Electrochemical Energy Systems Spring 2014 MIT, M. Z. Bazant. Midterm Exam

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1 10.66 Elctrochmical Enrgy Systms Spring 014 MIT, M. Z. Bazant Midtrm Exam Instructions. This is a tak-hom, opn-book xam du in Lctur. Lat xams will not b accptd. You may consult any books, handouts, or onlin matrial listd on th syllabus, but you must work indpndntly, without consulting any othr prson. 1. Discharg of a Raction-Limitd Battry. A battry has constant opn circuit voltag V O, constant intrnal sris rsistanc R int, and variabl Faradaic rsistanc at on lctrod, givn by th symmtric Butlr-Volmr quation I = I 0 η/kt η/kt Driv and sktch th voltag vrsus currnt, V (I), for battry discharg at constant currnt. W bgin with V = V O η int + η c η a (1) whr η int is th loss from th intrnal rsistanc and η a,c is th activation loss from th Faradaic ractions at th cathod and anod. W know that th loss from intrnal rsistanc is givn by and th raction loss, η act,isgivn. η int = IR int, () η act/kt ηact/kt I = I 0 = I 0 sinh (η act /kt) () η act = kt sinh 1 I I 0. (4) Not that way w hav writtn th currnt rlation is positiv whn nt rduction is occurring. Thus, for currnt dfind as positiv during discharg, w ar dscribing th cathod with this rlation. W would introduc a sign chang to dscrib th currnt at th anod whr nt oxidation is occurring. For simplicity, w assum th rlation applis to th cathod and assum no losss at th anod. Thus, V = V O IR int kt sinh 1 I I 0, () which is sktchd in Figur 1.. Voltag Hystrsis in a Li-ion Battry. Th homognous fr nrgy pr sit of a Liion battry cathod at filling fraction x is givn by th rgular solution modl. Th nthalpy of mixing is positiv h 0 > 0, and th tmpratur is blow th critical tmpratur for phas sparation. Nglct th intrfacial tnsion btwn phass and finit siz ffcts. Assum that nuclation is not possibl. Th anod and lctrolyt rmain at constant chmical potntials, and th opn circuit voltag is V 0 at half filling of th cathod. 1

2 Figur 1: Sktch of I-V curv for P1. Th dottd grn lin rprsnts th opn circuit voltag. (a) Writ down and plot th opn circuit voltag vrsus man filling fraction x, for both homognous and phas-sparatd stats. Kping th anod and lctrolyt at constant chmical potntial, th battry voltag is givn by th chmical potntial of th rducd stat in th cathod, μ, μ V = V O. (6) W bgin with th rgular solution modl as a function of th local concntration, nondimnsionalizd to th max concntration, c c μ h (c) = kt log +Ω a (1 c). (7) 1 c Thus, for a homognously filling cathod, c = x, μ = μ h (x), and μ h (x) V = V O (homognously filling). (8) This is plottd in Figur a. In th phas sparating stat, w accpt two diffrnt rsponss. First, at tru quilibrium, whnvr th systm is btwn th two fr nrgy minima, it should b in th fr nrgy minimum th phas sparatd stat. Th two fr nrgy minima occur at th binodal, which w will rfr to as x b,± for th uppr and lowr spinodal points. Bcaus of th symmtry in our fr nrgy modl, w can solv for ths points by stting μ h (c) = 0 and picking th solutions nar 0 and 1 (i.. c =0.), which can b solvd numrically. V O x [x b,,x b,+ ] V =. (9) μ h (x) ls This is plottd in Figur b. A scond accptabl rspons follows: If a (vry small) finit currnt is assumd without nuclation, phas sparation will occur at th spinodal, which occurs whn g "" (c) =0 or μ " h (c) =0. μ " h (c) = kt Ω a. (10) c 1 c

