ap chem week 29 unit 10 ch 18 part bflemingapchem.weebly.com/uploads/2/4/6/5/24658308/ap... ·...
TRANSCRIPT
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Electrochemistry
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ElectrochemicalCells
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TheVoltaicCell• ElectrochemicalCell=devicethatgenerateselectricitythroughredoxrxns
① Voltaic(Galvanic)CellAnelectrochemicalcellthatproducesanelectricalcurrentfromaspontaneousrxn
② ElectrolyticCellAnelectrochemicalcellthatconsumesanelectricalcurrenttodriveanonspontaneousrxn
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CellNotation• Ex/Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(aq)
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Half-CellPotentials• E°cell=E°oxidation+E°reduction– E°oxidation=-E°reduction– +E°cellmeanstherxnisspontaneous– whenaddingE°valuesforthehalf-cells,donotmultiplythehalf-cellE°values,evenifyouneedtomultiplythehalf-rxnstobalancetheequation
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PredictingtheSpontaneousDirection
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PredictingtheSpontaneousDirection² UseE°celltodeterminewhichwayaredoxrxnwillproceed
² Thehalf-rxnwiththemore(+)electrodepotentialattractse-smorestronglyandundergoesreduction• Goodoxidizingagents
² Thehalf-rxnwiththemore(-)electrodepotentialrepelse-smorestronglyandundergoesoxidation• Goodreducingagents
² Anyreductionreactioninthetableisspontaneouswhenpairedwiththereverseofanyoftherxnslisted
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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
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PredictingtheSpontaneousDirection• Dometalsdissolveinacids?• Yes,mostacidsdissolvemetalsbythereductionofH+ionstohydrogengasandthecorrespondingoxidationofthemetaltoitsion.
Zn(s)+2H+(aq)àZn2+(aq)+H2(g)• Canwepredictwhetherametalwilldissolveinacid?
• What’stheoxidation½rxnandthereduction½rxn?
• Metalswhosereductionhalf-rxnsarelistedbelowthereductionofH+toH2dissolveinacids,whilemetalslistedaboveitdonot.
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PredictingtheSpontaneousDirection• Whichmetal(s)dissolvesinHNO3butnotinHCl?• Hint:thecombinationofhalf-cellpotentialsshouldbepositive
a. Nib. Auc. Cu• Solution:• Cu
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CellPotential,FreeEnergy,andK
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E°cell,ΔG°andK• SincewehaverelatedE°celltowhetherornotarxnisspontaneousornot,canwerelateE°celltoΔG°?
• WhatabouttoK?
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E°cell,ΔG°andK• Foraspontaneousreaction(onethatproceedsintheforwarddirectionwiththechemicalsintheirstandardstates)– ΔG°<0(negative)– E°>0(positive)– K>1
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E°cell,ΔG°andK• Foranonspontaneousreaction(onethatproceedsinthereversedirectionwiththechemicalsintheirstandardstates)– ΔG°>0(positive)– E°<0(negative)– K<1
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Recall• Termstoknow:• ElectricCurrent=theflowofelectriccharge– Current(I)isoftenrepresentedbythenumberofelectronspassingthroughpersecond(Amperes)
• Voltage=thepotentialenergyperelectron
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E°cell,ΔG°andK• Areweabletoconvertchargetonumbersofelectrons?
• Why,yes!• Faraday’sconstant(F)representsthechargeincoulombsof1molofelectrons:
F=96,485coulombs/mole-s(onyourreferencesheet)
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E°cell,ΔG°andK• Mathematically,wecanderivearelationshipbetweenΔG°andE°cell:
• ΔG°=−RTlnK• ΔG°=−nFE°cell– nisthenumberofelectrons– F=Faraday’sConstant=96,485C/mole−
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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
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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
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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
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Note:• Atthispoint,wewouldnormallylookatnonstandardconditionsandthustheNernstEquation
• IamskippingallnotesontheNernstequation,sincetheyremovedallAPstandardsrelatedtoitwiththenewtest.
• However,IDOrecommendreviewingitbeforetakingcollegechemistry.
