10.626 Electrochemical Energy Systems

10.626 ElectrochemicalEnergySystems Spring2014 MIT,M.Z.Bazant

Midterm Exam

Instructions. Thisis atake-home, open-bookexamdue in Late examswill notbeaccepted.You mayconsult any books,handouts, oronlinemateriallistedon the syllabus,butyou must workindependently,without consultingany other person.

1. Discharge of a Reaction-LimitedBattery. Abatteryhasconstantopencircuitvoltage V_O,constantinternalseriesresistance R_int,andvariableFaradaicresistanceatoneelectrode, given by thesymmetric Butler-Volmerequation

−eη/2kT − e^eη/2kT I =I₀ e

Deriveandsketchthevoltageversuscurrent,V(I),forbatterydischargeatconstantcurrent.

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Webegin with

V =VO− ηint+ηc− ηa (1)

where η_int is the loss from the internal resistance and η_a,c is the activation loss from the Faradaic reactionsatthecathodeand anode. Weknow thatthelossfrominternalresistance is given by

ηint=IRint, (2)

and thereaction loss, η_act, is given.

−eη_act/2kT − e^eηact/2kT

I =I₀ e = −2I₀ sinh(eη_act/2kT) (3) ηact= − 2kT

e sinh⁻¹ I

2I₀ . (4)

Notethatwaywehavewrittenthecurrentrelationispositivewhennetreductionisoccurring.

Thus, for current defined as positive during discharge, we are describing the cathode with this relation. We would introduce a sign change todescribe thecurrent at theanode where netoxidation is occurring. For simplicity,weassume therelationapplies tothecathodeand assume no lossesatthe anode. Thus,

V =VO− IRint− 2kT

e sinh⁻¹ I

2I₀ , (5)

which is sketchedinFigure1.

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2. Voltage Hysteresis in a Li-ion Battery. The homogeneousfreeenergyper siteofa Li­

ionbattery cathodeatfillingfractionx isgivenby theregularsolution model. Theenthalpy of mixingispositiveh₀ >0,andthetemperatureis belowthecriticaltemperatureforphase separation. Neglecttheinterfacialtensionbetweenphasesandfinitesizeeffects. Assume that nucleation is not possible. Theanodeand electrolyteremainatconstantchemical potentials, and theopen circuit voltageis V⁰ athalf fillingof thecathode.

Lecture 22.

Figure 1: Sketchof I-VcurveforP1. The dotted green line representstheopencircuit voltage.

(a) Write down and plot the open circuit voltage versus mean filling fraction x, for both homogeneousand phase-separatedstates.

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Keepingtheanodeand electrolyteatconstantchemicalpotential,thebatteryvoltageis given by thechemical potentialofthe reduced stateinthecathode, μ,

V =V_O − μ. (6)

e

Webeginwith theregularsolution model asafunctionof thelocal concentration,non­

dimensionalized tothemax concentration,c

μ_h(c) =kTlog c + Ω_a(1 − 2c). (7) 1 − c

Thus,fora homogeneouslyfilling cathode, c=x,μ=μ_h(x), and μ_h(x)

V =V_O − (homogeneouslyfilling). (8) e

This isplottedin Figure2a.

In the phaseseparating state, we accept two different responses. First, attrue equilib­

rium, whenever the system is betweenthe two free energyminima, it should be in the free energy minimum – the phase separated state. The two free energy minima occur atthe“binodal”,which wewill refer toasx_b,± fortheupperand lowerspinodalpoints.

Because of the symmetry in our free energy model, we can solve for these points by setting μ_h(c) = 0 and picking the solutions near 0 and 1 (i.e. c = 0 .5), which can be solvednumerically.

V_O x ∈ [x_b,− , x_b,+]

V = . (9)

μ_h(x) else This isplottedin Figure2b.

Asecondacceptableresponsefollows: Ifa(verysmall)finitecurrentisassumedwithout nucleation,phase separationwill occuratthespinodal, which occurswhen g"" (c) = 0 or μ^"_h(c) = 0.

