ = ?
2[Cr 2+ (aq) ~ Cr 3 + (aq) + e-]
1
' ...''.1
(b)
Solve.
anode Er:d = -0.41 V
•
E C: lI = Er:d (cathode) - E r:d (anode); 1.19 V = E r:d -(-0.41 V);
Er : d = 1.19 V -0.41 V = 0.78 V
(c)
Anode(-)
Inert
Electrode (Pt)
Solution Contains
-
Salt
Bridge
aruons
•cations
I-
Cathode (+)
Inert
Electrode (Pt)
Solution Contains
Note that because Cr 2+(aq) is readily oxidized, it would be necessary to keep
oxygen out of the left-hand cell compartment.
556
tn
20 a
20.32
tEn ds li y of tIre "bIt eif 6liID@Nt·· .
(a)
PdCl/-(aq)+2e- --1-Pd(s)+4Cl-
cathode (ed =?
Cd(s) --1- Cd 2+ (aq) + 2e(b)
Solutions to Exercises
anode
E;ed = -0.403 V
E C: l1 = E r:d (cathode) - Er~ (anode); 1.03 V = E r:d - (-0.403 V);
E r:d = 1.03 V - 0.403 = 0.63 V
(c)
Anode(-)
Cd
Solution Contains
Cdz+(aq)
~I
I
Salt
Bridge
Cathode (+)
Pd
f-
I
Solution Contains
PdCl/-(aq), Cqaq)
anions
•
cations
•
20.33
Analyze/plan. Follow the logic in Sample Exercise 20.6.
(a)
CI 2(g) --1- 2Cr(aq) +2e-
Solve.
E r:d = 1.359 V
12(s) + 2e - --1- 2r (aq)
E r:d = 0.536 V
P = 1.359 V - 0.536 V = 0.823 V
Ni(s) --1- Ni 2+(aq) +2e­
(b)
2[Ce
P
(c)
4+(aq)
+ Le" --1-
Ce 3+(aq)]
Er~ = -0.28 V
E r:d
= 1.61 V
= 1.61 V - (-0.28 V) = 1.89 V
Fe(s) --1- Fe 2+ (aq) + 2e2[Fe 3+ (aq) + Le" --1- Fe 2+ (aq)]
E r:d
= -0.440 V
E r:d =
0.771 V
P = 0.771 V - (-0.440 V) = 1.211 V
(d)
3[Ca(s) --1- Ca 2+ (aq) + Ze"]
2[AI
P
20.34
(a)
=
3+
(aq) + 3e - --1- Al(s)]
Er~ = -2.87 V
E r:d = -1.66 V
-1.66 V - (-2.87 V)] = 1.21 V
F2(g) +2e- --1- 2P-(aq)
E r:d =2.87V
Er~ =O.OOV
H 2(g) --1- 2H+(aq) +2e-
P = 2.87 V - 0.00 V = 2.87 V
(b)
Cu(s) --1- Cu 2+ (aq) + 2e2+(aq)+2e--1- Ba(s)
Ba
EO = -2.90 V - (0.337 V) = -3.24 V
557
E r:d = 0.337 V
E r:d
= -2.90 V
'liT'"
20 QI'ft!£L icky 9' &9 8£2 bitmLIIlLnt
(c)
2[Fe (aq) --+
EO
Fe 3+ (aq)
+ Ie"]
Er : d = 0.771 V
= -0.440 V - 0.771 V = -1.211 V
Hg/+(aq)+2e- --+2Hg(l)
2[Cu+ (aq) --+
EO
20.35
E,~ =-0.440V
Fe 2+(aq)+2e- --+ Fe(s)
2+
(d)
Solutions to Exercises
Cu 2+(aq)
E':d = 0.789 V
+ Le-]
E r : d = 0.153 V
= 0.789 V - 0.153 V = 0.636 V
Analyze/plan. Given four half-reactions, find E':d from Appendix E and combine them
to obtain a desired Ec ell ' (a) The largest E c ell will combine the half-reaction with the most
positive E';d as the cathode reaction and th; one with the most negative E';d as the
anode reaction. (b) The smallest positive E eell will combine two half-reactions whose
E,~ values are closest in magnitude and sign.
Solve.
(a)
3{Ag+(aq)+le---+Ag(s)j
Cr(s) --+ Cr
3+
E,~ = -0.74
(aq) + 3e-
3Ag +(aq) + Cr(s) --+ 3Ag(s) + Cr
(b)
Er:d = 0.799
3+
E' =0.799-(-0.74) = 1.54 V
(aq)
Two of the combinations have essentially equal Eo values.
Er~ = 0.799 V
2[Ag +(aq) + Ie- --+ Ag(s)]
Cu(s)--+ Cu
2Ag+(aq) +Cu(s) --+ 2Ag(s) + Cu
3[Ni
2+
20.36
2+(aq)
(aq) + 2e- --+ Ni(s)]
2[Cr(s)
3Ni
E,~ = 0.337 V
2+(aq)+2e-
2+(aq)+2Cr(s)
(a)
--+ Cr 3+ (aq)
P = 0.799 V - 0.337 V = 0.462 V
Er : d = -0.28 V
Er~ =-0.74 V
+ 3e-]
--+ 3Ni(s)+2Cr
3+(aq)
p= -0.28 V -(-0.74 V) =0.46 V
2[Au(s)+4Br-(aq) --+AuBr4-(aq)+3e-]
E;ed
=
-0.858 V
3[2e- + IO-(aq)+H 20(l) --+ l-(aq)+ 20H-(aq)] E:ed = 0.49V
2Au(s)+8Br (aq)+3IO (aq)+3H 20(l)--+2AuBr4 (aq)+3I (aq)+60H (aq)
£0
(b)
2[Eu 2+(aq)--+ Eu 3+(aq)+1e-]
Sn 2+(aq)+2e- --+Sn(s)
= 0.49 - (-0.858) = 1.35 V
E:ed = -0.43 V
E:ed = -0.14 V
p= -0.14-(-0.43) = 0.29 V
20.37
Analyze/plan. Given the description of a voltaic cell, answer questions about this cell.
Combine ideas in Sample Exercises 20.4 and 20.7. The reduction half-reactions are:
Cu 2+(aq) + 2e- --+ Cu(s)
Sn
2+(aq)
+ 2e- --+ Sn(s)
EO
=
O.337V
EO = -0.136 V
Solve.
558
20 C12emistry HI ~hlliI2J it Jowel"t
(a)
Solutions to Exercises
It is evident that Cu 2+ is more readily reduced. Therefore, Cu serves as the
cathode, Sn as the anode.
20.38
(b)
The copper electrode gains mass as Cu is plated out/ the Sn electrode loses mass
as Sn is oxidized.
(c)
The overall cell reaction is Cu 2+(aq) + Sn(s) 4 Cu(s) + Sn 2+(aq)
(d)
£0 = 0.337 V - (-0.136 V) = 0.473 V
(a)
The two half-reactions are:
Pb 2+(aq)+2e- 4Pb(s)
£2= -0.126 V
£2= 1.359V
Cl 2(g)+2e- 42Cqaq)
Because £0 for the reduction of Cl 2 is higher, the reduction of Cl 2 occurs at the Pt
cathode. The Pb electrode is the anode.
(b)
The Pb anode loses mass as Pb 2+(aq) is produced.
(c)
CI 2(g) + Pb(s) 4 Pb 2+(aq) + 2Cl-(aq)
(d)
EO = 1.359 V - (-0.126 V) = 1.485 V
Strengths of Oxidizing and Reducing Agents
20.39
20.40
20.41
Analyze/plan. Follow the logic in Sample Exercise 20.8. In each case, choose the half­
reaction with the more positive reduction potential and with the given substance on the
Solve.
left.
(a)
CI 2(g) (1.359 V vs. 1.065 V)
(b)
Ni 2+(aq) (-0.28V vs. -0.403 V)
(c)
Br03 -(aq) (1.52 V vs. 1.195 V)
(d)
03(g) (2.07 V vs. 1.776 V)
The more readily a substance is oxidized, the stronger it is as a reducing agent. In each
case choose the half-reaction with the more negative reduction potential and the given
substance on the right.
(a)
Mg(s) (-2.37 V vs. -0.440 V)
(b)
Ca(s) (-2.87 V vs. -1.66 V)
(c)
H 2(g, acidic) (0.000 V vs. 0.141 V)
(d)
H 2C 20 4(aq) (-0.49 V vs. 0.17 V)
III
Analyze/plan. If the substance is on the left of a reduction half-reaction, it will be an
oxidant; if it is on the right, it will be a reductant. The sign and magnitude of the Er~
determines whether it is strong or weak.
Solve.
(a)
Cl 2(aq): strong oxidant (on the left, E r:d = 1.359V )
(b)
Mn0 4 "(aq, acidic): strong oxidant (on the left, E r:d
559
= 1.51 V)
20 Chcnjsh}
20.42
Solutions to Exercises
8i th@ EILCitOldIl€tft:
(c)
Ba(s): strong reductant (on the right, E r: d = -2.90 V)
(d)
Zn(s): reductant (on the right, E':d = -0.763 V)
If the substance is on the left of a reduction half-reaction, it will be an oxidant; if it is on
the right, it will be a reductant. The sign and magnitude of the E r: d determine whether
it is strong or weak.
