Electrochemistry

ElectrochemicalCells

TheVoltaicCell•  ElectrochemicalCell=devicethatgenerateselectricitythroughredoxrxns

① Voltaic(Galvanic)CellAnelectrochemicalcellthatproducesanelectricalcurrentfromaspontaneousrxn

② ElectrolyticCellAnelectrochemicalcellthatconsumesanelectricalcurrenttodriveanonspontaneousrxn

10

CellNotation•  Ex/Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(aq)

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

PredictingtheSpontaneousDirection

PredictingtheSpontaneousDirection² UseE°celltodeterminewhichwayaredoxrxnwillproceed

² Thehalf-rxnwiththemore(+)electrodepotentialattractse-smorestronglyandundergoesreduction•  Goodoxidizingagents

² Thehalf-rxnwiththemore(-)electrodepotentialrepelse-smorestronglyandundergoesoxidation•  Goodreducingagents

² Anyreductionreactioninthetableisspontaneouswhenpairedwiththereverseofanyoftherxnslisted

PredictingtheSpontaneousDirection•  WithoutcalculatingE°cell,predictwhethertherxnsbelowarespontaneousornot:

a.  Fe(s)+Mg2+(aq)àFe2+(aq)+Mg(s)b.  Fe(s)+Pb2+(aq)àFe2+(aq)+Pb(s)•  AsolutioncontainsbothNaIandNaBr.WhichoxidizingagentcouldyouaddtothesolutiontoselectivelyoxidizeI-(aq)butnotBr-(aq)

a.Cl2 b.H2O2 c.CuCl2 d.HNO3

PredictingtheSpontaneousDirection•  Dometalsdissolveinacids?•  Yes,mostacidsdissolvemetalsbythereductionofH+ionstohydrogengasandthecorrespondingoxidationofthemetaltoitsion.

Zn(s)+2H+(aq)àZn2+(aq)+H2(g)•  Canwepredictwhetherametalwilldissolveinacid?

•  What’stheoxidation½rxnandthereduction½rxn?

•  Metalswhosereductionhalf-rxnsarelistedbelowthereductionofH+toH2dissolveinacids,whilemetalslistedaboveitdonot.

17

PredictingtheSpontaneousDirection• Whichmetal(s)dissolvesinHNO3butnotinHCl?•  Hint:thecombinationofhalf-cellpotentialsshouldbepositive

a.  Nib.  Auc.  Cu•  Solution:•  Cu

CellPotential,FreeEnergy,andK

E°cell,ΔG°andK•  SincewehaverelatedE°celltowhetherornotarxnisspontaneousornot,canwerelateE°celltoΔG°?

•  WhatabouttoK?

E°cell,ΔG°andK•  Foraspontaneousreaction(onethatproceedsintheforwarddirectionwiththechemicalsintheirstandardstates)– ΔG°<0(negative)– E°>0(positive)– K>1

E°cell,ΔG°andK•  Foranonspontaneousreaction(onethatproceedsinthereversedirectionwiththechemicalsintheirstandardstates)– ΔG°>0(positive)– E°<0(negative)– K<1

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

•  Voltage=thepotentialenergyperelectron

E°cell,ΔG°andK•  Areweabletoconvertchargetonumbersofelectrons?

•  Why,yes!•  Faraday’sconstant(F)representsthechargeincoulombsof1molofelectrons:

F=96,485coulombs/mole-s(onyourreferencesheet)

E°cell,ΔG°andK•  Mathematically,wecanderivearelationshipbetweenΔG°andE°cell:

•  ΔG°=−RTlnK•  ΔG°=−nFE°cell– nisthenumberofelectrons– F=Faraday’sConstant=96,485C/mole−

26

E°cell,ΔG°andK•  Ex/UsethetabulatedelectrodepotentialvaluestocalculateΔG°forthereaction:

