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Physical Chemistry I for BiochemistsChem340

Lecture 38 (4/20/11)

Yoshitaka IshiiCh. 9.7‐9.12Ch. 9.7 9.12

Announcement

• HW10 due date is 4/27 (Wed)

• Exam 3 will be returned probably this Friday• Exam 3 will be returned probably this Friday

• Final Exam 5/4 (Wed) 1‐3 pm

• Quiz 5 on 4/29 (Fri)

• We will cover Ch 9 and Ch 10. Sec 1‐5

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9.6 The Electrochemical Potential• Assume that a Zn electrod is partially 

immersed in an aqueous solution of ZnSO4. 

• Zn(s)  Zn2+(aq) + 2e‐

‐Whereas Zn2+ goes into solution, e‐ stays on the electrode.  

‐ Only small Zn dissolves into ions (~10‐14 

mol), producing ~ 1V potential between Zn and electrolyte. 

To transfer a charge dQ from the potential To transfer a charge dQ from the potential 1 to 2, the required work is 

dG = dw = (2 ‐ 1)dQ ,where dQ = zFdn  and z is a charge number of an ion (z = 0, 1, 2 ..) and F is a charge of 1 mol of an electron in absolute value .    

(Faraday constant)

Electrochemical Potential (continued)

• dG = dw = (2‐ 1)dQ ,

where dQ = zFdn.

dG ( )dQ zF( ) dn (9 44) dG = (2‐ 1)dQ = zF(2‐ 1) dn (9.44)

We define now electrochemical potential as

(9.46 text incorrect)

So the difference             is given as

zF~ dndG 12~~

F~~ 12

~~

F~~

In electrochemical reaction, the equilibrium is reached when 

Fz 1212 Fz 1212

0~ ii

ireactionG 0 ii

ireactionG

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We choose (sol) =0 for solution. So

9.7 Electrochemical Cells and Half‐Cells (Correction) zF~

CellHalf Cell

)solutioninion(~

For electron in a metal electrode,(z = ‐1ele=0)

Now, we consider an equilibrium,Mz+ + ze‐ M,

Identicalelectrolyte

)solutioninion(ii

FFeleele ~

where M denotes metal.  For this,

Zn  Zn2+ + 2e‐ Cu2+ + 2e‐ Cu

eleZMM z ~~~

)()(0 MzFszFF ZMM

zFz ZMeleZMM ~~

Correct text

Equilibrium at a half cell

(9.53)

• For a metal (made of a pure element) at 1 bar in a d d i h i l i l h ld b 0 If 0

zF

MzFz

ZM

ZMeleZMM

~

)(~~

standard state, its chemical potential should be 0.  If =0,(In text MZ+

0 = 0 @1bar Not exact)

• Using (9.53) and 

)298,1(0 00 KbarZMM

zFz ZMeleZMM ~~~

zFMM ~

Fele ~

In a standard condition at 298K and 1 bar 

Cf.                           and  

zFMZM

zFzFMZM 0~

0~ 0 MM Fele ~

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Standard Hydrogen Electrode

H+(aq) + e‐ 1/2H2(g)

Calculating Potential for a Hydrogen Electrode

H+(aq) + e‐ 1/2H2(g)

22/1~)(~HeleH aq 0

0 /ln)( ffRTpgas

(Reference electrode)

Potential of the hydrogen electrodef: fugacity ~ kP

Using a standard state ( =0 in electrolyte, 298K, 1 bar),

)ln(2/12/1)ln( 20

2/0

HHHHHH fRTFaRT

0/ln)( ffRTp gasgas

)ln(

2/1 2/12

02

0

2/

H

HHH

HH a

f

F

RT

F

For unit activities of all species aH+ = f = 1, the cell (electrode) has its standard potential:

0

2/1 002

0

2/0

FF

HHHHH

Convention for H+

0H+ = 0

[Q1]

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9.8 Redox Reaction in Electrochemical Cells and the Nernst Equation

• Left(anode):

• Right (cathode) :

Zn(s)  Zn2+ (aq) + 2e‐

Cu2+ (aq) + 2e‐ Cu(s) Daniel Cell

• Overall:

• Anode: Red1Ox1+1e‐

• Cathode: Ox2 +2e‐Red2• Overall:

Zn(s) + Cu2+ (aq)  Zn2+(aq) + Cu(s)

Zn  Zn2+ + 2e‐ Cu2+ + 2e‐ Cu

• Overall:

2Red1 + 1Ox2 2Ox1 +1Red2Q. How do you get Greaction

using chemical potential?