• Iamalsoskippingoverbatteries.Theyaregoodtoreviewasexampleproblems,butfocusonthehalf-rxnsmorethanthedefinitions
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Electrolysis
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Electrolysis• Recall:anelectrolyticcellrequireselectricalenergytodriveanonspontaneousreaction
• Forexample,thereactionofhydrogenwithoxygentoformwaterisspontaneousandcanbeusedtoproduceanelectricalcurrentinafuelcell
• Conversely,bysupplyinganelectricalcurrent,wecancausethereversereactiontooccur,separatingwaterintohydrogenandoxygen
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Electrolysis• Electrolysisliterallymeanstosplitwithelectricity
• Electrolyticcells– Nonspontaneous–therevrserxncanoccurifwesupplyavoltagegreaterthanE°cellwecalculateforthespontaneousdirection– Usedtoseparateoresorplateoutmetals– WestillfollowANOX&REDCAT
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Electrolyticvs.VoltaicCells• Spontaneousvs.Nonspontaneous• Avoltaiccellisseparatedintotwohalf-cellstogenerateelectricity;anelectrolyticcelloftenoccursinasinglecontainer
• Avoltaiccellisabattery,whileanelectrolyticcellneedsabattery
• ANOX,REDCAT• However...• Forelectrolyticcells:EPA–electrolyticpositiveanode
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ElectrochemicalCells• inallelectrochemicalcells,oxidationoccursattheanode,reductionoccursatthecathode
• involtaiccells,– anodeisthesourceofelectronsandhasa(−)charge– cathodedrawselectronsandhasa(+)charge
• inelectrolyticcells– electronsaredrawnofftheanode,soitmusthaveaplacetoreleasetheelectrons,the+terminalofthebattery
– electronsareforcedtowardtheanode,soitmusthaveasourceofelectrons,the−terminalofthebattery
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electroplatingInelectroplating,theworkpieceisthecathode.
Cationsarereducedatcathodeandplatetothesurfaceoftheworkpiece.Theanodeismadeoftheplatemetal.Theanodeoxidizesandreplacesthemetalcationsinthesolution
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Electrolysis
• Electrolysisistheprocessofusingelectricitytobreakacompoundapart
• electrolysisisdoneinanelectrolyticcell
• electrolyticcellscanbeusedtoseparateelementsfromtheircompounds– generateH2fromwaterforfuelcells– recovermetalsfromtheirores
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ElectrolysisofWater
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ElectrolysisProducts
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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)
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ElectrolysisofNaCl(l)
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MixturesofIons• whenmorethanonecationispresent,thecationthatiseasiesttoreducewillbereducedfirstatthecathode– leastnegativeormostpositiveE°red
• whenmorethanoneanionispresent,theanionthatiseasiesttooxidizewillbeoxidizedfirstattheanode– leastnegativeormostpositiveE°ox
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AqueousSolutions• Aqueoussolutionscancomplicatethingsbecauseofthepresenceofwater
• Thereisthepossibilityoftheelectrolysisofwateritself
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SummingUp• Ifthereisnowaterpresentandyouhaveapuremoltenioniccompound,thecationwillbereduced,andtheanionwillbeoxidized
• Ifwaterispresent&youhaveanaqueoussolutionoftheioniccompound,thenwaterwillbeoxidizedinstead
• Nogroup1AorIIAmetalwillbereducedinanaqueoussolutionàwaterwillbeoxidizedinstead
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StoichiometryofElectrolysis
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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-
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Faraday’sLaw• Theamountofmetaldepositedduringelectrolysisisdirectlyproportionaltothechargeonthecation,thecurrent,andthelengthoftimethecellruns
chargethatflowsthroughthecell=currentxtime
1A=1C/s
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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
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Faraday’sLaw• TryThis:• Howlong(inminutes)mustacurrentof5.00AbeappliedtoasolutionofAg+toproduce10.5gsilvermetal?
• Hint:thisisareversalofwhatwasbeingaskedintheexample
• Solution:• 31.3min