Electrochemistry
ElectrodePotentialsandTheirMeasurement
Cu(s) + 2Ag+(aq)
Cu2+(aq) + 2 Ag(s)
Cu(s) + Zn2+(aq)
No reaction
Zn(s) + Cu2+(aq)
Cu(s) + Zn2+(aq)
In this reaction: Zn(s) g Zn2+
(aq) Oxidation Cu2+
(aq) g Cu(s) Reduction
• IfZn(s)andCu2+(aq)isinthesamesolution,thentheelectronisatransferreddirectlybetweentheZnandCu.
No useful work is obtained. However if the reactants are separated and the electrons shuttle through an external path...
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) ΔEcell = 1.103 V
AnElectrochemicalCell/2(Daniell)
Anode (-)
Negative electrode generates electrons
Oxidation occurs
Cathode (+)
Positive electrode accepts electrons
Reduction occurs
AnElectrochemicalCell/1
Anode (-)
Negative electrode generates electrons
Oxidation occurs
Cathode (+)
Positive electrode accepts electrons
Reduction occurs
Cu(s) | Cu2+(aq) || Ag+(aq) | Ag(s) ΔEcell = 0.460 V
ElectronTransferattheElectrodes
Anode
Cathode
Terminology
• Electromotiveforce,ΔEcell.– Thecellvoltageorcellpotential.
• Celldiagram.– Showsthecomponentsofthecellinasymbolicway.– Anode(whereoxidationoccurs)ontheleft.– Cathode(wherereductionoccurs)ontheright.
• Boundarybetweenphasesshownby|.• Boundarybetweenhalfcells(usuallyasaltbridge)shownby||.
• Couple,M|Mn+
Apairofspeciesrelatedbyachangeinnumberofe-.
Terminology
• Galvaniccells.– Produceelectricityasaresultofspontaneousreactions.
• Electrolyticcells.– Non-spontaneouschemicalchangedrivenbyelectricity.
StandardElectrodePotentials• Cellvoltages,thepotentialdifferencesbetween electrodes, are among them o s t p r e c i s e s c i e n t i f i cmeasurements.
• The potential of an individualelectrodeisdifficulttoestablish.
• Arbitraryzeroischosen.
The Standard Hydrogen Electrode (SHE)
StandardHydrogenElectrode2 H+(a = 1) + 2 e- D H2(g, 1 bar) E° = 0 V
Pt|H2(g, 1 bar)|H+(a = 1)
StandardElectrodePotential,E°
• E°definedbyinternationalagreement.• Thetendencyforareductionprocesstooccuratan
electrode.– Allionicspeciespresentata=1(approximately1M).– Allgasesareat1bar(approximately1atm).– Wherenometallicsubstanceisindicated,thepotentialis
establishedonaninertmetallicelectrode(ex.Pt).
ReductionCouplesCu2+(1M) + 2 e- D Cu(s) E°Cu2+/Cu = ?
Pt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) ΔE°cell = 0.340 V
Standard cell potential: the potential difference of a cell formed from two standard electrodes.
ΔE°cell = E°cathode - E°anode
cathode anode
StandardCellPotentialPt|H2(g, 1 bar)|H+(a = 1) || Cu2+(1 M)|Cu(s) ΔE°cell = 0.340 V
ΔE°cell = E°cathode - E°anode
ΔE°cell = E°Cu2+/Cu - E°H+/H2
0.340 V = E°Cu2+/Cu - 0 V
E°Cu2+/Cu = +0.340 V
H2(g, 1 atm) + Cu2+(1 M) D 2H+(1 M) + Cu(s) ΔE°cell = 0.340 V
MeasuringStandardReductionPotential
cathode cathode anode anode
StandardReductionPotentials
Most spontaneous <Reduction occurs> Oxidizing Agent
Most non-spontaneous Spontaneous in the reverse direction. <Oxidation occurs> Reducing Agent
ΔEcell,ΔG,andKeq
• Cellsdoelectricalwork.– Movingelectriccharge.
