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2.625

- Electrochemical

Systems Fall2013

Lecture

2

- Thermodynamics

Overview

Dr.Yang

Shao-Horn

Reading: Chapter 1 & 2 of Newman, Chapter 1 & 2 of Bard & Faulkner, Chapters 9 & 10 of PhysicalChemistry

I. LectureTopics:

A. Review1stand2nd lawsofthermodynamics

Firstlaw: dU =QW

Secondlaw: IS 0

B. ChemicalEquilibrium

II

iaprod i

Gr = G

+

nRT

ln

Ka =0;whereKa = Ir iareact i

i = +RTlnai; iin= 0i

C. ElectrochemicalEquilibrium

Gr

= G +nRTlnKa

+ nzrF Er

= 0r

NernstEquation: Er =E +

RT

lnKar zrF

II. Review1stand2ndlawsofthermodynamics

A. 1stLawofThermodynamics

For

a

closed

system

of

constant

mass,

i.internalenergyisconstant: IU = 0

ii.external energychangesthrough workand heat: EU =QW, where Q isheattransferred

intothesystem,andW isworkdonebythesystem. Note: inmost(fluid)systems,

W is PdV; for electrochemical systems, we will expand this definition to include electro

chemicalwork.

iii.the total energy change of the system is the sum the internal and external energy changes:dU

=IU +EU =QW

IU andEU areintensiveproperties whereasQandW areextensiveproperties

A.Intensive property: A physical property whichdoesnot depend on the size (volume,mass,number,etc.) ofthesystem

B.

Extensive

property:

A

physical

property

which

does

depend

on

the

size

(volume,

mass,

number,etc.) ofthesystem

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Lecture2 ThermodynamicsOverview 2.625,Fall13

IU

EU

System

Environment

Figure1: Thecanonicalthermodynamicsystem,withinternalenergychangeIU andexternalenergychangeEU.

B. 2ndLawofThermodynamics

i. changein internalentropy isequaltogeneratedentropy,whichisnever lessthanzero:

IS =gS 0

ii. externalentropychange isthechangeinheatscaledbythetemperatureofthesystem:QES =T

iii.

the

total

entropy

change

is,

similarly,

the

sum

of

the

internal

and

external

entropy

changes:

QdS=IS +ES =IS + T

MultiplyingthroughbyT,wehave: TdS=T IS +QAndsinceIS isalwaysgreaterthanorequaltozero,TdSQ.Theseexpressionscanbemappedintostatespace. ThechangeinentropywhileUandVareheldconstant,orHandPareheldconstant,areshownbelow.

C. Combiningthe1stand2ndLaws,Q=dU+PdV (whereW isgivenasPdV)

Subbing inTdSIS forQandrearranging,wehave

T IS =dU+PdV TdS0 (1)

For reversible processes, IS = 0, so TdS = dU +PdV. This is Gibbs fundamental equation.

Further,

a

closed

system

approaches

equilibrium

by

maximizing

entropy,

S.

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States

SAt constant U, V

equilibrium

Figure2: Equilibrium isachievedwhenentropy ismaximizedforasystemwhereUandVareconstant.

D. Alternatively,wecandefineequilibriumbyusingGibbsfreeenergy.

i. First,wedefineenthalpy,H:H =U+P V

ii. ThendefineGibbsfreeenergy,G:G=HT S

iii. G=U+P V T S,anddG=dU+d(P V)d(T S)

Byconsideringthecombinedfirstandsecondlawequation,wecannowwrite:

IS =

dG

V

dP

+

SdT

0.

Thus,asystemapproachesequilibriumbyminimizingGsuchthat dG0 :

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States

G At constant P, T

equilibrium

Figure3: Withtheproperchoiceofvariablesheldconstant,theHelmholtzandGibbsfunctionsshowsimilarprofiles. EquilibriumisachievedwhenHelmholtzorGibbsfunctionsareminimized.

Note: forareversiblesystem,dG=VdP SdT

III. Gibbsfreeenergyofmulticomponentsystems

ThegeneralformoftheGibbsfreeenergyforamulticomponentsystemorphaseisG(T,P,n1, n2,....ni),whereni

isthenumberofmolesofspeciesiinthesystemorphase.

G G GdG(P,T ,ni) = dT + dP + dni (2)

T P niP,n T,n T,P,nji

Wecandefinethepartialdifferentialsintheaboveequationasfollows: G S=T P,n

G V =P T,n

g r

Gi = ni T,P,nj=ni

ThenwecanwritethedifferentialfromoftheGibbsfreeenergyforamulticomponentsystemas:

dG(P,T ,ni) =SdT+VdP+ idni

(3)i

A. ComparingthistotheexpressionfordGabove,weseethattheinternalentropyisexpressedhereasthesumofthechemicalpotentials: T IS = iidniThisentropyisgeneratedwhenanewcomponentiisintroducedintothesystem,withachemicalg r

Gpotential,i = ni T,P,nj=niLookingback,recallthatatconstantTandP,T I

S=dG= i

idni

0

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B. Thermodynamicquantitiesaretypicallytabulatedonamolarbasis,meaningthatthequantityisprovided inunitspermoleofreactant/product. Examplesare:

g : molarGibbsfreeenergy

s

:

molar

entropy

h : molarenthalpy

C. Wedefinemolefraction: Xi =ini

ni

G= nigi;wheregi

= Xigi;andn= ni

D.WedefinethepartialmolarGibbsfreeenergywithrespecttocompositionas:g r

g r

g Ggi

= = =iXi niT,P,Xj=Xi T,P,nj=ni

E.Foran idealgas,wedefinethepartialpressureas:

Pi =XiP

F.We can define a more convenient form of the chemical potential, where we write the chemicalpotentialofaspeciesiintermsofitschemicalpotentialinthereferencestate, anditsactivity,i

ai:

i

=

+

RT

ln

aii

Defining the chemical potential of a species in terms of its activity is slightly more convenientbecausetheactivitycanbedescribedbysimplemodels,whichwillbediscussednext.

G.Activitymodels

Ideal(Raoultian)Solution: ai =XiAnidealsolutionassumesthattheactivityofspeciesiisequaltothemolefractionofspeciesi.

IdealGas: ai =PiForan idealgas,theactivityofspecies i isequaltothepartialpressureofspecies i. This isanextensionofthe idealsolutionmodel.

i.Figure ?? demonstrates how activities can be calculated from the partial pressures of gases

in

a

mixture.

ii.Theenthalpyofmixingcanbefoundbycalculatingtheenthalpybeforeandaftermixing:hinit

= Xihi(T, Pi)ihfinal = iXihi(T,Pi) = iXihi

hmix =hfinal hinit = 0

iii. Theentropyofmixingcanbefound inasimilarway:sinit = iXisi = iXi s

i RlnP

Pi

sfinal = si = iXi s RlnPi

iXi i P

smix =sfinal sinit = iXiRlnXiNote that sinceSmix isgreaterthanzero, mixing is irreversible. Thisshould lineupwithyour intuitiveunderstandingofgasbehavior.

iv.Next,theGibbsfreeenergyofamixture:

ginit =

Xigi =

Xi g +

RT

ln

Pi

i

i i P

gfinal

= i

Xigi

= i

Xi

gi +RTlnPP

i

gmix

=gfinal

ginit

= XiRTlnXi 0iWhengmix

0,mixingwilloccurspontaneouslyandworkwillbegenerated.

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System

Before Mixing: P1= P

2= P

3= P

i= P

After Mixing: P = p1+ p

2+ p

3+ p

i

P1

n1

P2

n2

P3

n3

P4

n4

p1, p

2, p

3, ... p

i

n1, n

2, n

3, ... n

i

Figure4: Initially,aboxcontainsseveralspeciesofagas,eachhavingthesamepressurePi,thesameastheoverallboxpressureP.Afterthepartitionsareremoved,thegasesmixandPisthesumofeachgaspartialpressurePi.

IV. ChemicalEquilibrium

A. Conditionforchemicalequilibrium

idni

= 0 ,or Gr

= 0 atconstantTandP.

Where

Gr =

iin,

and

i is

the

stoichiometric

coefficient

of

species

i.

6

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States

idn

i At constant T, P

equilibrium

Figure5: Equilibrium isachievedatconstantTandPwhenthesumofchemicalpotentialsofcomponentsinthesystemisminimized.

B. Example1i. Considertheformationofwater: H2(g) + O2(g) H2O2

= 0.5,andH2O = 1ii. H2(g) =1,O2(g)

C. Derivingtheconditionforchemicalequilibrium:i +RTlnai)i. Gr = iin=n i(

ii.

