Physicl Chemistry I for Biochemists Chem34 Lecture 38 (4//11) Yoshitk Ishii Ch. 9.7 9.1 9.1 Announcement HW1 due dte is 4/7 (Wed) Exm 3 will be returned probbly this Fridy Finl Exm 5/4 (Wed) 1 3 pm Quiz 5 on 4/9 (Fri) We will cover Ch 9 nd Ch 1. Sec 1 5 1
9.6 The Electrochemicl Potentil Assume tht Zn electrod is prtilly immersed in n queous solution of ZnSO 4. Zn(s) Zn + (q) + e Wheres Zn + goes into solution, e stys on the electrode. Only smll Zn dissolves into ions (~1 14 mol), producing ~ 1V potentil between Zn nd electrolyte. To trnsfer chrge dq from the potentil 1 to, the required work is dg = dw = ( 1 )dq, where dq = zfdn nd z is chrge number of n ion (z =, 1,..) nd F is chrge of 1 mol of n electron in bsolute vlue. (Frdy constnt) Electrochemicl Potentil (continued) dg = dw = ( 1 )dq, where dq = zfdn. dg = ( 1 )dq = zf( 1 ) dn (9.44) We define now electrochemicl potentil s zf (9.46 text incorrect) dg ~ ~ 1 So the difference ~ ~ is given s ~ dn ~ ~ F ~ ~ F 1 z 1 In electrochemicl rection, the equilibrium is reched when G ~ Grection ii rection i i i 1 1 z 1 i
9.7 Electrochemicl Cells nd Hlf Cells (Correction) ~ zf We choose (sol) = for solution. So ~ (ion in solution) i i For electron in metl electrode, ~ (z = 1 ele ele F F ele =) Now, we consider n equilibrium, z+ + ze, where denotes metl. For this, ~ ~ z~ Correct text Z ele Cell ) F zf( s) zf( Z ~ z~ ele zf Z Z Zn Zn + + e Identicl electrolyte Hlf Cell Cu + + e Cu Equilibrium t hlf cell ~ z ~ ele zf ( ) Z Z ~ zf Z (9.53) For metl (mde of pure element) t 1 br in stndrd dstte, its chemicl potentil ilshould ldbe. If =, (In text Z+ = @1br (1 br,98k) Z ~ Not exct) Using (9.53) nd ele F ~ z~ ele ~ zf Z Z ~ zf Z In stndrd condition t 98K nd 1 br ~ zf zf Z Cf. ~ nd ~ F ele 3
Stndrd Hydrogen Electrode H + (q) + e 1/H (g) Clculting Potentil for Hydrogen Electrode H + (q) + e 1/H (g) ~ ( ) ~ H q ele 1/ H (Reference electrode) f: fugcity ~ kp Potentil of the hydrogen electrode gs ( p ) RT ln f / f Using stndrd stte ( = in electrolyte, 98K, 1 br), H RT ln( H ) FH / H 1/ H 1/ RT ln( f H H / H H 1/ H F ) RT F f H ln( For unit ctivities of ll species H+ = f = 1, the cell (electrode) hs its stndrd potentil: Convention for H + H 1/ H H H+ = H / H [Q1] F F 1/ H ) gs 4
9.8 Redox Rection in Electrochemicl Cells nd the Nernst Eqution Left(node): Zn(s) Zn + (q) + e Right (cthode) : Cu + (q) + e Cu(s) Overll: Zn(s) + Cu + (q) Zn + (q) + Cu(s) Anode: Red 1 Ox 1 + 1 e Dniel Cell Cthode: Ox + e Red Overll: Red 1 + 1 Ox Ox 1 + 1 Red Q. How do you get G rection using chemicl potentil? Zn Zn + + e Cu + + e Cu G Nernst Eqution for Cell (Derivtion) Zn(s) + Cu + (q) Zn + (q) + Cu(s) ~ ~ ~ ~ G rection Zn Cu Zn Cu ~ ~ ln( / ) (9.69) Zn Cu RT Zn Cu For rection, n moles (n=) of electron re trnsferred from the cthode to node. The potentil difference is = cthode node. G rection ( nzf ) nf (z= 1 for electron) (9.7) By combining (9.69) & (9.79), we obtin rection ~ Zn ~ Zn Cu RT ln( ) nfe Zn RT ln( ) ) Cu Cu where we defined electromotive force (emf) s E =. When Zn+ = Cu+ = 1, E is defined s nfe = G re ction. E = E (RT/nF)ln( zn+ / Cu+ ) (9.7) E = E (RT/nF)lnQ (9.73) 5
How to obtin E from the hlf cell rection? For the rection Fe + + e Fe, E = -.45 (V) Q1. How much is E for Fe Fe + + e? E =.45 (V) Q. How much is E for Fe + + 4e Fe? E = -.45 (V) Q3. How much is E for Fe Fe + + 4e? E =.45 (V) G rection = -nfe E = E (RT/nF)lnQ 6
Nernst s Eqution for Whole nd Hlf Cells Blnce the chrges For whole cell, Red 1 Ox 1 +ne ; 1 Ox +ne 1 Red ( OX Q 1 Re d E = E (RT/nF)lnQ [ ] At 98.15K, we obtin the eqution clled Nernst s eqution E = E (RT/nF)lnQ= E (.5916/n)logQ (9.74) For hlf cell rection: Ox n+ + ne Red G G ~ n ~ ~ ~ F ~ Ox rection ~ Re d Ox RT ln( Ox ele / Re d Re d ) nf Ox / Re d 1 ( 1 Re d 1 Ox E ox/red =( OXn+ Red)/nF - (RT/nF)ln( Red / OXn+ ) (9.77) = E Ox/Re (RT/nF)ln( Red / Oxn+ ) ele ) Activity of electrons is not involved P9.4) By finding pproprite hlf-cell rections, clculte the equilibrium constnt t 98.15 K for the following rections:. Cd(OH) Cd + O + H O ) The hlf cell rections re Cd(OH) + e Cd + OH E red Tble 9.5 4OH O +H O + 4e E ox -E red Q. How mny is n in ()? n = 4 7