• Faradayconstant,F=96,488Cmol-1=q✕ NA=1.6022✕ 10-19C✕6.022045✕ 1023mol-1=chargeofonemoleofelectrons.
welec, rev = ΔG = -QΔE
ΔG = -nFΔE
ΔG° = -nFΔE°
SpontaneousChange
• ΔG<0forspontaneouschange.• ThereforeΔEcell>0becauseΔGcell=-nFΔEcell• ΔEcell>0
– Reactionproceedsspontaneouslyaswritten.• ΔEcell=0
– Reactionisatequilibrium.• ΔEcell<0
– Reactionproceedsinthereversedirectionspontaneously.
TheBehaviororMetalsTowardAcidsM(s) D M2+(aq) + 2 e- E° = -E°M2+/M
2 H+(aq) + 2 e- D H2(g) E°H+/H2 = 0 V
2 H+(aq) + M(s) D H2(g) + M2+(aq)
ΔE°cell = E°H+/H2 - E°M2+/M = -E°M2+/M
When E°M2+/M < 0, E°cell > 0. Therefore ΔG° < 0.
Metals with negative reduction potentials react with acids
ΔEcell,ΔG,andKeq
• Cellsdoelectricalwork.– Movingelectriccharge.
• Faradayconstant,F=96,488Cmol-1=q✕ NA=1.6022✕ 10-19C✕6.022045✕ 1023mol-1=chargeofonemoleofelectrons.
welec, rev = ΔG = -QΔE
ΔG = -nFΔE
ΔG° = -nFΔE°
RelationshipBetweenΔE°cellandKeq
ΔG° = -RT ln Keq = -nFΔE°cell
ΔEcell
0 = RTnF
ln Keq
SummaryofThermodynamic,EquilibriumandElectrochemicalRelationships.
ΔEcellasaFunctionofConcentrationΔG = ΔG° +RT ln Q
-nFΔEcell = -nFΔEcell° +RT ln Q
Convert to log10 and calculate constants
The Nernst Equation:
ΔEcell = ΔEcell
0 − RTnF
lnQ
R = 8.314472J × K−1 × mol−1
F = 96488 C mol−1
T = 298K
ΔEcell = ΔEcell
0 − 0.0592n
logQ
Applying the Nernst Equation for Determining ΔEcell.
What is the value of ΔEcell for the voltaic cell pictured below and diagrammed as follows?
Example
Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)
Pt|Fe2+(0.10 M),Fe3+(0.20 M)||Ag+(1.0 M)|Ag(s)
Fe2+(aq) + Ag+(aq) D Fe3+(aq) + Ag (s)
ΔEcell = 0.029 V – 0.018 V = 0.011 V
ΔEcell = ΔEcell
0 − 0.0592n
logQ
ΔEcell = ΔEcell0 − 0.0592
nlog
Fe3+⎡⎣ ⎤⎦Fe2+⎡⎣ ⎤⎦ Ag+⎡⎣ ⎤⎦
ΔEcellasaFunctionofConcentration:anAlternativeRoute
Cathode:Ox1èRed1Anode:Red2èOx2
AlternativeRoute
C
A
CombiningHalf-CellReactions/1
Reaction1:Cu2+(aq)+2e-DCu(s)
Reaction2:Cu+(aq)+e-DCu(s)
Reaction3:Cu2+(aq)+e-DCu+(aq)
SinceReaction3=Reaction1-Reaction2
NO!!
Reaction1:Cu2+(aq)+2e-DCu(s)
Reaction2:Cu+(aq)+e-DCu(s)
Reaction3:Cu2+(aq)+e-DCu+(aq)
CombiningHalfReactions/2
Fe3+(aq) + 3e- D Fe(s) E°Fe3+/Fe = ?
Fe2+(aq) + 2e- D Fe(s) E°Fe2+/Fe = -0.440 V
Fe3+(aq) + e- D Fe2+(aq) E°Fe3+/Fe2+ = 0.771 V
Fe3+(aq) + 3e- D Fe(s) E°Fe3+/Fe = +0.331 V
Equation3=Equation1+Equation2
1
2
3
Dismutation/1
Spontaneous
Dismutation/2
Non-spontaneous
ConcentrationCellsTwo half cells with identical electrodes
but different ion concentrations.