(a)
Na(s): strong reductant (on the right, E':d = -2.71 V)
(b)
03(g): strong oxidant (on the left, E':d = 2.07 V)
(c)
Ce 3+(aq): very weak reductant (on the right, E r: d = 1.61 V)
(d)
Sn 2+(aq): reductant (on the right, Er;d = 0.l54 V) or weak oxidant (on the left,
-0.136 V)
20.43
Analyze/plan. Follow the logic in Sample Exercise 20.8.
(a)
Solve.
Arranged in order of increasing strength as oxidizing agents (and increasing
reduction potential):
Cu 2+(aq) < 02(g) < Cr2072-(aq) < Cl 2(g) < H 20 2(aq)
(b)
Arranged in order of increasing strength as reducing agents (and decreasing
reduction potential):
H 202(aq) < l{aq) < Sn 2+(aq) < Zn(s) < AI(s)
20.44
20.45
(a)
The strongest oxidizing agent is the species most readily reduced, as evidenced
by a large, positive reduction potential. That species is H 202. The weakest
oxidizing agent is the species that least readily accepts an electron. We expect
that it will be very difficult to reduce Zn(s); indeed, Zn(s) acts as a comparatively
strong reducing agent. No potential is listed for reduction of Zn(s), but we can
safely assume that it is less readily reduced than any of the other species present.
(b)
The strongest reducing agent is the species most easily oxidized (the largest
negative reduction potential). Zn, E r: d = -0.76 V, is the strongest reducing agent
andF-, E r: d =2.87V, is the weakest.
Analyze/plan. In order to reduce Eu 3+ to Eu 2+, we need an oxidizing agent, one of the
reduced species from Table 20.1 or Appendix E. It must have a greater tendency to be
oxidized than Eu 3+ has to be reduced. That is, E : must be more negative than -0.43 V.
r d
Solve.
Any of the reduced species in Table 20.1 or Appendix E from a half-reaction with a
reduction potential more negative than -0.43 V will reduce Eu 3+ to Eu 2+. From the list
of possible reductants in the exercise, Al and H 2C20 4 will reduce Eu 3+ to Eu 2+.
Free Energy and Redox Reactions
20.46
Any oxidized species from Table 20.1 or Appendix E with a reduction potential greater
than 0.59 V will oxidize Ru0 4 2 - to Ru0 4 - . From the list of possible oxidants in the
exercise, Cr2072-(aq) and ClO-(aq) will oxidize RU0 4 2- to Ru0 4 -.
560
•• ~
20 Chpmjjtry of LliC EILsit6I&
20.47
Solutions to Exercises
AnalyzejPlan. In each reaction, Fe 2+ ~ Fe 3+ will be the oxidation half-reaction and one
of the other given half-reactions will be the reduction half-reaction. Follow the logic in
Sample Exercise 20.10 to calculate P and LlGo for each reaction.
Solve.
(a)
2Fe 2+(aq) + S2062-(aq) + 4H+(aq)
2Fe 3+(aq) + 2H 2S0 3(aq)
P = 0.60 V - 0.77 V = -O.17V
2Fe 2+(aq) + N 20 (aq) + 2H+(aq)
~
2Fe3+(aq) + N 2(g) + H 20(I)
EO = -1.77 V - 0.77 V = -2.54 V
Fe 2+(aq) + V0 2+(aq) + 2H+ (aq)
~
Fe 3+(aq) + V0 2+(aq) + H 20(l)
P
(b)
~
=
LlGO
1.00 V - 0.77 V = +0.23 V
= -nFP For the first reaction,
LlGQ = -2 mol x 96,500 J x (-0.17 V) = 3.281 x 10 4 = 3.3 x 10 4 J or 33 k]
1 V s mol
For the second reaction, LlGo = -2(96,500)(-2.54) = 4.902 x 10 5 = 4.90 x 10 2 k]
For the third reaction, LlGo = -1(96,500)(0.23) = -2.22
(c)
LlGo = -RT lnK; lnK
x
10 4 J = -22 k]
= -LlGo /RT; K = e-6.c°/RT
For the first reaction,
InK =
4
-3.281 x 10 J
= -13.243 = -13; K = e-13.2428 = 1.78 x 10- 6 = 2 x 10- 6
(8.314J/mol. K)(298 K)
[Convert In to log; the number of decimal places in the log is the number of sig
figs in the result.]
For the second reaction,
5
InK =
-4.902 x 10 J
= -197.86 = -198; K = e- 198 = 1.23 x 10- 86 = lxl0- 86
8.314 J/mol. K x 298 K
For the third reaction,
InK =
-(-2 22xl0 4 J)
.
=8.958=9.0; K=e9. 0 =7.77xl0 3 =8x10 3
8.314J/mol.K x 298K
Check. The equilibrium constants calculated here are indicators of equilibrium
position, but are not particularly precise numerical values.
20.48
(a)
2r(aq) ~ I 2(s)+ 2e­
Hg/+(aq)+2e- ~ 2Hg(l)
Er: d = 0.536 V
Er~ = 0.789 V
P = 0.789 - 0.536 = 0.253 V
96.5 kJ
x 0.253V = -48.829 = -48.8 kJ
V e mol e"
- (-4.8829 x 104 J)
InK =
= 19.708 = 19.7' K = e 19
.7 = 3.61 x 108 = 3.6 x 108
(8.314 J/mol e K)(298 K)
,
LlGQ = -nFP = -2 mol e- x
561
Solutions to Exercises
3{Cu+(aq) ~ Cu z+(aq)+le-:-]
(b)
Er'ed
=0.153V
N0 3 -(aq)+4H+(aq)+3e- ~ NO(g)+HzO(I)
E r: d = 0.96 V
3Cu+(aq)+N0 3 (aq)+4H+(aq) ~ 3Cu 2+(aq) + NO(g) + 2H zO(I)
EO = 0.96 - 0.153 = 0.81 V; L'.GO = -3(96.5)(0.81) = -2.345 x 10 2 kJ = -2.3 x 10 5 J
InK =
- (-2.345 x 105 J)
= 94.65 = 95; K = e 95 = 1.3 x 10 41 = 1041
(8.314 J/mol 0 K)(298 K)
2[Cr(OHh(s)+ 50H-(aq) ~ CrO/-(aq)+4H 20(I)+3e-] Er: d = -0.13 V
(c)
3[ClO-(aq)+H zO(I)+2e- ~ Cqaq)+20H-(aq)]
Er: d = 0.89V
2Cr(OHh(s)+3ClO (aq)+40H (aq)~2CrO/ (aq)+3Cl (aq)+5H zO(I)
EO = 0.89 - (-0.13)
InK =
=
1.02 V; L'.Go = -6(96.5)(1.02) = -590.58 kJ = -5.91 x 10 5 J
-(-5.9058 x 105J)
= 238.37 = 238; K = 3.3 x 10103 = 10103
(8.314 J/molo K)(298 K)
This is an unimaginably large number.
20.49
Analyze/plan. Given K, calculate L'.GO and EO. Reverse the logic in Sample Exercise 20.10.
According to Equation [19.22], L'.Go = -RT InK. According to Equation [20.12],
L'.Go = -nf'E", EO = - L'.Go/nF. Solve.
K = 1.5 X 10- 4
L'.GO = -RT inK = -(8.314 J/moloK)(298) in (1.5 x 10- 4) = 2.181 X 10 4 J = 21.8 kJ
EO = -L'.GO I nF; n = 2; F = 96.5 kJI mol ep=
-21.81kJ
=-0.113V
2mole- x 96.5kJ/Vomole-
Check. The unit of L'.Go is actually kl /mol, which means 1<J per 'mole of reaction', or for
the reaction as written. Since we don't have a specific reaction, we interpret the unit as
referring to the overall reaction.
20.50
K = 3.7 X 10 6 ; L'.Go = -RT inK; £0 = -L'.Go InF; n = 1; T = 298 K
L'.Go = -8.314 J/moloK x 298 K x in(3.7
EQ = -L'.GQ InF =
20.51
X
10 6 ) = -3.747
X
10 4 J = -37.5 kJ
-(-37.47 kJ)
= 0.388 V
le- x 96.5 kJ/V -rnol e­
Analyze. Given E r: d values for half reactions, calculate the value of K for a given redox
reaction.
Plan. Combine the relationships involving £0, L'.GO and K to get a direct relationship
between P and K. For each reaction, calculate P
relationship to calculate K.
562
from E r: d
,
then apply the
20 O} LCwC1iiy of toe EpIliroft~int
Solutions to Exercises
Solve. .1.Go = -nF£O, .1.Go = -RT inK; lnK = 2.303 log K
RT
2.303RT
-nFEQ=-RTlnK,P=-lnK=
logK
nF
nF
From Equation [20.15J and [20.16J, 2.303 RTIF = 0.0592.
