FLT• Iwillbeableto:– Identifyredoxreactionsandjustifytheidentificationintermsofelectrontransfer– Designand/orinterprettheresultsofanexperimentinvolvingaredoxtitration– Makequalitativeorquantitativepredictionsaboutgalvanicorelectrolyticreactionsbasedonhalf-cellreactionsandpotentialsand/orFaraday’slaws– Analyzedataregardinggalvanicorelectrolyticcellstoidentifypropertiesoftheunderlyingredoxreactions
bycompletingElectrochemistryNotes
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
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)
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.
Titrations• Ex/A16.42mLvolumeof0.1327MKMnO4solutionisneededtooxidize25.00mLofaFeSO4solutioninanacidicmedium.WhatistheconcentrationoftheFeSO4solutionifthenetionicequationis5Fe2++MnO4
-+8H+àMn2++5Fe3++4H2O• First,findmolesofKMnO4,andthenusethebalancedequationtofindthemolesofFeSO4
• Second,dividebythevolumeoftheFeSO4solutiontogetmolarity
• 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
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?
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
TheVoltaicCell• ElectrochemicalCell=devicethatgenerateselectricitythroughredoxrxns
① Voltaic(Galvanic)CellAnelectrochemicalcellthatproducesanelectricalcurrentfromaspontaneousrxn
② ElectrolyticCellAnelectrochemicalcellthatconsumesanelectricalcurrenttodriveanonspontaneousrxn
TheVoltaicCell• Twohalf-cells• SaltBridge– Connectstwohalf-cellsandneutralizeschargebuildup,allowingtherxntocontinue
• Electrodes=conductivesurfaces– Cathode=wherereductiontakesplace– Anode=whereoxidationtakesplace
Electrodes• Anode– electrodewhereoxidationoccurs– anionsattractedtoit– connectedtopositiveendofbatteryinelectrolyticcell– losesweightinelectrolyticcell
• Cathode– electrodewherereductionoccurs– cationsattractedtoit– connectedtonegativeendofbatteryinelectrolyticcell– gainsweightinelectrolyticcell• electrodewhereplatingtakesplaceinelectroplating
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(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)
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
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.