RateofelectrochemicalreactionsThehydrogen oxidation reactionisfastatPt(seewhy in a minute)sotheoverpotential fortheanodicreaction canbeignored.InPEMFCstheoxygenreduction reaction(ORR)is sluggish-requires significantoverpotential at the cathode,n°.ORRmechanismcomplex,butassumeitfollowsthisatlowoverpotentials:Step1:O,→Oad+OadrateconstantkStep 2: Oad +e-→O-adk2and k4 =10-1kg = 10-8kz~ 10-12kStep 3: O-ad + H+→ HOad k3Step 4: HOad + H+ + e- → H,O k4Sostep4isrdsandotherelectronsaretransferred"forfree"withnegligibleoverpotential.The question is:How does the applied potential change the rateof the rds (step4)?Answer-Butler-Volmerkinetics!
Rate of electrochemical reacGons • The hydrogen oxidaGon reacGon is fast at Pt (see why in a minute) so the overpotenGal for the anodic reacGon can be ignored. • In PEMFCs the oxygen reducGon reacGon (ORR) is sluggish – requires significant overpotenGal at the cathode, ηc . • ORR mechanism complex, but assume it follows this at low overpotenGals: • Step 1: O2 à Oad + Oad rate constant k1 • Step 2: Oad + e– à O– ad k2 • Step 3: O– ad + H+ à HOad k3 and k4 ≈10-1k3 ≈ 10-8k2 ≈ 10-12k1 • Step 4: HOad + H+ + e– à H2O k4 • So step 4 is rds and other electrons are transferred “for free” with negligible overpotenGal. • The quesGon is: How does the applied potenGal change the rate of the rds (step 4)? Answer – Butler-Volmer kineGcs!
Rateofelectrochemical reactionsCurrent is simply a measure of the rate of an electrochemical reaction so we writeitasasimple1oratelaw:i,=k(n)[O,l and k(n)is a rate constant that varies as a function of overpotetnial,nWecanexpresskinanArrheniusform:Eawhere E is the activation energy,A is the pre-exponential Arrhenius termk,=AexpRTAt EQM E~AG=-nFEo.Byapplyingapotential, E, todrivethe reaction awayfrom equilibrium we lower the activation barrier by a fraction -aFn where n=E-E@E?AGdAGOx↑ (I-α)F(E-E)(E-EeαF(E-EO)AG=FEOxRedRedrxn co-ordinaterxnco-ordinate
• Current is simply a measure of the rate of an electrochemical reacGon so we write it as a simple 10 rate law: • We can express k in an Arrhenius form: • At EQM Ea ≈ ΔG0 = -nFE0. By applying a poten:al, E, to drive the reac:on away from equilibrium we lower the ac:va:on barrier by a frac:on Rate of electrochemical reacGons if = k (η)[O2 ] and k (η) is a rate constant that varies as a function of overpotetnial, η kf = Aexp − Ea RT " # $ % & ' where Ea is the activation energy, A is the pre-exponential Arrhenius term −αFη where η=E - E0
Rate of electrochemical reactionsNow ouractivation energy is simply E-αFn plugback into Arrhenius:EEaFtheAexpterm is simplythestandardrateconstantatEQM,kk,=AexpRTRTRTαFSok,=k'expRTAndourcurrentdensity(volumetricchargepersecond)isnow:ai=i.[O,Texpnote:[O,Jis surface conc.of O,/eqm conc ofO,RTi,is the"exchange current density"a measure of the catalyst performance.ClosetoEQMthereversereactionoccurs:2H,O→O,+4H++4eIt is a simplematterto derive the Butler-Volmerformulation for this oxidation,combining with O, reduction we getthe full Butler-Volmer (note stoichiometrycoefficients in rateequations appearin B-V:i-510,Texp(cFn)-H,or"exp(-L-)人(Eq. 1.4)RTWhatthis tells us isthatthecurrent (rate)varies logarithmicallywithoverpotential!NB:here reduction currents arepositive (Ipreferotherway around)