EQ = 0.0592 log K; log K = ~; K = 10 1
0g K
n
0.0592
(a)
£0 = -0.28 - (-0.440) = 0.16 V, n = 2 (Ni 2+ + Ze: ~ Ni)
log K = 2(0.16) = 5.4054 = 5.4; K = 2.54 x 105 = 3 x 105
0.0592
(b)
£0 = 0 - (-0.277) = 0.277 V; n = 2 (2H+ + Ze"
~
H 2)
log K = 2(0.277) = 9.358 = 9.36' K = 2.3 x 10 9
0.0592
'
(c)
£0 = 1.51-1.065 = 0.445 = 0.45 V; n = 10 (2Mn0 4 - + 10e- ~ 2Mn+ 2)
log K = 10(0.445) = 75.169'" 75' K = 1.5 x 1075 = 10 75 '
0.0592
'.
Check. Note that small differences in EO values lead to large changes in the magnitude
of K. Sig fig rules limit precision of K values; using log instead of lnleads to more sig
figs in the K value. This result is strictly numerical and does not indicate any greater
precision in the data.
20.52
EQ = 0.0592 V log K; log K = nEQ
n
0.0592V.
explanation.
(a)
See Solution 20.51 for a more complete
EO = 1.00 V - 0.799 V = 0.201 = 0.20 V; n = 1 (V02 + + te
~
V02+)
log K = 1(0.210 V) = 3.3953 = 3.40' K = 2.48 x 103 = 2.5 x 103
0.0592 V
'
Note: The correctly balanced chemical equation for this reaction is
V0 2+(aq) + 2Ir (aq) + Ag(s) ~ 2 V02+(aq) + H 20(I) + Ag ' (aq), for which n = 1.
£0 = 1.61 V - 0.32 V = 1.29 V; n = 3 (3Ce 4+ + 3e-
~
3Ce 3+)
log K = 3(1.29) = 65.372 = 65.4-K = 2.35 x 10 65 = 2 x 10 65
0.0592
'
EO = 0.36 V - (-0.23 V) = 0.59 V; n = 4 (4Fe(CN)6 3- + 4elog K = 4(0.59)
0.0592
20.53
= 39.865 = 40;
K =7.3x 10 39
~
4Fe(CN)6 4-)
=1040
Analyze/plan. P = 0.0592 V log K. See Solution 20.51 for a more complete development.
n
log K =
(a)
nEo .
0.0592 V
log K =
1(0.177V)
0.0592 V
Solve.
=2.9899 = 2.99;
563
K =9.8 x102
1'1'
20 Qten dBh J of t}z@ 1Sfl!I}iI 6I dI[Hlft
20.54
Solutions to Exercises
2(0.177 V)
5
= 5.9797 = 5.98; K = 9.5x10
0.0592 V
(b)
log K =
(c)
log K = 3(0.177 V) = 8.9696 = 8.97; K = 9.32 x 108 = 9.3 x 108
0.0592 V
P = 0.0592 V log K; n = 0.0592 V log K. See Solution 20.51 for a more complete
EQ
n
development.
n=
0.0592 V
log (5.5x10 5);n=2
0.17V
Cell EMF under Nonstandard Conditions
20.55
20.56
20.57
20.58
(a)
The Nernst equation is applicable when the components of an electrochemical cell
are at nonstandard conditions.
(b)
Q = 1 if all reactants and products are at standard conditions.
(c)
If concentration of reactants increases, Q decreases, and E increases.
(a)
No. As the spontaneous chemical reaction of the voltaic cell proceeds, the
concentrations of products increase and the concentrations of reactants decrease,
so standard conditions are not maintained.
(b)
Yes. The Nernst equation is applicable to cell EMF at nonstandard conditions, so
it must be applicable at temperatures other than 298 K. There are two terms in the
Nernst Equation. First, values of P at temperatures other than 298 K are required.
Then, in the form of Equation [20.14], there is a variable for T in the second term.
In the short-hand form of Equation [20.16], the value 0.0592 assumes 298 K. A
different coefficient would apply to cells at temperatures other than 298 K.
(c)
If concentration of products increases, Q increases, and E decreases.
Analyze/plan. Given a circumstance, determine its effect on cell emf. Each circumstance
changes the value of Q. An increase in Q reduces emf; a decrease in Q increases emf.
Solve.
(a)
PH, increases, Q increases, E decreases
(b)
[2n 2 +] increases, Q increases, E decreases
(c)
[H+] decreases, Q increases, E decreases
(d)
No effect; does not appear in the Nernst equation
AI(s) + 3Ag + (aq) ~ A1
3
+ (aq)
+ 3Ag(s);
00592
[AI3+]
E = P - -'-n- log Q; Q = [Ag + ]3
Any change that causes the reaction to be less spontaneous (that causes Q to increase
and ultimately shifts the equilibrium to the left) will result in a less positive value for E.
564
,.
, 'I
20 ctr:enzisvy ot.the EngiropwsJilt
20.59
Solutions to Exercises
(a)
Increases E by decreasing [AP+] on the right side of the equation, which
decreases Q.
(b)
No effect; the "concentrations" of pure solids and liquids do not influence the
value of K for a heterogeneous equilibrium.
(c)
No effect; the concentration of Ag ' and the value of Q are unchanged.
(d)
Decreases E; forming AgCl(s) decreases the concentration of Ag", which
increases Q.
Analyze/plan. Follow the logic in Sample Exercise 20.11.
(a)
(b)
Ni 2+(aq)+2e­ ~ Ni(s)
Zn(s) ~ Zn 2+ (aq) + 2e-
Solve.
E r: d = -0.28 V
E r: d = -0.763 V
£2= -0.28 - (-0.763) = 0.483 = 0.48 V
2+].
E = £2 - 0.0592 10 [Zn
n = 2
2
n
g [Ni +j'
E = 0.483 - 0.0592 log (0.100) = 0.483 _ 0.0592 log (0.0333)
2
(3.00)
2
E = 0.483 - 0.0592 (-1.477) = 0.483 + 0.0437 = 0.527 = 0.53 V
2
(c)
20.60
(a)
(b)
E = 0.483 - 0.0592 10 (0.900) = 0.483 - 0.0193 = 0.464 = 0.46 V
2
g (0.200)
3[Ce 4+(aq)+le- ~ Ce 3+(aq)]
Cr(s) ~ Cr 3+(aq)+3e­
E r: d = 1.61 V
E r: d = -0.74 V
EQ= 1.61-(-0.74) = 2.35 V
0.0592
[Ce 3+]3 [Cr 3+]
E=£2---10g
; n=3
n
[Ce 4+]3
.
3
E = 2.35- 0.0592 log (0.010) (0.010)
3
(2.0)3
2.35- 0.0592 log (1.250 x 10- 9 )
3
E = 2.35 - 0.0592 (-8.903) = 2.35 + 0.176 = 2.53 V
3
(c)
E = 2.35- 0.0592 10 (0.85)3 (1.2)
3
g (0.35)3
= 2.35-0.0244 = 2.33 V
1'1'"
20.61
Analyze/plan. Follow the logic in Sample Exercise 20.11.
(a)
4[Fe 2+ (aq) ~ Fe 3+ (aq) + Ie "]
02(g)+4H+(aq)+4e­
~
2H 20(I)
565
Solve.
Er: d = 0.771 V
E;ed =1.23 V
£2 = 1.23 - 0.771 = 0.459 = 0.46 '
Solutions to Exercises
(b)
3+
[Fe ]4
; n = 4, [H+] = 10- 3 50 = 3.2 x 10- 4 M
E = P - 0.0592 10
n
g [Fe 2+]4[H+] 4P0 2
E = 0.459 V _ 0.0592 10
(0.010)4
= 0.459 - 0.0592 log (7.0x105 )
4
g(1.3)4(3.2x10-4)4(0.50)
4
E = 0.459 - 0.0592 (5.845) = 0.459 -0.0865 = 0.3725 = 0.37 V
4
20.62
(a)
2[Fe 3+(aq)+ Le" ~ Fe 2+(aq)]
~
H 2(g)
2H+(aq)+2e­
2+
f
E -- EQ-0.0592
- ­ Iog [Fe 3 f[H+
2
n
[Fe +] PH 2
(b)
Er~ =0.771 V
Er:d = 0.000 V
P = 0.771- 0.000 = 0.771 V
pH -10
. [H+] -10.
'
E = 0.771- 0.0592 10 (0.0010)2 (1.0 x 10­
2
g
(1.50)2(0.50)
E = 0.771
20.63
0.0592(-16.05)
2
=
5
)2
X
10- 5 ,n -- 2
= 0.771- 0.0592 10 (8.9 x 10- 17 )
2
g
0.771+ 0.4751 = 1.246 V
Analyze/plan. We are given a concentration cell with Zn electrodes. Use the definition of
a concentration cell in Section 20.6 to answer the stated questions. Use Equation [20.16]
to calculate the cell emf. For a concentration cell, Q = [diluteJI[concentrated].
Solve.
(a)
The compartment with the more dilute solution will be the anode. That is, the
compartment with [Zn 2+] = 1.00 x 10- 2 M is the anode.
(b)
Since the oxidation half-reaction is the opposite of the reduction half-reaction, EO
is zero.