μ^"_h(c) =kT 1

+ 1

− 2Ωa. (10) c 1 − c

6

�

(a) Homogeneous cathode volt- (b) Equilibrium phase separat­

age. ingcathodevoltage.

Figure2: Voltageofhomogeneousand phase separatingcathodes.

Thus,denotingthelowerandupperspinodalhomogeneouscompositionsasx_s− andx_s+, they aregiven by

x_s,± = 0.5 1 ± 1 − 2/Ω0_a (11)

where Ω0_a = Ωa/kT. After phase separation, because our model for the free energy is symmetric around c = 0.5, the chemical potential will be 0 until the filling fraction reaches the other free energy minimum and returns to a homogeneous state. Again because thesymmetry inour model forthe free energy, we cansolve for this point by setting μ_h(c) = 0, which can be solved numerically. We will denote these two limits x_b,±. Thus,with spinodalphase separation, there will bedifferent voltages whenfilling (discharging) oremptying (charging) thecathode. Forfilling,

μ_h(x) else

V = V_O x∈ [x_s,−, x_b,+]

μ_h(x) else . (12)

And foremptying,

V = VO x∈ [xb,−, xs,+]

. (13)

These areplottedinFigures3a and 3b.

(a) Phase separating cathode (b) Phase separating cathode voltageuponfilling(discharge). voltageuponemptying(charge).

Figure3: Voltageof phaseseparating cathodes assumingvery slow(dis)chargecurrents.

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(

(

(b) Onthis plot,also sketcha closedcurve torepresent slowcyclicvoltammetry,where the voltageissweptveryslowlybackandforthbetweenlargeandsmallvalues. Explainwhy there is hysteresis,i.e. different curvesfordischargingand charging.

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A representative CV curve for very slow scan rates involves following the curves in Figures 3a and 3b with some modifications. Asthe voltage is loweredfrom some large value, the cathode will be nearly empty, and the voltage will track the left part of Figure 3a. Then, when the spinodal is reached, because we are linearly sweeping the voltage, thefillingfractionwillrapidlygofromthelowspinodallimittotheintersection of the homogeneous curve at high filling fraction and the current applied voltage. In reverse (starting at low voltages and high filling fractions), the opposite occurs. The voltage will initially track Figure3b as thecathode empties,then when thespinodalis reached, the cathode will quickly empty until it reaches the intersection at low filling fractions. Thisisdepicted inFigure 4.

Figure 4: Slow CVscan ofa singeparticlewithphase transformationsby nucleation.

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(c) Deriveaformulaforthe“voltagegap”betweencharginganddischargingplateausinthe limitof zero current.

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Thevoltagegapisrelatedtothedifferenceinchemicalpotentialsatthelowerandupper spinodals. Thus,nondimensionlizing by thethermal voltage/energy

ΔV0_gap =μ0h(x_s,−) − μ0h(x_s,+) (14)

= 2 Ω0 1 − 2

− 2tanh⁻¹ 1 − 2

. (15)

0 0

Ω Ω

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3. Hydrogen-Bromine Flow Battery: Water Electrochemistry. Duringdischarge, the battery converts hydrogen gas (H₂) and liquid bromine (Br₂) to hydrobromic acid (HBr).

The half-cellreactionsare

anode: H₂ → 2H⁺ + 2e− E^Θ = 0

cathode: Br₂ + 2e⁻ → 2Br⁻ E^Θ = 1.087V

The electrolyte is 1M HBr(aq) with 1M Br₂(aq)added near the cathode and 1 atm H₂ gas attheanode, atroomtemperature.

(a) How doesthe cellvoltagevarywith pH?

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We will denote the reactions as anode = 1, cathode = 2, oxygen evolution = 3. The Nernst equation for each reaction requires (assuming room temperature, taken from class, and given inVolts)

E₁ = −0.06pH E₂ = 1.087

E₃ = 1.229 − 0.06pH.

