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OCRChemistryAH432 RedoxandElectrodePotentials

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RedoxReactionsReducingandoxidizingagentsInanyredoxreactiontherewillbeanoxidizingagentandareducingagent.

• Theoxidizingagentcontainsthespeciesbeingreducedandthereforedecreasinginoxidationnumber(itisthewholemolecule/ioncontainingthatspecies)

• Theoxidizingagenttakeselectronsfromthespeciesbeingoxidized

• Thereducingagentcontainsthespeciesbeingoxidizedandthereforeincreasinginoxidationnumber(itisthewholemolecule/ioncontainingthatspecies)

• ThereducingagentdonateselectronstothespeciesbeingreducedConstructingequationsforredoxreactionsMethod1:combininghalf-equations(whichshownumbersofelectronstransferred)Example:Acidifiedmanganate(VII)ionsareapowerfuloxidizingagent.TheycanoxidizeFe2+ionstoFe3+ions.Atthesametimethemanganate(VII)ionsarereducedtoMn2+ions.Wecanrepresentwhathappenstotheironionswithahalf-equation.Thenumberofelectronstoincludecomesfromthechangeinoxidationnumberofthespeciesbeingoxidizedorreduced:Oxidationislossofelectrons: Fe2+(aq) ! Fe3+(aq)+e- oxidationnumberofFe: +2 +3Wecanconstructahalfequationforthereductionofacidifiedmanganate(VII)ionsReductionisgainofelectrons: MnO4

-(aq)+5e-! Mn2+(aq)

oxidationnumberofMn +7 +2Thisisnotbalanced,however.Thechargesaredifferentoneachside,andweneedtodealwiththefouroxygensintheMnO4

-ion.Thecluehereisthereagentisacidifiedmanaganate(VII).IfweincludeH+ionsonthelefthandside,wecanproducewaterwiththeoxygensasaproduct.FouroxygenswillrequireeightH+ions:Reductionisgainofelectrons: MnO4

-(aq)+8H++5e-! Mn2+(aq)+4H2O(l)

Thisisbalancedforbothchargesandatomspresent.Tocombinethetwohalfequationsintoanoverallequation,wenowneedtohavethesamenumberofelectronslostintheoxidationasaregainedinthereduction,soweneedtomultiplythefirsthalf-equationby5throughout: 5Fe2+(aq) ! 5Fe3+(aq)+5e- Thenwecansimplyaddthehalfequationstogether,andcancelthefiveelectronsoneachside:


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MnO4-(aq)+8H+

(aq)+5e-! Mn2+(aq) +4H2O(l)

5Fe2+(aq) ! 5Fe3+(aq)+5e-

So MnO4-(aq)+8H+

(aq)+5Fe2+(aq)! 5Fe3+(aq)+Mn2+(aq) +4H2O(l) Example:Wearealsofamiliarwithacidifiedpotassiumdichromateasapowerfuloxidizingagent.Thedichromateionsactinasimilarwaytothemanganate(VII)ions.Forexample,theycanoxidizeSn2+ionstoSn4+ions.Youshouldrecallthatdichromateionsareanintenseorangecolour,andthatwhentheyactasanoxidizingagent,theircolourchangestodarkgreen.ThisisbecausetheyhavebeenreducedtoCr3+ions.Againwecanrepresentwhathappenswithhalf-equations Sn2+(aq)!Sn4+(aq)+2e- Oxidation Cr2O7

2-(aq)+6e-!2Cr3+(aq) Reduction

+6 +3 (Numberofelectronstoincludefrom +6 +3 changeinoxidationnumbers) Cr2O7

2-(aq)+14H+

(aq)+6e-!2Cr3+(aq)+7H2O(l) BalancingbyaddingH+andH2OAndagainwecancombinetheseintoasingleredoxequationbygettingthesamenumberofelectronsineachhalfequation–i.e.multiplyingtheoxidationequationby3throughout,andaddingthetwohalfequationssotheelectronscanceloneachside. 3Sn2+(aq)+Cr2O7

2-(aq)+14H+

(aq)+6e-! 3Sn4+(aq)+6e-+2Cr3+(aq)+7H2O(l)

N.B.wecanalsobeaskedtoworkouttheoxidationorreductionhalfequationifwearegiventheoverallequationandtheotherhalf-equation:Usingthesameexampleasabove,wemightbetoldthattheoverallequationis: 3Sn2+(aq)+Cr2O7

2-(aq)+14H+

(aq)!3Sn4+(aq)+2Cr3+(aq)+7H2O(l)andthattheoxidationgoingonisSn2+!Sn4+.Wearerequiredtoworkoutthereductionhalf-equation:Firstlymultiplythegivenhalf-equationuptogetthesamestochiometryasintheoverallequation: 3Sn2++6e-!3Sn4+Nowpositionthisabovetheoverallequationandworkoutwhatismissingoneachside:Oxidation 3Sn2++6e-!3Sn4+Reduction Cr2O7

2-+14H+!6e-+2Cr3++7H2OOverall 3Sn2+(aq)+Cr2O7

2-(aq)+14H+

(aq)+6e-!3Sn4+(aq)+6e-+2Cr3+(aq)+7H2O(l)