(c)
E = EO- 0.0592 log Q; Q = [Zn 2+, dilute]/[Zn 2+, conc.]
n
E
=0 _ 0.0592 10
2
(d)
(1.00 x 10­
g
(1.8)
2)
= 0.0668 V
In the anode compartment, Zn(s) ~ Zn 2+(aq), so [Zn 2+] increases from 1.00 x 10- 2
M. In the cathode compartment, Zn 2+(aq) ~ Zn(s), so [Zn 2+] decreases from
1.8M.
20.64
(a)
The compartment with 0.0150 M Cl (aq) is the cathode.
(b)
P = 0V
(c)
E = Eo - - - l o g Q; Q = [Cl", dilute]/[Cl-, conc.]
0.0592
n
E = 0 - 0.0592 10 (0.0150) = -0.13204 = -0.1320 V
1
g (2.55)
(d)
In the anode compartment, [Cl'] will decrease from 2.55 M. In the cathode, [Cl']
will increase from 0.0150 M.
566
20Cf!zcnrisky @f ihQ Fugirouw-t
20.65
Solutions to Exercises
Analyze/plan. Follow the logic in Sample Exercise 20.12.
Solve.
E == P _ 0.0592 10 [PH,] [Zn 2+] . P = 0.0 V _ (-0.763 V) = 0.763 V
2
g
[H+f
'
0.684 = 0.763- 0.0~92 x (log [PH, ][Zn 2+]-210g [H+])
== 0.763 - 0.0592 x (-0.5686 - 2 log [H+])
2
0.684 == 0.763 + 0.0168 + 0.0592 log [H+ ]; log [H+] =
0.684-0.0168-0.763
0.0592
log [H+] == -1.6188 == -1.6; [H+] = 0.0241 = 0.02 M; pH = 1.6
20.66
(a)
P
=
-0.136 V - (-0.126 V) = -0.010 V; n
=
2
2+]
2+]
0.22 == -0.010- 0.0592 10 [Pb
= -0.010- 0.0592 10 [Pb
2+]
2
g 1.00
2
g [Sn
log [Pb 2+] = -0.23 (2) = -7.770 = -7.8; [Pb 2+] == 1.7 x 10-8
0.0592
(b)
For Pb50 4(s),
x., = [Pb2+] [50
2
4 -]
=2
x 10- 8 M
= (1.0)(1.7 x 10- 8) = 1.7 X 10- 8
Batteries and Fuel Cells
20.67
(a)
The emf of a battery decreases as it is used. This happens because the
concentrations of products increase and the concentrations of reactants decrease.
According to the Nernst equation, these changes increase Q and decrease E eell•
(b)
The major difference between AA- and D-size batteries is the amount of reactants
present. The additional reactants in a D-size battery enable it to provide power
for a longer time.
20.68
First, H 20 is a reactant in the cathodic half-reaction, so it must be present in some form.
Additionally, liquid water enhances mobility of' the hydroxide ion in the alkaline
battery. OH- is produced in the cathode compartment and consumed in the anode
compartment. It must be available at all points where Zn(s) is being oxidized. If the
Zn(s) near the separator is mostly reacted, OH- must diffuse through the gel until it
reaches fresh Zn(s). A small amount of H 20(I) mobilizes OH- so that redox can
continue until reactants throughout the battery are depleted.
20.69
Analyze/plan. Given mass of a reactant (Pb), calculate mass of product (Pb0 2 ) . This is a
stoichiometry problem; we need the balanced equation for the chemical reaction that
occurs in the lead-acid battery. Then, g Pb ~ mol Pb ~ mol Pb0 2 ~ g Pb0 2. Solve.
The overall cell reaction (Equation [20.19]) is:
Pb(s) + Pb0 2(s) + 2H+ (aq) + 2H50 4 - (aq) ~ 2PbS0 4(s) + 2H 20(I)
239.2 g Pb0 2
402 Pb x 1 mol Pb x 1 mol Pb0 2
b
g
207.2 g Pb
1 mol Pb x 1 mol Pb0 2 = 464 g P O 2
567
Solutions to Exercises
20.70
The overall cell reaction is:
2Mn02(S) + Zn(s) + 2H 20(I)
12.6 g Zn x
20.71
~
2MnO(OH)(s) + Zn(OHh(s)
86.94 g Mn0 2 33 ~ MnO
1 mol Zn
2 mol Mn0 2
d
d
=
.o g
2 re uce
x
x
65.39 g Zn
1 mol Zn
1 mol Mn0 2
Analyze/plan. We are given a redox reaction and asked to write half-reactions, calculate
£0, and indicate whether Li(s) is the anode or cathode. Determine which reactant is
oxidized and which is reduced. Separate into half-reactions, find E r : d for the half­
reactions from Appendix E and calculate £O.
(a)
Li(s) is oxidized at the anode.
(b)
Ag2Cr04(S)+ 2e- ~ 2Ag(s) + CrO/- (aq)
Solve.
Er: d = 0.446 V
Er~ = -3.05 V
2[Li(s) ~ Li+(aq) + Ie"]
Ag 2Cr04(s) + 2Li(s) ~ 2Ag(s) + Cr0 4 (aq) +2Li+(aq)
£0
20.72
= 0.446 V - (-3.05 V) = 3.496 = 3.50 V
(c)
The emf of the battery, 3.5 V, is exactly the standard cell potential calculated in
part (b).
(d)
For this battery at ambient conditions, E ", £0, so log Q ", O. This makes sense
because all reactants and products in the battery are solids and thus present in
their standard states. Assuming that £0 is relatively constant with temperature,
the value of the second term in the Nernst equation is ", 0 at 37°C, and E ", 3.5 V.
(a)
HgO(s) + Zn(s)
(b)
EC:
ll =
~
Hg(l) + ZnO(s)
E r: d (cathode) - Er: d (anode)
E r: d (anode) = Er: d
(c)
-
EC: ll = 0.098-1.35 = -1.25 V
E r: d is different from Zn 2+ (aq) + 2e- ~ Zn(s) (-0.76 V) because in the battery the
process happens in the presence of base and Zn 2+ is stabilized as ZnO(s).
Stabilization of a reactant in a half-reaction decreases the driving force, so Er~ is
more negative.
20.73
20.74
Analyze/plan. (a) Consider the function of Zn in an alkaline battery. What effect would
it have on the redox reaction and cell emf if Cd replaces Zn? (b) Both batteries contain
Ni. What is the difference in environmental impact between Cd and the metal hydride?
Solve.
(a)
E r: d for Cd (-0.40 V) is less negative than Er: d for Zn (-0.76 V), so E ceU will have
a smaller (less positive) value.
(b)
NiMH batteries use an alloy such as ZrNi 2 as the anode material. This eliminates
the use and concomitant disposal problems associated with Cd, a toxic heavy
metal.
(a)
The alkali metal Li has much greater metallic character than Zn, Cd, Pb or Ni.
The reduction potential for Li is thus more negative, leading to greater overall
568
Solutions to Exercises
cell emf for the battery. Also, Li is less dense than the other metals, so greater
total energy for a battery can be achieved for a given total mass of material. One
disadvantage is that Li is very reactive and the cell reactions are difficult to
control.
(b)
20.75
Li has a much smaller molar mass (6.94 g/mol) than Ni (58.69 g/mol). ALi-ion
battery can have many more charge-carrying particles than a Ni-based battery
with the same mass. That is, Li-ion batteries have a greater energy density than Ni­
based batteries.
The main advantage of a H 2-0 2 fuel cell over an alkaline battery is that the fuel cell is
not a closed system. Fuel, H 2, and oxidant, O 2 are continuously supplied to the fuel
cell, so that it can produce electrical current for a time limited only by the amount of
available fuel. An alkaline battery contains a finite amount of reactant and produces
current only until the reactants are spent, or reach equilibrium.
Alkaline batteries are much more convenient, because they are self-contained. Fuel cells
require a means to acquire and store volatile and explosive H 2(g). Disposal of spent
.' alkaline batteries, which contain zinc and manganese solids, is much more problematic.
H 2-02 fuel cells produce only H 20(I), which is not a disposal problem.
20.76
No. The fuel in a fuel cell must be fluid, either gas or liquid. Because fuel must be
continuously supplied to the fuel cell, it must be capable of flow; the fuel cannot be
solid.
Corrosion
20.77
Analyze/plan. (a) Decide which reactant is oxidized and which is reduced. Write the
balanced half-reactions and assign the appropriate one as anode and cathode. (b) Write
the balanced half-reaction for Fez+(aq) -+ Fe Z0 3 • 3H zO. Use the reduction halfSolve.
reaction from part (a) to obtain the overall reaction.
(a)
anode: Fe(s) -+ Fez+(aq) + 2e­
cathode: Oz(g) + 4H+ (aq) + 4e- -+ 2H zO(I)
(b)
2Fe z+(aq) + 6H zO(I) -+ FeZ03 • 3H zO(s) + 6H+(aq) + 2e­
Oz(g) + 4H+(aq) + 4e- -+ 2H zO(I)
(Multiply the oxidation half-reaction by two to balance electrons and obtain the
overall balanced reaction.)
20.78
(a)
Calculate E C: 1l for the given reactants at standard conditions.
