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Physical Chemistry I for BiochemistsChem340Chem340
Lecture 37 (4/18/11)
Yoshitaka IshiiYoshitaka IshiiCh. 9.4‐8
Announcement
• HW10 due date is 4/25
• Exam 3 will be returned probably this Friday• Exam 3 will be returned probably this Friday
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Definition of ionic activities (Correction)
• a+ = (m+/m0)+ = (m+/m0)+where m+ is molality of positive ion, m is the molality of the compound dissolved, & m0 = 1mol kg‐1.
• Same for a‐ =(m‐/m0)‐
a+ = (c+/c0)+,
where c+ is molarity of positive ion & c0 = 1mol /L
Alternatively,
Chemical potential of electrolyteµ = µ
0 + RT ln(a), where a = {a+
+ a‐‐ }
Now using molality of ions,a+ = x++ = (m+/m0)+ and a‐ =(m‐/m0)‐
h ( / ) ( / )• Thus, a =(m+/m0)
+ (m‐/m0)‐++ ‐ ‐
• We define mean ionic molality m and mean ionic activity constant as
m (m+)
+(m‐)‐ = (m+) +(m‐)‐ & ++ ‐ ‐
• µsolute= µ = µ0 + RT ln(a) = [ µ 0 + RT ln( + ‐)] + RT ln(m/m )= [ µ0 + RT ln(++‐ )] + RT ln(m/m 0)
+ RT ln( )= µsolute
00 + RT ln(m/m 0) + RT ln( )
Deviation from ideal ioncsolution
“Normal” standard state
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Coulomb Interactions• Coulomb Energy E
)(4
),( 1221
21 qqqq
qqE q1 q2
Q. How much is r in vacuum?
• Coulomb Potential
4 120
21 rr q1 q2r
r
r 0
11 4)(
q1 1
r
• Energy (w’) required to fill a charge q1 on the surface of a sphere (radius R)
q1
R
q
R
qdqw
rr
q
0
2
1
00 84
1
R
(P215)
Gibbs Energy for Solvation (correction)
q2
Energy to accumulate a charge of q in a sphere of a radium r
Permittivity for free space
r
qw
08
Energy for the same sphere but in a solvent
qw
2
8'
+q
)1( rrr08
1
1
8'
0
20
rSolvation r
qwwG
The difference due to solvation is
< 0
+q
‐
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Coulomb Interaction in Three Medium
r
q
4
+qIn vacuum
r04
r
q
OH 024
+q
In water
I i i l ti ith i t th I
H2O ~80
Q. For the same r, how much isthe interaction in H2O?
x 1/80
)exp(4
)exp(4
2/1
02
02
rAIr
q
rr
q
OH
OH
+q
‐
‐
+
+ ‐
+
In ionic aqueous solution with ion strength I
9.4 Calculating from the Debey‐Hückel Theory – Ionic Screening and Ionic Strength (correction)
The potential of an isolated ion having a charge of ze in an dielectric medium at a location r from the charge is
On the other hand, the potential of an isolated ion having a charge of ze in an dielectric medium is
,
r
zer
rionisolated
04
)(
)exp(4
)( rze
rsolution
Ionic Strength,
where
)p(4
)(0 rr
solution
solvent
rA kT
zzmLmNe
0
22322 2/)1000(2
Ionic Strength
molecules/kg solvent
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)exp()(/)()( rrrrfionisolatedsolution
Q. So which is greater? solution or isolated ion?
zzmLmNe
22
32 2/)1000(2
m=0
m=0.001
Q. Find rDH for 3 solutions
rDH = 1/ is called Debye‐Hückel screening length f(1/ ) =e‐1 ~0.368
It is convenient to combine the
solventr
A kTLmNe
0
)1000(2
m=0.01
m=0 1concentration dependent terms as Ionicstrength:
i
iii
ii zmzmzzm
I 2222
2
1
2
m=0.1
Ionic Strength (Ex 9.2)
h h h f l/k f l
i
iii
ii zmzmzzm
I 2222
2
1
2
• What is the ion strength of 1 mol/kg of NaCl and Na2SO4 ?
