GeorgeMasonUniversity
GeneralChemistry212
Chapter21
Electrochemistry
Acknowledgements
CourseText:Chemistry:theMolecularNatureofMatterand
Change,6thed,2011,MartinS.Silberberg,McGrawHill
04/22/12
TheChemistry211/212GeneralChemistrycoursestaughtat
GeorgeMasonareintendedforthosestudentsenrolledina
science/engineeringorientedcurricula,withparticularemphasis
onchemistry,biochemistry,andbiologyThematerialonthese
slidesistakenprimarilyfromthecoursetextbuttheinstructor
hasmodified,condensed,orotherwisereorganizedselected
material.
Additionalmaterialfromothersourcesmayalsobeincluded.
Interpretationofcoursematerialtoclarifyconceptsandsolutions
1
Electrochemistry
RedoxReactionsandElectrochemicalCells
HalfReactionMethodforBalancingRedox
Reactions
ReviewofOxidationReductionConcepts
ElectrochemicalCells
VoltaicCell:UsingSpontaneousReactionsto
GenerateElectricalEnergy
CellNotation
04/22/12
ConstructionandOperation
WhyDoestheCellWork
2
Electrochemistry
CellPotential:OutputofaVoltaicCell
StandardCellPotentials
StrengthsofOxidizingandReducingAgents
FreeEnergyandElectricalWork
EffectofConcentrationofEcell
ChangesinEcellDuringCellOperation
StandardCellPotential
ConcentrationCells
ElectrochemicalProcessesinBatteries
04/22/12
Primary(NonrechargeableBatteries)
3
Electrochemistry
Corrosion:AcaseofEnvironmental
Electrochemistry
CorrosionofIron
ProtectingAgainstCorrosion
ElectrolyticCells:Usingelectricalenergytodrive
NonspontaneousReactions
PredictingElectrolysisProducts
04/22/12
ConstructionandOperation
StoichiometryofElectrolytes
4
Electrochemistry
Electrochemistry
Thestudyoftherelationshipbetween
chemicalchange(reactions)andtheflowof
electrons(electricalwork)
ElectrochemicalSystems
04/22/12
ElectrolyticWorkdonebyabsorbingfree
energyfromasource(passageofanelectrical
currentthroughasolution)todrivea
nonspontaneousreaction
Voltaic/GalvanicReleaseoffreeenergyfrom
aspontaneousreactiontoproduceelectricity
(Batteries)
5
Electrochemistry
OxidationReductionConceptsReview
OxidationLossofElectrons
ReductionGainofElectrons
OxidizingAgentSpeciesthatcausesanotherspecies
tobeoxidized(loseelectrons)
Oxidizingagentisreduced(gainse)
ReducingAgentSpeciesthatcauseanotherspecies
tobereduced(gainelectrons)
Reducingagentisoxidized(losese)
04/22/12
Oxidation(eloss)alwaysaccompanies
Reduction(egain)
Totalnumberofelectronsgainedbytheatoms/ionsof
6
Electrochemistry
04/22/12
7
Electrochemistry
OxidationNumber
Theoxidationnumberinabinaryioniccompound
equalstheioniccharge
Theoxidationnumberforeachelementinacovalent
compound(orpolyatomicion)areassignedaccording
totherelativeattractionofanatomforelectrons
04/22/12
Anumberequaltothemagnitudeofthechargean
atomwouldhaveifitssharedelectronswereheld
completelybytheatomthatattractsthemmore
strongly
Seenextslideforasummaryoftherulesforassigning
oxidationnumbers
8
Electrochemistry
04/22/12
9
Electrochemistry
BalancingRedoxReactions
OxidationNumberMethod
HalfReactionMethod
Thebalancingprocessmustinsurethat:
Thenumberofelectronslost
bythereducingagentequals
thenumberofelectrons
gainedbytheoxidizingagent
04/22/12
10
Electrochemistry
OxidationNumberMethod
Fromchangesinoxidationnumberofgiven
elements,identifyoxidizedandreducedspecies
Foreachelementthatundergoesachangeof
oxidationnumber,computethenumberof
electronslostintheoxidationandgainedinthe
reductionfromtheoxidationnumberchange(Draw
tielinesbetweentheseatoms)
04/22/12
Assignoxidationnumberstoallelementsinthe
reaction
Multiplyoneorboththesenumberbyappropriate
factorstomaketheelectronslostequaltothe
electronsgained
11
PracticeProblem
BalanceequationwithOxidationNumbermethod:
Cu(s) + HNO 3 (aq) Cu(NO 3 )2 + NO 2 (g) + H 2O(l)
0
+5
+1 -2
+5
+2
-2
-2
+4
+1 -2
Cu(s) + HNO 3 (aq) Cu(NO 3 )2 + NO 2 (g) + H 2O(l)
Loses 2e-
Cu(s) + HNO 3 (aq) Cu(NO 3 )2 + NO 2 (g) + H 2O(l)
Gains 1e-
Balance the Equation
Cu(s) + 2HNO 3 (aq) Cu(NO 3 )2 + 2NO 2 (g) + H 2O(l)
04/22/12
Cu(s) + 4HNO 3 (aq) Cu(NO 3 )2 + 2NO 2 (g) + 2H 2O(l)
12
Electrochemistry
HalfReactionMethod
ApplicabletoAcidorBasesolutions
DoesnotusuallyrequireOxidationNumbers(ON)
Procedure
Dividetheoverallreactioninto:
OxidationHalfReaction
ReductionHalfReaction
Multiplyoneorbothreactionsbysomeintegertomake
electronsgainedequaltoelectronslost
04/22/12
Balanceeachhalfreactionforatoms&charge
Recombinetogivenbalancedredoxequation
13
Electrochemistry
RedoxHalfReactionMethodExample
Cr2O 7 2- (aq) + I - (aq) Cr +3 (aq) + I 2 (s)
DividestepsintoHalfReactions
(
)
6 + 2 2Cr2 O7
I- I 2
04/22/12
Cr 3+ (Cr gains e - - reduction)
(Iodine loses e - - oxidation)
14
Electrochemistry
BalanceAtoms&ChargesforCr2O72/Cr3+
Cr2O 7 2- 2Cr 3+
Cr2O 7 2- 2Cr 3+ + 7H 2O
Add 7 Water molecules
to balance Oxygen
14H + + Cr2O 7 2- 2Cr 3+ + 7H 2O
Add 14 H+ ions on left to
balance 14 H on right
6e
-
+ 14H
+
+ Cr2O 7
2-
2Cr
3+
+ 7H 2O
(6 electrons gained this is the reduction reaction
Add 6 electrons (e-) on left
to balance reaction charges
2I - I 2
BalanceAtoms&ChargesforI/I2 No need to add H2O or H+
2I - I 2 + 2e-
Add 2 electrons (e-) on right
to balance reaction charges
(2 electrons lost this is the oxidation reaction
04/22/12
15
Electrochemistry
