( ) Lecture 14: E cell and ΔG

Transcription

1 Lecture 14: and ΔG Reading: Zumdahl: 11.3 (and 10.12) Outline: The vltage f a battery is determined by the Gibbs free energy f the chemical reactin that happens inside the battery. and wrev and ΔG and K wrk ( ) lectrchemistry is the avatar f Thermdynamics Prblems (Zumdahl Ch 11 5 th d.) 33(!), 34b,c, 36 (Is ClO a prblem?), 38, 39, 46,

2 A Cell (r Battery) MF This is the symbl we use t represent the vltage f a battery r MF (in Vlts). The naught (superscript) means under standard thermdynamic cnditins. We saw (previus lecture) hw t cnstruct a galvanic electrchemical (i.e. battery), capable f generating a flw f electrns. This flw f electrns (current) can perfrm wrk n the surrundings. 2

3 and wrk Frm the definitin f electrmtive frce (MF): Vlt = energy (,J)/charge (Q,C) 1Vlt = 1Jule/Culmb In nglish: 1 J f energy is available when 1 C f charge is transferred (r mves between) a ptential difference f 1 V. Think f gravity: P.. = m ( gh) Q If all f this energy is cnverted t wrk that wuld be the maximum electrical wrk available due t the ptential. Q is the symbl fr the amunt f charge in Culmbs V ( ) elect max elect wmax J = Δ Q C nnpdv rev ( ) ( ) w = w = ΔQ Cnventin: If the battery ptential is psitive the system (battery) des wrk s (inside) it lses energy, 3 hence the minus sign.

4 example Hw much wrk is dne by a galvanic in which 1.5 mles f electrns are passed between a ptential f 2 V (i.e. We have a 2V battery) under standard thermdynamic cnditins? Δ Q = w elect max 2V 1.5 mles =?? What d we d with this? We need t knw hw much charge this is 4

5 The Faraday Hw much charge des a mle f electrns have? The Faraday (F) is the amunt f charge cntained in 1 mle f electrns. 5 5 F = 96, 485 C / ml 10 C / ml = 10 J / ml / V e = ( ) C (Charge n a single electrn) F = NAe= = C/ ml Nw we can finish ur prblem and cmpute the wrk: Δ = = 5 Q n F 1.5 mles (0.96)10 C / mle w = Δ Q= n F elect max 5 5 2V C = 3 10 J = 300kJ A 1V battery can give abut 100 kj f wrk per mle f reactant. 5

6 Gibbs nergy and Wrk (at fixed P and T) Preparatin t talk abut batteries (next chapter) Frm Calrimetry: Δ H =ΔH X t actually get heat (r an enthalpy change) yu actually have t change the number f mles f smething (X). Nw t get wrk yu need t cnsume sme amunt f reactant (X) : w = Δ G =ΔG X electrical max This shws that t d wrk yu need a (chemical) ptential times a (chemical) distance. And the wrk yu get is prprtinal t the amunt f stuff yu use up. The useful wrk (text) is nn-pdv wrk, and is electrical wrk; like a battery Calling the wrk useful and nt useful (r PdV) wrk, which the bk des, is nt useful. (Sec ). 6

7 ΔG and Wrk (Fixed T and P) Cnsider a reversible reactin, and wrk is the sum f the PdV part and extra wrk. Then Δ G =ΔH Δ TS =Δ +Δ PV TΔS ( ) ( ) = q PΔ V + w + PΔ V q = w Δ G = w = w ( ) x x rev rev rev rev x rev nnpdv rev We previusly shwed that the reversible wrk is the maximum wrk (as a negative number) yu can get ut f a system. Nw see Gibbs energy is the same as the maximum nn-pdv wrk (i.e. electrical) we can get frm a chemical reactin when run at cnstant T 7 and P.

8 and ΔG Since there is a relatinship between ΔG and wrk, there is als a relatinship between ΔG and. Δ G = w = Δ Q= n F elect max The abve relatinship states that there is a direct relatinship between Gibbs free energy and ptential. Fr a galvanic (under standard cnditins): 0 Therefre, Δ < 0 (spntaneus) > G 8

9 and ΔG We have a chemical reactin, and sme f the reactin ccurs (mles f each species are used up r created): Δ ni = c ΔX Δ G=ΔG ΔX ( ), and ss i Sme electrns are transferred in the reactin. n(e) is the stichimetric cefficient fr the number f electrns transferred fr the reactin as written (even thugh the electrns are nt written ut in the net reactin) Putting all these cnnectins tgether in the equatin n the previus slide: Δ n e = n e ΔX and Δ Q= FΔn e ( ) ( ) ( ) 9

