Answers to Problems – Electrochemical Methods 3rd Edition by Allen J. Bard, Larry R. Faulkner, and Henry S. White

The Answers to Problems – Electrochemical Methods: Fundamentals and Applications (3rd Edition) provides clear and comprehensive step-by-step explanations for one of the most advanced and respected textbooks in electrochemistry.

Written by pioneers Allen J. Bard, Larry R. Faulkner, and Henry S. White, this resource simplifies complex electrochemical theories into easy-to-understand worked examples, helping students, researchers, and instructors develop a deep understanding of electroanalytical principles.

Each problem solution explains the reasoning, mathematical derivation, and practical meaning of concepts such as electrode kinetics, diffusion, voltammetry, impedance spectroscopy, electrolysis, and current-potential relationships. The guide balances theoretical depth with practical laboratory insight, making it an indispensable companion for students of chemistry, materials science, and electrochemical engineering.

It is ideal for graduate and advanced undergraduate learners who need structured guidance through numerical and conceptual electrochemistry problems. The solutions are carefully formatted to align with textbook chapters, ensuring effortless navigation.

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Mass Transport and Diffusion

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1. Electrochemical
Methods
Fundamentals and ApplicationsTHIRD EDITION
INSTRUCTOR SOLUTIONS MANUAL
Cynthia G. Zoski  Johna Leddy
ALLEN J. BARD
LARRY R. FAULKNER
HENRY S. WHITE
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2. to accompany
ELECTROCHEMICAL
METHODS
Fundamentals and Applications
Third Edition
INSTRUCTOR SOLUTIONS MANUAL
Cynthia G. Zoski
Johna Leddy
Department of Chemistry
Department of Chemistry
The University of Texas at Austin
The Center for Electrochemistry
Austin, TX USA
The University of Iowa
Iowa City, IA USA
WhatsApp: https://wa.me/message/2H3BV2L5TTSUF1 - Telegram: https://t.me/solutionmanual
[email protected]

3. TABLEOFCONTENTS
1.OVERVIEWOFELECTRODEPROCESSES 1
2.POTENTIALSANDTHERMODYNAMICSOFCELLS 21
3.BASICKINETICSOFELECTRODEREACTIONS 51
4.MASSTRANSFERBYMIGRATIONANDDIFFUSION 71
5.STEADY-STATEVOLTAMMETRYATULTRAMICROELECTRODES 81
6.TRANSIENTMETHODSBASEDONPOTENTIALSTEPS 101
7.LINEARSWEEPANDCYCLICVOLTAMMETRY 131
8.POLAROGRAPHY,PULSEVOLTAMMETRY,ANDSQUARE-WAVE
VOLTAMMETRY 155
9.CONTROLLED-CURRENTTECHNIQUES 165
10.METHODSINVOLVINGFORCEDCONVECTION-HYDRODYNAMIC
METHODS 181
11.ELECTROCHEMICALIMPEDANCESPECTROSCOPYAND
ACVOLTAMMETRY 197
12.BULKELECTROLYSIS 215
13.ELECTRODEREACTIONSWITHCOUPLEDHOMOGENEOUS
CHEMICALREACTIONS 239
14.DOUBLE-LAYERSTRUCTUREANDADSORPTION 259
15.INNER-SPHEREELECTRODEREACTIONSANDELECTROCATALYSIS 283
16.ELECTROCHEMICALINSTRUMENTATION 299
17.ELECTROACTIVELAYERSANDMODIFIEDELECTRODES 319
18.SCANNINGELECTROCHEMICALMICROSCOPY 329
19.SINGLE-PARTICLEELECTROCHEMISTRY 337
20.PHOTOELECTROCHEMISTRYANDELECTROGENERATED
CHEMILUMINESCENCE 353
A.MATHEMATICALMETHODS 367
B.BASICCONCEPTSOFSIMULATIONS 389
v
PREFACE vi

