Electrochemical Equilibrium Electromotive Force. Relation between chemical and electric driving forces

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1 C465/865, 26-3, Lctur 7, 2 th Sp., 26 lctrochmcal qulbrum lctromotv Forc Rlaton btwn chmcal and lctrc drvng forcs lctrochmcal systm at constant T and p: consdr G Consdr lctrochmcal racton (nvolvng transfr of ): ν A + ν B+ ν ν C+ ν D+ ν (1) A B R C D L ractants products Not: Stochomtrc coffcnts ( ν ) of ractants (A, B) hav ngatv sgn, thos of products (C,D) ar postv lctrons ar wrttn xplctly n ths quaton snc thy wll appar n th condton of lctrochmcal qulbrum! Objctv: us thrmodynamc argumnts to drv lctrcal potntal (th so-calld lctromotv forc, F) of a cll and rlat ths F to th composton of lctrochmcal clls Thrmodynamcs Composton lctrcal potntal, F

2 Racton (1) advancs by small amount dξ from lft to rght amount of A changs by dn A = ν Adξ < amount of B changs by dn B = νbdξ < amount of C changs by dnc = νcdξ > amount of D changs by dn D = ν Ddξ > xampl: 2Al O () s + 3S() s SO () s + 4Al() s R 2 ν = 2, ν = 3, ν = 3, ν = 4 A B C D L Chang n lctrochmcal Gbbs fr nrgy: dg = =, α and, thus, α µ dn α (dstngush spcs and phass) νµ dξ (: all spcs and all phass) G G = = ν µ ξ r T, p (gnral rlaton, nvolvs all spcs, all phass) Condton of lctrochmcal qulbrum: = r G Systm s not n chmcal qulbrum! < (spontanous racton) r G ths s th chmcal drvng forc of th racton

3 Lt s consdr som xampls frst: 1. tal soluton ntrfac, soluton contans ons of th mtal Ion-transfr: z+ + z - (cathodc racton, rducton) S Wrt qulbrum condton: µ z µ µ z+ = S ϕ ϕ whr ( ) S,S S µ = µ + RT ln a + zfϕ z+ z+ z+ µ µ ϕ, = F, µ = µ ( standard valu, unt actvty) r, µ s th standard chmcal potntal of spcs n α, phas α, rfrrng to standard condtons. Convntons: actvts of solvatd spcs: partal prssurs of gasous componnts: a = p = 1 1 bar concntratons of dlut solutons: c = 1 mol/l mtals, solds, lquds: for practcal purposs always at standard condtons!

4 Insrt all th rlatons nto condton of lctrochmcal qulbrum and solv for ϕ S. Rsult s an xprsson for th lctrod potntal or F of th z+ rdox coupl: and, thus,,s,, S z+ = ϕ = ϕ ϕ = + S µ + zµ µ zf ( ) RT = + z+ zf ln a RT ln zf dpnds on actvts of potntal dtrmnng ons. ( a z+ ) s th standard F: µ + zµ µ = zf,s,, z+

5 2. Non-mtal n contact wth ts ons on surfac of an nrt, conductng substanc,.g. 2 + on Pt 2 (g) (anodc racton, oxdaton) r G = S t,g ( a ) 2Fϕ + 2 2F RT ln( p ),S,Pt P 2µ + + 2RT ln + + µ ϕ µ = 2 2 g,g chmcal potntal of hydrogn gas: µ = µ + RT ln( p ) lctrod potntal: 2 2 2,g,S,Pt 1/2 µ 2µ S S Pt + 2µ 2 RT p 2 = Ptϕ = ϕ ϕ = + ln zf F a + RT F p 1/2 2 = + ln a +