3 (b) Equilibrium phas sparat ing cathod voltag. (a) Homognous cathod voltag. Figur : Voltag of homognous and phas sparating cathods. Thus, dnoting th lowr and uppr spinodal homognous compositions as x s and x s+, thy ar givn by ( ) x s,± =0. 1 ± 1 /Ω 0 a (11) whr Ω 0 a =Ω a /kt. Aftr phas sparation, bcaus our modl for th fr nrgy is symmtric around c =0., th chmical potntial will b 0 until th filling fraction rachs th othr fr nrgy minimum and rturns to a homognous stat. Again bcaus th symmtry in our modl for th fr nrgy, w can solv for this point by stting μ h (c) = 0, which can b solvd numrically. W will dnot ths two limits x b,±. Thus, with spinodal phas sparation, thr will b diffrnt voltags whn filling (discharging) or mptying (charging) th cathod. For filling, { V O x [x s,,x b,+ ] V =. (1) μ h (x) ls And for mptying, { V O x [x b,,x s,+ ] V =. (1) μ h (x) ls Ths ar plottd in Figurs a and b. (a) Phas sparating cathod voltag upon filling (discharg). (b) Phas sparating cathod voltag upon mptying (charg). Figur : Voltag of phas sparating cathods assuming vry slow (dis)charg currnts.

4 (b) On this plot, also sktch a closd curv to rprsnt slow cyclic voltammtry, whr th voltag is swpt vry slowly back and forth btwn larg and small valus. Explain why thr is hystrsis, i.. diffrnt curvs for discharging and charging. A rprsntativ CV curv for vry slow scan rats involvs following th curvs in Figurs a and b with som modifications. As th voltag is lowrd from som larg valu, th cathod will b narly mpty, and th voltag will track th lft part of Figur a. Thn, whn th spinodal is rachd, bcaus w ar linarly swping th voltag, th filling fraction will rapidly go from th low spinodal limit to th intrsction of th homognous curv at high filling fraction and th currnt applid voltag. In rvrs (starting at low voltags and high filling fractions), th opposit occurs. Th voltag will initially track Figur b as th cathod mptis, thn whn th spinodal is rachd, th cathod will quickly mpty until it rachs th intrsction at low filling fractions. This is dpictd in Figur 4. Figur 4: Slow CV scan of a sing particl with phas transformations by nuclation. (c) Driv a formula for th voltag gap btwn charging and discharging plataus in th limit of zro currnt. Th voltag gap is rlatd to th diffrnc in chmical potntials at th lowr and uppr spinodals. Thus, nondimnsionlizing by th thrmal voltag/nrgy ΔV 0 gap = μ0 h (x s, ) μ0 h (x s,+ ) (14) ( = Ω 0 1 ( tanh 1 1 )). (1) Ω 0 Ω 0. Hydrogn-Bromin Flow Battry: Watr Elctrochmistry. During discharg, th battry convrts hydrogn gas (H ) and liquid bromin (Br ) to hydrobromic acid (HBr). Th half-cll ractions ar anod: H H + + E Θ =0 cathod: Br + E Θ =1.087V Thlctrolytis1MHBr(aq) with 1M Br (aq) addd nar th cathod and 1 atm H gas at th anod, at room tmpratur. 4

5 (a) How dos th cll voltag vary with ph? W will dnot th ractions as anod = 1, cathod =, oxygn volution =. Th Nrnst quation for ach raction rquirs (assuming room tmpratur, takn from class, and givn in Volts) E 1 = 0.06 ph E =1.087 E = ph. Thus, th voltag is E c E a = E E 1 = ph, (b) Mak a Pourbaix diagram for th half-cll ractions, as wll as th oxygn volution raction (i.. lctrolysis, or watr splitting). W plot th abov quations in Figur. Figur : Pourbaix diagram for HBr Flow Battry. (c) What is th uppr bound for ph to avoid oxygn volution at th cathod nar opn circuit conditions? Oxygn volution will occur whnvr an lctrod potntial lis abov th E curv. Nar opn circuit conditions, this only occurs whn E >E,or ph >.67. (d) What is th uppr bound for cathodic ovrpotntial to avoid oxygn volution during battry rcharging? Whn charging, nt oxidiation is occurring at th cathod, so th lctrod potntial thr is highr than th quilibrium curvs. In ordr to driv raction in th oxidation dirction, th lctrod potntial, E c must b abov E. Similarly, for th oxygn to b volvd, th E c must b abov E. Thuswrquir that E <E c <E. Noting that th ovrpotntial for th bromin raction is η c = E c E <E E = ph. Bcaus w bgin with 1M HBr, w can assum that th ph is initially zro, lading to 0.14 V maximum ovrpotntial.

