FLT•  Iwillbeableto:– Identifyredoxreactionsandjustifytheidentificationintermsofelectrontransfer– Designand/orinterprettheresultsofanexperimentinvolvingaredoxtitration– Makequalitativeorquantitativepredictionsaboutgalvanicorelectrolyticreactionsbasedonhalf-cellreactionsandpotentialsand/orFaraday’slaws– Analyzedataregardinggalvanicorelectrolyticcellstoidentifypropertiesoftheunderlyingredoxreactions

bycompletingElectrochemistryNotes

Electrochemistry

Review:Oxidation-ReductionReactions

Oxidation-Reduction•  Oxidation-Reduction(redox)reactionscanbeconsideredelectron-transferreactions

•  Overallreaction:2Mg(s)+O2(g)à2MgO(s)

2Mg(s)+O2(g)à2MgO(s)

EachOgains2e-tobecomeO2-

EachMgloses2e-tobecomeMg2+

•  Half-Reactions(showtheelectronsinvolved):

2Mgà2Mg2++4e- O2+4e-à2O2-

Oxidation-Reduction•  Oxidation–involvesthelossofelectrons•  Reduction–involvesthegainofelectrons•  LEOthelionsaysGER

Oxidation-Reduction•  Whenlookingatreactions,wecanassignoxidationstatestoseewhichsubstanceisoxidized,andwhichisreduced

CH4(g)+2O2(g)àCO2(g)+2H2O(g)

Loste-s(oxidized)

Gainede-s(reduced)

C:-4H:+1

O:0 C:+4O:-2

H:+1O:-2

Review:BalancingRedoxReactions

Balancing1.  Assignoxidationstatesandlabel2.  Separatetherxnintohalf-reactions3.  BalanceallatomsexceptHandO4.  BalanceOatomsbyaddingoneH2Omoleculeforeach

Oatomneeded5.  AcidicsolutionsàbalancehydrogenbyaddingH+.

BasicsolutionsàaddoneH2OmoleculeforeachHatomneeded,andthenaddthesamenumberofOH-atomsontheoppositeside.

6.  Balancethechargesbyaddinge-stothepositiveside.7.  Multiplerxnsbytheappropriatenumbertogetthe

electronstobalanceoutandcancel.8.  Addreactionstogether,cancelingouttermsthat

appearonoppositesides

Balancing•  Ex/Al(s)+Cu2+(aq)àAl3+(aq)+Cu(s)

•  First,assignoxidationstatesandlabel

Al(s)+Cu2+(aq)àAl3+(aq)+Cu(s) 0

+2

+3

0

Oxidized(loste-s)

Reduced(gainede-s)

Balancing•  Next,separateintohalfreactions:

Al(s)+Cu2+(aq)àAl3+(aq)+Cu(s)

Oxidationhalf-rxnAl(s) à Al3+(aq)

Reductionhalf-rxnCu2+(aq) à Cu(s)

Al(s) àAl3+(aq)+3e-

2e-+Cu2+(aq)àCu(s)

2Al(s)à2Al3+(aq)+6e-

6e-+3Cu2+(aq)à3Cu(s)

2Al(s)+3Cu2+(aq)à2Al3+(aq)+3Cu(s)

Trythis:Fe2+(aq)+MnO4

-(aq)àFe3+(aq)+Mn2+(aq)

•  Answer:

5Fe2+(aq)+8H+(aq)+MnO4

-(aq)à5Fe3+(aq)+Mn2+(aq)+4H2O(l)

Trythis:BASICsolutionI-(aq)+MnO4

-(aq)àI2(aq)+MnO2(s)

AlsoReview:Titrations

Titrations•  Recall:Titrationsareavolumetricanalysiswecanusetodeterminetheamountofacertainsubstance

•  Inatitration,astandardsolution(ofKNOWNconcentration)isaddedgraduallytoasolutionofunknownconcentrationuntilthechemicalreactioniscomplete.

Titrations•  Acid-BaseTitrationMethod:1.  Analytesolution(ofunknownM)isplacedina

flaskorbeaker2.  Asmallamountofindicatorisadded3.  Titrantisplacedinaburetteandslowlyaddedto

theanalyteandindicatormixture4.  Theprocessisstoppedwhentheindicatorcauses

achangeinthecolorofthesolution5.  Thechangeinvolumeisusedtodeterminethe

volumeoftheanalytesolution

Titrations•  Justasanacidcanbetitratedagainstabase,wecantitrateanoxidizingagentagainstareducingagent.

•  Wecancarefullyaddasolutioncontaininganoxidizingagenttoasolutioncontainingareducingagent

•  Theequivalencepointisreachedwhenthereducingagentiscompletelyoxidizedbytheoxidizingagent

•  Westilluseanindicatorthatchangescolor.•  Theindicatorhasacharacteristiccolorofthereducedformandoxidizedform.Atorneartheequivalencepoint,asharpchangeincolorwilloccur.