4. 1OVERVIEWOFELECTRODE
PROCESSES
Problem1.1
c
(a).Inapproachingthiskindofproblem,itisusefultolistallthecouplesinTable
C.1thatarerelevanttothesystem.
E
0
vs.NHE(V) Reaction
1.229 O2+4H
+
+4eH2O
1.188 Pt
2+
+2ePt
0.340 Cu
2+
+2eCu
0.159 Cu
2+
+eCu
+
0.000 2H
+
+2eH2
-0.4025 Cd
2+
+2eCd
Alternatively,agraphicalrepresentationmayproveuseful.Here,thestandardorformalpotentials
foreachredoxcoupleareplottedonapotentialaxis.Thespeciespresentinsolutionareunderlined.
Notethereducedhalfofthecoupleisnotedtowardmorenegativepotentials.Theverticalline
indicatestheapproximatepotentialrangewherebothhalvesoftheredoxcouplecanexist.For
electrodepotentialspositiveofagivenline,theoxidizedhalfofthecoupleisstableattheelectrode
surface;forelectrodepotentialsnegativeoftheline,thereducedformisstable.Notethatforn= 1,
electrodepotentialswithin118mVofE
0
requirenolessthan1%ofeithertheoxidizedorreduced
halvesofthecoupleasgivenbylog
[O]
[R]
=n

EE
0

=0:059.Cu
2+
Cu
1+
Cd
2+
CdH
+
H2Cu
2+
  Cu
Pt
2+
Pt
O2H2O
Rest Potential
Range
­0.6­0.10.40.91.4
E (V vs. NHE)
Thecompositionofthesystemdictatesthattherest(zerocurrent)potentialbemorepositivethan
E
0
Cu
2+
=Cu
andmorenegativethanE
0
O2=H2O
orE
0
P t
2+
=P t
,i.e.,betweenabout0.34Vand1.2V
vs.NHE.Graphically,thisisapparentbecausethisisthevoltagerangeoverwhichtheoxidized
(Cu
2+
)andreducedspecies(PtorH2O)presentinthesolutionaremostadjacentonthegraph.
Thisdenesazoneofstabilitysetbytheoxidizedandreducedspecies.(Notethatthecellwould
notbeatequilibriumifoxidizedandreducedspeciesoftwoormorecoupleswerepresentsuchthat
theywereontheoutersidesofthelines.Forexample,ifthesolutioncontainedCuandO2,there
wouldbeathermodynamicdrivingforceforthesespeciestoreactspontaneouslytoformwater
andCu
2+
.)Intherestpotentialrange,thepotentialisnotwelldenedinathermodynamicsense;
theelectrodeisnotwellpoised,becausenocouplehasbothoxidizedandreducedformspresent.
CalculationoftheequilibriumpotentialbytheNernstequationcannotbemade.
Electrochemical Methods: Fundamentals and Applications, Third Edition, Instructor Solutions Manual. Cynthia G. Zoski and Johna Leddy.
© 2025 John Wiley & Sons Ltd. Published 2025 by John Wiley & Sons Ltd.

5. Chapter1OVERVIEWOFELECTRODEPROCESSES
Currentwillowwhenthepotentialismovednegativelyfromtherestpotentialtoabout0.340V
(or0:340 + (0:2412) = 0:099Vvs.SCE)sothatCu
2+
isreducedattheelectrodesurfacerst.
Cu
2+
+2eCu (rstreduction,0:1Vvs.SCE)
Apositivemovementfromtherestpotentialrstcausessignicantcurrentowwhenplatinumand
waterareoxidized.
PtPt
2+
+2e (rstoxidations,1.0Vvs.SCE)
2H2OO2+4H
+
+4e
Actually,Ptwouldformathinoxidelm,thenitwouldstabilize,andonlytheoxygenevolution
reactionwouldoccur,withasignicantnegative(anodic)currentowduetotheoxidationof
water,whichmarksthepositivebackgroundlimit.Thecurrent-potentialcurvewouldlooklikethe
followingwithatransitionfromalimitingpositive(cathodic)currentowduetoCu
2+
reductionto
CumetalthatplatesontothePtUMEtoasignicantcathodiccurrentbeginningat-0.1VvsSCE
duetoH
+
(2Minsolution)reduction,whichmarksthenegativebackgroundlimit.AlthoughCd
2+
ispresentinthesolution,itsreductiontoCdcannotbeobserved,becausethisreductionoccursat
about-0.6Vvs.SCE,farbeyondthenegativebackgroundlimit.
Thereadermaybepuzzledregardingthelocationsofthebackgroundlimits,whicharelessextreme
thantherelatedstandardpotentials.Onemustrememberthatabackgroundcurverepresentsjustthe
footofanenormouswavesupportedbyahighlyavailableelectroreactant(inthiscase,H
+
orH2O).
Thehalf-wavepotentialforthatwavemightbenearthestandardpotentialofthecouple,butonthe
currentscalethatallowsobservationofvoltammetricfeaturesofinterest,oneneverapproachesthe
standardpotentialbeforethebackgroundcurrentbecomesprohibitive.Ingeneral,abackground
limitbasedonareversibleorquasireversiblecoupleislessextremethanthecorrespondingstandard
potentialbyabout0.1V.
2