10 Reactin nergy The relatin between the actual energy and the reactin energy in terms f the balanced equatins gives a relatin between the reactin Gibbs energy and the vltage: Δ G = Δ Q= FΔ n e = FnΔX ( ) Δ G =ΔG Δ X = FnΔX Δ G = n F The stichimetric cefficient n(e). The Reactin Gibbs free energy is nt the electrical wrk. Cannt equate reactin energy and wrk. T d wrk yu actually have t have sme amunt f chemical reactin (specified as the number f mles f smething that reacted). Identify the intensive and extensive quantities abve. 10

11 Determine and ΔG 0 fr a Battery Fr the fllwing reactin, determine the verall standard ptential and determine ΔG Nte: The net reactin des nt give n, and des nt shw hw many electrns are used per reactin. T determine n(e) we must break the reactin int its half reactins. Cd +2 (aq) + Cu(s) Cd(s) + Cu +2 (aq) Cd e - Cd 1/2 = V Cu e - Cu 1/2 = V Cu Cu e - 1/2 = V 11

12 an example Cd e - Cd 1 2 = 0.40V Cu Cu e - Cd +2 (aq) + Cu(s) = 0.34V Cd(s) + Cu +2 (aq) = 0.74V Δ G = nf = (2 e / ) (96000 C/ ml e ) ( 0.74 J / C) 0 - ( ) = 150 kj / (nt spntaneus, nt galvanic) When adding half reactin t get a net reactin (where n(e) drps ut) then yu can add the vltages. Warning: If n(e) des nt drp ut, yu cannt just add vltages. 12

13 Δ G = nf and Κ Under Standard cnditins then: Δ G = nf Previusly: Therefre: RT = ln K nf Δ G = RTln K 0 nf = Δ G = RT ln K 0 r.t. T = K RT = 2.48 kj / mle RT 2 J / mle = = 25.7mV F Culmbs/ mle ln10 = mV ln10 = 59mV = ln lg n n 0 K = 10 K mv 13

14 and Κ = ln K = lg 10 K mv n n The abve relatinship states that by measuring, we can determine K. The abve relatinship illustrates that electrchemical s are a venue in which thermdynamics is readily evident (an avatar). 14

15 and ΔG 0 and K Develping the full picture Δ G = RTln K 0 Δ G = n F V = lg n RT = ln K nf It is imprtant t see hw all f these ideas interrelate. We can measure K directly under Standard State cnditins; very very far away frm quilbirium K

16 An xample: determine and K Balance, determine and K fr the fllwing: S 4 O 6 2- (aq) + Cr 2+ (aq) Cr 3+ (aq) + S 2 O 3 2- (aq) Break the net reactin int its half reactins 2e - + S 4 O S 2 O 2-3 Cr 2+ Cr 3+ + e - x 2 S 4 O Cr 2+ 2Cr S 2 O

17 A Battery xample Determining 2e- + S 4 O S 2 O =+ 0.17V 2Cr 2+ 2Cr e - S 4 O Cr 2+ 2Cr S 2 O =+ 0.50V =+ 0.67V Frm the tw half reactins we see: n=n(e)=2 This vltage is NOT dubled 17

18 Determining K fr a Battery =+ 0.67V S 4 O Cr 2+ 2Cr S 2 O = ln K n n ln K = = = K = e = 10 = 5 10 The larger the Battery Vltage; the larger the K; the larger the tendency t g t prducts; the greater the driving frce fr the reactin 18

19 Gibbs nergy frm K Clsing the lp (frm the triangle relatin f the full picture) Cmpute the Gibbs free energy f reactin: Δ G = RTln K 0 ( 22 ) kj mle ( ) = 2.47 ln 5 10 / = = 129 kj / Δ G = nf = J / 0 0 = 129 kj / The numbers are the same by the tw methds (with sme rund ff differences). The reactin is highly spntaneus. 19

20 Z11.33 Wrk and a Battery Hw much wrk will yu get frm the Battery shwn in 20b? See the Cd/V Battery, all species at standard state. Develp the half reactins, find half- Vltages (MFs) in Table and cmpute the MF. Determine n (fr electrns) and cmpute ΔG Nw cmpute the wrk. Insufficient Infrmatin!!!!! If yu use 0.1 mles f Cd (metal) hw much wrk will yu get? 20