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Method2:ConstructingredoxequationsusingoxidationnumbersWecanbalancearedoxequationusingoxidationnumbers,withouthavingtoconstructseparatehalf-equationsifwechoose.Thisworksbecausethetotalincreaseinoxidationnumbersinthebalancedequationmustbeequaltothetotaldecreaseinoxidationnumbers–aconsequenceofOILRIG!Example:Hydrogeniodide(HI)isoxidizedtoiodinebyconcentratedsulphuricacid.Thesulphuricacidisreducedtohydrogensulphide.Constructabalancedequationforthis.Step1:Writeanequationusingtheformulaeforreactantsandproductsyouknowabout HI+H2SO4!I2+H2S -don'tworryaboutbalancingormissingatomsyet!Step2:Usetheoxidationnumberstoseewhatisoxidizedandwhatisreduced HI + H2SO4 ! I2 + H2S +1-1 +1+6-2-2 0 +1-2 +1-2-2 0 +1Keeponlytheoxidationnumbersthathavechangedduringthereaction: HI + H2SO4 ! I2 + H2S -1 +6 0 -2 0Step3:BalancetheequationFORONLYTHOSEATOMSBEINGOXIDISEDORREDUCED.Don'tworryaboutanythingelseintheequationbeingbalancedyet. 2HI + H2SO4! I2 + H2S -1 +60-2 -1 0Step4:Calculatethetotalincreaseanddecreaseinoxidationnumbers: Reduction(S)is+6to-2=-8 Oxidation(I)is-1,-1to0=+2Step5:Multiplyupthespeciesintheoxidationand/orreductiontogetthesameincreaseanddecreaseinoxidationnumbers–inthiscaseweneedtomultiplythe2HI!I2by4 8HI + H2SO4! 4I2 + H2SStep6:Completethebalancing.Ifwearemissingoxygenatoms,addwatertothatside.Ifwearemissinghydrogenatoms,addH+tothatside(lookfor'acidicconditions'inthequestionasanotherclueastowhentoincludeH+asareactant).UnderalkalineconditionswemightaddOH-. 8HI + H2SO4! 4I2 + H2S + 4H2OFinally: Checkitisbalancedforallelements Checkitisbalancedforcharge–totalchargeoneachsidethesame.


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Practice:Hydrogensulphide,H2Sisoxidizedtosulphur,S,bynitricacid,HNO3,whichisitselfreducedtonitrogenmonoxide,NO.Step1 H2S + HNO3 ! S + NOStep2 -2+5 0+2Step3,4 nofurtherbalancingneeded oxidation(S)-2to0=+2 reduction(N)+5to+2=-3Step5 need3x(H2S!S)and2x(HNO3!NO)tobalance 3H2S + 2HNO3 ! 3S + 2NOStep6 missing4xOonRHSsoadd4xH2O(thennoneedtoaddH+) 3H2S + 2HNO3 ! 3S + 2NO +4H2OInacidicconditions,silvermetal,Ag,isoxidizedtosilver(I)ions,Ag+byNO3

-ions,whicharereducedtonitrogenmonoxide(NO).Step1 Ag + NO3

- + H+ ! Ag+ + NOStep2 0+5 +1 +2Step3,4 nofurtherbalancingneeded oxidation(Ag)0to+1=+1 reduction(N)+5to+2=-3Step5 Needx3(Ag+!Ag) 3Ag + NO3

- + H+ ! 3Ag+ + NOStep6 Need2moreOonRHS,soadd2H2O.Balancingthenneeds4H+onLHS. 3Ag + NO3

- + 4H+ ! 3Ag+ + NO +2H2O Nowprovetoyourselfthatthehalf-equationmethodusedformanganate(VII)ionsandiron(II)canbedonethiswayaswell,withthesameresult.Acidifiedmanaganate(VII)ionsarecapableofoxidizingiron(II)ionstoiron(III)ions.Themanganite(VII)isreducedtoMn2+ions:Step1 MnO4

- + Fe2+ +H+ ! Mn2+ + Fe3+Step2 +7+2 +2+3Step3,4 nofurtherbalancing oxidation+2to+3=+1 reduction+7to+2=-5Step5 Need5x(Fe2+!Fe3+) MnO4

- + 5Fe2+ +H+ ! Mn2+ + 5Fe3+Step6 NeedfourOonRHS,soadd4H2O.Balancingthenrequires8H+onLHS MnO4

- + 5Fe2+ +8H+ ! Mn2+ + 5Fe3+ +4H2O


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RedoxtitrationsInaredoxtitrationwearetitratingonesolutioninwhichsomethingisgoingtobeoxidized,againstanotherinwhichsomethingisgoingtobereduced.Wedon'tuseapHindicatorinaredoxtitration.Thechangesincolourofthesubstanceswhentheyareoxidizedandreducedtelluswhenwehavereachedtheendpoint.Ifweknowtheconcentrationofonesubstance,wecanworkouttheconcentrationoftheotherfromabalancedequationfortheredoxreaction.Therearetwoexamplesweshouldunderstandfully,althoughwemaymeetotherexamplesofredoxtitrationswhereinformationaboutthetitrationisprovidedinthequestion.Titrationofiron(II)ionsusingmanganate(VII)ionsunderacidicconditionsPurpose:TodeterminetheconcentrationofFe2+(aq)inasolution. e.g. determiningthemassofironinirontablets determiningthe%purityofironinanalloycontainingiron determiningtheconcentrationofiron(II)ionsincontaminatedwater determiningthemolarmassandformulaofaniron(II)saltHowtodoit:Ifthesampletobeinvestigatedisn'talreadyasolutionofiron(II)ions,thesamplemayneedtobecrushedanddissolvedinwaterandsulphuricacid.We'llneedtomeasurethemassofthesamplebeforedissolving.Theresultingsolutionwillbepalegreen,almostcolourlessiftheconcentrationofiron(II)isfairlylow.Wetitrateaknownvolumeofthesamplesolutioninaconicalflaskagainstpotassiummanagate(VII)solutionofknownconcentrationintheburette.Asthedeeppurplemanganate(VII)ionsreactwiththeiron(II)ionstheyaredecolourised.Thiscontinuesuntilalltheiron(II)ionshavereacted.Thenextdropofpotassiummanganate(VII)stayspurple,sotheendpointiswhenthefirsthintofpink/purplepersistsinthesolution.N.B.iftheiron(II)solutionisintheburette,theendpointwillbewhenthepurplecolourofthemanganate(VII)intheflaskdisappears.Thechemistry:Themanaganate(VII)ionsarereducedtomanagese(II),whiletheiron(II)isoxidizedtoiron(III)ions.Wehavealreadyconstructedanequationforthis: MnO4

-(aq)+8H+

(aq)+5e-! Mn2+(aq) +4H2O(l)

5Fe2+(aq) ! 5Fe3+(aq)+5e-

MnO4-(aq)+8H+

(aq)+5Fe2+(aq)! 5Fe3+(aq)+Mn2+(aq) +4H2O(l) Moleratio: 1:5OnemoleofMnO4

-ionsreactswithfivemolesofFe2+sowehavethemoleratioof1:5whenwedothetitrationcalculation.