6. Chapter1
(b).ThecouplestobeconsideredfromTableC.1areasfollows:
E
0
vs.NHE(V) Reaction
1.3583 Cl
2(g)+2e2Cl

1.229 O2+4H
+
+4eH2O
1.188 Pt
2+
+2ePt
0.15 Sn
4+
+2eSn
2+
0.000 2H
+
+2eH2
-0.1375 Sn
2+
+2eSn
Thegraphicalrepresentationisasfollows.Sn
4+
Sn
2+
Sn
2+
Sn
H
+
H
2
Cl
2Cl
­
Pt
2+
Pt
O
2H
2O
­0.50.00.51.01.5
E  ( V  vs . NHE )
BecausebothSn
4+
andSn
2+
arepresent,thesystemispoisedandawell-denedthermodynamic
equilibriumpotentialexists.FromtheNernstequation(1.3.13),theequilibriumpotentialis0.15V
vs.NHEor-0.09Vvs.SCE.MovingnegativelyfromthispotentialfavorsSn
2+
attheelectrode
andrequiresreductionofSn
4+
.
Sn
4+
+2eSn
2+
(rstnegativeprocess,at-0.09Vvs.SCE)
Likewise,amovepositiveoftherestpotentialfavorsSn
4+
attheelectrodeanddrivesoxidation.
Sn
2+
Sn
4+
+2e (rstpositiveprocess,alsoat-0.09Vvs.SCE)
Thecurrentpotentialcurveresemblesthefollowingwheretheequilibriumpotential(-0.09Vvs
SCE)fortheSn
4+
/Sn
2+
coupleisshownbecausebothtincationsarepresentintheelectrochemical
cellatthesameconcentration.MovingnegativefromthispotentialforSn
4+
/Sn
2+
andpositive
fromthispotentialforSn
2+
/Sn
4+
leadstoacathodicandanodiclimitingcurrentrespectivelythat
areequalbutoppositeinsign.Movingyetmorenegativeinpotentialfromthecathodiclimiting
currentleadstoasignicantincreaseincathodiccurrentat-0.1VvsSCEduetoproton(1M)
reduction,whichcorrespondstothenegativebackgroundlimit.Thelimitingcurrentforreduction
ofSn
4+
isnotfullyresolvedfromthecurrentriseatthislimit.Movingmorepositiveinpotential
fromabout+0.9Vfromtheanodiclimitingcurrentleadstoasignicantexcursionoftheanodic
currentduetowateroxidation,whichcorrespondstothepositivebackgroundlimit.
3

7. Chapter1OVERVIEWOFELECTRODEPROCESSES
(c).ThecouplestobeconsideredfromTableC.1areasfollows:
E
0
vs.NHE(V) Reaction
1.3583 Cl
2(g)+2e2Cl

1.229 O2+4H
+
+4eH2O
0.7960 Hg
2+
2
+2e2Hg
0.26816 Hg2Cl2+2e2Hg+2Cl

0.000 2H
+
+2eH2
-0.3515 Cd
2+
+2eCd(Hg)
-0.7656 Zn
2+
+2eZn(Hg)
Thegraphicalrepresentationisshown.Zn
2+
  Zn
Hg
2Cl
2Hg,Cl
­
H
+
H
2Cl
2Cl
­
Hg
2
2+
Hg
O
2H2O
R e s t P o t.
 R a nge
­1.0­0.50.00.51.01.5
E  (V  vs N H E )
Asin(a),thesystemisunpoisedandtherestpotentialisnotwelldened,butexistsbetween
E
0
H
+
=H
2=0.0Vvs.NHEandE
0
Hg2Cl2=Hg
at0.26816Vvs.NHE;thatis,between-0.2412Vand
0.02696Vvs.SCE.Therstoxidationoccurswhenthepotentialisdrawntomorepositivevalues
than 0.1Vvs.SCE,wherethefollowingreactionbegins.
4
Cd
2+
,HgCd(Hg)