6 3. Consdr th followng galvanc cll: Pt L 2 (g) Cl(m) AgCl(s) Ag(s) Pt R (m: molalty of th lctrolyt soluton) Lft lctrod: ANOD, oxdaton 2 (g) (at Pt L ) Rght lctrod: CATOD, rducton 2 AgCl(s) (at Pt R ) 2 Ag(s) + 2 Cl - ovrall: 2 (g) + 2 AgCl(s) (at Pt R ) 2 Cl (m) + 2 Ag(s) (at Pt L ) lctrochmcal qulbrum (all spcs, all phass): aq s PtL g S PtR 2µ Cl + 2µ Ag + 2µ µ 2µ 2 AgCl 2µ = Agan: solds, mtals, lquds unt actvty Only lctrochmcal potntals of lctrons n Pt-wrs ar potntal dpndnt. Why can aq µ Cl b consdrd as on lctronutral spcs? Insrt xprssons for lctrochmcal potntals: 2µ + 2µ µ µ + µ µ Fϕ + Fϕ = 2 PtR PtL ( ϕ ϕ ) PtL PtR PtL PtR aq s g S Cl Ag AgCl G+ 2F = r G a Cl = ϕ ϕ =, wth r = r + 2 T ln 1/2 2F p 2 PtR PtL r G G R

7 Rlaton btwn F and racton Gbbs fr nrgy r G ow s F dtrmnd by th chmcal composton of th systm? lads to Nrnst s quaton Agan: consdr lctrochmcal cll wth ovrall racton ν A + ν B+ ν ν C+ ν D+ ν A B R C D (could b for nstanc a Danll lmnt) Racton Gbbs fr nrgy vs. composton of th mxtur: = + rg rg RTlnQ L Th rlaton btwn F and G r s R L rg rg RT = ϕ ϕ = = ν F ν F ν F RT = ν F lnq or lnq Ths s Nrnst quaton! r, Q s th so-calld racton quotnt: Q = νc [ C] [ D] νa [ A] [ C] Not: nstad of concntratons, thr could b actvts or partal prssurs n th racton quotnt as wll dpnds on typ of spcs. ν D νb

8 Th Nrnst quaton s just anothr way to formulat th condton of lctrochmcal qulbrum. Now: what would b th quvalnt xprsson to Nrnstquaton for chmcal qulbrum??? Conncton btwn two lctrods through xtrnal mtal wr: systm wll go to chmcal qulbrum (by lctrons flowng spontanously from anod to cathod). Dschargd battry: chmcal qulbrum Ful cll: cannot rach chmcal qulbrum du to contnuous supply of ractants (opn systm)

9 lctrod confguratons and rfrnc lctrods Nrnst quaton: n prncpl possbl to calculat and masur Fs for half ractons and lctrochmcal clls (xampls n problm sts) In gnral, th F RT ν F = lnq has to parts: Standard F standard condtons ar: 25 C partal prssurs of gasous spcs: 1 bar 1 solutons unt actvts of ons composton-dpndnt part du to dvatons from standard condtons (nvolvng tmpratur varaton)! If standard potntals and composton of th systm ar known, thn n prncpl Fs of all systms could b dtrmnd! owvr: thr s no absolut scal, only dffrncs n lctrod potntals can b dtrmnd. Is that bad?

10 lctrod potntal of sngl lctrod confguraton: tal soluton ntrfac lctrod potntal: S S l = ϕ ϕ (cathodc) ϕ ϕ S Rmmbr: ths s proportonal to th amount of work rqurd to mov a tst charg across th mtal soluton ntrfac asurmnt: at last two lctrods ar rqurd lctrochmcal Cll F: C A cll = ϕ ϕ ϕ A Anod ϕ S ϕ C Cathod lctrcal work that th systm can prform n brngng sngl lctron from anod to cathod. Ths valu can b calculatd f th lctrod potntals ar known at th two dstnct S ntrfacs = C A cll l l Ths s masurabl (mtal conncton btwn lctrods, voltmtr wth hgh lctronc rsstanc, that allows only a vry small currnt to flow - masurmnt prformd practcally undr condtons of lctrochmcal qulbrum.

11 Znc lctrod mmrsd n watr Schmatc dagram of Danll cll Small amount of Zn 2+ ons go nto soluton, lavng ngatv charg on lctrod. Potntal dffrnc: ϕ ϕ S ~ 1V Zn 2+ Zn and Cu 2+ Cu halfclls ar connctd through salt brdg (ntrnal crcut). Voltmtr n xtrnal crcut.