RedoxHalfReactionMethodExample(cont)
Multiplyeachhalfreaction,ifnecessary,byaninteger
tobalanceelectronslost/gained
2elostinoxidationreactionand6egainedin
reduction
Multiplyoxidationhalfreactionby3
3(2I - ) 3I + 3(2e- )
2
6I - 3I 2 + 6e Add2halfreactionstogether 2Cr 3+ + 7H O
6e - + 14H + + Cr2O 7 2-
2
6I 6I 04/22/12
+ 14H +
3I 2 + 6e -
+ Cr2O 7 2- 3I 2
+ 2Cr 3+ + 7H 2O
16
Electrochemistry
HalfReactionMethodinaBasicsolution
Sodium Permanganate & Sodium Oxalate
( Mn
+7
O4 2
)
-
NaMnO4
(aq) +
(
)
+3 2 2C2 O 4
Mn
Na2C2O4
+4
O 2 (s)
2
(
+4
+ C O3
)
2-
(aq)
Half-Reactions
MnO4- MnO 2
C2O4 2- CO 3 2-
MnO 4- MnO 2 + 2H 2O
2 H 2O + C2O4 2- 2CO 3 2-
4H + + MnO 4- MnO 2 + 2H 2O
2 H 2O + C2O 4 2- 2CO 3 2- + 4H +
3e- + 4H + + MnO 4- MnO 2 + 2H 2O
(reduction)
2 H 2O + C2O 4 2- 2CO 3 2- + 4H + + 2e (oxidation)
Multiply each reaction by appropriate integer
6e- + 8H + + 2MnO 4- 2MnO 2 + 4H 2O
04/22/12
6H 2O + 3 C2O 4 2- 6CO 3 2- + 12H + + 6e17
Electrochemistry
SodiumPermanganate&SodiumOxalate(cont)
Add reactions
6e- + 8H + + 2MnO 4- 2MnO 2 + 4H 2O
6H 2O + 3 C2O 4 2- 6CO 3 2- + 12H + + 6e2MnO 4- + 2H 2O + 3 C 2O 4 2- 2MnO 2
+ 6CO 3 2- + 4H +
Add OH- to neutralize H+ , balance H2O, and form basic solution
2MnO 4- + 2H 2O + 3 C 2O 4 2- + 4OH - 2MnO 2 + 6CO 3 2- + 4H + + 4OH 2MnO 4- + 2H 2O + 3 C2O 4 2- + 4OH - 2MnO 2 + 6CO 3 2- + 4H 2O
2MnO 4- + 3 C2O 4 2- + 4OH - 2MnO 2 + 6CO 3 2- + 2H 2O
04/22/12
18
Electrochemistry
ElectrochemicalCells
Voltaic(Galvanic)Cells
DifferenceinChemicalPotentialenergybetween
higherenergyreactantsandlowerenergyproducts
isconvertedtoelectricalenergytopowerelectrical
devices
04/22/12
Usespontaneousreaction( G<0)togenerate
electricalenergy
ThermodynamicallyThesystemdoesworkonthe
surroundings
19
Electrochemistry
ElectrochemicalCells
ElectrolyticCells
Electricalenergyfromanexternalpowersupply
convertslowerenergyreactantstohigherenergy
products
ThermodynamicallyThesurroundingsdoworkon
thesystem
04/22/12
Useselectricalenergytodrivenonspontaneous
reaction( G>0)
ExamplesElectroplatingandrecoveringmetals
fromores
20
Electrochemistry
04/22/12
21
Electrochemistry
ElectrochemicalCells
Cellnotationisusedtodescribethestructureofa
voltaic(galvanic)cell
FortheZn/Cucell,thecellnotationis:
Zn(s) 2+(aq)Cu2+(aq)
Zn
Cu(s)
=phaseboundary(solidZnvs.AqueousZn2+)
=saltbridge
Anodereaction(oxidation)isleftofthesaltbridge
Cathodereaction(reduction)isrightofthesaltbridge
Halfcellcomponentsusuallyappearinthesameorder
asinthehalfreactions(Zn(s)+2eZn2+).
04/22/12
Zincsolidloses2e(oxidized)toproducezinc(II)atthe
22
Electrochemistry
Voltaic(Galvanic)Cells
Zincmetal(Zn)insolutionofCu++ions
Cu 2+ (aq) + 2e- Cu(s) [reduction]
Zn(s) Zn 2+ (aq) + 2e- [oxidation]
Zn(s) + Cu 2+ (aq) Zn 2+ + Cu(s)
ConstructionofaVoltaicCell
04/22/12
Theoxidizingagent(Zn)andreducingagent
(Cu2+)inthesamebeakerwillnotgenerate
electricalenergy
Separatethehalfreactionsbyabarrierand
connectthemviaanexternalcircuit(wire)
Setupsaltbridgebetweenchamberstomaintain
23
Electrochemistry
OxidationHalfCell
ZincmetalinsolutionofZn2+electrolyte(ZnSO4)
Znisreactantinoxidationhalfreaction
AnodeCompartmentOxidationofZinc(AnOx)
Conductsreleasedelectrons(e)outofitshalfcell
ReductionHalfCell
CathodeCompartmentReductionofCopper
(RedCat)
CopperbarinsolutionofCu2+electrolyte(CuSO4)
04/22/12
Coppermetalisproductinreductionhalfcell
reaction
Conductselectronsintoitshalfcell
24
Electrochemistry
ZincCopperVoltaicCell
04/22/12
25
Electrochemistry
RelativeChargesontheAnode/Cathode
electrodes
Electrodechargesaredeterminedbythesource
oftheelectronsandthedirectionofelectronflow
Zincatomsareoxidized(lose2e)toformZn 2+
attheanode
Anodenegativecharge(erich)
Releasedelectronsflowtorighttowardcathode
tobeacceptedbyCu2+toformCu(s)
04/22/12
Cathodepositivecharge(edeficient)
26
Electrochemistry
urposeofSaltBridge
ElectronsfromoxidationofZnleaveneutralZnSO4
solutionproducingnetpositivecharge
IncomingelectronstoCuSO4solutionwould
producenetnegativechargeinsolutionascopper
ionsarereducedtocoppermetal
Saltbridgeprovidesliquidwireallowingionsto
flowthroughbothcompartmentscompletingcircuit
04/22/12
Resultingchargeimbalancewouldstopreaction
SaltbridgeconstructedofaninvertedUtube
containingasolutionofnonreacting
27
Electrochemistry
ActivevsInactiveElectrodes
ActiveElectrodes
ElectrodesinZn/Cu2+cellareactive
Zinc&Copperbarsarecomponentsofthe
cellreactions
MassofZnbardecreasesasZn2+ionsin
cellsolutionincrease
MassofCopperbarincreasesasCu2+ions
acceptelectrontoformmorecoppermetal
04/22/12
28
Electrochemistry
ActivevsInactiveElectrodes
InactiveElectrodes
InmanyRedoxreactions,oneortheother
reactant/productisnotcapableofserving
asanelectrode
InactiveelectrodesGraphiteorPlatinum
Canconductelectronsintoandoutof
halfcells
Cannottakepartinthehalfreactions
04/22/12
29
Electrochemistry
VoltaicCell
with
InactiveGraphite
Electrodes
04/22/12
30
PracticeProblem
A mercury battery, used for hearing aids and electric
watches, delivers a constant voltage (1.35 V) for long
periods. The half reactions are given below. Which
half reaction occurs at the Anode and which occurs at
the Cathode? What is the overall cell reaction?