1"
02(g)+4H+(aq)+4e- -+2H 20(l)
2[Cu(s) -+ Cu 2+ (aq) + 2e-j
2Cu(s) + 02(g) +4H+ (aq) -+ 2Cu 2+(aq) +2H 20(l)
Er: d = 1.23 V
E r: d = 0.337 V
P =1.23-0.337 =0.89 V
At standard conditions with Oz(g) and H+(aq) present, the oxidation of Cu(s) has
a positive P value and is spontaneous. Cu(s) will oxidize (corrode) in air in the
presence of acid.
569
Solutions to Exercises
20.79
(b)
Fe 2+ has a more negative reduction potential (-0.440 V) than Cu 2+ (+0.337 V), so
Fe(s) is more readily oxidized than Cu(s). If the two metals are in contact, Fe(s)
would act as a sacrificial anode and oxidize (corrode) in preference to Cu(s); this
would weaken the iron support skeleton of the statue. The teflon spacers prevent
contact between the two metals and insure that the iron skeleton doesn't corrode
when the Cu(s) skin comes in contact with atmospheric 02(g) and H+ (aq).
(a)
A "sacrificial anode" is a metal that is oxidized in preference to another when the
two metals are coupled in an electrochemical cell; the sacrificial anode has a more
negative Er;d than the other metal. In this case, Mg acts as a sacrificial anode
because it is oxidized in preference to the pipe metal; it is sacrificed to preserve
the pipe.
Er~ for Mg 2+ is -2.37 V, more negative than most metals present in pipes,
(b)
including Fe (Er~
= -0.44
V) and Zn (Er;d
= -0.763 V).
20.80
No. To afford cathodic protection, a metal must be more difficult to reduce (have a
more negative reduction potential) than Fe 2+. E r:d C0 2+ = -0.28 V, Er~ Fe 2+ = -0.44 V.
20.81
Analyze/plan. Given the materials bras's, composed of Zn and Cu, and galvanized steel,
determine the possible spontaneous redox reactions that could occur when the
materials come in contact. Calculate £0 values for these reactions.
Solve. The main metallic component of steel is Fe. Galvanized steel is steel plated with
Zn. The three metals in question are Fe, Zn, and Cu; their Er~ values are shown below.
E r: d Fe 2+(aq) = -0.440 V
Er~ Zn 2+ (aq) = -0.763 V
•
2+
E red Cu (aq) = 0.337 V
Zn, with the most negative Er~ value, can act as a sacrificial anode for either Fe or Cu.
That is, Zn(s) will be preferentially oxidized when in contact with Fe(s) or Cu(s). For
environmental corrosion, the oxidizing agent is usually 02(g) in acidic solution,
Er~
=
1.23 V. The pertinent reactions and their £0 values are:
2Zn(s) + 02(g) + 4H+(aq)
£0
= 1.23 V - (-0.763 V) = 1.99 V
2Fe(s) + 02(g) + 4H+ (aq)
£0
=
=
~
2Fe 2+(aq) + 2H 20(I)
1.23 V - (-0.440 V) = 1.67 V
2Cu(s) + 02(g) + 4H+(aq)
£0
2Zn 2+(aq) + 2H 20(l)
~
~
2Cu 2+(aq) + 2H 20(l)
1.23 V- (0.337 V) = 0.893 V
Note, however, that Fe has a more negative E r: d than Cu so when the two are in
contact Fe acts as the sacrificial anode, and corrosion (of Fe) occurs preferentially. This
is verified by the larger £0 value for the corrosion of Fe, 1.67 V, relative to the corrosion
of Cu, 0.893 V. When the three metals Zn, Fe, and Cu are in contact, oxidation of Zn will
happen first, followed by oxidation of Fe, and finally Cu.
570
20 QlrNtri3h, ef ~e FoviniHament
20.82
Solutions to Exercises
The principal metallic component of steel is Fe. Er: d for Fe, -0.763 V, is more negative
than that of Cu, 0.337 V. When the two are in contact, Fe acts as the sacrificial anode and
corrodes (oxidizes) preferentially in the presence of 02(g).
2Fe(s) + 02(g) + 4H"r (aq) -+ 2Fe 2+(aq) + 2H 20 (1)
EO = 1.23 V - (-0.440 V) = 1.67 V
2Cu(s) + 02(g) + 4H+(aq) -+ 2Cu 2+(aq) + 2H 20 (1)
£0 = 1.23 V - (0.337 V) = 0.893 V
Both reactions are spontaneous, but the corrosion of Fe has the larger EO value and
happens preferentially.
Electrolysis; Electrical Work
20.83
20.84
(a)
Electrolysis is an electrochemical process driven by an outside energy source.
(b)
Electrolysis reactions are, by definition, nonspontaneous.
(c)
20-(1) -+ 02(g) + 2e­
(a)
An electrolytic cell is the vessel in which electrolysis occurs. It consists of a power
source and two electrodes in a molten salt or aqueous solution.
(b)
It is the cathode. In an electrolysis cell, as in a voltaic cell, electrons are consumed
(via reduction) at the cathode. Electrons flow from the negative terminal of the
voltage source and then to the cathode. .
(c)
20.85
A small amount of H 2S0 4(aq) present during the electrolysis of water acts as a
change carrier, or supporting electrolyte. This facilitates transfer of electrons
through the solution and at the electrodes, speeding up the reaction.
(Considering H+ (aq) as the substance reduced at the cathode changes the details
of the half-reactions, but not the overall £0 for the electrolysis. S04 2-(aq) cannot
be oxidized.)
Analyze/Plan. Follow the logic in Sample Exercise 20.14, paying close attention to units.
Coulombs = amps -s: since this is a 3e- reduction, each mole of Cr(s) requires
Solve.
3 Faradays.
(a)
IF _
7.60A x 2.00d x 24hr x 60min x ~ x - lC
-- x
Id
1 hr
1 min
96,500C
1 amp-s
x
(b)
1 mol Cr
52.00 g Cr
3F
x 1 mol Cr
=
236 g Cr(s)
96,500C
3F
1 amp - s
1
1 hr
1 min
0.250 mol Cr x - - - x
x
x---x---x-­
1 mol Cr
F
1C
8.00 hr
60 min
60 s
=
20.86
Coulombs = amps .s; since this is a 2e- reduction, each mole of Mg(s) requires
2 Faradays.
S71
2.51 A
20 C,JaewjoLty of the lSilOitbl went
(a)
5.25A
x
2.50d
24 hr
Id
60 min
1 hr
x -- x ---
Solutions to Exercises
60 s I C
x --1 min
1 ampes
X
--
x
(b)
10.00 g Mg x
2F
1 mol Mg
x ---1 mol Mg
24.31g Mg
x
IF
96/500 C
X --­
1 mol Mg x _24_._31--"g....M-=g = 143 g Mg
2F
1 mol Mg
96/500 C
1 ampes
1 min
1
x
x--x-F
C
60 s
3.50 A
=378 min
20.87
Analyze/plan. Given a spontaneous chemical reaction, calculate the maximum possible
work for a given amount of reactant at standard conditions. Separate the equation into
half-reactions and calculate cell emf. Use Equation [20.19L w max = -nFE, to calculate
maximum work. At standard conditions, E = EO.
Solve.
12 (s) + 2e-
~
2r (aq)
Sn(s) ~ Sn
W max
E r: d = 0.536 V
Er~
2+(aq)+2e­
= -0.136 V
P = 0.536 - (-0.136) == 0.672 V
= -2(96.5)(0.672) = -129.7 = -130 kJI mol Sn
. -129.7 k]
1 mol Sn
1000 J
4
---~ x
x 75.0 gSn x -~ = -8.19 x 10 J
k]
mol Sn(s)
118.71 gSn
Check. The (-) sign indicates that work is done by the cell.
20.88
For this cell at standard conditions, EO = 1.10 V.
W max = ~Go
= -nF£O = -2(96.5)(1.10)
=
-212.3 = -212 kJImol Cu
_
.
1 mol Cu
-212.3 kJ
5
oO.OgCu x
x
=-167kJ=-1.67xlO J
63.55 g Cu
mol Cu
20.89
Analyze/plan. Follow the logic in Sample Exercise 20.15/ paying close attention to units.
Solve.