Gr i +

nRT

iln

ai=

n i

iii. LetG i=n irII

iaprod iI

iv. Gr = G +nRTlnr

iareact iII

iprod ia

I

iareact i

v. RememberthatKa

=

vi. Atequilibrium: Gr = Gr +nRTlnKa = 0

G =nRTlnKarvii. Thus,atwecandefinethechemicalequilibriumcondition:

D. Twointerestingrelations

i. Thenaturallogarithmoftheequilibriumconstant,Ka,canbewrittenasfollows:h h G

H

r rKa

=

exp

=

A

expnRT nRT

HrlnKa =lnAnRT

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1/T

ln Ka

(high T) (low T)

H2+ 0.5O

2 H

2O

CO2 CO + 0.5O

2

1 1Figure6: ExampleplotoflnKa vs. lnKa variesapproximately linearlywith. .T T

ii. Assuming that Hrrange of T), the following relationship can be developed:

is nota functionof temperature(this is a good assumption for awide

r

rrSinceG(T, P) = H TS

r

G=S ,assumingthatH

T

r

r

,

=f(T)

1Example: H2(g) + O2(g) H2O2

T

(K)

hr (kJ/mol) g

r (kJ/mol) E

r

300 -286 -237 1.23500 -242.4 -219 1.13

1000 -245 -192 0.99

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T

Gr

0

-Sr

Figure7: Visualrepresentationof howthechange inGibbs freeenergywithtemperature isapproximatelyequaltothechangeinentropy(i.e. theslopeofG asafunctionoftemperatureisequaltoS).r r

E. ElectrochemicalEquilibrium

i.Relative to a chemical system, equilibrium for an electrochemical system has an additionalworkterm.

ii.W =Edq forchargedparticleswithachargeofdq

iii. Conditionforelectrochemicalequilibrium(atconstantTandP): ii

+zrF Er

= 0

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MIT OpenCourseWarehttp://ocw.mit.edu

2.625 / 10.625JElectrochemical Energy Conversion and Storage: Fundamentals, Materials, and Applica

Fall2013

For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
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2.625 - Electrochemical Systems Fall 2013

Lecture 3 - Chemical Potential and Equilibrium

Dr.Yang Shao-Horn

I. Review

A. 1st and 2nd laws of thermodynamics

B. Chemical Equilibrium: Gr =

iin= 0, at constant T and P

II. Lecture Topics

A. Electrochemical Equilibrium

Electrochemical Gibbs free energy (including charged species): Gr = iin+nzrF Er = 0,at constant T and P

B. Nernst equation that governs the cell/reaction potential

i

E =E + RT

ln proda

i

r r zF a ireact i

Er = f(T,P,Xi)

Example reactions and concentration cells

C. Electrochemical series to determine the cell/reaction potential

III. Electrochemical Equilibrium

A. The work term for charged particles under an applied potential is: W = E dq, where particleshave chargedq

B. dG(P , T , n , E ) = SdT+ V dP+

idni+ Edq, at constant T and P

C. At electrochemical equilibrium (under constant T and P): dG=

idni+ Edq= 0

D. Alternatively, we can write that for electrochemical equilibirum (Gibbs-Faraday Equilibrium Equa-tion): idni= Edq

Where F is Faradays constant: F =N 23Aqe = 6.022 10 1.6 1019 = 96500C/mol

zr is the number of charges/electrons transferred in the reaction (per mole of reactant orproduct).Example: H2(g)+

1 O2 2(g) H2O;zr = 2 per mole of H2(g)zr = 4 per mole of O2(g)

IV. Nernst Equation that governs reaction potential

A.ii= zrF Er, at (T, P

)

B. i i = T

zrF Er, at ( , P)

C. ii

i i+ zr(Er Er)F= 0

a iD. Er = E

Rr

T lnzr F

prod i

a ireact i

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Lecture 3 Chemical Potential and Equilibrium 2.625, Fall 13

RT a iE. Nernst Equation: Er = E + ln

zrF

react

r

i

prod a i

i

V. Nernst Equation for ideal reacting gases

A. Activity of gaseous species, i: ai = Pi =P

XiP P

B. Er = Er RT lnzr F

a ireact i

a iprod i

( P

i ) iC. E =E RT react P

r r lnzr F prod(

P i

P ) i

T P

R i RT i

D. Er = Er + ln

+ ln react

zrF P

Xi

zrFprod X

i

i

Where i= i i

E. The reaction

poten

tial, Er

is ultimately a function of a temperature dependent term, system

pressure dependent term, and a concentration dependent term.g

i. Temperature dependent term: E rr = Er =

=f(T, P)zr F

Er

s= rT zr F

The temperature dependent term can be significant. For example consider the reaction:H2(g)+

1 O2 2(g) H2O.At 300K, Er

= 1.23VAt 1000K, Er

= 0.99V

ii. Pressure dependent term: Er = RT ln P2zr F P1

Varying the pressure only can only provide a small gain when increasing from very low to veryhigh pressures in terms of thermodynamics (pressure variations can, however, have a largeimpact on kinetics). For example consider T= 300K, zr = 4, P1 = 1atm, and P2 = 10atm.Then: Er = .056V.

iii. Concentration dependent term: E = RTr ln Pb

zr F PaThe concentration dependence also does not greatly affect the reaction potential, but again,it can have important kinetic effects.For example, consider T= 300K, zr = 4, P1 = .21atm, andP2 = 1atm.Then: Er = .01V.

RT RT iF. Nernst Equation for ideal liquids/solids: Er = Er

+ + ln iz F z F

react X

r rprod X

i

i

VI. Chemical Equilibrium vs. Electrochemical Equilibrium

A. Chemical Equilibrium

Under chemical equilibrium, no work can be generated: Gr = 0 = G

r(P = 1atm) +

nRTln Ka Example: Gr = zrF Er

Er = 0 andKa is very large, so mostly H2O is present on right side

B. Electrochemical Equilibrium

Gr = iin= 0, where i= i+ ziF E. i is the electrochemical potential.

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Lecture 3 ChemicalPotential and Equilibrium 2.625, Fall 13 Thus, we can write: Gr = iin + nzrF Er = 0

Example: H2(g)+ 1 O2 2(g)

a H2O(l)

Gr = G

r+ nRTln H aO2 2 , where aH2 =aO2 =aHa 2O = 1H O

Gr = Gr 2

= nzrF ErTherefore: Er

= 1.229V

VII. Concentration Cells

A. Concentration cells are used for desalination, hydrogen sensors, oxygen sensors, etc.

P (100atm)B. Example: g100atmH =g2 H

+ RTln H22 P

g108atm (

8 )H =g2 H

P

ln H 10 atm

+ RT 22 P

8 P (108atm)gr = g

10 a tm 100atmH =RTln

H2

H2 g

2 PH (100atm)2

C. Physcial Origin of the electrical potential and work generated: entropy change of mixing

D. Alternatively, desalination and gas separation requires work input due to the entropy change ofmixing

VIII. Electrochemical Series

A. The purpose of the electrochemical series is to determine the potential of individual electrodes(half-cell reactions) with respect to a standard electrode, such as the Stanard Hydrogen Electrode(S.H.E.).

B. Standard Hydrogen Electrode:

aH+ =aH2 = 1, T = 298K, and P = 1atm.

We define E+ = 0V for 2H+ + 2e H2.H /H2C. Example 1

Consider the reaction O2+ 4H+ + 4e 2H2O, whereEO

2,H+ = 1.229V/H2O

Combine:

2H2 4H+ + 4e

+

O2+ 4H+ + 4e 2H2O (1)

O2+ 2H2 2H2O

EH

+ =E/H2 ano

de = 0V

E + =E =O2,H /H2O catho

de 1.229V

Ecell =Ecatho de Eano de = 1.229V

D. Example 2: Consider the thermodynamic stability of the following (T= 298K and P= 1atm)

i. Fe/H2O

ii. Fe/H2O + H+

iii. Fe/O2/H2O

iv. Fe/O2/H2O/H+

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Lecture 3 Chemical Potential and Equilibrium 2.625, Fall 13

First check the electrochemical series and find relevant redox reactions:

Fe2+ + 2e Fe = EF

e2+ = 0.409V/Fe

O2+ 4H+ + 4e 2H2O = EO 2,H+ = 1.229V/H2O

O2+ 2H2O + 4e 4OH = E + = 0.401VO2,H /OH

2H+ + 2e H2= E =H

+ 0V/H2

i. anodic: Fe Fe2+ + 2e

no cathodic reaction can be found

ii. anodic: Fe Fe2+ + 2e

cathodic: 2H+ + 2e H2

Eano

dic= 0.409V; Ecatho

dic= 0VEcell =Ecatho

dic Eano

dic= 0.409V> 0

This reaction will occur spontaneously. Iron is not stable.

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2.625 - Electrochemical Systems Fall 2013Lecture 4 - Electrochemical Cells

Dr.Yang Shao-Horn

Nernst Equation: = or+ RT ln

a i

reactants i i r = f(T , P , xi)ZrF aprod iElectrochemical series of half cell reactions (Po, To, oO/R)Electrochemical series of half cell potentials can help us...

1. Estimate the Nernst Voltage of aqueous electrochemical reactions under standard conditions(To = 298K, Po = 1atm, ai= 1)

2. Understand the stability of metals in aqueous environment

Example: Metal plating vs H2 productionIn an acid solution (aH+ = 1), can we plate Cu spontaneously? Can we plate Zn spontaneously?

Figure 1: Electrochemical series of copper, hydrogen, and zinc

Cu Case

CathodeC u2+ + 2e Cu, ocathode= 0.340V

AnodeH2 2H+ + 2e, oanode= 0V

o ocell= cathode oanode= 0.340V >0 Will Plate

Zn Case

CathodeZ n2+ + 2e Zn, ocathode= 0.763V

AnodeH2 2H+ + 2e, oanode= 0V

o

cell

= o

cathode

o

anode

= 0.763V >0 Will NOT Plate

All metals with half cell potentials more positive than SHE can be plated out spontaneously inaqueous electrolytes

A quick question: changing from acidic to basic solution, What/How would the driving forcechange?