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Thecalculations:i)25.0cm3ofasolutionofiron(II)sulphaterequired23.0cm3of0.0020moldm-3potassiummanganate(VII)tocompletelyoxidizeitinacidicsolution.Whatwastheconcentrationoftheiron(II)sulphatesolution?Step1:Calculatemolesofthepermanganate moles =conc.xvol =0.00200x(23/1000)=4.60x10-5molesStep2:Usethemoleratiofromthebalancedequationtogetmolesofiron(II)ions MnO4

-(aq)+8H+

(aq)+5Fe2+(aq)!5Fe3+(aq) +Mn2+(aq) +4H2O(l) 1:5 4.60x10-5 2.30x10-4molesofFe(II)Step3:Findconcentrationofiron(II)solution ConcofFe2+=moles/vol(indm3) =2.3x10-4/0.0025=0.00920moldm-3ii)Thesolutionwasmadebydissolvingatabletcontaininghydratediron(II)sulphateandsucrose(whichdoesnotreactwithpermanganateions)in250cm3ofwater.CalculatethemassofFeSO4inthetablet.MolesofFeSO4=concxvolindm3 =0.00920x0.25=0.0023mol(or10xthemolesin25cm3)MassofFeSO4=0.0023xMr =0.0023x(55.8+32.1+(16.0x4))=0.0023x151.9 =0.350giii)TheformulaforthehydratedironsulphateusedinthetabletisFeSO4.7H2O.Ifthemassofthetabletwas8.5g,calculatethe%bymassofhydratediron(II)sulphateinthetablet.MrofFeSO4.7H2O=151.9+(7x18)=277.9Massintablet=molesxMr=0.023x277.9 =0.6392g%bymass=(0.6392/8.5)x100 =7.5%Practical:Nuffieldexpt19.1cp.458TitrationofiodineinsolutionusingthiosuphateionsPurpose:Todeterminetheconcentrationofanoxidizingagentinasolutione.g.Cu2+(aq)ions. e.g. determiningthe%ofcopperinanalloyssuchasbrassorbronze determiningconcentrationofcopperionsinasolution determiningconcentrationofdichromateionsinasolution determiningtheconcentrationofchlorate(I)ionsClO-inbleach determiningtheconcentrationofhydrogenperoxidesolutionHowtodoit:Theanalysisisdoneintwostages.


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Inthefirststage,theoxidizingagenttobedeterminedisreactedwithexcessiodideions.Thisresultsiniodinebeingformedinthesolution,theamountofiodinebeingdirectlyrelatedtotheamountoftheoxidizingagentpresent.e.g. 2Cu2+

(aq)+4I-(aq)!2CuI(s)+I2(aq) sointhiscase2molesofCu2+ionsproduce1moleofiodine.Inthesecondstage,theconcentrationofiodineinthesolutionisdeterminedbytitratingitagainstsodiumthiosulphateofknownconcentration.Thered-browncolouroftheiodineinsolutionfadestoapalestrawcolour,andtocolourlesswhenalltheiodinehasbeenreducedtoiodideions.Thismakestheendpointdifficulttosee,soclosetotheendpointanamountofstarchisaddedasanindicator.Starchisblue-blackwheniodineispresent,butattheendpointthestarchbecomescolourless.Thechemistry: I2(aq)+2e-!2I-(aq) reductionAtthesametimethethiosulphateionsareoxidizedtotetrathionateions: 2S2O3

2-(aq)!S4O6

2-(aq)+2e- oxidation

Theoverallequationistherefore: I2(aq)+2S2O3

2-(aq)! 2I-(aq)+S4O6

2-(aq)

MoleRatio: 1:2Thismeansthat2molesofthiosulphateionsreactwith1moleofiodine.ThecalculationThecalculationisalsodoneintwostages.Inthefirststage,thetitreforthethioulphateanditsconcentrationareusedalongwiththemoleratioabovetoworkoutthemolesofiodineintheportionofsolutionthatwastitrated.Thisisthenscaleduptogetmolesofiodineinthewholesolution.Inthesecondstage,themolesofiodineinthewholesolutionareusedalongwiththemoleratioofiodinetooxidizingagenttoworkouttheconcentrationofoxidizingagentinthewholesolution.Thiscanthenbeusedfurtherifneededtoworkoutthemassoftheoxidizingagentgivenaformulaetc.e.g.30.00cm3ofbleachwasreactedcompletelywithiodideionsinacidicsolution.Theiodineformedwastitratedagainst0.2000moldm-3sodiumthiosulphate,requiring29.45cm3ofthethiosulphatesolutiontoreachtheendpoint.Thechlorate(I)ionsinthebleachreactwithiodideionsaccordingtotheequation: ClO-

(aq)+2I-(aq)+2H+(aq)!Cl-(aq)+I2(aq)+H2O(l)

Calculatetheconcentrationofthechlorateionsinthebleach.Step1:Calculatemolesofthethiosulphateionsintitration moles=concxvol(indm3)=0.2000x(29.45/1000)=5.890x10-3moles


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Step2:Usetheequationforthetitrationtogetmolesofiodineinthetitre I2(aq) + 2S2O3

2-(aq) !2I-(aq)+S4O6

2-(aq)

1 : 2 2.945x10-3 : 5.89x10-3 Thewholebleachsolutionwastitrated,somolesofI2releasedbytheClO-ionsinthebleach=2.945x10-3mol.Usingthemolesofiodinetofindmolesofchlorateions: ClO-

(aq)+2I-(aq)+2H+(aq)!Cl-(aq)+I2(aq)+H2O(l)