HgO(s) + H2O(l) + 2e- Hg(l) + 2 OH-(aq)
Zn(s) + 2 OH-(aq) Zn(OH)2(s) + 2eAns: ReductionoccursatCathode(RedCat)
Hg2+gains2e(reduced)toformHg
OxidationoccursattheAnode(AnOx)
04/22/12
HgO(s) + Zn(s) + H2O Zn(OH)2 + Hg(l)
2+
Znloses2e (oxidized)toformZn
31
PracticeProblem
Write the cell notation for a voltaic cell with the
following cell reaction
Ni(s) + Pb 2+ (aq)
Ans:
Pb 2+ (aq) + 2e - Pb(s)
Ni(s) Ni 2+ + 2e-
Ni 2+ (aq) + Pb(s)
Reduction @ Cathode (Red Cat)
Oxidation @ Anode (An Ox))
Ni(s) I Ni 2+ (aq) P Pb 2+ (aq) I Pb(s)
Anodeisrepresentedon
leftsideofCellnotation
04/22/12
Cathodeisrepresentedon
rightsideofcellnotation
32
PracticeProblem
Write the cell reaction for the following voltaic cell
Pt|H2(g) | H+(aq) Br2(l) | Br-(aq)|Pt
Note: Platinum (Pt) serves as a reaction site at the anode,
but does not participate in the reaction
Br2 (l) + 2e - 2Br - (aq)
Ans:
gain 2 e -
Reduction @ Cathode (Red Cat)
H 2 (g) 2H + (aq) + 2e -
lose 2 e-
Oxidation @ Anode (An Ox)
PtIH 2 (g) + Br2 (l) 2H + (aq) + 2Br - (aq)IPt
04/22/12
33
Electrochemistry
CellPotential
Themovementofelectronsisanalogousto
thepumpingofwaterfromonepointto
another
Water
moves from a point of high pressure
to a point of lower pressure. Thus, a
pressure difference is required
The
work expended in moving the water
through a pipe depends on the volume of
water and the pressure difference
04/22/12
34
Electrochemistry
CellPotential
MovementofElectrons
An
electric charge moves from a point of
high electrical potential (high electrical
pressure) to one of lower electrical potential
The
work expended in moving the electrical
charge through a conductor depends on the
potential difference and the amount of
charge
Work(w) = Potential difference (E) Charge
w = Ecell Charge
04/22/12
35
Electrochemistry
CellPotential
Purposeofavoltaiccellistoconvertthefree
energyofaspontaneousreactionintothe
kineticenergyofelectronsmovingthroughan
externalcircuit(electricalenergy)
Electricalenergyisproportionaltothe
differenceintheelectricalpotentialbetween
thetwocellelectrodes
CellPotential
04/22/12
36
Electrochemistry
CellPotential
Negativecellpotentialisassociatedwitha
nonspontaneouscellreaction
Ecell < 0 for a nonspontaneous process
04/22/12
PositiveCellPotentialElectronsflowspontaneously
fromthenegativeelectrode(Anode)tothepositive
electrode(Cathode)
Ecell > 0 for a spontaneous process
Cellpotentialforacellreactionatequilibriumwouldbe
Ecell = 0 for an equilibrium process
0
AswithEntropy,thereisaclearrelationshipbetween
37
Electrochemistry
UnitsofCellPotential
TheSI(metric)unitofelectricalchargeisthe:
Coulomb(C)
TheSI(metric)unitofcurrentisthe:
Ampere(A)
1 coulomb
1 ampere =
second
1A = 1 C / s
TheSI(metric)unitofelectricalpotentialisthe:
Volt(V)
Bydefinition,theenergyreleasedbyapotentialdifference
ofonevoltmovingbetweentheanodeandcathodeofa
voltaiccellreleases1jouleofworkpercoulombofcharge
1 volt =
04/22/12
1J
C
1C =
1J
V
38
Electrochemistry
Thecharge(F)thatflowsthroughacellequalsthe
numberofmolesofelectrons(n)transferredtimesthe
chargeof1molofelectrons
charge
Charge = moles of e
mol eMoles e- = n
Charge
Mol e-
96, 485 C
= F = Farday Constant =
mol e-
Charge
= nF
96, 485 C
F=
mol e-
F = 96, 485
04/22/12
J
C - Coulomb , SI unit of charge
V
J
J
= 9.6104
V mol e V mol e 39
Electrochemistry
StandardCellPotential
EocellThepotentialmeasuredataspecific
temperature(298K)withnocurrentflowing
andallconcentrationsintheirStandard
States
1atmforgases
1Mforsolutions
Puresolidsforelectrodes
Zn(s) + Cu 2+ (aq; 1M) Zn 2+ (aq; 1M) + Cu(s)
04/22/12
Eocell = 1.10V
40
Electrochemistry
StandardElectrodeHalfCellPotentials
EohalfcellPotentialassociatedwithagivenhalfcell
reaction(electrodecompartment)whenall
componentsareinStandardStates
StandardElectrodePotentialforahalfcellreaction,
whetheranode(oxidation)orcathode(reduction)is
writtenasareduction
Cu 2+ (aq) + 2e- Cu(s) [reduction - cathode]
Ex.