6 4. Hydrogn-Bromin Flow Battry: Polybromid complxs. In hydrobromic acid, bromin can form tribromid and pntabromid ion complxs Br + K =16.7 Br + K =7.7 whr K and K ar th quilibrium constants (Molar). Assum room tmpratur, dilut solution approximations (activity = molar concntration) and hydrogn gas at 1 atm. (a) What is th quilibrium constant of th scond complxation raction, +Br K =? Whn in quilibrium, q q q q q Br. a = K a Br a (16) a = K (a ) a (17) Th quilibrium constant of th givn raction is, by dfinition, q a Br K = q q (18) a Br a Br K q q (a Br ) abr = q q (19) K (a Br ) a K 7.7 = =.6. (0) K 16.7 (b) What ar th standard potntials of bromin rduction to th polybromid complxs? Br + E Θ =? Br + E Θ =? Th standard potntial is obtaind by using th Nrnst quation. kt abr a E = E Θ + ln (1) a Br Br kt a a = E Θ + ln () (K a Br a ) ( kt kt a Br a ) = E Θ ln K + ln. () a Now, w not that w can rlat th last trm to th simpl Bromin rduction raction from P, which w will now dnot as raction B kt a Br a E B = EB Θ + ln. (4) a 6

7 If ths ractions ar both in quilibrium at th sam lctrod, thr is only on mtal potntial, so E = E B = E, and (in Volts) Similarly, E = E Θ kt ln K +(E E Θ B ) () E Θ = kt ln K + E Θ B = ln 16.7 = (6) E = E Θ + kt ln a Br a a = E Θ kt ln K + kt ln a Br a a (7) E Θ = kt ln K + EB Θ = (8) (c) What ar th concntrations of and in quilibrium with a rsrvoir of 1M HBr +1MBr? Can this quilibrium vr b rachd? Using th rlations from part (a), q q q Br q q Br a = K a a = K =16.7 M (9) a = K (a ) a = K =7.7 M, (0) which cannot b achivd bcaus ths valus xcd th solubility limits. (d) If instad th total concntration of bromin (in all forms: Br,, )isfixdat th initial Br concntration of 1M (prior to complxation ractions) and th systm quilibrats in contact with a rsrvoir of 1M HBr, what is th opn circuit voltag? First w not that th first two givn spcis ach hav on quivalnt of Br,whras Br contains quivalnts of Br. Fixing th total concntration of Br and using that to obtain th OCV, w ar also in quilibrium c Br + c +c = 1 (1) a Br + a +a = 1 () a Br (1 + K a +K a Br a ) = 1 () a Br (1 + K +K a Br )=1 (a = 1 from rsrvoir) (4) K a +(1+ K ) a Br 1 = 0 () Br (1 + K )+ (1 + K ) +8K a Br = = (6) 4K Thn, assuming unit activity for lctrons, w can us th Nrnst quation to dtrmin th OCV, V O = kt ln a Br a a (7) kt = ln (a Br ) (8) =1.047 V, (9) 7

8 which dmonstrats that occupying som of th Br via th complxs, whil kping th fixd rducs th OCV. () Extra crdit: If th total concntrations of all forms of bromin (Br,, )and of bromid (,, ) ar ach fixd at 1M (for an initial solution of 1M Br + 1M HBr, prior to complxation ractions) and allowd to rach quilibrium in a closd systm (no rsrvoir), what is th opn circuit voltag? Hr, rathr than having a =1, w hav that a and from th constraint on bromin, + a + a = 1 (40) a + K a a Br + K a Br a = 1 (41) a Br + K a a Br +K a Br a =1. (4) Ths two quations can b solvd numrically to obtain a Br =0.1 a =0.8. (4) Thus, 0.1 V O = ln =1.09 V, (44) 0.8 which is quit clos to th standard potntial bcaus both Br and concntrations wr rducd. 8

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