1: 1 2.945x10-3 2.945x10-3

Thereforethe30cm3ofbleachcontained2.945x10-3molesofchlorate(I)ions.Step3:Calculatetheconcentrationofthechlorate(I)ionsinthebleach concofchlorate(I)=moles/volindm3=2.945x10-3/0.030=0.0982moldm-3Practical:Nuffieldexpt.6.8c (potassiumiodate(V)with10%potassiumiodidebymass)ElectrodePotentialsElectrontransfersinredoxreactionsIfwedipazincstripintocopperIIsulphate,weobservealayerofcopperdepositedonthezinc.Wemightalsoobservethatthebluecolourofthecoppersulphatesolutionfadestowardscolourlessaszincsulphateisformedinsolution. CuSO4(aq)+Zn(s)!Cu(s)+ZnSO4(aq)

+200+2Theoxidationofzincisproducingelectrons,whilethereductionofcopperionsisconsumingthem.Ifwecouldseparatethesetworeactions,wecouldhavetheelectronsflowthoughanelectricalcircuittogetfromwheretheywereproducedtowheretheyareconsumed.Thisiswhathappensinthecellsofabattery.CellsandhalfcellsAnelectricalcellcomprisestwohalf-cells,oneinwhichoxidationistakingplacetosupplyelectrons,andoneinwhichreductionistakingplacetoconsumethem.Eachhalfcellthereforecontainsanelementwhichchangesoxidationstate.

electrical cell

half-cell in which oxidation releases electrons

half-cell in which reduction consumes electrons

flow of electrons

external circuit


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Thesimplesttypeofhalf-cellcomprisesametalplacedinanaqueoussolutionofitsions.Anequilibriumexistsatthesurfaceofthemetalbetweentheoxidationstateofthemetalandtheoxidationstateinitsion.Byconvention,theequilibriumisalwayswrittenwiththeelectronsontheleftside:e.g.forcopperandcopperIIions: Cu2+(aq)+2e-⇌Cu(s)andforzincandzincions: Zn2+(aq)+2e-⇌Zn(s)ThisconventionmeansthattheforwardreactionisalwaysREDUCTION(gainingelectrons)andthereversereactionisalwaysOXIDATION(losingelectrons).Wehavewrittentheseasreversiblereactionsbecauseeitheroxidationorreductioncouldtakeplaceinanyhalfcell,itisonlywhenitisconnectedtoanotherhalfcellthatonegoesintheoxidationdirectionandonegoesinthereductiondirection.Todecidewhichwayeachreactionwillgo,weneedtoknowtheelectrodepotentialofeachhalfcell.ElectrodePotentialThetendencyofahalfcelltogainelectronsismeasuredusingavaluecalledstandardelectrodepotential.Thelarger(morepositive)theelectrodepotential,thegreaterthetendencyofthathalf-celltogainelectrons.ItismeasuredinVolts,V.Wecomparetheelectrodepotentialsofthetwohalfcells:Inthehalf-cellwiththemostpositiveelectrodepotential,

• thereactionwillgointheforwarddirection• reductionwilltakeplace• electronswillbegained

Thiswillbethepositiveterminalofthecell.Inthehalf-cellwiththesmallest(mostnegativeorleastpositive)electrodepotential

• thereactionwillgointhereversedirection• oxidationwilltakeplace• electronswillbereleasedtoflowaroundthecircuit

Thiswillbethenegativeterminalofthecell.Theelectrodepotentialsforthetwohalfcellswehaveconsideredsofarare:

Cu2+(aq)+2e-⇌Cu(s) +0.34VZn2+(aq)+2e-⇌Zn(s) -0.76V

HerewecanseethattheCu2+/Cuhalf-cellhasthemorepositiveelectrodepotential: -sothiswillbethepositiveterminalofthecell -thereactionwillbeCu2+(aq)+2e-!Cu(s) -copperionswillbereduced -theelectronstodothiswillflowINfromtheexternalcircuit


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TheZn2+/Znhalf-cellhastheleastpositiveelectrodepotential -sothiswillbethenegativeterminalofthecell -thereactionwillbeZn2+(aq)+2e-"Zn(s) -zincwillbeoxidizedtozincions -electronsaregivenuptoflowOUTintotheexternalcircuitMetal/MetalionhalfcellsTheexperimentalsetupforthehalfcellsdescribedabovewouldrequiretwohalfcellsinwhicheachmetalisincontactwithasolutioncontainingionsofthesamemetal:Itisthedifferencebetweentheelectrodepotentialsforthetwohalf-cellswhichdrivesonehalf-celltoproduceelectronsandtheothertoconsumethem–wecanmeasurethiswithavoltmeter.Itisreferredtoasthepotentialdifference(sameisinphysics).Asthereactionsproceed,bothhalf-cellendsupwithanimbalancebetweenpositiveandnegativeions.TheZn2+/Znhalf-cellendsupwithextrapositiveions,andtheCu2+/Cuhalf-cellendsupwithfewerpositiveions.Thesaltbridgeallowsionstomovefromonehalf-celltotheotherinordertocorrectthisimbalance,otherwisetheelectronswouldstopflowingintheexternalcircuit.Thesaltbridgecouldbeassimpleasastripoffilterpapersoakedinanaqueoussolutionofanioniccompound,howevertheioniccompoundchosenmustn'treactwiththesolutionsineitherofthehalfcells.OftenaqueousKNO3orNH4NO3isused.Non-metal/non-metalionhalfcellsAhalf-cellsimplyhastosupportanequilibriumwhereanelementisintwodifferentoxidationstates.Wecouldthereforehaveanon-metalelementincontactwithasolutioncontainingionsofthatelement.e.g.ahydrogenhalf-cellcompriseshydrogengas,H2,incontactwithaqueoushydrogenions: 2H+

(aq)+2e-⇌H2(g)

Aninertplatinumelectrodeisusedtotransfertheelectronsintooroutofthehalf-cellandmaketheconnectiontotherestoftheelectricalcircuit.Theplatinumdoesnotreactatall.Theplatinum