(a)
4
3600 s
lC
IF
Imol Li
x - __
7.5 x 10 A x 24 hr x x --- x
Lhr
Larnp s s
96/500C
IF
x 6.94 g Li
1 mol Li
(b)
ff
96,500 C x
If th e ceu:IS 85 01
/0 e icient,
F
x
0.85 = 3.961
x
105 = 4.0
x
105 g Li
IF
= 1.135 x 105
0.85 mol
=
1.1 x 105 C/mol Li required
_
1.135x105 C
IJ
lkWh
Energy = 7.0 Vx
x--x
= 0.24 kWh/mol Li
molLi
1CeV 3.6x106J
20.90
(a)
6.5
X
103 A x 48hr x 3600s
1 hr
x
1C
IF x -Larnp s s
96,500C
x ---
x
1 mol Ca
2F
40.08 gCa
x 0.68 = 1.586 x 105 = 1.6 x 105 g Ca
1mol Ca
572
20 Ci'rEnristry of th@ j;iM?lJii9liUigCilnt
(b)
If the cell is 68% efficient,
Solutions to Exercises
96,500 C
2 F
5
x - - - - = 2.838 x 10
F
0.68 mol Ca
=
2.8 x 105 C/mol Ca required
2.838 x 105 C
1J
1 kWh
Energy = 5.00 V x - - - - - x - - x ------..,.6- = 0.3942 = 0.39 kWh
mol Ca
C. V
3.6 x 10 J
Additional Exercises
20.91
(a)
Ni+(aq)+le- -+Ni(s)
Nnaq)
-+ Ni 2+(aq) + 1e­
2Ni+ (aq)
-+ Ni(s) + Ni 2+ (aq)
(b)
MnO/-(aq)+4H+(aq)+2e- -+Mn02(s)+2H20(l)
2[MnO/-(aq) -+Mn04-(aq)+le-]
(c)
(d)
H 2S0 3(aq)+4H+(aq)+4e- -+S(s)+3H 20(l)
2[H 2S0 3 (aq) + H 20(I) -+ HSO 4- (aq) + 3H+(aq) + 2e-]
Cl 2(aq) +2H 20(l) -+ 2ClO-(aq) + 4H+ (aq) +2e­
40H- (aq)
+ 40H- (aq)
Cl 2(aq) + 40H- (aq) -+ 2ClO- (aq) + 2H 20(I) + 2e­
Cl 2(aq)+2e- -+ 2Cqaq)
1/2[2Cl 2(aq) +40H-(aq) -+ 2Cl-(aq) + 2ClO-(aq) +2H 20(l)]
Cl 2(aq) + 20H-(aq) -+ Cl{aq) + ClO-(aq) +H 20(I)
20.92
(a)
Anode (-)
Cathode (+)
573
20 aasmjstrY·G{.'hi.)2Jrjl:Qp~.t
(c)
Solutions to Exercises
Mn0 4-(aq)+8H+(aq)+5e- ~Mn2+(aq)+4H20(l)
Er: d =1.51 V
5[Fe2+(aq)~Fe3+(aq)+le-]
Er: d =0.771 V
£0 = 1.51 V - 0.771 V = 0.74 V
(d)
3+]5
2+]
E = p_ 0.0592 10
[Fe
[Mn
; pH = 0.0, [H+] =1.0
5
g [Fe 2+]5[Mn0 4-][H+]8
4)5
E = 0.74 V - 0.0592 10 (2.5 x 10(0.010) ; Q = 6.510 x
5
g (0.10)5(1.50)(1.0)8
10-16
= 6.5 x
10-16
E = 0.74 V 0.0592(-15.1864) = 0.74 V + 0.18 V = 0.92 V
20.93
en
(a)
5
V
It e
Anode
Fe(s) ~ Fe 2+(aq)+2e­
2Ag+(aq)+2e- ~ 2Ag(s)
Fe(s) +2Ag+(aq) ~ Fe 2+(aq)+ 2Ag(s)
Cathode
Fe(s)
Ag(s)
Porous
separator
(b)
Zn(s) ~ Zn 2+(s)+2e­
2H+(aq)+2e- ~ H2(g)
err
+
0
0
0
0
Zn(s)
n
e
V
Pt
Zn2+
Porous
separator
(c)
Cuj Cu 2+11 CIO 3 -, C\-I Pt Here, both the oxidized and reduced forms of the
cathode solution are in the same phase, so we separate them by a comma, and
then indicate an inert electrode.
'Anode (-)
Cathode (+)
......
v-'
»>
Cu
Salt
IBridge
Cu h
anions
Inert
Electrode
"'~I------l.~
20.94
CJ~-,
Cl-
cations
We need in each case to determine whether P is positive (spontaneous) or negative
(nonspontaneous).
574
20 Gfl~lJilirwy 'it' the Enyjropmest
(a)
(b)
12(s) + 2e­ ~ 21- (aq)
Sn(s) ~ Sn 2+ (aq) + 2e­
E r: d = 0.536 V
E r: d == -0.136
£0 = 0.536-(-0.136) = 0.672 V, spontaneous
Ni 2+(aq)+2e­ ~ Ni(s)
Er: d = -0.28 V
2r-(aq) ~ 12(s)+2e­
(c)
Solutions to Exercises
r:d
E = 0.536 V
EO = -0.28-0.536 = -0.82 V, nonspontaneous
2[Ce 4+(aq)+le- ~Ce3+(aq)]
E r: d = 1.61 V
H 20 2(aq) ~ 02(g)+ 2H+ (aq) + 2e­
(d)
Cu 2+(aq)+2e­ ~ Cu(s)
E r: d = 0.68 V
P = 1.61- 0.68 = 0.93 V, spontaneous
E r: d = 0.337 V
+2e­
Sn 2+ (aq) ~ Sn
E r: d = 0.154 V
2+
4+
Cu2+(aq) +Sn (aq) ~ Cu(s)+Sn (aq) P = 0.337 -1.54 = 0.183 V, spontaneous
4+(aq)
20.95
(a)
The reduction potential for 02(g) in the presence of acid is 1.23 V. 02(g) cannot
oxidize Au(s) to Au +(aq) or Au 3+(aq),even in the presence of acid.
(b)
The possible oxidizing agents need a reduction potential greater than 1.50 V.
These include C 03+(aq), F 2(g), H 20 2(aq), and 03(g). Marginal oxidizing agents
(those with reduction potential near 1.50 V) from Appendix E are Br0 3"(aq),
Cev'(aq), HClO(aq), Mn04 "(aq), and Pb0 2(s).
(c)
4Au(s) + 8NaCN(aq) + 2H 20(I) + 02(g)
~
4Na[Au(CNh](aq) + 4NaOH(aq)
Au(s) + 2CN-(aq)
~
[Au(CNh]- + le­
02(g) + 2H 20(I) + 4e­
~
40H-(aq)
Au(s) is being oxidized and 02(g) is being reduced.
(d)
2[Na[Au(CNh](aq)+ Ie" ~ Au(s)+ 2CN-(aq) + Na+ (aq)]
Zn(s) ~ Zn 2+(aq) + 2e­
2Na[Au(CNh](aq) + Zn(s) ~ 2Au(s)+ Zn 2+(aq)+ 2Na+(aq) + 4CN (aq)
Zn(s) is being oxidized and [Au(CNh]-(aq) is being reduced. While OH-(aq) is
not included in the redox reaction above; its presence in the reaction mixture
probably causes Zn(OHh (s) to form as the product. This increases the driving
force (and £0) for the overall reaction.
20.96
(a)
(b)
2[Ag+(aq)+ Ie"
~
Ag(s)]
E r: d = 0.80 V
Ni(s) ~ Ni 2+ (aq) + 2e­
Er~ =-0.28 V
2Ag+ (aq)+ Ni(s) ~ 2Ag(s) + Ni 2+(aq) P = 0.80 - (-0.28) = 1.08 V
As the reaction proceeds, Ni 2+(aq) is produced, so [Ni 2+] increases as the cell
operates.
(c)
575
20 Glil @l1isEI' o£
tIlf
RRvin d" !iJU"'f
Solutions to Exercises
_ 0.04(2) = log(0.0100) -log[Ag +]2; log[Ag +]2 = log(0.0100)+ 0.04(2)
0.0592
0.0592
10g[Ag'P = -2.000 + 1.351 = -0.649; [Ag+j2 = 0.255 M; [Ag "] = 0.474 = 0.5 M
(Strictly speaking, [E - P] having only one sig fig leads (after several steps) to the
answer having only one sig fig. This is not a very precise or useful result.)
20.97
(a)
12(s) + 2e- 4' 2r (aq)
E r: d =0.536 V
2[Cu(s)4'Cu+(aq)+le-]
Er: d =0.521 V
12(s) + 2Cu(s) 4' 2Cu +(aq) + 21 (aq) P = 0.536-0.521 = 0.015 V
E = EO - 0.0592 log Q
.
=
n
0.015- 0.0592 log [Cu +f [r]2
2
E = +0.015- 0.0592 log (2.5)2 (3.5)2 = +0.015- 0.056 = -0.041 V
2
(b)
Since the cell potential is negative at these concentration conditions, the cell
would be spontaneous in the opposite direction and the inert electrode in the
12 / 1- compartment would be the anode; Cu(s) would be the cathode.
(c)
No. At standard conditions the cell reaction is as written in part (a) and Cu(s) is
the anode.
(d)
E = 0, + 0.Q15 = 0.0592 log (1.4)2 [I" f; 2(0.015) = log (1.4)2 + 2 log [r];
2
0.0592
10g[I-] = 0.107 = 0.11; [1-] = 10°.107 = 1.28 = 1.3 M 1­
1
:
20.98
Both P and K are related to ~Go. (See Solution 20.51.)
I
,I
~GO
= -nFP; ~Go = -RT InK
- nFE O = -RT InK, P
=
RT
-InK
nF
In terms of base 10 logs, InK = 2.303 log K.
o
E =
2.303RTI
ogK
nF
From the development of the Nernst equation,
2.303 RT
F
20.99
=0.0592 EO = 0.0592 log K
'n
Use the relationship developed in Solution 20.98 to calculate K from P. Use data from
Appendix E to calculate P for the disproportionation.