Thus,thevoltage isE_c − E_a =E₂ − E₁ = 1.087+0.06pH,

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(b) Make a Pourbaix diagram for the half-cell reactions, as well as the oxygen evolution reaction (i.e. electrolysis, orwatersplitting).

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Weplot theabove equationsinFigure 5.

Figure 5: Pourbaixdiagram forHBrFlow Battery.

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(c) What is the upper bound for pH to avoid oxygen evolution at the cathode near open circuit conditions?

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Oxygen evolution will occur whenever an electrode potential lies above the E₃ curve.

Near opencircuit conditions, this only occurs whenE₂ > E₃, or pH>2.367.

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(d) What is the upper bound for cathodic overpotential to avoid oxygen evolution during battery recharging?

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When charging, net oxidiation is occurring at the cathode, so the electrode potential there ishigherthantheequilibriumcurves. Inordertodrivereaction 2intheoxidation direction, theelectrode potential,E_c mustbe aboveE₂. Similarly, fortheoxygentobe evolved, the E_c must be above E₃. Thus we require that E₂ < E_c < E₃. Notingthat theoverpotentialforthebrominereactionisηc =Ec− E₂ < E₃ − E₂ = 0.142 − 0.06pH.

Because webeginwith1M HBr,wecanassume thatthe pHis initiallyzero,leading to 0.142 V maximumoverpotential.

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4. Hydrogen-Bromine Flow Battery: Polybromide complexes. In hydrobromic acid, bromine canform tribromideandpentabromide ioncomplexes

Br₂ + Br⁻ → Br⁻ ₃ K₃ = 16.7

2Br₂ + Br⁻ → Br⁻ ₅ K₅ = 37.7

where K₃ and K₅ are theequilibrium constants (Molar). Assume room temperature, dilute solution approximations (activity =molarconcentration) and hydrogengasat1 atm.

(a) What is theequilibrium constantofthesecond complexation reaction, Br⁻ ₃ + Br₂ → Br⁻ ₅ K=?

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When inequilibrium,

eq eq eq

a =K₃a_Br

2 a (16)

Br⁻ ₃ Br⁻ eq eq

a_Br− =K₅(aBr₂ )² a Br⁻ . (17)

5

The equilibrium constantofthe givenreaction is,by definition, aeq_Br−

K = _eq ⁵ _eq (18)

aBr⁻ a_Br

3 2

eq eq

K₅(a_Br2 )²a_Br−

= _{eq eq} (19)

K₃(a_Br )²a

2 Br⁻

K₅ 37.7

= = ≈ 2.26. (20)

K₃ 16.7

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(b) What are thestandardpotentials ofbromine reduction tothepolybromidecomplexes?

3Br₂ + 2e⁻ → 2Br⁻ ₃ E₃ ^Θ =?

5Br₂ + 2e⁻ → 2Br⁻ ₅ E₅ ^Θ =?

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The standardpotentialis obtainedby using theNernstequation.

⎛ ⎞

3 2

kT a_Br2 a_e

E₃ =E₃ ^Θ + ln⎝ ₂ ⎠ (21)

2e a_Br− 3

3 2

kT a_Br2 a_e

=E₃ ^Θ + ln (22)

2e (K₃a_Br2 a_Br− )² kT kT a_Br2 a² _e

=E₃ ^Θ − lnK₃ + ln ₂ . (23)

e 2e a_Br−

Now, wenotethatwecanrelatethelasttermtothesimpleBrominereductionreaction from P3,which wewill nowdenote asreaction B

kT aBr₂ a_e ²

EB=E_B ^Θ + ln ₂ . (24)