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electrodeisimmersedinasolutioncontainingtheH+ions(i.e.anacid),andhydrogengasisbubbledovertheelectrodesurface.Non-metal/non-metalionhalf-cellsarenotlimitedtohydrogenandpositivelychargedions.Wecouldjustaswellhavee.g.chlorinegasincontactwithaqueouschlorideions: Cl2(g)+2e-⇌2Cl-(aq) Ametalion/metalionhalfcellThistypeofhalf-cellcontainsanaqueoussolutionwithionsofthesameelementintwodifferentoxidationstates,e.g.Iron(II)andiron(III). Fe3+(aq)+e-!Fe2+(aq) Asbefore,aninertplatinumelectrodeisneededtotransfertheelectronsintooroutofthehalf-cell.StandardelectrodepotentialsWeneedtomeasureandcompareelectrodepotentialsundercontrolledandreproducibleconditions.Wethereforechoosetocomparetheelectrodepotentialsofdifferenthalf-cellstotheelectrodepotentialofahydrogenhalf-cell(ourreferencestandard),whichwedefineashavinganelectrodepotentialofzerovoltswhenunderstandardconditionsof298K(25°C)andwithaH2gaspressureof101kPa(1atmosphere)andanH+solutionconcentrationof1moldm-3.Standardelectrodepotentialsofotherhalf-cellsalsoneedtobemeasuredwiththesamestandardconditions:298K,101kPagaspressure,and1moldm-3solutionconcentrations,orwithequalconcentrationsofeachioninametalion/metalionhalfcell.Tomeasurethestandardelectrodepotentialofotherhalf-cells,wecombinethemwithastandardhydrogenelectrodetoformacompletecell,thenmeasuretheoverallcellpotential,(thepotentialdifferenceacrossthecell)withahighresistancevoltmeter.Thisgivesadirectreadingofthestandardelectrodepotentialofthehalf-cellbeingmeasured.Definition:Thestandardelectrodepotentialofahalf-cell,Eө,istheelectrodepotentialofahalf-cellcomparedwithastandardhydrogenhalfcell,measuredat298Kwithsolutionconcentrationsof1moldm-3andagaspressureof101kPa.


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CellpotentialElectrodepotentialscanbeusedtopredictthecellpotential(potentialdifference)producedbyanycombinationofhalf-cells.Forexample,ifweusedourZn2+/ZnandCu2+/Cuhalfcellstogethertomakeacell:

Cu2+(aq)+2e-⇌Cu(s) Eө=+0.34V mostpositiveelectrodepotential=>reductionZn2+(aq)+2e-⇌Zn(s) Eө=-0.76V leastpositiveelectrodepotential=>oxidation

Wecanthencalculatethestandardcellpotential,Eөcell. Eөcell=Eө(reductionreaction)-Eө(oxidationreaction)So Eөcell=0.34-(-0.76)=+1.10V SHOWTHE+SIGN!N.B.Cellpotentialsshouldbepositive–ifitcomesoutnegative,you'vegotthewronghalf-cellsdoingoxidationandreduction! Workedexample:Asilver/coppercellismadebyconnectingtogetheranAg+/AghalfcellandaCu2+/Cuhalfcell.Thehalf-cellequilibriaandstandardelectrodepotentialsare: Ag+(aq)+e-⇌Ag(s) Eө=+0.80V reductionreaction Cu2+(aq)+2e-⇌Cu(s) Eө=+0.34V oxidationreaction Eөcell=0.80-0.34 =+0.46VPractice:Foreachofthehalf-cellcombinationsbelow,givethevalueofEөcellCa2+/Ca | 2H+/H2 Given Eөvalues: Ca2+/Ca -2.87V Ans:+2.87V 2H+/H2 0.00VMn2+/Mn | Cl2/2Cl- Mn2+/Mn -1.19VAns:+2.55V Cl2/2Cl- +1.36VCr3+/Cr2+ | Mn2+/Mn Cr3+/Cr2+ -0.41VAns:+0.78VCellreactionsWemayalsobeaskedtoderivetheoverallequationfortheredoxreactiontakingplaceinthetwohalf-cells.Thisisjustthesameascombiningareductionandanoxidationhalf-equationtogettheoverallequationforaredoxreaction,whichwe'vedonebefore.Step1:Writethetwohalfequations Ag+(aq)+e-⇌Ag(s) Cu2+(aq)+2e-⇌Cu(s) Step2:Changethe⇌to!toshowwhichwayeachisgoing,basedontheelectrodepotentals


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Ag+(aq)+e-!Ag(s) Cu2+(aq)+2e-"Cu(s) betterwrittenasCu(s)!Cu2+(aq)+2e-

Step3:Getthesamenumberofelectronsineachhalf-equation 2Ag+(aq)+2e-!2Ag(s) Cu(s)!Cu2+(aq)+2e-Step4:Addthetwoequationsandcanceltheelectronsoneachside Cu(s)+2Ag+(aq)+2e-!Cu2+(aq)+2e-+2Ag(s)

Practice:Foreachofthehalf-cellcombinationsbelow,givetheoverallcellreaction:Ca2+/Ca | 2H+/H2 Given Eөvalues: Ca2+/Ca -2.87V Ans:Ca(s)+2H+

(aq)!Ca2+(aq)+H2(g) 2H+/H2 0.00VMn2+/Mn | Cl2/2Cl- Mn2+/Mn -1.19VAns:Cl2(g)+Mn(s)!Mn2+(aq)+2Cl-(aq) Cl2/2Cl- +1.36VCr3+/Cr2+ | Mn2+/Mn Cr3+/Cr2+ -0.41VAns:2Cr3+(aq)+Mn(s)!2Cr2+(aq)+Mn2+(aq)Wemightalsobegiventheoverallequationandoneofthehalfequations,andaskedtoworkoutwhatisgoingonintheotherhalf-cell:e.g.Inacell,thefollowingoverallreactionoccurs: Cu(s)+2Ag+(aq)!Cu2+(aq)+2Ag(s) Thehalf-equationfortheAg/Ag+electrodeisAg++e-!Ag Workoutthehalf-equationforwhatistakingplaceintheotherhalf-cell:Step1:Calculatethechangesinoxidationnumbertakingplaceintheoverallequation,andmultiplyuptheknownhalfequationtothecorrespondingnumberofelectrons.OverallCu+ 2Ag+! Cu2++2Ag so2electronsarebeingtransferred 0 +1 2+ 0 +1 2Ag++2e-!2Ag multiplyinguptheknownhalf-equation