I2(s)+2Br-(aq)à2I-(aq)+Br2(l).Istherxnspontaneous?Steps:① Labelandseparatetherxninto½reactions② FindE°cellusingyourtableofvalues③ Thevalueofncorrespondstothenumberofe-s

canceledinthehalf-rxns.④ PlugintoΔG°=−nFE°cell.Note:1V=J/C

Example - Calculate ΔG° for the reaction I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq)

since ΔG° is +, the reaction is not spontaneous in the forward direction under standard conditions

Answer:

Solve:

Concept Plan:

Relationships:

I2(s) + 2 Br−(aq) → Br2(l) + 2 I−(aq) ΔG°, (J)

Given: Find:

E°ox, E°red E°cell ΔG° !!!

redoxcell EEE += !! cellFEG n=Δ

ox: 2 Br−(aq) → Br2(l) + 2 e− E° = −1.09 v

red: I2(l) + 2 e− → 2 I−(aq) E° = +0.54 v

tot: I2(l) + 2Br−(aq) → 2I−(aq) + Br2(l) E° = −0.55 v

!!

cellFEG n=Δ( )( )( )

J 101.1G

55.0485,96 mol 2G5

CJ

mol

C

×+=Δ

−=Δ −−

!

!

ee

E°cell,ΔG°andK•  Ex/UsethetabulatedelectrodepotentialvaluestocalculateΔG°forthereaction:

2Na(s)+2H2O(l)àH2(g)+2OH-(aq)+2Na+(aq).Istherxnspontaneous?Steps:① Labelandseparatetherxninto½reactions② FindE°cellusingyourtableofvalues③ Thevalueofncorrespondstothenumberofe-s

canceledinthehalf-rxns.④ PlugintoΔG°=−nFE°cell.Note:1V=J/C

Note:•  Atthispoint,wewouldnormallylookatnonstandardconditionsandthustheNernstEquation

•  IamskippingallnotesontheNernstequation,sincetheyremovedallAPstandardsrelatedtoitwiththenewtest.

•  However,IDOrecommendreviewingitbeforetakingcollegechemistry.

•  Iamalsoskippingoverbatteries.Theyaregoodtoreviewasexampleproblems,butfocusonthehalf-rxnsmorethanthedefinitions

Electrolysis

Electrolysis•  Recall:anelectrolyticcellrequireselectricalenergytodriveanonspontaneousreaction

•  Forexample,thereactionofhydrogenwithoxygentoformwaterisspontaneousandcanbeusedtoproduceanelectricalcurrentinafuelcell

•  Conversely,bysupplyinganelectricalcurrent,wecancausethereversereactiontooccur,separatingwaterintohydrogenandoxygen

Electrolysis•  Electrolysisliterallymeanstosplitwithelectricity

•  Electrolyticcells– Nonspontaneous–therevrserxncanoccurifwesupplyavoltagegreaterthanE°cellwecalculateforthespontaneousdirection– Usedtoseparateoresorplateoutmetals– WestillfollowANOX&REDCAT

Electrolyticvs.VoltaicCells•  Spontaneousvs.Nonspontaneous•  Avoltaiccellisseparatedintotwohalf-cellstogenerateelectricity;anelectrolyticcelloftenoccursinasinglecontainer

•  Avoltaiccellisabattery,whileanelectrolyticcellneedsabattery

•  ANOX,REDCAT•  However...•  Forelectrolyticcells:EPA–electrolyticpositiveanode

ElectrochemicalCells•  inallelectrochemicalcells,oxidationoccursattheanode,reductionoccursatthecathode

•  involtaiccells,–  anodeisthesourceofelectronsandhasa(−)charge–  cathodedrawselectronsandhasa(+)charge

•  inelectrolyticcells–  electronsaredrawnofftheanode,soitmusthaveaplacetoreleasetheelectrons,the+terminalofthebattery

–  electronsareforcedtowardtheanode,soitmusthaveasourceofelectrons,the−terminalofthebattery

36

electroplatingInelectroplating,theworkpieceisthecathode.