Nernst Equation for a Cell (Derivation)

(9.69)• For a reaction, n moles (n=2) of electron are transferred

)/ln(~~

~~~~

2220

20

22

CuZnCuZn

CuZnCuZnreaction

aaRT

G

Zn(s) + Cu2+ (aq)  Zn2+(aq) + Cu(s)

For a reaction, n moles (n 2) of electron are transferred from the cathode to anode.  The potential difference is  =cathode‐anode.  

(z= ‐1 for electron) (9.70)

By combining (9.69) & (9.79), we obtain

nFnzFG reaction )(

nFEa

RTa

RTG ZnCuZn

Znti

)ln(~~)ln( 2

20

2020

where we defined  electromotive force (emf) as E = .  When aZn2+ = aCu2+ = 1, E

0 is defined as ‐nFE0 = G0re action.

E = E0 –(RT/nF)ln(azn2+/aCu2+) (9.72)

(9.73)

nFEa

RTa

RTGCu

CuZn

Cureaction

)ln()ln(2

22

2

E = E0 – (RT/nF)lnQ

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How to obtain E0 from the half cell reaction?

• For  the reaction “Fe2+ + 2e‐ Fe”,

E0 = -0.45 (V)

Q1. How much is E0 for “Fe Fe2+ + 2e‐”?

E0 = 0.45 (V)

Q2. How much is E0 for “2Fe2+ + 4e‐ 2Fe”?

E0 = -0.45 (V)

Q3. How much is E0 for “2Fe2Fe2+ + 4e‐”?

E0 = 0.45 (V) G0reaction = -nFE0

E = E0 – (RT/nF)lnQ

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Nernst’s Equation for Whole and Half Cells

For a whole cell,  2Red1 2Ox1 +ne‐ ; 1Ox2 +ne‐ 1Red2

• E = E0 (RT/nF)lnQ [ ])(1

2Re2

1

dOX aaQ

Balance the charges

• E =  E0 – (RT/nF)lnQ [                           ]

At 298.15K, we obtain the equation called Nernst’s equation

E = E0 – (RT/nF)lnQ = E0 – (0.05916/n)logQ (9.74)

For a half cell reaction:  Oxn+ + ne‐ Red~~~0 G F~

)(1

22

1ReOxd aa

Q

• Eox/Red =(0OXn+  0

Red)/nF - (RT/nF)ln(aRed/aOXn+) (9.77)

= E0Ox/Re– (RT/nF)ln(aRed/aOxn+)

0)/ln(~~0

Re/ReRe00

Re

dOxdOxdOx

deleOxreaction

nFaaRT

nG

Fele

Activity of electrons is not involved

• P9.40) By finding appropriate half-cell reactions, calculate the equilibrium constant at 298.15 K for the following reactions:

2Cd(OH) 2Cd O 2H O• a. 2Cd(OH)2 2Cd + O2 + 2H2O

a) The half cell reactions are

• Cd(OH)2 + 2e– → Cd + 2 OH– E 0 red Table 9.5

• 4OH– → O2 + 2H2O + 4e–

• Q.  How many is n in (a)?                    n = 4

E 0 ox -E0

red

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9.9 Combining Standard Electrode Potential to Determine Cell Potential

• By convention standard potentials are listed as reduction potentials as (see Table 9.5 in appendix B)

Reduction E0(V) Q Which is spontaneous reaction inReduction               E0(V)      

Cu2+ + 2e‐ Cu 0.674

Zn2+ + 2e‐ Zn ‐0.7618

Whether the reaction is spontaneous at the standard state is determined by E0.   Because G0 = ‐nFE0 and G0

ox = ‐G0red ,

E0reduction = E0oxidization.   (i.e. Zn  Zn2+ + 2e‐ E0 ox = 0.7618V) 

Q. Which is spontaneous reaction in a standard state?

Cu2+ + 2e- Cu

Because G0 = -nFE0 <0

reduction oxidization ( ox )

The potential of whole cell E0cell is related to those of the half cells E0red and E

0ox as

Q. How much is E0cell for the above

half cells? E0cell = 0.674 – (-0.7618) V.

E0cell = E0

red + E0ox

G0reaction = -nFE0 S0

reaction= -(G0/T)P=nF(E0 /T)P

• P9.32) Calculate G0reaction and the equilibrium

constant at 298 15K for the reactionconstant at 298.15K for the reaction

• Cr2O72-(aq) + 3H2(g) + 8H+(aq)

2Cr3+(aq) + 7H2O(l)

• 3H2 6H+ + 6e‐ Eox = ?

• Cr2O72-(aq) + 6e- 2Cr3+(aq) Ered = Find in text

UseG0

reaction = -nFE0cell & K = exp(- G0

reaction /RT)

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9.10 The Relationship between the Cell emf and Equilibrium Constant