2 H+(1 M) D 2 H+(x M)
Pt|H2 (1 atm)|H+(x M)||H+(1.0 M)|H2(1 atm)|Pt(s)
2 H+(1 M) + 2 e- D H2(g, 1 atm)
H2(g, 1 atm) D 2 H+(x M) + 2 e-
Concentration Cells
2 H+(1 M) D 2 H+(x M)
ΔEcell = ΔEcell0 − 0.0592
nlogQ
ΔEcell = ΔEcell0 − 0.0592
2log
x2
12
ΔEcell = 0− 0.0592 log x = 0.0592 × pH
MeasurementofKsp
Ag+(0.100 M) D Ag+(sat’d M)
Ag|Ag+(sat’d AgI)||Ag+(0.10 M)|Ag(s)
Ag+(0.100 M) + e- D Ag(s)
Ag(s) D Ag+(sat’d) + e-
Using a Voltaic Cell to Determine Ksp of a Slightly Soluble Solute.
With the date given for the reaction on the previous slide, calculate Ksp for AgI.
Example
AgI(s) D Ag+(aq) + I-(aq)
Let [Ag+] in a saturated Ag+ solution be x:
ΔEcell = ΔEcell0 + 0.0592
1log Ag+⎡⎣ ⎤⎦C
− 0.05921
log Ag+⎡⎣ ⎤⎦A
0.417 = − 0.05921
logAg+⎡⎣ ⎤⎦A
Ag+⎡⎣ ⎤⎦C
= 0.05921
logAg+⎡⎣ ⎤⎦A
0.1=
0.417 = −0.0592 log Ag+⎡⎣ ⎤⎦A+ 1( )
−7.04− 1 = log Ag+⎡⎣ ⎤⎦A
Ag+⎡⎣ ⎤⎦A= 10−8.04
KS0 = 10−8.04( )2
= 8.31× 10−17
Batteries:ProducingElectricityThroughChemicalReactions
• PrimaryCells(orbatteries).– Cellreactionisnotreversible.
• SecondaryCells.– Cellreactioncanbereversedbypassingelectricitythroughthecell
(charging).
• FlowBatteriesandFuelCells.– Materialspassthroughthebatterywhichconvertschemicalenergy
intoelectricenergy.
TheLeclanché(Dry)Cell
DryCellZn(s) D Zn2+(aq) + 2 e- Oxidation:
2 MnO2(s) + H2O(l) + 2 e- D Mn2O3(s) + 2 OH- Reduction:
NH4+ + OH- D NH3(g) + H2O(l) Acid-base reaction:
NH3 + Zn2+(aq) + Cl- → [Zn(NH3)2]Cl2(s) Precipitation reaction:
AlkalineDryCell
Zn2+(aq) + 2 OH- D Zn (OH)2(s)
Zn(s) D Zn2+(aq) + 2 e-
Oxidation reaction can be thought of in two steps:
2 MnO2(s) + H2O(l) + 2 e- D Mn2O3(s) + 2 OH- Reduction:
Zn (s) + 2 OH- D Zn (OH)2(s) + 2 e-
Lead-Acid(Storage)Battery• Themostcommonsecondarybattery
PbO2(s) + 3 H+(aq) + HSO4-(aq) + 2 e- D PbSO4(s) + 2 H2O(l)
Lead-AcidBattery
Oxidation:
Reduction:
Pb (s) + HSO4-(aq) D PbSO4(s) + H+(aq) + 2 e-
PbO2(s) + Pb(s) + 2 H+(aq) + HSO4-(aq) D 2 PbSO4(s) + 2 H2O(l)
ΔE°cell = E°PbO2/PbSO4 - E°PbSO4/Pb = 1.74 V – (–0.28 V) = 2.02 V
TheSilver-ZincCell:AButtonBattery
Zn(s),ZnO(s)|KOH(sat’d)|Ag2O(s),Ag(s)
Zn(s) + Ag2O(s) D ZnO(s) + 2 Ag(s) Ecell = 1.8 V
TheNickel-CadmiumCell
Cd(s) + 2 NiO(OH)(s) + 2 H2O(L) D 2 Ni(OH)2(s) + Cd(OH)2(s)
FuelCellsO2(g) + 2 H2O(l) + 4 e- D 4 OH-(aq)
2{H2(g) + 2 OH-(aq) D 2 H2O(l) + 2 e-}
2H2(g) + O2(g) D 2 H2O(l)
ΔE°cell = ΔE°O2/OH- - ΔE°H2O/H2
= 0.401 V – (-0.828 V) = 1.229 V