Cationsarereducedatcathodeandplatetothesurfaceoftheworkpiece.Theanodeismadeoftheplatemetal.Theanodeoxidizesandreplacesthemetalcationsinthesolution

Electrolysis

•  Electrolysisistheprocessofusingelectricitytobreakacompoundapart

•  electrolysisisdoneinanelectrolyticcell

•  electrolyticcellscanbeusedtoseparateelementsfromtheircompounds–  generateH2fromwaterforfuelcells–  recovermetalsfromtheirores

ElectrolysisofWater

ElectrolysisProducts

ElectrolysisofPureCompounds•  Puresaltsmustbeinmolten(liquid)state•  electrodesnormallygraphite•  cationsarereducedatthecathodetometalelement

•  anionsoxidizedatanodetononmetalelement•  Ex/

Oxidation:2Cl-(l)àCl2(g)+2e- Reduction:2Na+(l)+2e-à2Na(s) 2Na+(l)+2Cl-(l)àCl2(g)+2Na(s)

ElectrolysisofNaCl(l)

MixturesofIons•  whenmorethanonecationispresent,thecationthatiseasiesttoreducewillbereducedfirstatthecathode–  leastnegativeormostpositiveE°red

•  whenmorethanoneanionispresent,theanionthatiseasiesttooxidizewillbeoxidizedfirstattheanode–  leastnegativeormostpositiveE°ox

AqueousSolutions•  Aqueoussolutionscancomplicatethingsbecauseofthepresenceofwater

•  Thereisthepossibilityoftheelectrolysisofwateritself

SummingUp•  Ifthereisnowaterpresentandyouhaveapuremoltenioniccompound,thecationwillbereduced,andtheanionwillbeoxidized

•  Ifwaterispresent&youhaveanaqueoussolutionoftheioniccompound,thenwaterwillbeoxidizedinstead

•  Nogroup1AorIIAmetalwillbereducedinanaqueoussolutionàwaterwillbeoxidizedinstead

StoichiometryofElectrolysis

StoichiometryofElectrolysis•  Recall:Inanelectrolyticcell,electricalcurrentisusedtodrivearxn

•  Wecanlookate-sasareactantandusestoichiometricrelationships

•  Fore-s,wemeasurequantityascharge•  Forexample,givenCu2+(aq)+2e-àCu(s)•  Wecansee2mole-:1molCu(s)•  Wecanalsorelatemolofe-tochargeusingFaraday’sconstant:

F=96,485C/mole-

Faraday’sLaw•  Theamountofmetaldepositedduringelectrolysisisdirectlyproportionaltothechargeonthecation,thecurrent,andthelengthoftimethecellruns

chargethatflowsthroughthecell=currentxtime

1A=1C/s

Example - Calculate the mass of Au that can be plated in 25 min using 5.5 A for the half-reaction

Au3+(aq) + 3 e− → Au(s)

units are correct, answer is reasonable since 10 A running for 1 hr ~ 1/3 mol e−

Check:

Solve:

Concept Plan:

Relationships:

3 mol e− : 1 mol Au, current = 5.5 amps, time = 25 min mass Au, g

Given:

Find:

s 1C 5.5

Au g 6.5Au mol 1

g 196.97 mol 3Au mol 1

C 96,485 mol 1

s 1C 5.5

min 1s 60min 25

=

×××××−

−

ee

t(s), amp charge (C) mol e− mol Au g Au

C 6,4859 mol 1 −e

−e mol 3Au mol 1

Au mol 1g 196.97

Faraday’sLaw•  TryThis:•  Howlong(inminutes)mustacurrentof5.00AbeappliedtoasolutionofAg+toproduce10.5gsilvermetal?

•  Hint:thisisareversalofwhatwasbeingaskedintheexample

•  Solution:•  31.3min