Zn(s) Zn 2+ (aq) + 2e- [oxidation - anode]
Cu 2+ (aq) + 2e- Cu(s) EoCopper (Eocathode ) [reduction]
wouldbewritten:
Zn 2+ (aq) + 2e- Zn(s) Eo Zinc (Eoanode ) [reduction]
04/22/12
41
Electrochemistry
StandardElectrodeHalfCellPotentials
ElectronsflowspontaneouslyfromAnode(negative)to
Cathode(positive)
CathodemusthaveamorePositiveEohalfcellthanthe
Anode
ForapositiveEocell
Eocell = (Eocathode (reduction) - Eoanode (oxidation) ) > 0
Eocell = Eocopper - Eo Zinc
04/22/12
Thestandardcellpotentialisthedifferencebetweenthe
standardelectrodepotentialoftheCathode(reduction)
halfcellandthestandardelectrodepotentialofthe
Anode(oxidation)halfcell
Standardhalfcellpotentialsareintensiveproperties,
42
PracticeProblem
Write out the overall equation for the cell reaction and
determine the standard cell potential for the following
galvanic cell
[Eo (Ag+/Ag) = 0.80 V; Eo (Ni2+/Ni) = - 0.26 V]
Ni(s) I Ni 2+ (aq) P Ag + (aq) I Ag(s)
2Ag + (aq) + 2e- 2Ag(s)
Ni(s) Ni 2+ (aq) + 2e -
[reduction (cathode)]
[oxidation (anode)]
Ni(s) + 2Ag + (aq) Ni 2+ + 2Ag(s)
Writebothreactionsin"reduction"form
2Ag + (aq) + 2e- 2Ag(s)
Ni 2+ (aq) + 2e -
Ni(s)
[reduction (cathode)]
[reduction (anode)]
Eocell = EoSilver - Eo Nickel
Eocell = 0.80
04/22/12
- (-0.26) = 1.06 V
43
Electrochemistry
TheStandardHydrogenElectrode
Halfcellpotentialsarenotabsolutequantities
Thevaluesfoundintablesaredeterminedrelativetoa
Standard
TheStandardElectrodepotentialisdefinedaszero
(Eoreference)=0.00
Thestandardreferencehalfcellisastandard
Hydrogenelectrode
04/22/12
SpeciallypreparedPlatinumelectrodeimmersedina
1Maqueoussolutionofastrongacidthroughwhich
2H +gasat1atmisbubbled 2 (g); 1 atm)
(aq; 1 M) + 2e - H
Eo reference = 0.00V
H2
44
Electrochemistry
ReferenceHalfCellandUnknownHalfCell
TheStandardelectrodecanactaseitherthe
AnodeortheCathode
OxidationofH2(losee)atanodehalfcelland
reductionofunknownatcathodehalfcell
Eocell = Eocathode - Eoanode Eounknown - Eoreference
Eocell = Eounknown - 0.00 V = Eounknown
ReductionofH+(gaine)atcathodehalfcelland
Eocell = Eocathode - Eoanode Eoreference - Eounknown
oxidationofunknownatanodehalfcell
Eocell = 0.00 V - Eounknown
04/22/12
= - Eounknown
45
PracticeProblem
Determinethestandardelectrodepotential,Eozinc,usinga
voltaiccellconsistingoftheZn/Zn2+halfreactionandthe
H+/H2halfreaction.
Eocell=+0.76V
Ans:Zincisbeingoxidized(loses2e)producingelectronsat
2H +thenegativeanode;H+gainsereference = 0.00V
(aq) + 2e- H 2 (g)
Eo atpositivecathode
Zn(s) Zn 2+ (aq) + 2e -
Eo zinc = ? V
Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g)
Eocell = Eocathode - Eoanode
Eocell = 0.76V
Eoreference - Eo Zinc
Eo Zinc = Eoreference - Eocell = 0.00 V - 0.76 V = - 0.76 V
04/22/12
46
Electrochemistry
RelativeStrengthofOxidizingandReducingAgents
Cu 2+ (aq) + 2e- = Cu(s)
2H + (aq) + 2e - = H 2 (g)
Eo = 0.00 V
Zn 2+ (aq) + 2e - = Zn(s)
Eo = 0.34 V
Eo = - 0.76 V
ThemorepositivetheEovalue,themorereadilythe
reactionoccurs
StrengthofOxidizingAgentsCu2+>H+>Zn2+
StrengthofReducingAgentsZn>H2>Cu
OxidizingagentsdecreaseinstrengthasthevalueofEo
decreases,whilethestrengthofthereducingagents
increasesasthevalueofEodecreases
Cu2+isthestrongerOxidizingagent
04/22/12
47
Electrochemistry
TableofStandardElectrodePotentials
(TheemfSeries)
Allreactionsarewrittenasreductions
04/22/12
48
Electrochemistry
EMFSeries
AllValuesarerelativetothestandardhydrogen
(reference)electrode
Allreactionsarewrittenasreductions
F2isstrongestoxi AllValuesarerelativetothestandardhydrogen(reference)electrode
dizingagent(high,positiveEo)
Fluorineisveryelectronegative
Itiseasilyreduced(gaine)toformweakreducing
agent,F(reluctanttoloseelectrons)
Limetalisstrongestreducingagent(low,negativeEo)
04/22/12
haslowionizationpotential
easilyoxidized(losese)toformstrongoxidizing
agent,Li+(reluctanttogainelectrons)
49
Electrochemistry
WritingSpontaneousRedoxReactions
SimilaritiesAcid/BasevsRedox
AcidStrengthvsBaseStrengthusingKa&Kb
values
Redox(OxidizingagentvsReducingagent)using
Eovalues
TablesofStandardElectrodePotentials(Eo)
Thestrongeroxidizingagent(speciesonleftsideof
table)hasahalfreactionwithalarger(more
positiveorlessnegative)Eo
04/22/12
Thestrongerreducingagent(speciesontheright