2e a_Br−

Ifthesereactionsarebothinequilibriumatthesameelectrode,thereisonlyone“metal”

potential, soE₃ =E_B =E,and (inVolts) E=E₃ ^Θ − kT

e lnK₃ + (E − E_B ^Θ ) (25) E₃ ^Θ = kT

e lnK₃ +E_B^Θ = 1.087+0.026ln16.7 = 1.160. (26) Similarly,

⎛ ⎞ E=E₅ ^Θ +kT

2e ln⎝ a⁵ _Br2 a² _e a²

Br⁻ ₅

⎠ =E₅ ^Θ − kT

e lnK₅ +kT

2e ln aBr₂ a² _e

a² _Br− (27) E^Θ = kT

lnK₅ +E_B ^Θ = 1.181. (28)

5 e

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(c) What aretheconcentrationsof Br⁻ ₃ and Br⁻ ₅ inequilibrium withareservoirof 1MHBr + 1M Br₂? Can this equilibriumever bereached?

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Usingtherelations from part(a),

eq eq eq

a_Br− =K₃a_Br2 a_Br−=K₃ = 16.7 M (29)

3 eq eq

a_Br− ₅ =K₅(aBr₂ )² aBr⁻ =K₅ = 37.7 M, (30)

which cannot beachieved because thesevalues exceedthe solubilitylimits.

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(d) If instead the total concentration of bromine (in all forms: Br₂, Br⁻ ₃ , Br⁻ ₅ ) is fixed at the initial Br₂ concentration of 1M (prior to complexation reactions) and the system equilibrates incontact witha reservoirof 1M HBr,what isthe opencircuit voltage?

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First we notethat the first two given specieseach have oneequivalent of Br₂, whereas Br−₅ contains2equivalentsofBr₂. FixingthetotalconcentrationofBr₂ andusingthat toobtain theOCV, weare alsoin equilibrium

c_Br2 +c_Br3− + 2cBr⁻ ₅ =1 (31) aBr₂ +a_Br− ₃ + 2aBr⁻ ₅ =1 (32) a_Br2 (1+K₃a_Br− + 2K₅a_Br2 a_Br− )=1 (33) a_Br2 (1+K₃ + 2K₅a_Br2 ) = 1 (a_Br− =1 from reservoir) (34) 2K₅a2_Br2 + (1 +K₃)a_Br2 − 1=0 (35)

−(1+K₃) + (1 +K₃)² + 8K₅

a_Br2 = = 0.047. (36)

4K₅

Then,assumingunitactivityforelectrons, wecanusetheNernstequationtodetermine theOCV,

VO = 1.087+kT

2e ln aBr₂ a² _e

a² _Br− (37)

= 1.087+kT

2e ln(a_Br2 ) (38)

= 1.047V, (39)

which demonstratesthat“occupying”someoftheBr₂ via thecomplexes,while keeping theBr⁻ fixedreducesthe OCV.

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(e) Extra credit: If the total concentrations of all forms of bromine (Br₂, Br⁻₃, Br⁻₅ ) and of bromide (Br⁻, Br⁻₃, Br⁻₅ ) are each fixed at 1M (for an initial solution of 1M Br₂ + 1M HBr, priorto complexationreactions) and allowedtoreach equilibrium in a closed system (noreservoir),what isthe opencircuit voltage?

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Here, ratherthanhaving a_Br− = 1, wehavethat

+a +a =1 (40)

a_Br− Br₃⁻ Br₅ ⁻

a_Br− +K₃a_Br− a_Br2 +K₅a2 _Br2 a_Br− =1 (41)

and from theconstraint on bromine,

a_Br2 +K₃a_Br− a_Br2 + 2K₅a2 _Br2 a_Br− = 1. (42)

These two equationscanbesolvednumericallytoobtain

a_Br2 = 0.12 a_Br− = 0.28. (43)

Thus,

V_O = 1.087+0.013 ln 0.12 = 1.093V, (44) 0.28²

whichisquiteclosetothestandardpotentialbecausebothBr₂ and Br⁻ concentrations were reduced.

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10.626 Electrochemical Energy Systems

Spring 2014

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