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Titrations•  Ex/A16.42mLvolumeof0.1327MKMnO4solutionisneededtooxidize25.00mLofaFeSO4solutioninanacidicmedium.WhatistheconcentrationoftheFeSO4solutionifthenetionicequationis5Fe2++MnO4

-+8H+àMn2++5Fe3++4H2O•  First,findmolesofKMnO4,andthenusethebalancedequationtofindthemolesofFeSO4

•  Second,dividebythevolumeoftheFeSO4solutiontogetmolarity

Mini-CW

MiniCW•  BalancingWS!•  AnyunfinishedequationswillbeyourHWinadditiontothebookproblems.

Electrochemistry

Referencebacktothisslideafterwefinishelectrochem:

•  LEOsaysGER•  Oxidizingagentscausereduction•  Reducingagentscauseoxidation•  Galvaniccellsarespontaneous(battery)•  Eletrolyticcellsrequireexternale-source(DC)•  ANOX,REDCAT(anode–oxidationoccurs;cathode–reductionoccurs)

•  ElectrochemistryandEquilibrium=Buddies•  Atequilibrium,voltaiccellshavezerovoltage•  VoltaiccellshaveverylargeKvalues(favorable)•  Q=1atstandardconditions•  IfQincreasestoapproachK,voltagedecreases•  IfQdecreasestoapproachK,voltageincreases

Electrochemistry

ElectricCurrentsandRedox•  Inelectrochemistry,westudytherelationshipsbetweenelectricalandchemicalprocesses

•  Thisincludesbatteries,electroplating,fuelcells,hydrogenproduction,biologicalprocesses

•  Redoxrxnsinvolvethetransferofelectrons•  Wecanlookatenergyreleasedbyaspontaneouschemicalrxnbeingconvertedintoelectricity(ex/battery)

•  Wecanalsolookatprocesseslikeelectrolysis,whereweuseelectricalenergytoforceanonspontaneousrxntooccur

ElectricCurrentsandRedox•  Termstoknow:•  ElectricCurrent=theflowofelectriccharge– Current(I)isoftenrepresentedbythenumberofelectronspassingthroughpersecond(Amperes)

•  Voltage=thepotentialenergyperelectron

ElectricCurrentsandRedox•  Let’slookatredoxrxns:•  Sinceredoxrxnsinvolvethetransferofe-s,wecanlookatthemintermsofgeneratingelectricalcurrents

Zn(s)+Cu2+(aq)àZn2+(aq)+Cu(s)•  What’shappeninginthisrxn?

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ElectricCurrentsandRedoxZn(s)+Cu2+(aq)àZn2+(aq)+Cu(s)

•  ZnmetalisplacedinasolutionofCu2+•  ThegreatertendencyofZntolosee-sresultsinZnbeingoxidizedandCu2+beingreduced

•  Let’sthink…•  Thee-sarebeingtransferreddirectlyfromZntoCu2+.ThecopperionacceptstheelectronsanddepositsontheZincassolidcopper.

ElectricCurrentsandRedoxZn(s)+Cu2+(aq)àZn2+(aq)+Cu(s)

•  Canweseparatethezincatomsandcopperions,andforcetheelectrontransfertooccurthroughawireconnectingthetwohalf-rxns?

•  Yes!•  Theflowofe-swouldconstituteanelectricalcurrentandcouldbeusedtodoelectricalwork

ElectrochemicalCells

ElectricCurrentFlowingIndirectlyb/tAtoms

TheVoltaicCell•  ElectrochemicalCell=devicethatgenerateselectricitythroughredoxrxns

① Voltaic(Galvanic)CellAnelectrochemicalcellthatproducesanelectricalcurrentfromaspontaneousrxn

② ElectrolyticCellAnelectrochemicalcellthatconsumesanelectricalcurrenttodriveanonspontaneousrxn

TheVoltaicCell

TheVoltaicCell•  Twohalf-cells•  SaltBridge– Connectstwohalf-cellsandneutralizeschargebuildup,allowingtherxntocontinue

•  Electrodes=conductivesurfaces– Cathode=wherereductiontakesplace– Anode=whereoxidationtakesplace

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Electrodes•  Anode– electrodewhereoxidationoccurs– anionsattractedtoit– connectedtopositiveendofbatteryinelectrolyticcell– losesweightinelectrolyticcell

•  Cathode– electrodewherereductionoccurs– cationsattractedtoit– connectedtonegativeendofbatteryinelectrolyticcell– gainsweightinelectrolyticcell• electrodewhereplatingtakesplaceinelectroplating

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CurrentandVoltage•  Thecurrent=thenumberofe-sthatflowthroughthesystempersecond–  unit=Ampere–  1Aofcurrent=1Coulombofchargeflowingbyeachsecond–  1A=6.242x1018electrons/second–  Electrodesurfaceareadictatesthenumberofelectronsthatcanflow