Cu+(aq)+le- 4'Cu(s)
Cu+ (aq) 4' Cu 2+ (aq) + le-
Er: d = 0.521 V
Er: d = 0.153 V
EO = 0.521 V -0.153 V = 0.368 V
nP
1 x 0.368
0.0592
EO =- - l o g K, log K = - - =
= 6.216 = 6.22
n
0.0592
0.0592
K = 10 6 .216 = 1.6 X 10 6
576
20(5ECfIl ieay o£ the Fuyiropwew
20.100
Solutions to Exercises
(a)
False. For standard cell potentials derived from Er~ values listed in Appendix E,
the maximum E C: ll value is 5.92 V. This reaction involves F 2(g), the element with
the largest positive Er: d , and Li, the element with the most negative Er: d • Other
species stable in aqueous solution will have E~ values between these two
limiting values, so other possible Ec: u values would be less than 5.92 V. Aqueous
E c: n values are not likely to exceed 5.92 V.
(b)
False. The effect of concentration of voltage is given by the Nernst equation;
E == £0 - 0.0592 log Q
where Q is the reaction quotient which inCOrporates
n
concentrations. Clearly E is not directly proportional to either individual
concentrations or Q.
(c)
True. Oxidizing agents are themselves reduced. The strength of an OXidizing
agent is measured by the magnitude of its reduction potential, Er~'
20.101
(a)
In discharge: Cd(s) + 2NiO(OH)(s) + 2H 20(l) -+ Cd(9Hh(s) + 2Ni(OHh(s)
In charging, the reverse reaction occurs.
(b)
£0 = 0.49 V - (-0.76 V) = 1.25 V
(c)
The 1.25 V calculated in part (b) is the standard cell potential, £0. The
concentrations of reactants and products inside the battery are adjusted so that
the cell output is greater than £0. Note that most of the reactants and products are
pure solids or liquids, which do not appear in the Q expression. It must be [OB:-]
that is other than 1.0 M, producing an emf of 1.30 rather than 1.25.
(d)
I K I k
nEo
Eo =0.0592
= - - og ; og = = - ­
n
log K ==
0.0592
2 x 1.25
42
42
== 42.23 == 42.2' K == 1.7 x 10 == 2 x 10
0.0592
'
20.102
The ship's hull should be made negative. By keeping an excess of electrons in the metal
of the ship, the tendency for iron to undergo oxidation, with release of electrons, is
diminished. The ship, as a negatively charged "electrode," becomes the site of
reduction, rather than oxidation, in an electrolytic process.
20.103
It is well established that corrosion occurs most readily when the metal surface is in
contact with water. Thus, moisture is a requirement for corrosion. Corrosion also occurs
more readily in acid solution, because O 2 has a more positive reduction potential in the
presence of H +(aq). S02 and its oxidation products dissolve in water to produce acidic
solutions, which encourage corrosion. The anodic and cathodic reactions for the
corrosion of Ni are:
Ni(s) -+ Ni 2+ (aq) + 2e­
02(g) + 4H+(aq) +4e- -+ 2H 20(l)
Er : d == -0.28 V
Er~
=
1.23 V
Nickel(ll) oxide, NiO(s), can form by the dry air oxidation of Ni. This NiO coating
serves to protect against further corrosion. However, NiO dissolves in acidic solutions
such as those produced by S02 or 503' according to the reaction:
NiO(s) + 2H+(aq) -+ Ni 2+(aq) + H 20(1). This exposes Ni(s) to further wet corrosion.
577
20 '¥emis1tY oltllil ~I [J •
Solutions to Exercises
20.104
A battery is a voltaic cell, so the cathode compartment contains the positive terminal of
the battery. In the alkaline, Ni-Cd and NiMH batteries, OH-(aq) is produced at the
cathode. The wire that turns the indicator pink is in contact with OH-(aq), so the
rightmost wire is connected to the positive terminal of the battery. [The battery could
not be a lead-acid battery because OH-(aq) is not present in either compartment, so
neither wire would turn the indicator pink.]
20.105
(a)
Total volume of Cr == 2.5 x 10- 4 m x 0.32 m Z == 8.0 x 10- 5 m 3
mol Cr == 8.0 x 10- 5 m 3 Cr x
_1_00_3_cm~_3
The electrode reaction is:
Cr0 4 Z-(aq) + 4H zO(I) + 6e-
Cr(s) + 80H-(aq)
~
1 m3
Coulombs required == 11.077 mol Cr x
x 7.20 g Cr x 1 mol Cr == 11.077
52.0 g Cr
lcm 3
== 11 mol Cr
6F
6
x 96,500 C = 6.41 x 10
IF
1 mol Cr
=6.4 x 106 C
1 amp.s
1
5
x --=6.4 x 10 amp
1 C
10.0 s
(b)
6.41 x 106 C x
(c)
If the cell is 65 efficient, (6.41 x 10 6/0.65) == 9.867 x 10 6 == 9.9 x 10 6 C are required
to plate the bumper.
IJ
1 kWh
6.0 V x 9.867 x 106 C x - - x
6
1 C. V 3.6 x 10
20.106
(a)
J
= 16.445 = 16 kWh
The work obtainable is given by the product of the voltage, which has units of
J/C, times the number of Coulombs of electricity produced:
3600s
lC
6J
lkWh
wm ax =300amp.hrx--x
x-x
6
=1.8kWh:==2kWh
1 hr
1 amp. sIC 3.6 x 10 J
(b)
20.107
(a)
(b)
This maximum amount of work is never realized because some of the electrical
energy is dissipated in overcoming the internal resistance of the battery, because
the cell voltage does not remain constant as the reaction proceeds, and because
the systems to which the electrical energy is delivered are not capable of
completely converting electrical energy into work.
7 x 108 mol Hz x
2F
I mol Hz
x 96,500 C = 1.35
IF
2H z0(1) ~ 0z(g) + 4H+(aq) +4e2[2H+(aq) +2e- ~ Hz(g)]
2H zO(I) ~ Oz(g) + 2H 2(g)
X
1014 = 1 x 1014 C
Er: d = 1.23V
Er~ =OV
P = 0.00 - 1.23 == -1.23 V
PI == 300 atm = POI + PH,' Since H 2(g) and Oz(g) are generated in a 2:1 mole ratio,
PH, ==200ann and Po,
=
100ann.
578
Solutions to Exercises
0.0592
2
0.0592
2
E=Eo---log(PO 2 x PH 2 )=-1.23V---log[100 x (200) 1
4
4
E = -1.23 V - 0.100 V = -1.33 V; E m in = 1.33 V
8
(c)
Energy = nFE = 2(7 x 10 mol) (1.33 V)
(d)
1.80
x
10 14 J
x
1 kWh
3.6 x 106 J
96,500 J
14
14
= 1.80 x 10 = 2 x 10 J
Vs mol
$0.85 = $4.24
kWh
x
x
107 = $4
x
107
It would cost more than $40 million for the electricity alone.
Integrative Exercises
20.108
2[N0 3-(aq) + 4H+(aq) +3e- ~NO(g)+2H20(l)1
E r:d
=0.96 V
3[Cu(s)~Cu2+(aq)+2e-l
Er~ =0.34V
3Cu(s) + 2N0 3 (aq) + 8H+ (aq) ~ 3Cu 2+ (aq) + 2NO(g) + 4H 20(I)
P = 0.96 - 0.34 = 0.62 V
2H+(aq)+2e-
~
H 2(g)
Cu(s) ~ Cu 2+(aq)+2e­
Er
: d = 0 V
E r: d =0.34 V
P = 0 - 0.34 = -D.34 V
The overall cell potential for the oxidation of Cu(s) by HN0 3 is positive and the
reaction is spontaneous. The cell potential for the oxidation of Cu(s) by HCl is negative
and the reaction is nonspontaneous. Note that in the reaction with HN0 3 , it is N0 3 ­
that is reduced, not H+; Cl from HCl cannot be further reduced.
20.109
N 2(g) + 3H 2(g) ~ 2NH 3(g)
(a)
The oxidation number of H 2(g) and N 2(g) is O. The oxidation number of N in
NH 3 is -3, H in NH 3 is +1. H 2 is being oxidized and N 2 is being reduced.
(b)
Calculate ~Go from ~G; values in Appendix C. Use ~Go = -RT InK to calculate
K.
~Go = 2~G; NH 3(g) -~G; N2(g)-3~G; H 2(g)
~Go =
°
~G
K
2(-16.66 kJ) - 0 - 3(0) = -33.32 kJ
- ~Go
- (-33.32 x 103 J)
= -RTlnK InK = - - =
= 13.4487 = 13.45
,
RT
(8.314 J/mol.K)(298 K)
= e 1 3.4487 = 6.9 x 105
(c)
n = ? 2 N atoms change from 0 to -3, or 6 H atoms change from 0 to +1. Either
way, n = 6.
P
= -(-33.32 kJ) = 0.05755 V
6 x 96.5 k]/V
579
20 •
20.110
J1iIHst7)' ~ K Li 2
sr. ,._,
Solutions to Exercises
The redox reaction is: 2Ag+(aq) + H 2(g) ~ 2Ag(s) + 2H+(aq). n
= 2 for this reaction.