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Step2:Subtracttheknownhalf-equationfromtheoverallequationtorevealtheotherhalf-equation: Cu+ 2Ag+ ! Cu2++ 2Ag minus 2Ag++2e- ! 2Ag = Cu -2e- ! Cu2+ = Cu! Cu2++2e-becausebyconventionweuse+not–inchemicalequations. Practice:Theexampleabovewasrathertrivial,andtheycanbequitealothardertofigureout,butthemethodisasdescribedabovee.g.Inamethanolfuelcell,theoverallequationisCH3OH+1½O2!CO2+2H2OAtoneelectrode,thereactiontakingplaceis4H++4e-+O2!2H2OWorkoutwhathappensattheotherelectrode:Oxidationnumberchanges: CH3OH+1½O2!CO2+2H2O -2+1-2+10+4-2+1-2 +1 -2 +1 +1Nowmultiplyupknownhalf-equationtoget6electronstransferred: 6H++6e-+1½O2!3H2O Subtractthishalfequationfromtheoverallequation(don'tworryabout–signsyet) CH3OH +1½O2 !CO2 +2H2O 6H++6e- +1½O2 ! 3H2O CH3OH–6H+-6e-!CO2-H2OFinallyrearrangetogetridofsubtractions(andcheckifthehalf-equationwillsimplify) CH3OH+H2O!CO2+6H++6e-

-2 to +4 = 6 electrons

0 to -2 (3x) = 6 electrons


p. 15

StorageCells(batteries)Half-cellsarecombinedtoproducetwotypesofcommercialelectricalstoragecell,orbattery.i)non-rechargeablecellse.g.alkalinecellsTheseprovideelectricalenergyuntiltheoxidationandreductionhasbeencarriedouttosuchanextentthatthepotentialdifferencecannolongerbemaintained.Thecellisthen"flat",andmustbedisposedof.Alkalinecellconstruction:

Detailsnotrequired

Thezincanodesuppliedelectrons,asthezincisoxidizedbythehydroxideions: ZnO(s)+H2O(l)+2e-⇌Zn(s)+2OH-

(aq)

Atthegraphitecathode,electronsaresuppliedtothemanganeseIVoxidewhichisreduced.Hydroxideionsareproduced,replacingthosewhichwereconsumedattheanode 2MnO2(s) +H2O(l)+2e-⇌Mn2O3(s)+2OH-Eөcellis+1.5V(typicalofAAcells)ii)Rechargeablecellse.g.NiCd,NiMH,Lithiumionbatteries(inlaptops),Lithiumionpolymerbatteries(iPodetc.)Thereductionandoxidationinthetwohalf-cellswhichsuppliestheelectricalcurrentcanbereversedduringrecharging.Thisreversesthetworedoxequilibria,recreatingthereactantssothatthecellcanbeusedagainandagain.Cadmium,alongwithothersubstancescommonlyusedinbatteries,istoxic.Thereareenvironmentalproblemsassociatedwithitsdisposal,henceitsuseonlyinrechargeableandthereforereusablecells.Lithiumishighlyreactive,andreactsexothermically,sothereisariskoffireifthecontentsofalithium-basedbatteryareexposedbybreakingopenthebattery.

zinc anode (-)

graphite cathode (+)

MnO2 and NaOH paste


p. 16

InNiCdbatteriesthepositiveandnegativeelectrodes,isolatedfromeachotherbytheseparator,arerolledinaspiralshapeinsidethecase,andsurroundedwithpotassiumhydroxideastheelectrolyte.ThisallowsaNiCdcelltodeliveramuchhighermaximumcurrentthananequivalentsizealkalinecell.Theinternalresistanceforanequivalentsizedalkalinecellishigherwhichlimitsthemaximumcurrentthatcanbedelivered.Theoxidationhalf-cellreactionatthenegativeterminalinaNiCdbatteryduringdischargeis:

andthereductionhalf-cellreactionatthepositiveterminalinaNiCdbatteryis:

DetailsnotrequiredThisgivesacellreactionof:

Whenthestoragecellisrecharged,thesetworeactionsgointheoppositedirection.FuelCellsAfuelcellreleasestheenergyfromthereactionofafuelwithoxygeninelectricalform,producingavoltageratherthanheat.Themostcommonfuelcellsusehydrogen,althoughotherhydrogen-richfuelssuchasmethanolandnaturalgascanbeused.Purehydrogenfuelemitsonlywaterastheproductofthereaction,whilsthydrogen-richfuelsproduceonlysmallamountsofCO2andothergaseouspollutants,sovehiclesfittedwithfuelcellshavethepotentialtobecauselessdamagetotheenvironmentthanthoseburningpetrolordiesel.Howafuelcellworks:• Thereactants(hydrogenandoxygen)flowinandthe

products(water)flowout,whiletheelectrolyteremainsinthecell

• Fuelcellsdonothavetoberechargedandcanoperatevirtuallycontinuouslyaslongasthefuelandoxygencontinuetoflowintothecell

• Theelectrodesaremadeofacatalystmaterialsuchasatitaniumspongecoatedinplatinum

• Theelectrolyteisanacidoralkalinemembranethatallowsionstomovefromonecompartmentofthecelltotheother(likeasaltbridge)

