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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/'
7
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