sideoftable)hasahalfreactionwithasmaller(less
positiveormorenegative)value
50
Electrochemistry
WritingSpontaneousRedoxReactions
Aspontaneousreaction(Eocell>0)willoccurbetween
anoxidizingagentandanyreducingagentthatlies
belowitinthetable
Aspontaneousreactionwilloccurwhenthehalf
reactionhigherinthelistoccursatthecathode
(reduction)aswrittenandthehalfreactionlowerinlist
occursattheanode(oxidation)inreverse
04/22/12
Zn(s)(reducingagentonrightsideoftablewith
Eo=0.76V)willreactspontaneouslywithCu2+
(oxidizingagentofleftsideoftablewithEo=+0.34V)
whichliesaboveZninthetable
Recallallhalfreactionsarewrittenasreductionsin
theelectrodepotentialtable
51
PracticeProblem
Consultthetableofstandardelectrodepotentialsinyour
textbookinordertodecidewhichoneofthefollowing
reagentsiscapableofreducingI2(s)toI(aq,1M)
I2(s)+2e2I(aq)+0.53V
ReductionForm
Br2(l)+2e2Br(aq)+1.07V
a.Br (aq)
b.Ag(s)
c.Sn(s)
d.Zn2+(aq,1M)
e.Sn4+(aq,1M)
04/22/12
Ag+(aq)+eAg(s)+0.80V
Sn2+(aq)+2eSn(s)0.14V
Zn2+(aq)+2eZn(s)0.76V
Sn4+(aq)+2eSn2+(aq)+0.13V
(cont)
52
PracticeProblem(cont)
Ans:C
TheI2isbeingreducedgainingelectronsatCathode(RedCat)
TheothersolutionmustbecapableofbeingoxidizedatAnode(AnOx)
a.TheBrsolutionundergoesoxidation(losee)attheAnode Eo
Eo
= Eo
- Eo
+ 0.53 - (+1.70) = -1.17
cell
Thus
Neg
Eo
cell
cathode
anode
Bromide (Br - ) will not reduce I 2
Eocell = Eocathode - Eoanode + 0.53 -)toformAg+ o
b.TheSilver(Ag(s)undergoesoxidation(losese (+0.80) = - 0.27
Neg Eo Ag(s) will not reduce I 2
cell
Thus:
Eocell = Eocathode - Eoanode
+ 0.53 - (-0.14) = + 0.67o
c.TheSn(s)willundergooxidationtoformSn2+
Pos Eo Sn(s) will reduce I
cell
2
Zn 2+ will not reduce I 2
d.TheZn2+solutionisalreadyoxidized(willnotloseanymoree)
Sn 4+ will not reduce I 2
Thus:
04/22/12
4+
53
PracticeProblem
Writeaspontaneousredoxreactionforthefollowing
reactantswhentableDisnotavailable
Ag + (aq) + e-
= Ag(s)
Sn 2+ (aq) + 2e- = Sn(s)
Eo = + 0.80 V
Eo = - 0.14 V
Steps:
Bothreactionsarewrittenasreductions(egain)
Reverseoneofthereactionssuchthatelectrode
potentials(cathode[reduction]minusanode
[oxidation])givesaPositiveEocell
ThesignofEoneednotbereversedforthereversed
reaction(adjustmentmadeinEocell=EocathodeEoanode)
04/22/12
Addrearrangedhalfreactionstoobtainbalanced
overallreaction
(Cont)
54
PracticeProblem(cont)
Pairstrongeroxidizingandreducingagentsas
reactants
Ag+hashigherEo,0.80V(gainsemoreeasily)
Sn(s)isstrongerreducingagent(loseseeasily)
ReverseTin(Sn)halfcellreaction
AddhalfcellreactionsandcomputeEocell
2Ag + (aq) + 2 e - = 2Ag(s)
Sn(s) =
Sn 2+ (aq) + 2e-
Eo = + 0.80 V (Cathode Reduction)
Eo = - 0.14 V (Anode Oxidaton)
Sn(s) + 2Ag + (aq) Sn 2+ (aq) + 2Ag(s)
Eo cell = Eocathode reduction(Ag) - Eoanode oxidation(Sn)
Eocell = 0.80 - (-0.14) = + 0.94 V
04/22/12
(Spontaneous)
55
PracticeProblem
Combinethefollowing3halfcellreactionsinto3balanced
spontaneousreactions;CalculateEocellforeach;rankthe
relativestrengthsoftheoxidizing&reducingagents
(A)
NO 3- (aq) + 4H + (aq) + 3 e - NO(g) + 2H 2O(l)
N 2 (g) + 5H + + 4e- N 2 H 5 +
(B)
MnO 2 (s) + 4H + (aq) + 2e- Mn 2+ (aq) + 2H 2o (l)
(C)
Eo = 0.96 V
Eo = - 0.23 V
Eo = 1.23V
1. Combine&BalancehalfreactionsA&B
ReverseB(Eoin(A)ismorepositivethanEoin(B)
(Acathodereduction;Banodeoxidation)
(A)
(B)
4NO 3- (aq) + 16H + (aq) + 12 e - 4NO(g) + 8H 2O(l)
3N 2 H 5
+
3N 2 (g) 15H + + 12e -
Eo = 0.96 V
Eo = -0.23 V
3N 2 H 5 +(aq) + 4NO 3 - (aq) + H + (aq) 3N 2 (g) + 4NO(g) + 8H 2O(l)
Eocell (A + B) = 0.96 V - (-0.23 V) = 1.19 V
04/22/12
Cont
56
PracticeProblem(cont)
2. Combine&BalancehalfreactionsA&C
(C)
(A)
Reversehalfreaction(A)(Eo1.23>Eo0.96)
(Ccathodereduction;Aanodeoxidation)
3MnO 2 (s) + 12H + (aq) + 6e - 3Mn 2+ (aq) + 6H 2o (l)
Eo = 1.23 V
2NO(g) + 4H 2O(l) 2NO 3 - (aq) + 8H + (aq) + 6 e-
Eo = 0.96 V
3MnO 2 (s) + 4H + (aq) + 2NO(g) 3Mn 2+ (aq) + 2H 2O(l) + 2NO 3
Eo cell (A + C) = (1.23 V - 0.96 V) = 0.27 V
3. CombineandBalancehalfreactionsB&C
ReversehalfreactionB(1.23>0.23)
(Ccathodereduction;Banodeoxidation)
(C)
(B)
04/22/12
2MnO 2 (s) + 8H + (aq) + 4e - 2Mn 2+ (aq) + 4H 2o(l)
N 2 H 5 + N 2 (g) + 5H + + 4e -
Eo = 1.23 V
Eo = -0.23 V
N 2H 5 + (aq) + 2MnO 2 (s) + 3H + (aq) N 2 (g) + 2Mn 2+ (aq) + 4H 2O(l)