9.9 Combining Stndrd Electrode Potentil to Determine Cell Potentil By convention stndrd potentils re listed s reduction potentils s (see Tble 9.5 in ppendix B) Reduction E (V) Q. Which is spontneous rection in stndrd stte? Cu + + e- Cu Cu + + e Cu.674 Zn + + e Zn.7618 Becuse G = -nfe < Whether the rection is spontneous t the stndrd stte is determined by E. Becuse G = nfe nd G ox = G red, E reduction = E oxidiztion. (i.e. Zn Zn + + e E ox =.7618V) The potentil of whole cell E cell is relted to those of the hlf cells E red nd E ox s Q. How much is E cell for the bove hlf cells? E cell =.674 (-.7618) V. E cell = E red + E ox G rection = -nfe S rection= -(G /T) P =nf(e /T) P P9.3) Clculte G rection nd the equilibrium constnt t 98.15K for the rection Cr O - 7 (q) + 3H (g) + 8H + (q) Cr 3+ (q) + 7H O(l) 3H 6H + + 6e E ox =? Cr O 7 - (q) + 6e - Cr 3+ (q) E red = Find in text Use G rection = -nfe cell & K = exp(- G rection /RT) 8
9.1 The Reltionship between the Cell emf nd Equilibrium Constnt If the redox rection is llowed to proceed until equilibrium is reched, G =, nd thus E =. For the equilibrium stte, the rection quotient Q = K. Therefore, E =(RT/nF)ln(K) K = exp(nfe /RT) P33 AgBr(s) + e Ag(s) + Br (q) E =.7133 V (1) Ag + (q) + e A() Ag(s) E =.7996 V () Obtin K sp for AgBr(s). Ag(s) Ag + (q) + e E =.7996 V ( ) By clculting (1) + ( ), AgBr(s) Ag + (q) (q) + Br (q) (q) E = (.7133.7996)) V K sp = exp( G rection/rt)= exp(nfe /RT) 9
9.11 The Determintion of E nd Activity Coefficients Using n Electrochemicl Cell The min problem in determining stndrd potentil E is knowing the ctivity constnt for given solute. Now ssume cell consisting of the Ag + /Ag nd SHE hlf cells t 98K. For Ag + /Ag, Ag+ rises from the dissocition of AgNO 3. Assume tht Ag+ = Cl. Recll ctivities of individul ions cnnot be mesured directly. + = + = Ag+ Cl. & = Ag+ = Cl Similrly, = Ag+ = Cl nd m = m Ag+ = m Cl. Then the cell potentil is E = E Ag+/Ag+(RT/F)ln( Ag+ ) = E Ag+/Ag +(RT/F){ln(m )+ln } For low m Ag+, log =-.596(m ) 1/ t 98K. E = E Ag+/Ag+(RT/F)ln( Ag+ ) = E Ag+/Ag +(RT/F){ln(m )+ln } E-.5916 log 1 (m ) = E Ag+/Ag -.313(m ) 1/ 1
9.1 Biochemicl Stndrd Stte Consider the rection, A A(q) + B B(q) D D(q) + E E(q)+xH + (q) Assuming i ~ 1 for ll the species, K D E c D / c D c E / c E c H / c H A c / c c / c B A For biochemistry, we tke ph =7 (rther thn c H+ = 1 ) s stndrd condition. Thus, c H = 1 1. x 1 77 mol/l A nd for other species c i =1. K ' K = K x 1 ( 7x) B B D E 7 c /1. c /1. c / 1. 1 D A c / 1. c / 1. B A E B x H x continued If we define G s G = RTln(K ), G = RTln(K ) = RTln(K1 7x ) = RTln(K ) 7x RTln(1) = G 7x RTln(1) E = G /nf = (RT/nF)ln(K ) = (RT/nF)ln(K1 7x ) = (RT/nF){lnK+7xln1} = (RT/nF){lnK+7xln1} = (RT/nF)lnK + (RT/nF){7xln1} = E + (RT/nF){7xln1} 11
9.13 The Donnn Potentil There re two comprtments seprted by membrne tht is freely permeble to N + nd Cl, but not for P z. Becuse the membrne is permeble to N + nd Cl, the system reches equilibrium s L = R. We ssume tht L = R. This leds to L = R. When i ~ 1, c L L = R R + c c + c. If P z were not present, c +L /c +R = c R /c L = 1 Since P z is present, c +L /c +R = c R /c L = r D 1, (r D : Donnn rtio) Net Chrge Net Chrge When z = 1, =1 (b+x)x = (-x)(-x) Net Chrge 1
At the equilibrium, (b+x)x = ( x)( x) x = /(b+) c L N+ eq = (b+x) = (+b) /(b+) c L Cl eq = x = /(b+) c R N+ eq = c R Cl eq = ( x) = (+b)/(b+) r D = c L N+ eq /c R N+ eq = (+ b)/ = L R = = D + (RT/F)ln(r D ) Donnn Potentil D = (RT/F)ln{(+b)/} Potentil due to difference of ion concentrtions Potentil due to polriztion of membrne 13