•  Potentialdifference=thedifferenceinpotentialenergybetweenthereactantsandproducts–  unit=Volt–  1Vofforce=1Jofenergy/Coulombofcharge–  thevoltageneededtodriveelectronsthroughtheexternalcircuit

–  amountofforcepushingtheelectronsthroughthewireiscalledtheelectromotiveforce,emf

CellPotential

CellPotential•  CellPotential(Ecell)=thedifferenceinpotentialenergybetweentheanodeandthecathodeinavoltaiccell

•  Dependsontherelativetendenciesofthereactantstoundergooxidationandreduction

•  Cellpotentialunderstandardconditionsiscalledthestandardemf,E°cell– 25°C,1atmforgases,1Mconcentrationofsolution– sumofthecellpotentialsforthehalf-reactions

•  Overall,thecellpotentialisameasureoftheoveralltendencyforaredoxrxntooccur.Thelowerthecellpotential,thelowerthetendency

CellNotation•  shorthanddescriptionofVoltaiccell•  electrode|electrolyte||electrolyte|electrode•  oxidationhalf-cellonleft,reductionhalf-cellontheright

•  single|=phasebarrier–  ifmultipleelectrolytesinsamephase,acommaisusedratherthan|

– oftenuseaninertelectrode•  doubleline||=saltbridge•  Ex/Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(aq)

Fe(s)|Fe2+(aq)||MnO4-(aq),Mn2+(aq),H+(aq)|Pt(s)

CellPotential•  Ahalf-rxnwithastrongtendencytooccurhasalarge+half-cellpotential

•  Whentwohalf-cellsareconnected,thee-sflowsothatthehalf-rxnwiththestrongertendencywilloccur

•  ThesevaluesarestandardizedbySHE,orthestandardhydrogenelectrode,whichisassignedapotentialdifferenceof0v

Half-CellPotentials•  SHEreductionpotentialisdefinedtobeexactly0v•  half-reactionswithastrongertendencytowardreductionthantheSHEhavea+valueforE°red

•  half-reactionswithastrongertendencytowardoxidationthantheSHEhavea-valueforE°red

•  E°cell=E°oxidation+E°reduction– E°oxidation=-E°reduction– +E°cellmeanstherxnisspontaneous– whenaddingE°valuesforthehalf-cells,donotmultiplythehalf-cellE°values,evenifyouneedtomultiplythehalf-rxnstobalancetheequation

Example•  CalculateE°cellforthereactionat25°C:Al(s)+NO3

-(aq)+4H+

(aq)àAl3+(aq)+NO(g)+2H2O(l)Steps:1.  Separatetherxnintotheoxidationand

reductionhalf-reactions2.  FindtheE°foreachhalf-reactionandsum

CalculateE°cellforthereactionat25°CAl(s)+NO3

−(aq)+4H+

(aq)→Al3+(aq)+NO(g)+2H2O(l)

Separate the reaction into the oxidation and reduction half-reactions

ox: Al(s) → Al3+(aq) + 3 e−

red: NO3

−(aq) + 4 H+

(aq) + 3 e− → NO(g) + 2 H2O(l)

find the E° for each half-reaction and sum to get E°cell

E°ox = −E°red = +1.66 v E°red = +0.96 v E°cell = (+1.66 v) + (+0.96 v) = +2.62 v

Example•  Willtherxnbelowbespontaneousunderstandardconditions?

Fe(s)+Mg2+(aq)àFe2+(aq)+Mg(s)

Predictifthefollowingreactionisspontaneousunderstandardconditions

Fe(s)+Mg2+(aq)→Fe2+(aq)+Mg(s)Separate the reaction into the oxidation and reduction half-reactions

ox: Fe(s) → Fe2+(aq) + 2 e−

red: Mg2+

(aq) + 2 e− → Mg(s)

look up the relative positions of the reduction half-reactions

red: Mg2+(aq) + 2 e− → Mg(s)

red: Fe2+(aq) + 2 e− → Fe(s)

since Mg2+ reduction is below Fe2+ reduction, the reaction is NOT spontaneous as written

the reaction is spontaneous in the reverse direction

Mg(s) + Fe2+(aq) → Mg2+

(aq) + Fe(s) ox: Mg(s) → Mg2+

(aq) + 2 e− red: Fe2+

(aq) + 2 e− → Fe(s)

sketch the cell and label the parts – oxidation occurs at the anode; electrons flow from anode to cathode

Practice-SketchandLabeltheVoltaicCellFe(s)⏐Fe2+(aq)⏐⏐ Pb2+(aq)⏐Pb(s),WritetheHalf-ReactionsandOverallReaction,andDeterminethe

CellPotentialunderStandardConditions.

ox:Fe(s)→Fe2+(aq)+2e−E°=+0.45V

red:Pb2+(aq)+2e−→Pb(s)E°=−0.13V

tot:Pb2+(aq)+Fe(s)→Fe2+(aq)+Pb(s)E°=+0.32V