H+/H2 =f(pH) means that H +

2 2H + 2e is pH dependent

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Lecture 4 Electrochemical Cells 2.625, Fall 13

Cu2+ + 2e Cu is pH independent

I. Nernstian Half Cell Potential

Given a redox coupleoO+ Ze RR (Reduction)

o and R are stoichiometric coefficients and Zis the # of electrons

Couple O/R redox with SHE to make a chemical reaction

OO+ Ze RR+ =OO+ H2 H2 RR+ H+ H

+

H2 H2 Ze + H+ H+

Ha o a 2

Nernst Voltage of this reaction is :cell= oR+ RT ln

o H2

ZF + a H a R+ RH

Also, the reaction voltage can be defined as: cell= cathode anode= O/R O= R

o RT ao

thus O/R = O/R+ ln oR is the Nernst half cell potential for O/RZF aR

Example- Consider H2 concentration cell

Overall reaction H2(1000atm) H2(108atm)

Figure 2: Hydrogen concentration cell, a high H2 pressure and low H2 pressure chambers areseparated, only allowing for H+ ion transport

a2

anode= o RT 2+ a

2H RT +H RT 1

H+ + ln = 0 + ln/H2 2F 2F Panode =aanode ln2F PanodeH H H2 2 2P

assuming ideal gas for H2 and Po=1atm aH2 = H2

Po

a2 a2R 2 =o + T ln + = 0 + R ln +H T H = RT ln 1cathode H+/H2 2F acathode 2F Pcathode 2F PcathodeH H H2 2 2

= Panode

= RT ln H2cell cathode anode =2F Pcathode 0.295VH2

II. Pourbaix Diagrams

Potential-pH diagrams established by Pourbaix in 1974

Example: Stability of water in aqueous electrolytes

a. Consider hydrogen electrode potential vs. pH

2H+ + 2e H2[H+]aH+ =H+

[H+]o

activity of solution species

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activity coefficient = 1 for ideal solution

[H+

] molar concentration ofH+

mol/L [H+] molar concentration ofH+ at 1 mol/L (1M)

Po = 1atm,To = 298K, oH+ = 0, a/H H+ = 12

Recall that ln x= 2.303 log xand pH= log[H+]

= 2.303RT log P + 2.303RT log[H+H+/H2 H2F 2 ]2F

= 2.303RTH+/H2 log PH2F 2 2.303RTpH

2F Hydrogen Electrode

At room temperature, 298 K, Hydrogen electrode potential as f(pH, PH2 ):

H+/H2 = 0.059 2log PH2 0.059pH

As pH , H+/H2 ; PH2 , H+/H2

b. Consider the oxygen electrode

1 O2+ 2H+ + 2e2 H2O(l) 298K1/2

a a2

O+/H2 = o

O O2,H+ + RT ln

H2 2

2,H = 1/H2 2F aH O2 (l)

303RO ,H+/H =o 2. T 1/2 2O ,H+ + +2 2 log PO a (ideal gas)2 /H2 2F 2 H

Oxygen electrode potential at room temperature

O2,H+/H2 = 1.229 + 0.0148 log PO2 0.059pH

As pH , O2,H+/H2 ; PO2 , O2,H+/H2

c. Pourbaix diagram of water

Figure 3: Pourbaix diagram of water, showing regions of gas evolution and water stability

Physical significance of Pourbaix Diagram

i. H+/H2 half cell reaction (from pH = 0(aH+ = 1) to pH = 14)voltage decreases by0.82V (0.059V 14). 2H+ + 2e H2 reduction less thermodynamically favorable.

ii. O2, H+/H2O half cell reaction voltage decreases by 0.82V from pH= 0 to 14. 2H+ +O2 H2O reduction less thermodynamically favorable.

III. Description of electrochemical cells

For any electrochemical cells, we need at least:

A. Two electrodes (one for oxidation, one for reduction) to provide electrons

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B. One ion conduction to provide ionic transport from one electrode to another. Must no

conduct electronsC. External electrical circuit to provide electron flow from one electrode to another

Generally speaking, we have two type of cells

A. Galvanic Cells: cell > 0 and GR < 0. Reactions can occur spontaneously to producework. Examples include primary batteries, fuel cells, secondary batteries upon discharge,flow batteries, and capacitors upon discharge.

B. Electrolytic Cells: cell< 0 and GR> 0. Electric work is needed to drive the reaction.Example is water electrolysis to produce H2.

i. Production of Al, Li, Na, K, Mg, and Zn by electrolysis

ii. Production of Chlorine and NaOH for water purification

Cathode 2H

+

aq+ 2e

H2,(g),

o

+ = 0VH /H2 Anode 2Claq

H2,(g)+ 2e, o =Cl/Cl2 1.358V

2NaCl+ H2O Cl2+ H2+ 2NaOH

Historical Note: Invented as a weapon in WWI by Fritz Haber (Nobel Laureate inChemistry). Haber-Bosch N2+ 3H2 2N H3.

Figure 4: Schematic of cell types. Galvanic cells produce work during the reaction. Electrolyticcells require work or energy to drive the reaction.

IV. Faraday law i m, i in C/s, m in g/s

Michael Faraday (1791-1867). Discovered water electrolysis and coined the terms anode, cath-ode, ion and electron.

m

i= M Zr F

in C/S or At

m weight of reactant or product (g)

M molecular weight of reactant or product (g/mol)

F Faraday Constant = 96485 C/mol 96500 C/mol

Zr # of electrons involved

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Example: To makeH2 at 1g/s, what current is needed?

96500C/moli= m Zr F = 1g 2 105C/s = 105At M s 2g/mol

Large current needed (1g/s 105A) due to large F

Consider H2/O2 electrolytic cells to make H2 using electricity at 1 A/cm2 and cell area of

105cm2, providing current at 1A and cell voltage is 1.6 V.

To make 1 kg H2 105A 1000s= 108C

1kg 2 96500C/molFromt = mZr F = iM 105 = 1000sC/s2g/mol

Energy needed 108C

3600J1.6V

/W h 44kW h

Assuming $0.15/kWh, electricity cost of $7/kg for H2.

CurrentH2, $10/kg and target cost of $3/kg.

V. Electrochemical energy conversion efficiencies

Maximum thermal efficiency- Thermodynamics

ideal= W = G = 1H H

TSH

Second Law efficiency (voltage efficiency)

= EmeasuredEo

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Lecture 8 - Kinetics of Chemical and Electrochemical Reactions

Dr.Yang Shao-Horn

I. Overview

A. For a given O + e R:

i. Electrode kinetics lead to activation losses. a is defined as the overpotential loss dueto kinetic limitations and is called the activation polarization (a ln i). Electrode kineticsinclude:

Electron transfer

Ion/molecule adsorption

Ion/molecule desorption

ii. Ion/mass transport gives rise to ohmic losses() andconcentration losses(c). ( i)and (c exp(i))

iii. Overpotential loss curves for electrolytic cells and galvanic cells a are shown below.

Electrolytic cell: Ecell= Eeq+ a+c+ Galvanic cell: Ecell= Eeq a c

iv. Further, schematic plots showing common losses in electrochemical capacitors, lithium bat-teries, fuel cells, and electrolytic cells are also shown below.

Electrode Electrolyte

e-

Oads

Rads

Osurface

Rsurface

Obulk

Rbulk

Bulk

Solution

Figure 1: Schematic of transport and kinetic steps which must occur during a single electron transfer reaction.

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Lecture 8 Kinetics of Chemical and Electrochemical Reactions 2.625, Fall 13

iE

eq

Electrolytic Cells:

a

c

E

Figure 2: Overpotential losses as a function of Figure 3: Overpotential losses as a function ofcurrent common to an electrolytic cell. current common to a galvanic cell.

i

Eeq

Galvanic Cells:

a

c

E

Charge

Electrochemical Capacitor

(typically limited)

E

Figure 4: Illustration of ohmic overpotential Figure 5: Lithium battery potential as a functionlosses in an electrochemical capacitor. of time during battery cycling.

Charge time

Li-Battery Cycling

(typically transport (c) limited)

E

Eeq

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E

Fuel Cells or

Electrolytic Cells

Kinetics (a) limited

Figure 6: Current as a function of potential for a kinetically limited device.

II. Key Concepts Covered

A. Chemical kinetics: basic concepts

i. Arrhenius expression

ii. Transition state theory

iii. Why catalysts?

B. Surface thermodynamics: basic concepts

i. Physisorption

ii. Chemisorption

iii. d-band center theory

iv. Langmuir isotherms

C. Electron transfer kinetics

i. Addinge transfer to transition state theoryii. Butler-Volmer Equation

III. Basic Concepts in Chemical Kinetics

A. The rate of chemical reactions

kf

[A] [B]kb

for a homogeneous reaction (single phase) is given by:

i. RHomo d[A] d[B]chem = = =kf[A] kb[B] mol/cm3dt dt This equation is valid for a single reaction in a closed system with constant volume.

[A], [B]: molar concentration of A or B in mol/cm3

kf, kb: forward/backward reaction rates in 1/s (kf and kb are functions of temperature,but not [A] or [B])

ii. When kf >> kb, then RHomochem =kf[A]

d[A]having =kf[A]dt

d[A]and = [A]

dt t=0exp ( kft)

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G

Reaction Coordinates

G#> 0

Gr< 0

(driving force)

H2+ 0.5O

2

H2O

Transition State Theory

Figure 7: Gibbs free energy as a function of reaction coordinate for the formation of water from oxygen andhydrogen. The free energy curve exhibits a reaction barrier which limits how fast the chemical reaction canoccur.


G

G#

Butler-Volmer Equation(electrochemical rxn)

O + e- R

~

~

Figure 8: Reaction barrier for a single electron transfer electrochemical reaction. This reaction governs theButler-Volmer equation, which will be discussed in a later lecture.

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Time

[A]

[A] @ t=0

Figure 9: Concentration of reactant A as a function of time as the chemical reaction proceeds.

iii. When the reaction reaches equilibrium, RHomochem = 0

A. kf[A]eq = kb[B]eqk [A]

B. f = eqkb [B]eq

Ka, where Ka is the equilibrium constant

i

C. Ka=

prodXi

reactX

ii

nB nB

D. [= X n +n

K B ]eq =

A B eq [ [a [XA] n = V ]eq B]

nA = eq

Aeq

+n

[B

V ] [A]eqn eqA eq

B. Law of mass action k

i. Reactants 1, 2, ... f

i P, for (kf >> kb)

ii. = dP

R =k

[i] ifdt react

e.g. aA +bB +cCdP a b

kf

Pc

iii. R= =kf[A] [B] [C] , wherea, b, and c are reaction ordersdtC. Arrhenius expression of chemical reactions

i. Remember:

kf[A] [B]

kb

d[B]ii. and: R = = kb[B] + kf [A], where kf, and kb are functions of temperature but aredtindependent of the concentrations of A and B.

iii. In 1889, Arrhenius introduced the concept of an activation barrier:

A. Recall from chemical equilibrium: Gr = 0 = G

r+ RTln Ka

S

ln Ka = rR Hr

RT

d lnKa H

= rdT RT 2

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T

kfor k

b

Figure 10: Temperature dependence ofkf andkb.