8. Chapter1
2Hg+2Cl

Hg2Cl2+2e (rstpositiveprocess,E
0
0.03Vvs.SCE)
Thechartanddiagrampredictthattherstreductionwillbetheevolutionofhydrogen.
2H
+
+2eH2 (predictedrstreduction,-0.24Vvs.SCE)
However,thisreactionisextremelyslowonmercury(i.e.,ithasahighoverpotential;seeSection
1.1.7(c)andChapters1and3)anddoesnotoccuratanappreciablerateuntilfarmorenegative
potentialsarereached.Thus,therstreductionofsignicanceisthedepositionofcadmiuminto
themercurytoformtheamalgam.
Cd
2+
+2eCd(Hg) (actualrstreduction,-0.59Vvs.SCE)
Thisisanexampleofkineticssupersedingthermodynamicexpectationduringdynamicperturba-
tionofanelectrochemicalsystem.Theoverpotentialofhydrogenonmercurysignicantlywidens
therangeofaccessiblepotentials(i.e.,the“potentialwindow”)inwater,andisonereasonmercury
waslong-favoredforelectrochemicalanalysisandthermodynamicevaluationsdespiteitstoxicity.
ThereductionwavesforCd
2+
andZn
2+
canbothbeobservedatamercuryelectrode.
ThereductionofCd
2+
willleadtoalimitingcurrentat-0.59VvsSCE,followedbyalimiting
currentofsimilarheightforthereductionofZn
2+
atabout-1.0Vvs.SCE.Finally,thereduction
ofH
+
beginsatapproximately-1.1Vvs.SCEwithacathodiccurrentthatappearsinniteonthe
currentscaleduetotheH
+
concentrationof1M.
Thecurrent-potentialcurveresemblesthefollowingwheretheanodiccurrentduetotheoxidation
ofHgat0.0VvsSCEappearspracticallyinniteonthecurrentscale.Thisoxidationdenesthe
anodicbackgroundlimit.
Thenegativebackgroundlimitinthiscaseisnotduetoareversibleorquasireversiblecouple,but,
rather,thehighlyirreversibledischargeofH2onamercuryelectrode.Itoccursatpotentialsabout
1VmoreextremethanE
0
forthecouplebecausethekineticshavetobeactivatedthroughalarge
overpotential.
5

9. Chapter1OVERVIEWOFELECTRODEPROCESSES
Problem1.2
c
Rotationrateof10revolutionspersecondisexpressedinradianspersecondas
!= 210s
1
= 62:8s
1
Fromtheequationgivenintheproblemforthemasstransportratetoarotatingdisk,
mO= 0:62D
2=3
O

1=6
!
1=2
= 0:62

5:210
6
cm
2
/s

2=3


62:8s
1

1=2


0:010cm
2
/s

1=6
= 3:210
3
cm/s
Fromequation(1.3.10),andn= 1,
il=nF AmOC

O
= 1(96485C/mol)(0:30cm
2
)(3:210
3
cm/s)(1:010
5
mol/cm
3
)
= 9:210
4
A= 920A
6

10. Chapter1
Problem1.3
c
Theimportantreactionsare
Fe
3+
+eFe
2+
E
0
=0.771Vvs.NHE
Sn
4+
+2eSn
2+
E
0
=0.15Vvs.NHE
(a).From(1.3.10)andn= 1,il=nF AmOC

O
= 580A.
(b).Becausetheconcentrationofstannicionishalfthatofferricionbutn= 2,andthemass
transfercoefcientsofthetwoionsarethesame,thelimitingcurrentforthereductionofSn
4+
is
also580A.Thehalfwavepotential,E
1=2,fortheferricreductionisnearE
0
=0.77Vvs.NHE,
whereasthatforthereductionofstannicionisnear0.15Vvs.NHE.TheiEcurveisasfollows:
7

11. Chapter1OVERVIEWOFELECTRODEPROCESSES
Problem1.4
c
(a).Fromequation(1.5.3):R= 2:3k
(b).Fromequation(1.5.3):
x(cm)0.050.10.51.0Ru()3877385769
(c).ForA=0.1cm
2
,r0=0.0892cm.Then,fromequation(1.5.4):
x(cm)0.050.10.51.0Ru()25365863
8