Eo cell (B + C) = 1.23 V - (-0.23) = 1.46 V
Cont
57
PracticeProblem(cont)
OverallRankingofOxidizing&ReducingAgents
(C) MnO 2 (s) + 4H + (aq) + 2e- Mn 2+ (aq) + 2H 2o (l) Eo = 1.23 V
(A) NO 3- (aq) + 4H + (aq) + 3 e- NO(g) + 2H 2O(l)
(B) N 2 (g)
+ 5H +
Eo = 0.96 V
+ 4e- N 2 H 5 +
Oxidizing Agents : MnO 2
> NO 3-
Reducing Agents : N 2 H 5 + > NO
Eo = -0.23 V
> N2
> Mn 2+
RankOxidizing&ReducingAgentsWithinEachEquation
04/22/12
Equation1(A+B):OxidizingAgentNO3>N2
ReducingAgentN2H5+>NO
Equation2(C+A):OxidizingAgentMnO2>NO3
ReducingAgentNO>Mn+2
Equation3(C+B):OxidizingAgentMnO2>N2
+
+2
58
Electrochemistry
RelativeReactivitiesofMetals
MetalsthatdisplaceH2fromacid
IftheEocellforthereactionofH+ismorepositivefor
metalAthanitisformetalB,metalAisastronger
reducingagentthanmetalBandamoreactive
metal +
2H (aq) + 2e H 2 (g)
Eo = 0.00 V (cathode - reduction)
Fe(s) Fe +2 (aq) + 2e Eo = - 0.44 V (anode - oxidation)
Fe(s) + 2H + (aq) H 2 (g) + Fe 2+ (aq)
Eo = 0.00 - (-0.44) = 0.44 V
cell
04/22/12
MetalsLithroughPb(includesFe)inthestandard
electrodepotentiallist(appendixD)liebelowH+
o
+
59
Electrochemistry
RelativeReactivitiesofMetals
MetalsthatcannotdisplaceH2fromacid
Metalsthatlieabovethestandardhydrogenreference
halfreactioncannotreduceH+fromacids
TheEocellforthereversedmetalhalfreactionis
negativeandthereactiondoesnotoccur
2H + (aq) + 2e - H (g)
Eo = 0.00 V (cathode - reduction)
+
2
2 Ag(s) 2Ag (aq) + 2e- Eo = 0.80 V (anode - oxidation)
2Ag(s) + 2H + (aq) H 2 (g) + 2Ag + (aq)
Eo = 0.00 V - 0.80 V = - 0.80 V (not spontaneous)
cell
04/22/12
Thehigherthemetalinthelist,themorenegativeisits
EocellforthereductionofH+toH2,thusitsreducing
strength(andreactivity)isless
60
Electrochemistry
RelativeReactivitiesofMetals
MetalsthatdisplaceH2fromwater
Metalsthatliebelowthehalfcellreactionpotential
forwatercandisplaceH2fromwater
InthereactionbelowtheEvalueforwaterisnotthe
standardstatevaluelistedinthetablebecausein
purewater,[OH]is1.0x107M,notthestandard
statevalueof1M(0.83V) + 2OH 2H O(l) + 2e - H (g)
E = -0.42V
2
2Na(s)
2
2Na + (aq) + 2e -
Eo = -2.71 V
2Na(s) + 2H 2O(l) 2Na + (aq) + H 2 + 2OH -
Eo = - 0.42 V - (-2.71 V) = + 2.29 V
cell
04/22/12
Ecell>0
SodiumdisplacesHydrogenfromwater
61
Electrochemistry
RelativeReactivitiesofMetals
Metalsthatcandisplaceothermetalsfromsolution
Anymetalthatislowerinthestandardelectrode
halfcelllistcanreducetheionofametalthatis
higherinthelist,thusdisplacingthatmetalfrom
solution
Fe 2+ (aq) + 2e - Fe(s)
Eo = -0.44V (cathod; reduction)
Zn(s) Zn 2+ (aq) + 2e -
E o = -0.76V (anode; oxidation)
Zn(s) + Fe 2+ Zn 2+ (aq) + Fe(s)
Eo = - 0.44 V - (-0.76 V) = 0.32 V
cell
04/22/12
Ecell>0Zincisthestrongerreducingagent
reducingFe2+toFeanddisplacingitfromsolution
62
Electrochemistry
FreeEnergyandElectricalWork
ElectricalWork
Potential(Ecell,involts)timesthecharge
Work(w) = Potential difference (E) Charge
w = Ecell charge
Ecellmeasuredwithnocurrentflowing
Noenergylosttoheating
Ecellvoltageismaximumpossibleforcell
Workismaximumpossible
Onlyreversibleprocesscandomaximumwork
Reversibleprocesswithnocurrentflow:
Forwardreactionifopposingpotentialissmaller
04/22/12
63
Electrochemistry
SpontaneousReaction G<0
SpontaneousReactionEcell>0
G - Ecell
Thevoltaiccelllosesenergyasitdoesworkonthe
surroundings;thustheworkterm(wmax)isnegative
w = - E charge
max
cell
G = w max
Recall Slide 70 from Chapter 20
w max = - Ecell charge = G
G = w max = - Ecell charge
G = w max = - Ecell nF
G o = - Eo n F
cell
Recall :
(Recall slide # 38)
(components in standard states)
n = moles
Coulombs
J
4
F = 9.65 10
= 9.65 10
mole
V mol e 4
04/22/12
64
Electrochemistry
G o = - RT lnK
(Slide 82 from Chapter 20)
G o = - Eo n F
cell
-Eo n F = - RT lnK
cell
J
298.15 K
RT
mol rxn K
=
lnK =
2.303(log K)
n mol e
J
nF
(9.65 104
)
mol rxn
V mol e
8.314
Eo
cell
E
04/22/12
o
cell
0.0592 V
=
log K
n
nEo
cell
log K =
0.0592 V
(at 298.15K)
65
Electrochemistry
SummaryRelationshipbetween
GoEocellK
04/22/12
66
Electrochemistry
EffectofConcentrationonCellPotential
Mostcellsdonotstartwithconcentrationsintheir
standardstates o
G = G + RTlnQ
(Slide 83 from Chapter 20)
Recall:
G = - nF E
cell
G o = - nF Eo
cell