B. Arrhenius proposed a similar expression for reaction rate constants

d lnkf E

f

= adT RT 2

d lnkb E

b

= adT RT 2

whereEfa >0 andEba > 0. These are called activation barriers for the forward/backward

reactions.

C. The activation barriers are related to the change in enthalpy of the reaction by:Hr

f=Eb Eba

Ef

kf=Af exp a

RT

f

E

kb= Ab exp b

RT

fD. Typically, Af,

Ab, Efa

, and Eb can be obtained from fitting experimental data:

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1/T

ln Ka

(high T) (low T)

ExothermicH

r0

o

o

Figure 11: Behavior of ln Ka for both exothermic and endothermic reactions.

E


A

B

Ea

b

Ea

f

Hr

o

Figure 12:

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1/T

ln kf

(high T) (low T)

f-Ea

R

Figure 13: Activation energy barrier as proposed by Arrhenius.

IV. Transition State Theory - 1935 (activated complex theory)

kA. We began with the model that a species transforms from reactant, A to product, B: A

f

B.#

B. Now consider the formation of an intermediate species called the Activated Complex (AB) :#

K# k

a #

A (AB) f

B

where K# #

a is the equilibrium constant between A and (AB) K# = exp G

#

a RT

# #and, kf is the reac

tion rate constant from (AB) to B

d[B] #

# #C. R= =k (AB)

= k K# [A]dt f f a

=kf[A]

## GD. kf=kf exp

RT

i. The jumping frequency of a species is given by: kB T 1013 1/s.h

where kB is the boltzmann constant, T is temperature, and h is Plancks constant.

#ii. kf

=kf

exp G#

RT

where G

# =

#

= kB T expG# H= Af

exph RT

RT

H# TS#

V. The influence of catalysts on the rate of chemical reactions

A. criteria for a catalyst in chemical reactions

i. increase reaction rates

ii. not consumed in the reaction

iii. a small quantity increases the reaction rate of a large amount of reactants

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E


A

B

G#

(AB)#

#Figure 14: Free energy barrier for a reaction to proceed from A to B, where a transition state species, (AB) ,is formed as an intermediate.

E


A

B

G#

(AB)#

(AB)#c

G#

f,c

f

without catalyst

with catalyst

Figure 15: When a catalyst is introduced, the free energy barrier for formation of the transition statedecreases.

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ln Kads

Rchem

Figure 16: Typical volcano curve showing how the rate of the chemical reaction changes a function ofequilibrium constant for adsorption.

iv. does not shift chemical thermodynamics ofKa

B. Transition State Theory with a catalyst

K#

i. A a k#

(AB) f

B

K# k#ii. A

a,c

(AB) f,c

c B

iii. The rate change between the reaction without a catalyst and the reaction with a catalyst isgiven by:

#G

exp f,c

k (AB)# RTkf,c [A] c]= f,c[

=

k [A] k [(AB)#]

#f Gf

exp fRT

#where Gf,c is dependent on the catalyst.

C. Sabatier Principle (Heterogeneous Catalysis)

i. Homogeneous reactions reactions occur in one phase: R= kf[A] kb[B] mol/cm3 s

ii. Heterogeneous reactions reactions occur in multiple phases

iii. Rchem= kf[A]ads kb[B]ads, where [A]ads is given in mol/cm2

k

A f

B, or

kA

f,cB

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2.625 - Electrochemical Systems Fall 2013Lecture 9 - Surface Reactivity and Electronic Structure

Dr.Yang Shao-Horn

Topics to be covered in the lecture

I. Sabatier principle for heterogenous reaction

II. Basic concepts of thermodynamics= Physisorption/Chemisorption

III. Langmuir Isotherm

Homogenous reactions: Reactions occur in one phase

Heterogenous reactions: reactions occur is multiple phasesExamples

N2+ 3H2 2N H3 (Haber Process)

H2+ 12

O2 H2O

CO + 12O2 CO2

I. Sabatier Principle for heterogenous reaction kinetics

The best catalysts are surfaces that bind reactant/product neither too weakly nor too strongly

Just right

Too weakly Too strongly

Activity/rates

ln Kads

Figure 1: Schematic of volcano relationship

Recall homogenous reactions

kfckf

=exp

G#

f,cRT

exp

G

#c

RT

>1

Where the reaction rate constant can be increased by catalysts via lowering the reactionbarrier.

A kfB (assuming kf >> kb for simplicity)

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Lecture 9 Surface Reactivity and Electronic Structure 2.625, Fall 13

!Gf#

(AB)#, without catalyst

(AB)C#, with catalyst

A

B

!Gf,c#

(a) catalyst vs no catalyst

!Gf#

(AB)#, without catalyst

(AB)C#, with catalyst

A

B

!Gadso

(b) absorbed catalyst vs no catalylst

C+A KadsAC

k#ac(AB)#Ck#f

BCB +Cwhere C represents the catalyst

or we write ifBc B +C is barrier-less

C+A KadsACk#

ac(AB)#Ck#f

B +C

Rchem for heterogenous reaction has a unit ofmol/cm2 s

Rchem = k#f [(AB

#C] = k

#f[AC]K

#ac, where [AC] is the moles of A adsorbed on the catalyst

surface inmol/cm2.

Rchem = k#fK

#ac[A]KadsLcky

Letting [AC] = [A]KadsLC, where Kads is the adsorption equation constant exp

Goads

RT

andLC is the characteristic length of adsorption.

In mol/cm2 [AC] = Jsurface, where Jsurface is the flux of molecules hitting the surface inmol/cm2s and is the residence time of an absorbed species in (s).

Jsurface= 14 [A]gas|vgas|

[A]gas = ngasV

= PRT

and|vgas| is the average velocity of gas in 3D.

=oexp

Goads

RT

[AC] = 14 [A]gas |vgas| oexp

G

oads

RT

= [A]gasLcKads, whereLC=

14 [A]gas |vgas| o

Rchem = k#fK

#a,c[A]KadsLC

kf,c = k#fK

#a,cKadsLC cm/s for heterogenous reaction rate constant

kf,c =kbTh

K#a,cKadsLC

kf,ckf

=

kbT

h

K#a,cKadsLC

kbT

h K#a LC

= K#a,cKads

K#a

catalyzed

Sabatier principle came from the fact that G#f,c = b agoads, where the reaction barrier scales

with adsorption energy and b and a are constants.

Rchem =kbTh

exp

b

RT

exp

aGoads

RT

exp

G

oads

RT

[A]Lc

where exp

bRT

exp

aGoads

RT

= K#a,c= exp

G#f,c

RT

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Eaads>0Ea

des

chemisorption

Physisorption

Figure 5: Activated molecular adsorption

Figure 6: Activated molecular adsorption for H2 and Pt

Hads = 2UHPt UHH < 0 where U represents the bond strength of H-Pt and H-Hrespectively. If this condition is met, the process is thermodynamically feasible.

Examples with be seen on next lecture

O2 onP t 370kJ/mol

H2 onP t 50-60 kJ/mol

CO onP t 140 kJ/mol

III. Langmuir Isotherm

1 =KadsP relates the coverage of adsorbates to gas pressure/concentration at a given

temperature.

Historical note: formulated in 1916, received 1932 Nobel Prize in Chemistry

Assumptions:

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adsorption on homogenous surface

terminated upon completion of one adsorbed monolayer adsorbed atoms do not interact with each other

A+C(1) adsorptionAc() monoatomic gas adsorption

Gads= Goads+RTln

aAC

aacPAPo

wherePo = 1

At equilibrium Gads= 0 Goads= RTln

aACaC(1)

+ RTln PA where aAC=

exp

GoadsRT

=

( 1 )PA

1 =KadsPA monoatomic gas adsorption

1= (KP))12 dissociative adsorption such as H2, O2, N2,...

Consider 1 =KP or = KP1+KP

!

P

!

log P

1

Figure 7: vs P and vs log P

At a given T,

!

log P

!" T2 T3

Figure 8: |Hads| , Kads . 1 at higher P

At a fixed Hads

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Lecture 13 - Butler Volmer Equation

Dr.Yang Shao-Horn

I. Lecture topics: Butler Volmer Equation

A. Application of Transition State Theory to O + e R (one-electron, one-step reactions)

B. Current is linearly proportional to potential at low overpotentials

C. Current is exponentially related to potential at high overpotentials (i.e. i exp(E))

II. Reading Materials: Chapter 3 of Bard & Faulkner

III. Pseudocapacitance

A. Recall the lab demo in which cyclic voltammetry of platinum in aqueous solutions was performed:B. Pt in 1M HClO4

C. RHE: reversible hydrogen electrode (H2 at 1 atm at the pH of the reaction of interest)

D. EDLC: electrical double layer capacitance

E. Pseudocapacitance: HPt + + Ptad H +e ande + H Had

F. Redox on the surface

G. What is the source of H+? Decomposition of H2O

H2O + e HPt+ OH

or H +2O O2+ 4H + 4e (counter)

2H+ + 2e 2HPt

IV. Transition State Theory for one-step and one-electron transfer reactions

A.

k

O + e f

Rkb

i. : electrochemical potential

ii. =i+ziF E

iii. At electrochemical equilibrium, Rel

ectrochem= 0

A. Rel

ectrochem= kf [O]

andkb

R

B. Where unlike chemical reaction

s,

kf

and kb

are f(T, [X] , P , E ) such that Rchem orkchem= f(T ,P, [X])

iv. Assuming [O]surface= [R]

surface

Then: k =k =k G#

= kbTf b exph f LcRT

#Where Gf are kinetic barri

ers and

typically very difficult to measure or compute.#Generally Gf = C Gads, where C is a constant and Gads applied to the rate

determining step.