• Now our acGvaGon energy is simply plug back into Arrhenius: • i 0 is the “exchange current density” a measure of the catalyst performance. • Close to EQM the reverse reacGon occurs: 2H2O à O2 + 4H+ + 4e– • It is a simple maber to derive the Butler-Volmer formulaGon for this oxidaGon, combining with O2 reducGon we get the full Butler-Volmer (note stoichiometry coefficients in rate equaGons appear in B-V: • What this tells us is that the current (rate) varies logarithmically with overpotenGal! NB: here reducGon currents are posiGve (I prefer other way around) Rate of electrochemical reacGons i = i 0 [O2 ] * exp αF RT η ! " # $ % &−[H2O] *2 exp − (1−α)F RT η ! " # $ % & ( ) * +* , - * .* Ea 0 −αFη kf = Aexp − Ea 0 RT " # $ % & 'exp αF RT η " # $ % & ' the Aexp − Ea 0 RT " # $ % & ' term is simply the standard rate constant at EQM, k 0 : So kf = k 0 exp αF RT η " # $ % & ' And our current density (volumetric charge per second) is now: i = i 0[O2 ] * exp αF RT η " # $ % & ' note: [O2 ] * is surface conc. of O2 / eqm conc of O2 (Eq. 1.4)
Butler-VolmeratEQM=NernstAtEQM anykinetic model should givethethermodynamic resultIt isa simple matterto showthatthe Butler-Volmerequation collapsestotheNernsteguation,ifwerememberthat:1.atEQM[]'=bulkconcentrationofthatspecies;2.AtEQMnonetcurrentisflowing,i=0.3.n=E-EOThisisleftasanexerciseforyoutofollow:provethatfortheORRprocess1-α)Fi=i[0,fexpl]-[H,"expRTRT[O,RTBecomesEoc=Eo[H,nF
• At EQM any kineGc model should give the thermodynamic result. • It is a simple maber to show that the Butler-Volmer equaGon collapses to the Nernst equaGon, if we remember that: 1. at EQM [ ]*= bulk concentraGon of that species; 2. At EQM no net current is flowing, i = 0. 3. η = E – E0 • This is leÅ as an exercise for you to follow: prove that for the ORR process Becomes Butler-Volmer at EQM = Nernst i = i 0 [O2 ] * exp αF RT η ! " # $ % &−[H2O] *2 exp − (1−α)F RT η ! " # $ % & ( ) * +* , - * .* EOC = EOC 0 − RT nF ln [O2 ] [H2O] 2 " # $ % &
SowhatdoestheB-V model tell us?K°=1E-5k°=100Overpotential/VThereare 2 limiting cases corresponding to small orlargevalues of i,(i.e.kostandard electron transfer rate constant):Largei(k):"reversible"(FAST)electrodekinetics.Onlyasmalloverpotentialisrequiredtodrivethereactionand currentreadilyflows ineitheroxidativeorreductivedirection.Note:- because i.=-red=ioxand i,is large, there is a significant contribution to the current from BOTH OxandRedprocessesatallbutthemostextremeoverpotentials.ii.Small ig(k):"irreversible"(SLOW)electrodekinetics.Need largeoverpotential,n,togetcurrentflowingwhichis useful for
So what does the B-V model tell us? • There are 2 limiGng cases corresponding to small or large values of i 0 (i.e. k0 standard electron transfer rate constant): i. Large i 0 (k0): “reversible” (FAST) electrode kineGcs. Only a small overpotenGal is required to drive the reacGon and current readily flows in either oxidaGve or reducGve direcGon. Note:- because i o= -i red = i ox and i o is large, there is a significant contribuGon to the current from BOTH Ox and Red processes at all but the most extreme overpotenGals. ii. Small i 0 (k0): “irreversible” (SLOW) electrode kineGcs. Need large overpotenGal, η, to get current flowing which is useful for.