(Standard State)
G = - nF Ecell = Go + RTlnQ = - n F Eo + RTlnQ
cell
o
-nF Ecell = - n F Ecell + RT lnQ
RTlnQ
Ecell = Eo (Nernst Equation)
cell
nF
J
8.314
298.15 K
o
mol rxn K
Ecell = Ecell 2.303 log Q
n mol eJ
(9.65 104
)
mol rxn
V mol e
Ecell = Eo cell
04/22/12
0.0592 V
log Q
n
(at 298.15 K)
67
Electrochemistry
ChangesinPotentialDuringCellOperation
Thepotentialofacellchangesastheconcentrationof
thecellcomponentschange
Zn(s) + Cu 2+ (aq) Zn 2+ (aq) + Cu(s)
[Zn 2+ ]
o
Ecell = 1.10 V
Q=
[Cu 2+ ]
0.0592 V
o
Ecell = Ecell log Q
(at 298.15 K)
n
Stage 1 When Q < 1
o
Ecell > Ecell
Stage 2 When Q = 1
o
Ecell = Ecell
Stage 3 When Q > 1
o
Ecell < Ecell
Stage 4 When Q = K Ecell = 0 (Equilibrium)
If Q / K < 1, Ecell is positive (+) (cell does work)
If Q / K = 1, Ecell = 0
04/22/12
(cell no longer does work)
If Q / K > 1, Ecell is negative (-) (reaction reverses until Q / K = 1)
68
Electrochemistry
If Q / K < 1, Ecell is positive (+)
(cell does work)
If Q / K = 1, Ecell = 0
(cell no longer does work)
If Q / K > 1, Ecell is negative (-)
(reaction reverses until Q / K = 1)
Ecell = E
04/22/12
o
cell
0.0592 V
log Q
n
69
Electrochemistry
ConcentrationCells
Inacellcomposedofthesamesubstance,but
differingconcentrationsinthetwohalfcells,thetwo
concentrationsmovetoequilibrateproducingelectrical
energy
Thecellreactionisthesumofidenticalhalfcell
reactionswritteninoppositedirections
TheStandardElectrodePotentials(Eocell)areboth
basedona1Msolution(standardconditions),sothey
canceleachother,i.e.,Eocell=0
04/22/12
Thenonstandardcellpotential,Ecell,dependsonthe
ratioofthetwoconcentrations[A]dil/[A]conc=Q
70
Electrochemistry
HowtheConcentrationCellWorks
ThedilutesolutionisintheAnodecompartment
(oxidation)andtheconcentratedsolutionisinthe
Cathodecompartment(Reduction)
2+
-
Cu(s) Cu (aq;0.10M) + 2e (Oxidation)
Cu 2+(aq;1.0M) + 2e -
Cu(s)
( Reduction )
IntheCathode(conc)halfcell,Cu2+ionsgain2
electronsandtheresultingCuatomsplateoutonthe
electrode,makingthesolutionlessconcentrated
04/22/12
IntheAnode(dilute)halfcell,Cuatomsgiveup2
electronsandtheresultingCu2+ionsenterthe
solutionandmakeitmoreconcentrated
InthistypeofVoltaiccell,thedilutioncontinuesuntil
equilibriumisattained,i.e.,
71
Electrochemistry
AlkalineBattery
MnO 2 (s) + 2H 2O(l) + 2e - Mn(OH) 2 (s) + 2O H - (aq)
Zn(s) + 2OH - (aq) ZnO(s) + H 2O(l) + 2e-
Zn(s) + MnO 2 (s) + H 2O(l) ZnO(s) + Mn(OH) 2 (s)
E cell = 1.5 V
NickelMetalHydride(NiMH)Battery
NiO(OH)(s) + H 2O(l) + e - Ni(OH) 2 (s) + OH - (aq)
MH(s) + OH - (aq) M(s) + H 2O(l) + e -
MH(s) + NIO(OH)(s) M(s) + Ni(OH) 2 (s)
Ecell = 1.4 V
LithiumIonBattery+
Li1-xMn 2O 4 (s)(s) + xLi + xe LiMn 2O 4 (s) + OH - (aq)
Li xC6
xLi +
+ xe - + C6 (s)
Li xC6 + Li1-xMn 2O 4 (s) LiMn 2O 4 (s) + C6(s)
04/22/12
[Cathode (Reduction]
[Anode (Oxidation]
Overall Cell Reaction
[Cathode (Reduction]
[Anode (Oxidation]
Overall Cell Reaction
[Cathode (Reduction]
[Anode (Oxidation]
Ecell = 3.7 V
Overall Cell Reaction
72
PracticeProblem
CalculateEcellforavoltaiccellcontainingthefollowinghalf
cells:
Zn(s) I Zn 2+ (aq) II H + I H 2 (g)
[Zn2+]=0.010M[H+]=2.5MPH2=0.30atm
ConstructEocell2e- H (g)
2H + (aq) +
2
Zn(s) Zn 2+
(Eo = 0.00 V) (red; cathode)
+ 2e -
(Eo = - 0.76 V) (ox; anode)
2H + (aq) + Zn(s) H 2 (g) + Zn 2+ (aq)
Eocell = 0.00 V - (-0.76 V) = 0.76 V
PH2 [Zn 2+ ] 0.30 0.010
Q=
=
= 4.8 10-4
[H + ]2
2.52
CalculateQ
RT
0.0592 V
Ecell = Eocell - 2.303
log Q = Eocell log Q
nF
n
04/22/12
0.0592 V
Ecell = 0.76V -
log(4.8 10 -4 ) = 0.76V - (-0.0982 V = 0.86 V
2
73
PracticeProblem
What is the equilibrium constant for the following reaction
[Eo(Ce4+/Ce3+) = 1.72 V
Eo(Cl2/Cl-) = 1.36 V
2 Cl-(aq) + 2 Ce4+(aq) Cl2(g) + 2 Ce3+(aq)
Cl - (aq) I Cl 2 (g) II Ce4+ (aq) I Ce 3+ (aq)
2Ce4+ (aq) + 2e2Cl - (aq) Cl 2
2Ce 3+ (aq)
(Eo = 1.36 V) (ox; anode)
+ 2e -
2Ce4+ (aq) + 2Cl -
(Eo = 1.72 V) (red; cathode)
Cl 2 (g) + 2Ce 3+ (aq)
Eocell = 1.72V - 1.36 V = 0.36 V
E
o
cell
0.0592 V
=
log K
n
nEo
2 0.36 V
cell
log K =
=
= 12.2
0.0592 V 0.0592 V
04/22/12
K = 1012.2 = 1.45 1012
(at 298.15 K)
74
PracticeProblem
Write out the overall equation for the cell reaction and
determine the standard cell potential for the following
galvanic cell.