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Lecture 13 Butler Volmer Equation 2.625, Fall 13

i

E vs. RHE

HadH++e-

H++e-Had

EDCL

EDCL

Only EDCL RemainsIncreasing

scan rate

Figure 1: Schematic representation of how the psuedocapacitance of platinum varies with scan rate. At low

scan rates, hydrogen adsorption and desorption peaks are visible. At high scan rates, on the electrochemicaldouble layer capacitance (EDCL) can be observed.

Ptele

ctrode

-

-

-

-

-

-

-

-

-

-

-

-

-

-

-

H+

H+

H+

H+

H+

H+

H+

H+

H+

H+

Figure 2: Diagram of the EDCL.

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i

E vs. RHE

Bulter-

VolmerBubbling Ar

H2transport limited

H2transport limited

Figure 3: Schematic showing i-E curve for platinum pseudocapacitance, Butler-Volmer dominated hydrogenevolution/reduction, and H2 transport limited currents.

Gf

~

o,#

R

O + e-

Figure 4: Graphical representation of the free energy of formation of the transition state.

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Gf+F

a

~

o,#

R

O + e-

Gf

o,#

Gf+(1-)F

a

o,#

Figure 5: Role of the transfer coefficient in reaction behavior.

B. At E = Eeq (or finite activation overpotential: a = E Eeq), where the potential loss is usedto promote the reaction rate, the energy of the electrons is lowered by |a|relative to Eeq as Eincreases, promoting the oxidation reaction rate.

i. Energy loss is given by a F .

ii. We define . which is the transfer coefficient. A typical value of the transfer coefficient is= 0.5.

iii. Energy loss term due to impeding the forward reaction:

aF

.iv. Energy loss term due to promotion of the backward reaction: (1 ) aF.

v. Derivation of the

A. k = kbTf exph

Butler-Volmer Equation:#

G

f Lc RT#

B. = k T exp G

b f F a

h

Lc

C. k exp

aF RT

L cRT#

D. kb = k T Gexp

h

b b LcRT

( ) FE. k exp

a

RT

Lc

F. Relectrochem= kf[O] kb[R] mol/cm2

sG. i= ZrF RAelectrode

Where ZR is the number of electrons transfered (Zr = 1)

Aelectrode is the electrode surface area

Fis Faradays constant

i is current

H. assuming [O] = [R] = [O]bulk= [R]bulk; (assumes no transport limitations)F a (1 ) F a

I. Butler-Volmer Equation for Electrode Kinetics: i= io exp

R

exp

T RT

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E vs. RHE

i

increasing io

Figure 6: Change in Butler-Volmer i-E curve with changing exchange current density (io .

J. io= F Aelectrodek [O]bulk= F Aelectrodek

[R]bulk

V. Special cases of Butler-Volmer Equation

A. When |a| 10mV, i-E curve exhibits linear behavior.

i. For small values ofx:

A. exp(1 +x) 1 +x

B. exp(1 x)

i

1 x

ii. i= oF aRT From the linear region, the current exchange current ( io) can be calculated.

B. Tafel relation

i. The Tafel relation applies at large overpotentials (i.e. |a| 120mV).ii. ib F= exp |a|

f RT

>> 1, (where ifcan be neglected)i

iii. i= (1ioexp

)F aRT

iv. a= RT ln io+

R

T ln ((1)F (1)F i)

v. Tafel relation: a= a+b log i

vi. Example:

A. 2H+

+ e

H2B. Tafel reaction: 2Had H2

1

(KP)ads= 2

11+(KP) 2

C. Volmer Re

action:

ad+ H+ + e Had.

The Volmer react

ion is the rate limitin

D. io= F Aelectrodekads n =F Aelectrodeexp

n: the number of adsorption sits per el

g step. G

#

adsRT kB T n

h C/s

ectrode

area.

G# =C Gads, where C is a constant.

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E vs. RHE

ln |i|

Eeq

Figure 7: Typical Tafel plot, showing E vs. ln |i|, for the Butler-Volmer equation.

i

E vs. RHE

10 mV

Figure 8: The linear region of Bulter-Volmer i-E curve occurs at small overpotentials.

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E vs. RHE

log |i|

Eeq

Tafel Slope

Figure 9: At high overpotentials the logarithm of the current is linearly proportional to the overpotential,as exhibited by the linear Tafel slope.

ads= exp

Gads

RT

P, for low ads

io exp Gads

RT

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2.625 - Electrochemical Systems Fall 2013Lecture 17 - Controlled Current Methods and Potential Step Methods

Dr.Yang Shao-Horn

Motivation of Transport

Current Collector

Pt H+

O2gas

Figure 1

H2/O

2

H2/Air

!

i

iL

1.5A/cm2

Figure 2: Fuel Cell. Reaction rate is transport limited [O2] surface limiting, e.g. electrodeflooding air vs. pure O2

!

Capacity

Increasingcurrent

Figure 3: Lithium ion battery. Limited by D+Li in liquid electrolyte.

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Lecture 17 Controlled Current Methods and Potential Step Methods 2.625, Fall 13

jL= nPo[O]F

Jo= 4965002.310

59.5107C/cm50106100cms = 1.69mA/cm

2 O+2H2O+ 4e 4OH

much slower than 1.5A/cm2 achieved in fuel cells.Mass TransferReading: Bard and Faulkner Chapter 4, Newman Chapter 11

1. Modes of mass transfer

1. Diffusion: movement of a species under the influence of a gradient of a chemical potential(i)

2. Migration: movement of a charged particle under the influence of an electric field E (a gradientof electric potential ).

3. Convection: stirring or hydrodynamic transport of fluid particles

2. Nernst-Plank Equation for mass transfer toward an electrode

Ji(x) = Di[i])x

x

ZiF

RTDi[i]

x + [i]uxa. Ji(x): flux of species i mol/s/cm

2

b. Di[i])xx : diffusion

c. ZiFRTDi[i]x : migration

d. [i]ux: convection

Electrode

Solution/ElectrolyteJi(x)

Figure 4: Schematic of electrolyte solution flux into electrode.

Di: the diffusivity of species i in the electrolyte cm2/s

x : electrical field V/m

ux: velocity of fluid/electrolyte particles in the x direction (cm/s)

1. Thermodynamic driving for for diffusion and migration

x

Point r

Points

J!" !i

!

Figure 5: Diffusion + migration driven by electrochemical potential

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i(r) = oi +RTln ai(r) +ZiF (r)

i(s) = oi +RTln ai(s) +ZiF (s)

Ji = Di[i]RTi

With convection Ji = Di[i]RT

2i + [i]u

assumingai= [i] for ideal solutions and 1D transport

Ji = [i]DiRT

x [RTln[i]] +

x [ZiF ]

+ [i]ux

Ji = Di[i]x

Di[i]ZiFRT

x + [i]ux Nernst-Plank Eq.

2. Current density associated with the flux to the electrode

ji= ziF J)i if species i is charged with valence ofzi

j =

iji =

i ziF Ji =

i ZiF Di[i]x

i

z2iF2

RT Di[i]x +

i ZiFi[i]ux total current

from all charged species

as

i zi[i] = 0 due to electroneutrality

j =

i ziF D[i]x

i

z2iF2[i]D

RTx =i

D +iM where

Diffusion currentiD =

i ziF D[i]x

Migration current iM =

iz2iF

2[i]DRT

x

3. Current density in the electrolyte isjM as concentration gradients are generally small

ji= jMi =

z2iF2[i]DiRT

x

We definite mobility /= |zi|FDi

RT = uiddx

Absolute mobility /Bi = DiRT cm

2mol/Js

ji= jMi = |zi|F/i[i]

x = i

x

ji= i x Ohms law

i = |zi|F/i[i]: The conductivity of species i in the solution

j =

ji=

i |zi|F/i[i]x =

i

x

Transference number: ti = jij =

|zi| /i[i]i|zi| /i[i]

= iiti is unitless 0 ti 1

Ion Conduction/Conductivity is much lower than electron conduction in metals (106S/cm)

Electrolyte Material Conductivity ti

AqueousKClHClH+ in Nafion

101 to 102 S/cm at 25 C

101 to 103S/cm at 80 C, high RH

tK+ = 0.5tH+ = 0.8tH+ = 1.0

Nonaqueous LiPF6 inEC:DEC

102 to 103S/cm at 25 C tLi+ 0.2 0.4

Solid State YSZ 102S/cm at 800 C tO2 = 1

4. Mixed migration and diffusion near an active electrode

j = jD +jM

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iftR= 1, [R] = 0. AstR, |[R]|

iL FZRDR[R]bulk

(1tr)o

tR, iL

Example: Lithium ion batteries

Carbon, 200!m LiCoO2, 200!mSeparator/Electrolyte

25 !miLi+

Figure 10: The Rate of lithium ion batteries is limited by DLi in the liquid phase of theelectrode (200m)

Concentration Overpotential

[O]

[O] Surface

[O] Bulk

Figure 11: Oxygen concentration surface to bulk

O+e R

surf,O/R

= oO/R

+ RTF

ln [O]surf

[R]

bulk = oO/R+

RTF ln

[O]bulk[R]

/c = surf+ bulk = RTF ln

[O]surf[O]bulk


/i= RTF ln

iLi

L

= RTF ln

1 iiL


!o

[O] Bulk

o+e!R

Figure 12

[O]surf[O]bulk

= 1 iiLj = FDo [[O]bulk [O]surf] A

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jL= FDoo

[O]bulkA the limiting current

Concentration overpotential/c = RTF ln

1 iiL

i

!