Theequilibriainvolvedinanalkalinefuelcellare:2H2O(l)+2e-⇌H2(g)+2OH-

(aq) Eө=-0.83V (negativeelectrode) ½O2(g)+H2O(l)+2e-⇌2OH-

(aq) Eө=+0.40V (positiveelectrode)


p. 17

Theequilibriainvolvedinanacidicfuelcellare: 2H+

(aq)+2e-⇌H2(g) Eө=+0.00V (negativeelectrode) ½O2(g)+2H+

(aq)+2e-⇌H2O(l) Eө=+1.23V (positiveelectrode)Theoverallreactionandcellpotentialinbothcasesisthesame: H2(g)+½O2(g)→H2O(l) Eөcell=1.23VFeasibilityofreactionsThebenefitsofusingelectrodepotentialsarenotlimitedtocalculatingthecellpotentialofacell.Moregenerally,wecanuseelectrodepotentialstopredictwhetheraredoxreactionwilltakeplace.Forexample,"Willasolutioncontainingiron(III)ionsoxidisenickeltonickel(II)ions,andbereducedtoiron(II)?”Weneedtolookattheelectrodepotentialsfortheoxidationandreductionsinvolved.# Weusethesameconventionofwritingeachhalfequationwiththeelectronsontheleft# Wechangetheequilibriumsignstoarrowstoshowwhichwaythereactionwouldhavetogoin

eachhalfcellforthereactiontogoasdescribed:inthiscasewewantNi!Ni2+andFe3+!Fe2+ Ni2+(aq)+2e-⇌Ni(s) Eө=-0.25V Fe3+(aq)+e-⇌Fe2+(aq) Eө=+0.77V• Nowcheckthatonearrowisgoingineachdirection.Ifnot,youdon'thavearedoxreaction,you

havetwospeciesthatbothwanttobeoxidized,orbothreducedsonoreactionwilloccur.• WorkoutthevalueEөcellwouldhaveifthereactiondidhappen. Eөcell=Eөreductionreaction(forwardarrow)-Eөoxidationreaction(backwardarrow)• IfthecalculatedEөcellhasapositivevalue,thereactionisfeasible.Ifthevalueisnegative,the

reactionisnotfeasible.InourexampleEөcell=0.77–(-0.25)=+1.02VSowecansaythatiron(III)ionsCANoxidizenickeltonickel(II)ions,butweneedtobeabletoexplainwhatthecalculationhasshown:Generalexplanationifthereactionisfeasible:Theelectrodepotentialforthe<REDUCTIONHALFCELL>ismorepositivethantheelectrodepotentialforthe<OXIDATIONHALFCELL>so<SPECIESBEINGREDUCED>isasufficientlypowerfuloxidizingagenttooxidise<SPECIESBEINGOXIDISED>to<SPECIESAFTEROXIDATION>,releasingtheelectronsneededtoreducethe<SPECIESBEINGREDUCED>to<SPECIESAFTERREDUCTION>.


p. 18

most likely to gain electrons and be reduced i.e. oxidising agents

most likely to lose electrons and be oxidised i.e. reducing agents

"WhydoesFe3+reactwithNi?"TheelectrodepotentialfortheFe3+/Fe2+halfcellismorepositivethantheelectrodepotentialforNi2+/Ni,soFe3+isasufficientlypowerfuloxidizingagenttooxidiseNitoNi2+,releasingtheelectronsneededtoreducetheFe3+toFe2+.Nowlet'slookattheoppositeeffect.Anotherpossibleredoxreactionmightbe"Willnickel(II)ionsoxidizeiron(II)ionstoiron(III),themselvesbeingreducedtonickel?"Weapproachthisthesameway:WewantFe2+!Fe3+andNi2+!Ni

Fe3+(aq)+e-⇌Fe2+(aq) Eө=+0.77V Ni2+(aq)+2e-⇌Ni(s) Eө=-0.25V

Eөcell=-0.25–0.77=-1.02VCheck:Yeswe'vegotonearrowgoingeachway SinceEөcellisnegativewecansaythatnickel(II)ionsCAN'Toxidizeiron(II)toiron(III).Againweneedtobeabletoexplainwhatthecalculationhasshown:Generalexplanationifreactionnotfeasible:Theelectrodepotentialforthe<OXIDATIONHALFCELL>ismorepositivethantheelectrodepotentialfor<REDUCTIONHALFCELL>,sothe<SPECIESBEFOREREDUCTION>isnotapowerfulenoughoxidizingagenttooxidise<SPECIESBEFOREOXIDATION>to<SPECIESAFTEROXIDATION>andthenecessaryelectronswillnotbeavailabletoreduce<SPECIESBEFOREREDUCTION>to<SPECIESAFTERREDUCTION>.“Whydonickel(II)ionsnotoxidiseiron(II)toiron(III)?”TheelectrodepotentialfortheFe3+/Fe2+,halfcellismorepositivethantheelectrodepotentialfor

Ni2+/Ni,sotheNi2+isnotapowerfulenoughoxidizingagenttooxidizeFe2+toFe3+andthenecessaryelectronswillnotbeavailabletoreduceNi2+toNi.Solongasweknowthestandardelectrodepotentials,wecandealwithquitecomplicatedredoxsystems. Zn2+(aq)+2e- ⇌ Zn(s) Eө=-0.77V Fe2+(aq)+2e- ⇌ Fe(s) Eө=-0.44V 2H+

(aq)+2e- ⇌ H2(g) Eө=0.00V Cu2+(aq)+2e- ⇌ Cu(s) Eө=+0.34V Fe3+(aq)+e- ⇌ Fe2+(aq) Eө=+0.77V Ag+(aq)+e- ⇌ Ag(s) Eө=+0.80V NO3

-(aq)+2H+

(aq)+e-⇌ NO2(g)+H2O(l) Eө=+0.80V O2(g)+4H+

(aq)+4e- ⇌ 2H2O(l) Eө=+1.23V Cl2(g)+2e- ⇌ 2Cl-(aq) Eө=+1.36V


p. 19

Example:Doescopperreactwithacids?Ifweconsidertheusualreactionofametalwithanacid,i.e.metal+acid!metalsalt+hydrogenthenwewantH+!H2andCu!Cu2+ 2H+

(aq)+2e- ⇌ H2(g) Eө=0.00V Reduction Cu2+(aq)+2e- ⇌ Cu(s) Eө=+0.34V OxidationEөcellforthisreaction=0.00–0.34=-0.34Vsocopperwon'treactwiththeH+ions.Explanation:TheelectrodepotentialfortheCu2+/Cuhalfcellismorepositivethantheelectrodepotentialfor