[Eo (Ag+/Ag) = 0.80
Eo (Ni2+/Ni) = -0.26]
Ni(s)|Ni2+(aq)Ag+(aq)|Ag(s)
2Ag + (aq) + 2e -
2Ag + (aq)
Ni(aq) Ni 2+
(Eo = 0.80 V) (red; cathode)
+ 2e -
(Eo = - 0.26 V) (ox; anode)
2Ag + (aq) + Ni(aq) Ni 2+ (aq) + 2Ag + (aq)
Eocell = 0.80V - (- 0.26) = 1.06 V
04/22/12
75
PracticeProblem
What is the maximum work you can obtain from 15.0 g of
Ni in the galvanic cell shown in the previous problem when
the Ecell is 0.97 V?
[Eo (Ag+/Ag) = 0.80
Eo (Ni2+/Ni) = -0.26]
Ni(s)|Ni2+(aq) Ni 2+ (aq) + 2Ag + (aq)
Ag+(aq)|Ag(s)
2Ag (aq) + Ni(aq)
+
w max = - Ecell charge
w max
charge = nF
F=
96,485 C
J
= 96,485
mol eV mol e -
mol eJ
= -Ecell n F = -0.97 V 2
96,485
= - 1.87 10 5 J / mol
mol Ni
V mol e-
Mass
15.0g
mol(Ni) =
=
= 0.256 mol
Mol Wgt
58.69 g / mol
w max = 0.256 mol (-1.87 105 J / mol = -4.7810 4 J
w max = - 4.78 104 J = - 4.78 kJ
04/22/12
76
PracticeProblem
What is the cell voltage (Ecell) for the following galvanic cell?
Cd(s)|Cd2+(0.026 M)Ni2+(0.00420 M)|Ni(s)
Ni 2+ (aq) + 2e- Ni(s)
Cd(s) Cd 2+ + 2e-
Eo = - 0.25 V
Eo = - 0.40 V
(Reduction)
(Oxidation)
Cd(s) + Ni 2+ (aq) Cd 2+ + Ni(s)
Eo = - 0.25 V - (-0.40 V) = 0.15 V
cell
[Cd 2+ ]
0.026M
Q=
=
= 6.19
2+
0.00420M
[Ni ]
o
Ecell = Ecell -
Ecell = 0.15 V 04/22/12
0.0592 V
log Q
n
(at 298.15 K)
0.0592 V
log 6.19 = 0.15 V - (0.0296 0.79) = 0.13 V
2
77
PracticeProblem
ConstructaVoltaicCelltodeterminethepHofanUnknown
Solution
Compartment#1CathodeconsistingofStandard
HydrogenElectrodebasedonH 2/H+
halfcellreactionatstandard
conditions(H +1M;H21atm)
Compartment#2Anodeconsistingofsameapparatus
butdippingintoasolutionof
unknownH +(pH)
2H + (aq; 1o +
M)
2e - H 2 (g; 1 atm)
[cathode; reduction]
AlthoughE cell=0,theindividualhalfcellsdifferin[H+]
H 2 (g; 1atm) 2H + (aq; unknown) + 2e - [anode; oxidation]
andEcellisnot=0
2H + (aq; 1M) 2H + (aq; unknown)
04/22/12
Ecell = ?
Cont
78
PracticeProblem(Cont)
Ecell = E
o
cell
2
0.0592 V
0.0592 V
[H + ]unknown
o
log Q = Ecell log + 2
n
n
]H ]standard
Substituting 1M for [H + ]standard and 0 V for Eocell gives :
Ecell
2
0.0592 V
[H + ]unknown
0.0592 V
2
= 0V log
= log[H + ]unknown
2
12
2
0.0592 V
2 log [H + ]unknown = - 0.0592V log[H + ]unknown
2
Ecell
-log[H + ]unknown =
0.0592
Ecell = -
Since pH = - log[H + ]
pH =
Ecell
0.0592
Measure the cell potential with a Voltmeter and calculate pH
04/22/12
79
PracticeProblem
What is the pH of the test solution when Ecell = 0.612 V at
25 oC?
Pt|H2(g)(1 atm)|H+(test soln)AgCl(s),Ag(s)|Cl-(2.80 M)
o
AgCl(s) + e Ag(s) + Cl (aq)
E = 0.22V
[cathode; reduction]
H 2 (g; 1atm) 2H + (aq; unknown) + 2e -
Eo = 0.0V
[anode; oxidation]
H 2 (g;1atm) + 2AgCl(s) 2Ag(s) + 2Cl - (aq) + 2H + (aq, unknown)
[Cl - ]2 [H + ]2
Q=
H 2 (g;1atm)
Ecell = E
o
cell
Eocell = 0.22 V - 0.0 V = 0.22 V
0.0592 V
0.0592 V
[Cl - ]2 [H + ]2
o
log Q = Ecell log 2
n
2
H (g;1atm)
0.0592 V
(2.80)2 [H + ]2
0.612 V = 0.22 V log
= 0.22 V - 0.0296(log(7.84) + 2log[H + ])
2
1
-0.0296 2log[H + ] = 0.612 - 0.22 + 0.0296 0.894
-log[H + ] = pH = 7.07
04/22/12
80
SummaryEquations
charge
Charge = moles of e
mol eCharge = nF
F - Faraday Constant
-
F=
96, 485 C
mol e-
(C - coulomb, SI unit of charge)
Eocell = Eocathode (reduction) - Eoanode (oxidation)
2H + (aq) + 2e - H 2 (g)
Eo = 0.00 V (cathode - reduction)
Ag(s) Ag + (aq) + e - Eo = 0.80 V (anode - oxidation)
2Ag(s) + 2H + (aq) H 2 (g) + 2Ag + (aq)
Eo = 0.00 V - 0.80 V = - 0.80 V
cell
04/22/12
81
SummaryEquations
G = w max = - Ecell charge
G = w max = - Ecell nF
G o = - Eo n F
cell
(components in standard states)
G o = - RT lnK
-Eo n F = - RT lnK
cell
G = G o + RTlnQ
-nF Ecell = - n F Eo + RTlnQ
cell
o
Ecell = Ecell -
Ecell = Eo cell
04/22/12
RTlnQ
nF
(Nernst Equation)
0.0592 V
log Q
n
(at 298.15 K)
82