1

2

3

Figure 13: 1. i = /

R (ohmic, electronic+ionic) 2. /c (concentration overpotential) 3. /a

(activation overpotential)

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Lecture 2 Basics of Impedance Measurements. 2.625, Fall 13

Z R

0 Z

The magnitude ofZ:

Z

R2

1

C

The phase shift associated with the circuit:

Ztan

Z

1C

R

1

C R

0 for resistors and for capacitors.2

Impedance data are displayed typically in Nyquist plots and Bode plots below.

Nyquist Plots:

3

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Bode Plots:

4

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Example:

1log Z log R2

2C2

, Z 100

0, Z 100

1 log Z 10Z Z0

10Z

capacitor resistor-like1

log Z log R2 C2

1R2

C2

1log Z log 6

C

5

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2.

Rct Charge transfer resistance.

Z Warburg Impedance element, a resistance to mass transfer.

Now consider the Faradaic impedance

jZ Rs

Csq

E iRs Cs

dE di i

dt Rs

dt

Cs

As i imaxsin t, we have:

E i

t Rsimax cos t

maxsin t

Cs

Consider a redox reaction: O ne R

E E

i, 0 0, t , R 0, t

d

E E di

dt

i dt

0, t

R

E

O

d 0 0, t E d t

dt

R 0,

R 0, t

dt

ct

0

d

d

R

E di d , t R

RRct

00

O 0, t

dt

dt

dt

dt

Rctimax cos t need to find O x, t andR x, t .

Assuming semi-infinite diffusion:

i O 0, t O

max

cos tnF A 2OO

sin t

1

2

i R 0, t R

maxsin t cos t

nF A R

2O 1 2

8

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Lecture 2 Basics of Impedance Measurements. 2.625, Fall 13d

O

0,t

d

R

0,tPlug and into the expression of d

dt dt dtE:

dE

Rct i ax 1

2

max cos t im sin tdt

1 2

1 Rwhere

nF

0

A 2

1

2O

1

2

O OR

Rs Rct

1 2

1Cs

1 2

j jZf Rct R

C

Rct

1 2

1 2

1

Rs

j

Cs

j j

Zf

Rct

R

C

Rct

1 2

1 2

Rct 1 2

j 1 2

9

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY2.625/3.625/10.625 FALL 2013

HOMEWORK 1 (ungraded)Solutions posted September 12, 2013

Chemical equilibrium questions warm up

Problem #1

Calculate the equilibrium constant K for ammonia synthesis reaction, N2+ 3H22NH3at 298

K, and show how K related to the partial pressure of the species at equilibrium when the overall

pressure is low enough for the gases to be treated as perfect.

Problem #2

Suppose that an Iron catalyst at a particle manufacturing plant produces ammonia in the mostcost effective manner at 450C when the pressure is such that Grfor the reaction

N2(g)+ 3/2H2(g)NH3(g)is equal to -500 J/mol.

(a)What pressure is needed?

(b)Now suppose that a new catalyst is developed that is most cost effective at 400C whenthe pressure gives the same value of Gr. What pressure is needed when the new catalyst

is used? What are the advantages of the new catalyst? Assume that all gases are perfect

gases. Isotherms of Gr(T,P) in the pressure range 100atm p 400atm are needed to

derive the answer.(c)Do the isotherms you plotted confirm Le Chateliers principle concerning the response of

equilibrium changes in temperature and pressure?

Hint: Le Chateliers principle: A system at equilibrium when subjected to a disturbance,

responds in a way that tends to minimize the effect of the disturbance.

Electrostatics questions warm up

Problem #3

The potential around a cylindrical shell. You have a hollow inner cylinder with radium a,surrounded by a concentric outer cylinder with radius a2>a1. Charges -1 and +1 per unit length

are distributed uniformly over the inner and outer cylinders, respectively. The arrangement is in amedium with dielectric constant D.

(a)What is the potential as a function of the axial distance r?

(b)What is the capacitance per unit length of the arrangement of cylinders?

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E&''+38 )*&('+236 !$ < (34 ; '28)'0)- 8+F)6 &6 ( 4+-)>' -),('+2360+@ %)'?))3

?0+>0 ?) .(H @,2' (' '?2 4+11)-)3' ').@)-('&-)6 '2 2%'(+3 2&- +62'0)-.6

!!!(34 E$

I' JKLL MN.2,$

E-)66&-) (' OKL P Q !R('(,H6' +6 +' )3(%,)6 '0) -)(>'+23 (' (%2&' 0(,1 '0) 2-+8+3(,

@-)66&-)" /0+6 ?+,, -)4&>) '0) (.2&3' 21 @&.@+38 (34 >2.@-)66+23 3))4)4 '2

@-)66&-+T) '0) -)(>'+23 >0(.%)-$ (34 .(U) (..23+( 6H3'0)6+6 >0)(@)-"

B>C V)6" !!!4)>-)(6)6 (6 @-)66&-) +3>-)(6)6$ ?0+>0 >2--)6@2346 '2 ( 60+1' '2 '0) -+80'

6+4) 21 '0) -)(>'+23$ ?0)-) '0)-) (-) 1)?)- .2,)6 21 8(6" I,62$ 6+3>) '0) 12-?(-4

-)(>'+23 +6 )A2'0)-.+>$

').@)-('&-) +3>-

!!!82)6 &@ B>2--)6@234+38 '2 ( 60+1' '2 '0) -)F)-6) -)(>'+23C (6

)(6)6$ 12- ( 8+F)3 @-)66&-)"

;" B(C

W) .(H (66&.) '0(' '0) ,)38'0 21 '0) >H,+34)-$ X YY ('-+> 1+),4 = +6 4)1+3)4 (6 !! !" ! !

!"!

$ ?0)-) \ Q ! @)- &3+' ,)38'0

]+3>) 1+),4 ,+3)6 (-) @)-@)34+>&,(- '2 2&- 7(&66+(3 6&-1(>)$ ! !!" ! !

!!!

$ (34

! ! !

!!"#!!

!

!!! ! ! !

!!!! !"

!!!! ! !

!!"!!

!" !

!!B-)1)-)3>)4 '2 '0) 6&-1(>) 21 '0) +33)- >H,+34)-C

B%C P(@(>+'(3>) @)- &3+' ,)38'0$ P Q \N^_

5-2. B(C$ !! ! !

!!"!!

!"!!

!!

!

(34 ?) U32? \ Q !

/0)-)12-) ! !!"!

!"!!

!

!!

3

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#

Problem #2 Short Questions

a. A steady current of 1.00 amp is passed through an electrolytic cell containing a 0.1 M aqueoussolution of CuF2 using inert graphite electrodes until 1.54 g of copper is deposited. At which

electrode is the copper deposited? (anode or cathode) and what is the half-cell reaction at the

other electrode?

b.Calculate the equilibrium constants for the following reactions at 25oC in aqueous solutions.

(10 pt)(a) Ni(s)+ CuSO4(aq)!Cu(s)+ NiSO4(aq)

(b) 2H2(g)+ O2(g)!2H2O(l)

c. Predict whether a reaction takes place spontaneously and explain why, and give a balancedreaction equation.

a. Au(s)+ HCl(aq)b. Mg(s)+ ZnSO4(aq)

c. Cu(s)+ AuCl3(aq)d. Sn(s)+ Al2(SO4)3(s)

d. Consider an electrochemical cell involving Zn(II) and Ni(II) and answer the following

questions:a). Identify the anode and the cathode for spontaneous reaction

b). Write the balanced overall reactionc). Which electrode will lose mass?

d). Which way do the electrons flow? toward Zn or Ni?

Problem #3

Considering the following electrochemical cell:

Pb/PbCl2, Hg2Cl2/Hg

At 25oC, the thermodynamic voltage of the cell is 0.4857 V and the temperature coefficient is

1.65x10-4

Volt/degree. Please calculate:

A:

(a) The maximum work available from the cell at 25oC per mole of Pb reacted.

(b) The entropy change of the cell operation.

(c) The heat adsorbed by the cell at 25oC per mole of Pb reacted when the cell is operating

reversibly.

B:

You replace the Hg electrode with an Hg-Y alloy electrode in which mole fraction XHg= 0.3.

The cell thermodynamic voltage at 25oC is found to increase by 0.0094 V. Calculate the activity

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of Hg in the alloy at 25oC.

Problem #4 - Desalination

You have a desalination cell with 0.1 M solution of NaCl on each side. The activity of NaCl can

be defined as the molar concentration of NaCl divided by a reference concentration

(a) Assuming no kinetic limitations, what is the minimum applied potential required tomake one side 10

8times more concentrated than the other?

(b) From the electrochemical series, devise a galvanic cell that would involve lithium

redox ( ), which will provide enough electromotive force orthermodynamic voltage for your desalination cell.

Problem #5

Efficient and economical water splitting is a key technology component of a hydrogen economy

since it utilizes the inexpensive natural resource, water, to produce hydrogen fuel.

(a) You are asked to do water splitting in an acidic electrolyte. Please write out the half cell

reactions at both hydrogen electrode and oxygen electrode. Please specify the cathode and anode

side. What is the overall reaction?

(b) Repeat (a) but with an alkaline electrolyte.

(c) Calculate the Grofor water splitting at standard condition (T = 298K, P=1atm).

(d) In the acidic electrolyte, derive the expression of the two half cell potential at cathode and

anode and the cell potential.