2H+/H2halfcell,sotheH+ionisnotapowerfulenoughoxidizingagenttooxidiseCutoCu2+andthenecessaryelectronswillnotbeavailabletoreduceH+toH2.Sowhydoweobservethatcopperdoesreactwithnitricacid?IftheH+ionsaren’ttheoxidizingagent,itmustbesomethingtodowiththeotherionsinthesolution:thenitrateions: Cu2+(aq)+2e- ⇌ Cu(s) Eө=+0.34V Oxidation NO3

-(aq)+2H+

(aq)+e- ⇌ NO2(g)+H2O(l) Eө=+0.80V Reduction Eөcell=0.80–0.34=+0.46VSocopperwillreactwhenH+andNO3

-ionsarebothpresent.Explanation:TheelectrodepotentialfortheNO3

-/NO2halfcellismorepositivethantheelectrodepotentialfortheCu2+/CuhalfcellsoNO3

-inthepresenceofH+isasufficientlypowerfuloxidizingagenttooxidiseCutoCu2+,releasingtheelectronsneededtoreducetheNO3

-toNO2.ComparingOxidisingAgentsandReducingAgentsWemightbeaskedtocomparethespeciesinaredoxreactionandsaywhichisthebetteroxidizingagent,orreducingagent.e.g.Whichspeciesisthebestoxidizingagent? Cu2+(aq)+2e- ⇌ Cu(s) Eө=+0.34V NO3

-(aq)+2H+

(aq)+e- ⇌ NO2(g)+H2O(l) Eө=+0.80VAnoxidizingagentcausessomethingelsetobeoxidized,soitisitselfreduced.Welookforthehalfequationthatisgoinginthereductiondirection(theonewiththemostpositiveelectrodepotential,sothehalfcellinwhichthereactiongoesintheforwarddirection),andthebetteroxidizingagentwillbethereactant(s)forthatreaction:inthiscase,acidifiednitrateions.


p. 20

Ifwewereaskedforthebestreducingagent,welookforthesubstancethatisitselfoxidized,inthisexampletheCu,becauseinthishalf-equationtheelectrodepotentialisleastpositive,sothereactionisgoinginthereverse(oxidation)direction.OverallequationsOncewehavedeterminedthataredoxreactionwillwork,andweknowinwhichdirectionthereactionineachhalf-cellisgoingwecanmultiplyuptobalancetheelectrons,andcombinethehalfequationstogetanoverallequationfortheredoxreactionEXACTLYaswehavedonebeforewithhalfequations.e.g.whencopperreactswithconcentratednitricacid: Cu2+(aq)+2e- ⇌ Cu(s) NO3

-(aq)+2H+

(aq)+e- ⇌ NO2(g)+H2O(l) Step1: Cu(s) ! Cu2+(aq)+2e- turningroundequation NO3

-(aq)+2H+

(aq)+e- ! NO2(g)+H2O(l)

Step2: Cu(s) ! Cu2+(aq)+2e- 2NO3

-(aq)+4H+

(aq)+2e- ! 2NO2(g)+2H2O(l) multiplyingupelectronsStep3: Cu(s)+2NO3

-(aq)+4H+

(aq) ! 2NO2(g)+2H2O(l)+Cu2+(aq)

Limitationsofpredictionsusingstandardelectrodepotentials1)Non-standardconditionsalterthevaluesoftheelectrodepotentials,soiftheredoxreactionistohappenunderanyconditionsotherthan298K,101kPaandconcentrationsof1moldm-3(orequalconcentrationsinametalion/metalionhalfcell)thentheelectrodepotentialswon'tbecorrect.e.g. Cu2+(aq)+2e- ⇌ Cu(s) Eө=+0.34VIfweincreasetheconcentrationofCu2+ions,LeChateliertellsusthatthepositionofequilibriummovesintheforwarddirectiontoopposethechangeanduseuptheCu2+ions.Theelectrodepotentialmeasuresthetendencyofthereactiontogointheforwarddirection(i.e.togainelectronsbyreduction),sotheelectrodepotentialincreasescomparedtothestandardEөvalue.2)Standardelectrodepotentialsapplytoaqueousequilibria.Manyredoxreactionsdon'thappenundertheseconditions,sotheelectrodepotentialscan'treflectthoseconditions.e.g. 2Mg(s)+O2(g)!2MgO(s)3)Eveniftheelectrodepotentialsindicatethataredoxreactionisfeasible,itsaysnothingabouttheRATEofthereaction,whichmaybesounobservablyslowduetoaveryhighactivationenergy.Asageneralrule,thelargerthedifferencebetweentheelectrodepotentialsofthetworedoxequilibria,themorelikelyitisthatareactionwilltakeplaceinpractice.


p. 21

Consideragainthereactionofcopperwithnitricacid,usingtheserules. Cu2+(aq)+2e- " Cu(s) Eө=+0.34V NO3

-(aq)+2H+

(aq)+e- ! NO2(g)+H2O(l) Eө=+0.80VStandardconditionsmeanstheconcentrationofnitrateionsmustbe1moldm-3.Soforcopperreactingwith1Mnitricacid,thedifferenceinelectrodepotentialsis0.80–0.34=+0.46V.Thisisafairlysmalldifference,sothereactionmighttakeplace.Whatifweuseconcentratedacid?Ifweincrease[H+]thenthepositionofequilibriumfortheNO3

-/NO2redoxequilibriumshiftstotherighttouseuptheH+.Thismeanstheelectrodepotentialofthereductionhalf-cellgetsmorepositive.RememberingthatEөcell=Eөreduction-Eөoxidation,ifEөreductionincreases,thenEөcellincreasesandthereactionismorelikelytobefeasible,andwecananticipateitismorelikelytakeplaceundertheseconditions(i.e.gofaster).WouldthereactiongofasterifweaddedCu2+ionstothesolution?Ifweincrease[Cu2+]thenthepositionofequilibriumfortheCu2+/Curedoxequilibriumshiftstotheright(forwarddirection)touseuptheCu2+,inaccordancewithLeChatelier’sprinciple.Thismeanstheelectrodepotentialoftheoxidationhalf-cellwillhavebecomemorepositive.SinceEөcell=Eөreduction-Eөoxidation,andincreaseinEөoxidationwillreduceEөcellandthereactionislesslikelytohappen.

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