12. Chapter1
Problem1.5
c
ThediscussionforthisproblemisfoundinSection1.6.4a.Fromequation(1.6.17),
i=
E
Ru
exp


t
RuCdA

(1.6.17)
AreaAisintroducedwhereCdisexpressedascapacitanceperarea(Fcm
2
).Thecelltime
constant=RuCdA.
i=
E
Ru
exp


t


ForpotentialsteppedfromE1toE2inthenonfaradaicregion,E=E2E1,thecapacitor
ischarged.Theinitialcurrent,whent= 0,isthemaximumcurrent,=E=Ru.Thecurrent
decaysto0:05i(t= 0)attimei(t
5 %)when
i(t
5%)
i(t= 0)
= 0:05 = exp


t
5%


ln (0:05) = 3:0 =
t
5%

Thecurrentdecaysto5%oftheinitialcurrentwhent
5%= 3:0.Atthistime,double-layer
chargingis95%complete.
ForCd= 20Fcm
2
andA= 0:1cm
2
,=RuCdAandt
5%= 3:0arefoundforRuof1,10,
and100.
Rs()(s)3(s)1 2 6
10 20 60
100 200 600
9

13. Chapter1OVERVIEWOFELECTRODEPROCESSES
Problem1.6
c
Note:Itwastheauthors’intentiontochangethereferenceelectrodeusedin
thisproblemtothemorecommonlyusedA
g/AgCl.Inpart(c)oftheproblemstatement,
"SCE"shouldread"Ag/AgCl."ThereisnopurposetobeservedbyusingtheSCEforpart
(c).alone.InstructorsandstudentsareurgedtousetheAg/AgClreferencethroughout.This
solutionisbaseduniformlyonAg/AgCl.
ItisnotedthatSCEis+0.244VvsNHEandAg/AgClis+0.197VvsNHE.SCEisthe+0.047V
vsAg/AgCl.SCEandAg/AgClareformedinsaturatedKCl.
FromProblem1.5,Cd= 20Fcm
2
ata0.1cm
2
electrode.Inthisproblem,thepotentialstepis
fromE1=0:1VtoE2=0:6VvsAg/AgClwithPZCEz=0:3VvsAg/AgCl.
(a).Fromequation(1.6.11),whereareaAappearsbecauseCdisexpressedascapacitanceperunit
area.ThechargeontheelectrodeattheinitialpotentialE1isdesignatedasq1.
q1=CdA(E1Ez)
q1=

20F/cm
2

0:1cm
2

(0:1V(0:3V))
= 0:410
6
C
Toappreciatetheunits,itishelpfultorecognizeafarad(F)asacoulombpervolt(C/V).The
chargeontheelectrodeoncesteppedtoE2isalsofoundfromequation(1.6.11).
q2=CdA(E2Ez)
=

20F/cm
2

0:1cm
2

(0:6V(0:3V))
=0:610
6
C
(b).ForEz= +0:1Vvs.Ag/AgCl,thesolutioniscarriedoutexactlyasin(a).Theresultsare
q1=0:4Candq2=1:4C.
(c).ForEz=0:1Vvs.Ag/AgCl,thesolutioniscarriedoutexactlyasin(a).Theresultsare
q1= 0:0Candq2=1:0C.
10

14. Chapter1
Problem1.7
c
Fromequation(1.6.23),
i=Cd

1exp


t
RuCd

(1.6.23)
Thetransienttermdecaysinafewtimeconstants.Here,itisnegligibleafter10s(whichcorre-
spondsto5inProblem1.5atRs= 1 )to2ms(atRs= 100 ).Atsteadystate,
i=vCdA
AappearsasCdisexpressedhereincapacitanceperunitarea.CurrentsareindependentofRu.
(Vs
1
)0.02120i(A)0.04240
11

15. Chapter 1 OVERVIEW OF ELECTRODE PROCESSES
Problem 1.8
c
(a).From equations (1.3.10) and (1.3.18) for the limiting currents,
il;c
il;a
=
nF AmOC

O
nF AmRC

R
=
4:00A
2:40A
= 1:67or
mO
mR
= 1:67
C

R
C

O
= 0:833. From equation (1.3.16),E
1=2=E
0
0

RT
nF
ln
mO
mR
=
0:498V vs. NHE.
(b).The wave has both cathodic and anodic parts, because bothA
3+
andA
+
are present in the
bulk. The potential where the current is midway between the plateaus isE
1=2. The potential where
i= 0is the equilibrium value calculable from the Nernst equation. TheiEcurve is shown.
(c).From (1.3.21), the plot islog [(il;ci)=(iil;a)]vs. NHE. The crossing of the abscissa is
E
1=2, and the slope magnitude isnF=(2:303RT) =n38:92V
1
at 25

C. The plot is shown.
12
T
The sample finish on this page. The last solved problem in chapter 1 is #13.