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY

2.625 FALL 2013

HOMEWORK 3

DUE 10/17/2013 in class (note this is a change of due date)

State Your Assumptions Clearly

1. The rate constant for the first-order decomposition of N2O5 in the reaction 2N2O5 (g)!

4NO2 (g) + O2 (g) is k = 7.48e-5 s-1

at 25 C.(a) What is the half-life of N2O5?

(b) What will be the pressure, initially 300 Torr, 20s after the initiation of the reaction?(c) How long will it take for the pressure, initially 300 Torr, to reach 100 Torr?

2. The separate adsorption isotherms for gases A and B on a certain solid obey the Langmuir

equation, and it may be assumed that the mixed or competitive adsorption obeys thecorresponding form of the equation. Gas A, by itself, adsorbs to a coverage theta of 0.03 at P

= 350 mm Hg and gas B, by itself, adsorbs a coverage of 0.03 at P = 35 mm Hg; T = 95 K inboth cases.

(a) Derive a Langmuir model expression for the competitive adsorption of gases A and B.(b) Calculate the difference in the heat of adsorption for A and B on the solid surface.

(c) Calculate the value for coverage of A when the solid, at 95 K, is equilibrated with amixture of A and B such that the final pressures are 350 mm Hg each.

3. This problem will walk you through the hydrogen oxidation reaction (H2 (g) ! 2H+

+ 2e) in

PEM fuel cells. The adsorption of CO can act as a poison, inhibiting the hydrogen oxidationreaction by site blocking.

(a) Derive an expression using the Langmuir model for dissociative adsorption of H2 on a Ptsurface: H

2 (g)! 2H

ads.

(b) The adsorption of CO can compete with this process. Modify your expression from 2a) toconsider the competition between the formation of COads and Hads starting from CO (g)and H2 (g) reactants.

(c) How does this equation simplify in the limit of strong CO binding? What is the reaction

order with respect to CO (g) and with respect to H2 (g)?(d) Plot the coverage of COads and Hads as a function of the concentration of CO, assuming a

CO adsorption energy of !HCO = -1.2 eV and a dissociative H2 adsorption energy !HH2 =-0.7 eV, at pH2 = 1 bar and 300K.

(e) At what concentration of CO (in ppm) will the coverage of CO, "CO, exceed that of "H?

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!A&66*42 6%& &()"6*043 90- /01&-"2& 09 E &()"+ 60 /01&-"2& 09 YZL 04& 0,6"*43 "

;-&33)-& 90- YZ 2"3 S%*/% /"4 ,& /041&-6&: 60 ;;. )3*42 6%& *:&"+ 2"3 &()"6*04

_@U@^ ;;. YZ

4

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AAC C

2.625 A 2013

4

10/17/2013 ( 3)

A

1. The initial rate of the reaction 2A + 2B ---> C + D is determined for different initial

conditions, with the results listed in the following table:

Run # [A]0, M [B]0, M Initial rate, M/s

1 0.266 0.185 1.35 X 10-3

2 0.133 0.370 6.75 X 10-4

3 0.266 0.370 2.70 X 10-3

4 0.133 0.185 3.35 x 10-4

Find the rate law and rate constant for this reaction.

2. (a) Show that for a first-order homogeneous reaction, , the average lifetime of

A is 1/kf .

() A ,

, - 7.21 . 0.100

8 .., 10 ..?

3. Consider the reaction: 2A + B 3C. In one experiment it was found that at 300 K therate constant is 0.268 L/(mol.s). A second experiment showed that at 450 K, the rate

constant was 1.138 L/(mol.s). Determine the activation energy for the reaction.

4. Consider the elementary reaction: A + BC AB + C

(a) Express the rate law (define rA) using the Arrhenius equation.

(b) Now utilize transition state theory (TST), where an activated molecule is formed

during the reaction at the transition state between forming products from reactants:

We can express the rate of reaction as the product of the frequency of the activated

complex crossing the barrier and the concentration of the transition state complex,with the transition state molecule and reactants in pseudo equilibrium at the top of the

energy barrier. Write the expression for the equilibrium constant of KC#.

(c) Express the rate law as a function of T, a change in free energy, and the

concentration of reactants. Draw a diagram specifying this change in free energy as a

function of reaction coordinate, and draw how the curve would change with the use ofa catalyst.

1

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1. This is an initial rates problem with more than on species present as a reactant. Wemust therefore isolate the effect of each of the reactants' concentrations on the overall rate

of reaction. Runs number 3 and 1 have the same concentration of A but the concentration

of B is half. The rate of the reaction is halved too, so the reaction is first order in B. Runsnumber 3 and 2 have the concentration of B the same but the concentration of A halves.

The rate is four times lessunder these conditions so the reaction is second order in A.

Thus the rate law is

rate = k[A]2[B]The rate constant may be evaluated with any set (row) of data. Try run #3

2.70 x 10-3M/s = k(0.266 M)2(0.370 M)

k = 0.103 M-2

s-1

As a check, lets use run #43.35 x 10-4M/s = k(0.133 M)2 (0.185 M)

k = 0.102 M-2s-1

(a)

For first order reactions,

, where is the number ofmoles of A at time, t = 0.

First we obtain a probability distribution of A with time, by normalizing:

!

, where a is the normalization constant

Average lifetime is therefore the expectation of the probability distribution:

(b) Half-lives are constant for first order reactions: ln2 = kthalfthalf= 7.21 hr, so k = ln2/7.21hr = 0.0961 hr

-1

ln(mfinal/minitial) = -ktln(mfinal/0.100 mg) = -(0.0961 hr

-1)(14 hr)

mfinal= 0.026 mgastatine remaining
http:///reader/full/ln2/7.21http:///reader/full/ln2/7.21

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3. Compare the two rate constants to solve for Ea.

, similar for k450

Ea= 10.8 kJ

where q is overall the partition function per unit volume and is the product of

translational, vibration, rotational and electric partition functions (this part is not

necessary for credit). To put it more simply:

-rA= kF#[A][BC]exp(-Gf

o/RT)where kF

#= kBT/h

4. (a)

(b)

(c) Combining with gives ,

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1) Calculate the equilibrium fuel cell voltage under the operating conditions.

2) Develop an analytical expression that relates the fuel cell operating voltage to the current

density obtainable from the fuel cells by considering activation overpotentials in theanode and in the cathode. Plot the effect of activation overpotentials to the fuel cell

voltage loss as a function of current density in mA/cm2

in the range from 0 to 2A/cm2.

2

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY

2.625/10.625 FALL 2011

EXAM I (OPEN BOOK, in class)

October 4, 2011

Please state your assumptions clearly and show your analysis step by step

Short Questions (total 30 points) feel-good questions

Question 1.Which of the following is the strongest oxidizing agent? (5 pt)

A. H2O2in acid

B. MnO4-in acid

C. MnO4-in base

D. CrO42-

in acid

Please state your rationale.

Question 2.In an electrolytic cell to generate H2from water splitting, which statement is always

true? (5 pt)

A. H2evolves at the cathode

B. Protons migrates from anode to cathode

C. O2evolves at the cathode

D. The electrode for H2evolution is the positive electrode

Question 3. A steady current of 1.00 amp is passed through an electrolytic cell containing a

0.1 M aqueous solution of CuF2 using inert graphite electrodes until 1.54 g of copper is

deposited. (10 pt)A. At which electrode is the copper deposited? (anode or cathode) and what is the half-cell

reaction at the other electrode? (5 pt)

B. How many minutes does the current flow to obtain this deposit? (5 pt)

Question 4.Calculate the equilibrium constants for the following reactions at 25oC in aqueous

solutions. (10 pt)

(a) Zn(s)+ CuSO4(aq)Cu(s)+ ZnSO4(aq)

(b) 2H2(g)+ O2(g)2H2O(l)

1

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Problem #1 (10 points)

Extraction of lithium from LiFePO4, first reported by Goodenough and co-workers in 1997,1

results in the formation of FePO4 ((1-x) LiFePO4 + xFePO4, where the solid solution of

LixFePO4 does not exist). The following figure shows the charge and discharge profiles of

LiFePO4in an Li/FePO4cell:1

FePO4LiFePO4LiClO4in PC:DME(1:1)Li

[Ref]: (1) J.B. Goodenough et al., JES, 144 (4), 1997

(a) What are the half-cell reactions at the two electrodes and the whole-cell reaction? (2 pt)

(b) Please compute the gravimetric energy (in Wh/gLiFePO4) of the cell (by only considering

the weight of LiFePO4) (4 pt)

(c) Please explain why lithium removal from LiFePO4 exhibits a voltage plateau but not a

slopping voltage profile? Hint: develop your rational using Nernst equation (4 pt)

Molecular weight of LiFePO4is 157.75 g/mol

2

The Electrochemical Society, Inc. All rights reserved. This content is excluded from our Creative Commons licen

For more information, see http://ocw.mit.edu/help/faq-fair-use/.

Source: Padhi, A. K., K. S. Nanjundaswamy, et al. "Phospho.olivines as Positive-Electrode Materials for Rechargea

Lithium Batteries." Journal of the Electrochemical Society 144, no. 4 (1997): 1188-94.
http://ocw.mit.edu/help/faq-fair-use/http://dx.doi.org/10.1149/1.1837571http://dx.doi.org/10.1149/1.1837571http://dx.doi.org/10.1149/1.1837571http://dx.doi.org/10.1149/1.1837571http://ocw.mit.edu/help/faq-fair-use/

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Problem #3 (30 points)

Pourbaix diagrams are graphical representations of thermodynamic and electrochemical

equilibria occurring in aqueous systems. They are thus the electrochemical analogues of the

chemical phase stability diagram. In the Pourbaix diagram of Aluminum, the species

participating in the various chemical and electrochemical equilibria are the solids Al and Al2O3-

and ions Al3+

and AlO2 .