• If the redox reaction is allowed to proceed until equilibrium is reached, G =0, and thusuntil equilibrium is reached, G  0, and thus E = 0.

For the equilibrium state, the reaction quotient Q = K.  Therefore, 

E0 = (RT/nF)ln(K) K = exp(nFE0/RT)E = (RT/nF)ln(K) K exp(nFE /RT)

P233

• AgBr(s) + e‐ Ag(s) + Br‐(aq)   E0 = 0.07133 V (1) 

A +( ) A ( ) E0 0 7996 V (2)• Ag+(aq) + e‐ Ag(s)  E0 = 0.7996 V    (2)

• Obtain Ksp for AgBr(s).

• Ag(s)  Ag+(aq) + e‐ E0 = ‐0.7996 V  (2’)

By calculating (1) + (2’),

• AgBr(s)  Ag+(aq) + Br‐(aq)    E0 = (0.07133 ‐0.7996) Vg ( ) g ( q) ( q) ( )

• Ksp = exp(‐G0reaction/RT)=  exp(nFE

0/RT)

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9.11 The Determination of E0 and Activity Coefficients Using an Electrochemical Cell

• The main problem in determining standard potential E0

is knowing the activity constant for a given soluteis knowing the activity constant  for a given solute.• Now assume a cell consisting of the Ag+/Ag and SHE half‐cells at 298K. For Ag+/Ag, Ag+ arises from the dissociation of AgNO3.  Assume that aAg+ = aCl‐.  Recall activities of individual ions cannot be measured directly. 

• a = a + a ‐ a 2 = aA aCl & a = aA = aCla = a+ a‐ a = aAg+ aCl‐. &  a = aAg+ = aCl‐• Similarly, = Ag+ = Cl‐ and m= mAg+ = mCl‐ .  Then the cell potential is 

•

• For low mAg+, 

E = E0Ag+/Ag+(RT/F)ln(aAg+) = E0

Ag+/Ag +(RT/F){ln(m)+ln }

log =-0.50926(m)1/2 at 298K.

E- 0.05916 log10(m) = E0Ag+/Ag -0.03013(m)1/2

E = E0Ag+/Ag+(RT/F)ln(aAg+) = E0

Ag+/Ag +(RT/F){ln(m)+ln }

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9.12 Biochemical Standard State• Consider the reaction, 

AA(aq) + BB(aq)  DD(aq) + EE(aq)+xH+(aq) 

Assuming i ~ 1 for all the species,Assuming i  1 for all the species, 

For biochemistry, we take pH =7 (rather than cH+ = 1 M) 

as a standard condition.  Thus, 0 1 00 10 7 l/L

B

BB

A

AA

x

HH

E

EE

D

DD

cccc

ccccccK

00

000

//

///

c0H = 1.00 x 10‐7 mol/L

and for other species c0i =1.00 M

K = K’ x 10(‐7x)

B

BA

A

x

HE

ED

D

McMc

MMcMcMcK

00.1/00.1/

1000.1/00.1/00.1/'

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continued

• If we define G0’ as G0’=‐RTln(K’),

• G0’ RTln(K’) RTln(K107x)• G0 = ‐ RTln(K’) = ‐ RTln(K107x)

= ‐ RTln(K’) ‐ 7x RTln(10)

= G0 – 7x RTln(10)

• E0’ = ‐ G0’/nF =  (RT/nF)ln(K’)

= (RT/nF)ln(K107x) = (RT/nF){lnK+7xln10}

= (RT/nF){lnK+7xln10}

= (RT/nF)lnK + (RT/nF){7xln10}

= E0 + (RT/nF){7xln10}

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9.13 The Donnan PotentialThere are two compartments separated by membrane that is freely permeable to Na+ and Cl‐, but not for Pz‐.  Because the membrane is permeable to Na+ andBecause the membrane is permeable to Na+ and Cl‐ , the system reaches equilibrium as

L = R.  We assume that 0L = 0R. This leads to 

aL = a

R.  When i ~ 1, 

c L c L = c R c Rc+ c‐  c+ c‐ .

If Pz‐ were not present, c+

L /c+R = c‐

R /c‐L = 1

Since Pz‐ is present, c+

L /c+R = c‐

R /c‐L = rD 1,  (rD: Donnan ratio)

Net Charge 0Net Charge 0

(b+x)x = (a-x)(a-x) Net Charge 0

When z = 1, =1

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• At the equilibrium,

(b+x)x = (a‐x)(a‐x)   x = a2/(b+2a)

cLNa+ eq = (b+x) = (a+b)2/(b+2a)

cLCl‐ eq = x = a2/(b+2a)

cRNa+ eq =  cRCl‐ eq = (a‐x) = a(a+b)/(b+2a)

rD = cLNa+ eq /c

RNa+ eq = (a+ b)/a Na eq Na eq

= L ‐ R = 0 = D + (RT/F)ln(rD)  

D =‐(RT/F)ln{(a+b)/a}Donnan Potential Potential due to difference

of ion concentrations

Potential due to polarization of membrane