Ece
°11 = E oed cathode - E °d anode = 0.799 V -0 V = 0.799
r
re
E = p_ 0.0592 10
[H+f
n
g [Ag+12 P
H z
[H+] in the cell is held essentially constant by the benzoate buffer.
C 6H sCOOH(aq) ':=:H+(aq) + C 6H sCOO-(aq)
K, =?
K = [H+][C 6H sCOO-l. [H+1= K a[C 6H sCOOHl = 0.10 M x K
a
[C6H sCOm"n'
[C6H sCOO-l
0.050 M
a
=2K
a
Solve the Nemst expression for [H+] and calculate K, and pK a as shown above.
1.030 V = 0.799 V _ 0.0592 10
[H+ f
n
g (1.00)2(1.00)
0.231 x _2_ = -Iog [H+ f
0.0592
=-2 log[H+ 1
0.231 = -log [H+1= pH; pH = 3.9020 = 3.90;[H+1= 10- 3.902 = 1.253 x 10- 4 = 1.3
0.0592
[H+] = 2K a ,
X
10- 4 ;
x, = [H+]j2 = 6.265 x 10- 5 = 6.3 x 10- 5; pK a = 4.20
Check. According to Appendix D, K, for benzoic acid is 6.3 x 10- 5 .
20.. 111
(a)
The oxidation potential of A is equal in magnitude but opposite in sign to the
reduction potential of A +.
(b)
Li(s) has the highest oxidation potential, Au(s) the lowest..
(c)
The relationship is reasonable because both oxidation potential and ionization
energy describe removing electrons from a substance. Ionization energy is a
property of gas phase atoms or ions, while oxidation potential is a property of the
bulk material.
20.112 (a)
N0 3-(aq)+4H+(aq)+3e- ~ NO(g)+ 2H 20(1)
Au(s)~Au3+(aq)+3e-
E r: d =0.96 V
E':d =1.498 V
EO = 0.96 - 1.498 = -0.54 V; EO is negative, the reaction is not spontaneous.
(b)
3[2H+(aq)+2e- ~H2(g)1
Er: d =0.000 V
E r: d = 1.002 V
2[Au(s)+ 4Cr(aq) ~ AuCl 4 - (aq) + 3e-l
2Au(s) + 6H+ (aq) + 8CI (aq) ~ 2AuCl 4 (aq) + 3H 2 (g)
EO = 0.000 -1.002 = -1.002 V; EO is negative, the reaction is not spontaneous.
(c)
N0 3-(aq) +4H+(aq) +3e- ~NO(g)+2H20(1)
Au(s)+ 4Cqaq) ~ AuCl 4-(aq)+3e-
E r: d =0.96V
E r: d = 1.002 V
EO = 0.96 -1.002 = -0.04; EO is small but negative, the process is not spontaneous.
580
20 GfrefirtSlij
II' *0 FPJTir9Rn£11l~
Solutions to Exercises
(d)
If [H+], [Cl"] and [N0 3 - ] are much greater than 1.0 M, the log term is negative
and the correction to EO is positive. If the correction term is greater than 0.042 V,
the value of E is positive and the reaction at nonstandard conditions is
spontaneous.
20.113
(a)
Ag+(aq)+eFe
2+
~
Er~ =0.799 V
Ag(s)
(aq) ~ Fe
3+
(aq) +le-
Er:d = 0.771 V
£0 = 0.799 V - 0.771 V = 0.028 V
(b)
Ag ' (aq) is reduced at the cathode and Fe 2+(aq) is oxidized at the anode.
(c)
~Go = -nFE· = -(1)(96.5)(0.028) = -2.7 k]
65° = S° Ag(s) + S° Fe 3+(aq) - SOAg+ (aq) - S° Fe 2+(aq)
= 42.55 J + 293.3 J- 73.93 J -113.4 J = 148.5 J
~GO
= ~Ho - T65° Since 65° is positive, ~Go will become more negative and ED
will become more positive as temperature is increased.
20.114
(a)
MI o = UHOH 20(I) - UW H 2(g) 65°
=
25° H 20(I) - 25° H 2(g) -
~So
~H002(g)
= 2(-285.83) - 2(0) - 0 = -571.66 kJ
02(g)
= 2(69.91) - 2(130.58) - (205.0) = -326.34 J
20.115
(b)
Since 65° is negative, -T~S is positive and the value of ~G will become more
positive as T increases. The reaction will become nonspontaneous at a fairly low
temperature, because the magnitude of ~So is large. .
(c)
~G
= w max' The larger the negative value of ~G, the more work the system is
capable of doing on the surroundings. As the magnitude of ~G decreases with
increasing temperature, the usefulness of H 2 as a fuel decreases..
(d)
The combustion method increases the temperature of the system, which quickly
decreases the magnitude of the work that can be done by the system. Even if the
effect of temperature on this reaction could be controlled, only about 40% of the
energy from any combustion can be converted to electrical energy, so combustion
is intrinsically less efficient than direct production of electrical energy via a fuel
cell.
First balance the equation:
4CyFe 2+(aq) + O 2(g) + 4H+ (aq) ~ 4CyFe 3+(aq) + 2H 20(I); E = +0.60 V; n
(a)
=4
From Equation [20.11] we can calculate ~G for the process under the conditions
specified for the measured potential E:
~G
_
= -nFE = -( 4 mol e ) x
96.5 k]
. _ (0.60V) =-231.6 =-232 k]
1 V cmol e
581
Solutions to Exercises
£L
The moles of ATP synthesized per mole of
(b)
2 is given by:
1 mol ATP formed
.
= approximately 6 mol ATP/mol O 2
37.7 kJ
231.6 kJ
02 molecule
20.116
°
x ~------
AgSCN(s)+e- ~Ag(s)+SCN-(aq)
Er: d = 0.0895 V
Ag(s)~Ag+(aq)+eEr~ =0.799 V
AgSCN(s) ~ Ag+(aq)+SCN (aq) EO = 0.0895-0.799 = -0.710 V
P = 0.0592 10 K . 10 K
n
g SP' g sp
K sp =
20.117
10- 11.993 =
1.02
x 10- 12
= (-0.710) (1) =-11.993 = -12.0
0.0592
= 1 x 10- 12
The reaction can be written as a sum of the steps:
Pb 2+(aq)+2e­ ~ Pb(s)
PbS(s) ~ Pb 2+ (aq) + S2- (aq)
E r: d = -0.126 V
PbS(s)+2e ~Pb(s)+S2 (aq)
Ered
Q
-?
- .
"Eon for the second step can be calculated from K sp.
EO = 0.0592 log Ksp
n
= 0.0592 10g (8.0
2
x 10- 28) = 0.0592 (-27.10) = -0.802 V
2
EO for the half-reaction = -0.126 V + (-0.802 V) = -0.928 V
Q
Calculating an imaginary E for a nonredox process like step 2 may be a disturbing
idea. Alternatively, one could calculate K for step 1 (5.4 x 10- 5) , K for the reaction in
question (K = K 1 x K s p = 4.4 x 10- 32) and then E for the half-reaction. The result is the
same.
Q
20.118
(a)
anode e - _
v
e - - cathode
H 2 (g), 1 atm - ­
_ - H 2 (g), 1 atm
H~
(aq, unknown)
W(aq, 1 M)
Porous
separator
- - - Anion
movement
(b)
2H+(aq, 1M)+2e- ~ H 2(g)
E r: d =0
H 2(g) ~ 2H+(aq, 1 M) + 2e(c)
At standard conditions, [H+] = 1 M, pH = 0
582
E r: d = 0
20 Glte tt:i:I~13
(d)
d
*9 &uiroawnzt
o
0.0592
[H+ (unknownjf PH,
E=E ---log----......"...------O...
[H+ (1 M)f PH,
2
E = 0- 0.0592 log [H+f
2
(e)
20.119
Solutions to Exercises
=
0.0592 x 2(-log [H+]) = 0.0592(pH) = 0.0592(5.0) = 0.30 V
2
E eell changes 0.0592(0.01) = 0.000592 = 0.0006 V for each 0.01 pH unit. The
voltmeter would have to be precise to at least 0.001 V to detect a change of
0.01 pH units.
The two half-reactions in the electrolysis of H 20(l) are:
2[2H 20(l)+2e-
~H2(g)+20H-]
2H 20(I) ~02(g)+4H+ +4e­
2H 20(l) ~ 2H 2(g) + 02 (g)
4 mol e- 12 mol H 2(g) or 2 mol e- I mol H 2(g)
Using partial pressures and the ideal-gas law, calculate the mol H 2(g) produced, and
the current required to do so.
0
P, = PH, + PH,o· From Appendix B, PH,o at 25.5 C is approximately 24.5 torr.
PH 2
= 768
torr - 24.5 torr
n = PV!RT =
4912
.
x 10
-4
= 743.5 = 744
torr
-4
-4
I
(743.5/760) atm x 0.0123 L
= 4.912 x 10 = 4.91 x 10 mo H 2
298.5 K x 0.08206 L 0 atm/mol s K
mo
IH
2x
2mole- 96,SOOC Lamp- s I min
1
7
x
x
x- - x
= O. 90 amp
mol H 2 1 mol e1C
60 s
2.00 min
583