•
• The principal ions of human blood plasma and
i
NaClI 22 11112
1 i
SONaI 2242 2112
2
1
• The principal ions of human blood plasma and their molal concentrations are
• Calculate the ionic strength of blood plasma. m
Na 0.14 m, m
Cl 0.10 m, m
HCO3 0.025 m.
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Debye‐Hückel Limiting Law
ezz
2
||)ln(
I1/2Derivation Raff (p416-428)
kTzz
r
08||)ln(
Q. Is > 1 or < 1 ?
Q. Why are the slopes different?
• µsolute= [ µ0 + RT ln(++‐‐)] + RT ln(m/m 0)
+ RT ln( )= µsolute
00 + RT ln(m/m 0) + RT ln( )
AgNO3
ZnBr2
Debey-Huckel “limiting” law is valid only for low ionic strength
Solvent is treated as a structure
CaCl2
-less dielectric medium Ions surrounded by structured water Assume point charge Ion has a volume Water coordinates to ion as complex Effective molality decreased
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• P9.14) Calculate the Debye–Hückel screening l th 1/ t 298 K i 0 00100 l ti flength 1/ at 298 K in a 0.00100 m solution of NaCl.
KatmkgL
molkgI solvent
r
298/
1091.2 11
110
Use this eq (9.31)to solve problems.(9.29) needs a factor m
The text is incorrect 108 1010
9.5 Chemical Eqilibrium in Electrolyte Solutions – Solubility ProductFrom Sec 8.13
• K =(aieq)i
• ai = i(ci/c0) kT
ezz
2
8||)ln( (*)
We next consider dissociation of MgF2 in an aqueous solution:
• MgF2(s) Mg2+(aq) + 2F‐(aq)
Because the activity of the pure solid can be set to equal 1,
Ksp =(aieq)i =aMg2+aF‐2 = (cMg2+/c
0) (cF‐/c0)2 3 =6.4x10‐9
kTr08
Q. Does the solubility depend on the ion strength? cMg2+ cF-
( )
sp i Mg2+ F Mg2+ F cF‐ = 2cMg2+. Two unknown (cF‐ and )
● First assume =1. Solve cF‐ cF‐ =0.565 x10‐3 molL‐1
From c, ionic strength I & use (*) =0.87.
Solve cF‐ with cF‐ = 0.68 x10‐3 molL‐1 (Repeat the process)
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• P9.21) Dichloroacetic acid has a dissociation constant of Ka = 3.32 x 10–2. Calculate the degree of dissociation for a 0.125 m solution of this acid (a) using the 0. 5 solutio of t is acid (a) usi g t eDebye–Hückel limiting law and (b) assuming that the mean ionic activity coefficient is one.
• HA H+ + A-
0.125 –m1 m1 m1
(a) Assume =1 Obtain m1.
Ksp =(aieq)i =aH+aA-
= (m1/m0) (m1/m0)2/{(0.125 – m1)/m0}= m1
2/(0.125 – m1) m1 =0.050 Then, revaluate ln new m1
Protein Salting in/Salting outP > P‐ + H+
Ksp = (c2/c0
2) 2 = s2 2
• When ionic strength I is low, < 1
Solubility of protein increases (Salting in)
• When ionic strength I is high, > 1
Solubility of protein decreases (Salting out)
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Solubility of hemoglobin in the presence of (NH4)2SO4
Solubility high for a lower ionic strength due to (NH ) SOdue to (NH4)2SO4
Adding some salts stabilizesprotein in a solution
Solubility lower for a higher ionic strength
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)
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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
We choose (sol) =0 for solution. So
9.7 Electrochemical Cells and Half‐Cells 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 ~~~
zFzFF ZMM 00
zFz ZMeleZMM ~~
Correct text
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Equilibrium at a half cell
(9.53)
• For a metal (made of a pure element)at 1 bar in a standard eleZMM z ~~
state, its chemical potential should be 0. (In text MZ+
0 = 0 Not exact)
• Using (9.53) and
)298,1(0 0 KbarZMM
zFz ZMeleZMM ~~~
zF ~
Fele ~
In a standard condition at 298K and 1 bar
Cf. and
zFMZM
zFzFMZM 0~
0~ 0 MM Fele ~
Standard Hydrogen Electrode
H+(aq) + e‐ 1/2H2(g)
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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 electrode
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
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?
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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
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