You are given:

Gfoo

(kJ/mol)

Al(s) 0

Al2O3(s) 1608.9

Al3+(aq) 481.2

AlO2

(aq) 839.8

H2O(l) 237.1

H+(aq) 0

(a) Write down the half-cell reaction between Al3+

and Al and plot it in a Pourbaix diagram for

[Al3+

] = 10-5

M and 1 M. (10 pt)

(b) Write down the half-cell reaction between Al2O3and Al and plot in the Pourbaix diagram for

PO2of 10-5

atm and 1 atm. (10 pt)-

(c) Plot the chemical reaction of Al3+

+ 2H2O AlO2 + 4H+ in the Pourbaix diagram, and

compute at what PH value, you would find [Al3+

] = [AlO2-]? (10 pt)

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Question 3. A steady current of 1.00 amp is passed through an electrolytic cell containing a

0.1 M aqueous solution of CuF2 using inert graphite electrodes until 1.54 g of copper is

deposited. (10 pt)

A. At which electrode is the copper deposited? (anode or cathode) and what is the half-cell

reaction at the other electrode? (5 pt)

B. How many minutes does the current flow to obtain this deposit? (5 pt)

Answer:

A: The copper is deposited at the cathode (Cu2+

+ 2e-Cu)

At the anode, we need to find the oxidation reaction:

Because 2F-F2+ 2e

--2.866V

2H2OO2+ 4H+

+ 4e-

-1.23V > -2.866V (water is easier to be oxidized than F-

)

So the reaction at anode is 2H2OO2+ 4H++ 4e

-

i t m (1.0) t 1.54B: Faradays law , t = 78 min

r 2 96485Z F 63.5 63.5

Question 4.Calculate the equilibrium constants for the following reactions at 25oC in aqueous

solutions. (10 pt)

(a) Zn(s)+ CuSO4(aq)Cu(s)+ ZnSO4(aq)

(b) 2H2(g)+ O2(g)2H2O(l)

Answer:

(a) The reaction can be written into two half cell reactions

ZnZn2+

+ 2e-

0.762V

Cu2+

+ 2e-Cu 0.342V

Overall: Zn + Cu2+Zn

2++ Cu 1.104V

oo kJG

Z F

(2)(96485)(1.104)

213.04r rmol

oo- G r 213040 37K= exp( ) exp( ) 2.210RT 8.314 298

2

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Problem #1

(a) What are the half-cell reactions at the two electrodes and the whole-cell reaction? (2 pt)

(b) Please compute the gravimetric energy (in Wh/gLiFePO4) of the cell (by only considering the

weight of LiFePO4) (4 pt)

(c) Please explain why lithium removal from LiFePO4 exhibits a voltage plateau but not aslopping voltage profile? Hint: develop your rational using Nernst equation (4 pt)

Molecular weight of LiFePO4is 157.75 g/mol

Answer:

(a) Positive electrode: LiFePO4FePO4+ Li++ e

-

Negative electrode: Li++ e

-Li

Overall: LiFePO4FePO4+ Li

(b) Energy = charge*voltage

r (1) CZ F FGravimetric charge = 611.63 170 mAh/gM.W. 157.75 g

Assume an average potential of 3.5 V

-3 AhE = 170 10 3.5 V = 0.595 Wh/g g

(c) According to Nernst equation

o

RT aLiFePO4

ln( ) Z F a ar Li FePO4

For a two-phase reaction, LiFePO4, FePO4are all pure substances (no mixing).

Hence, a a 1LiFePO FePO4 4

In addition, lithium is also pure substance, aLi 1

o

RT aLiFePO o ln( 4 ) , which is not a function of lithium content.

Z F a ar Li FePO4

Therefore, LiFePO4exhibits a voltage plateau upon lithium removal from LiFePO4.

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Problem #2

(a) Please write down the oxygen half-cell reaction and the net reaction of an Li-O2battery (right

graph), where the oxygen electrode is in an alkaline electrolyte (i.e, LiOH). What is the

equilibrium cell voltage if the cell operates at pH=14 and 1 atm O2for the oxygen electrode andat aLi+=1 for the lithium electrode (LiLi

++ e

-)? (5 pt)

Answer:

Positive electrode (cathode): O2(g)+ 4e-+ 2H2O(l)4OH

-(aq) 0.401V

Negative electrode (anode): 4Li(s)4Li+

(aq)+ 4e-

3.045V

Net reaction: 4Li(s)+ O2(s)+ 2H2O(l)4Li+

(aq)+ 4OH-(aq) 3.446V

RT a a2

a4

P 1 1 RTo

O2 H O2 Li o RT O2 o 1 1 1

ln( )

ln( )

ln( )

3.446 V,

cell

cell

4 cell

- cell4F a

+ aOH- 4F 4F1 [OH ] 1 1Li

assuming ideal gas for oxygen (aO2= PO2), and aOH- = [OH-] = 1M (pH = 14).

aLi(s)= aH2O(l)= 1 due to pure substance.

(b) Please write down the oxygen half-cell reaction and the net reaction for an Li-O2battery in an

acidic electrolyte (i.e., HClO4). What is the equilibrium cell voltage if the cell operate at 1M

HClO4, 1atm O2and at aLi+=1 for the lithium electrode (LiLi++ e

-)? (5 pt)

Answer:

Positive electrode (cathode): O2(g)+ 4H+

(aq)+ 4e-2H2O(l) 1.229V

Negative electrode (anode): 4Li(s)4Li+

(aq)+ 4e-

3.045V

Net reaction: 4Li(s)+ 4H+

(aq)+ O2(g)2H2O(l)+ 4Li+

(aq) 4.274V

4 4 + 4

o

RT aO2 aH+ aLi o RT PO2 [H ] 1 RT o 1 1 1cell

cell ln( 4 2 ) cell ln( ) cell ln( ) 4.274 V ,4F a + aH O 4F 1 1 4F 1 1Li 2

assuming ideal gas for oxygen (aO2= PO2), and aH+= [H+] = 1M

aLi(s)= aH2O(l)= 1 due to pure substance.

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(c) Please write down the oxygen half-cell reactions and the net reaction for an Li-O 2battery in

non-aqueous electrolytes (i.e., LiClO4) in the left graph. What is the equilibrium cell voltage

(1atm O2and 1M LiClO4) if the final product is Li2O2and Li2O at 298 K, respectively? (10 pt)

Answer:

If the final product is Li2O2

Positive electrode (cathode): O2(g)+ 2Li++ 2e

-Li2O2(s)

Negative electrode (anode): 2Li(s)2Li++ 2e

-

Net reaction: 2Li(s)+ O2(g)Li2O2(s)

RT aO2 a2

o

RT PO2 a2

o

RT 1 1 oo

Li Li ln( ) ln( ) ln( ) , assuming ideal gascell

cell

cell

cell

cell

2F a 2F a 2F 1Li O Li O2 2 2 2

for oxygen (aO2= PO2), and aLi+= [Li+] = 1M (1M LiClO4).

aLi(s)= aLi2O2(s)= 1 due to pure substance.

oo oo ooo 2 G ) ( 570900 0)-G ( G f, Li O Gf, Li f, O 0 o

r 2 2(s) (s) 2(g)cell

2.96 V

2F 2F 2(96485)

If the final product is Li2O

Positive electrode (cathode): O2(g)+ 4Li++ 4e

-2Li2O(s)

Negative electrode (anode): 4Li(s)4Li++ 4e

-

Net reaction: 4Li(s)+ O2(g)2Li2O(s)

RT aO2 a4

o

RT PO2 a4

o

RT 1 1 oo

Li Li ln( ) ln( ) ln( ) , assuming ideal gascell

cell 2 cell 2 cell cell4F a 4F a 4F 1Li O Li O2 2 2

for oxygen (aO2= PO2), and aLi+= [Li+] = 1M (1M LiClO4).

aLi(s)= aLi2O(s)= 1 due to pure substance.

oo oo oo

-G (2 Gf, Li O 4 Gf, Li f, O ) 0 o G ( 561200 0)o r 2 (s) (s) 2(g)

cell

2.91 V

4F 4F 4(96485)

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(c) Identify the metal(s), with corresponding metal ions in solution, that can plate

spontaneously vs SHE.

a. Ba/Ba2+

b. Pt/Pt2+

c. Ru/Ru2+

d.

Ca/Ca

2+

Does the driving force for metal plating increase or decrease as pH increases? Please

show why using a Pourbaix diagram.

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which describes the enthalpy per site in the lattice associated with interactions between

intercalated phases within the electrode.

The discharge voltage of the positive elec

trode

=

i

0

s

gi

ve

(

n b

)

y

We assume the negative electrode is lithium metal, and therefore the positive electrode voltage is

referenced to V vs. Li.

(a) Develop an expression for the Gibbs free energy, g, per site as a function of x and

given constants.

(b) What is the chemical potential, , as a function of x? What is the cell voltage, , as a

function of x?

(c) Consider that the positive electrode behaves as a solid solution during discharge.

Assuming that the enthalpic contributions, h, to the free energy, g, are negligible such

that ho = 0 (i.e., the free energy is determined predominantly by the entropy ofmixing). What equation results? Does this look familiar? Assuming a value of

0= 4

V vs. Li and kBT/e = 0.1, sketch (x) for the solid solution case below.

(d) Now assume that the enthalpic contributions are significant, such that kBT/e = 0.1 and

ho = 2kBT, which is representative of a two-phase intercalation process where mixing

between phases is not thermodynamically favorable. Sketch the shape of as a

function of x assuming 0

= 3.6 V below.

(e) Match your voltage profiles to the experimental data of two positive electrode

materials shown below. Which positive electrode material exhibits solid solution

behavior and which exhibits two-phase behavior? Briefly describe the physical

mechanisms happening during discharge for these

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