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