Transcript 7. Fundmentals of Charged Surfaces

8. Fundamentals of Charged Surfaces
Moving the reagents
Quickly and with
Little energy
Diffusion
electric fields
Yo
+
Charged Surface
Y*o
+
+
+
X=0
G
N*
 exp  kT
No
1. Cations distributed thermally
with respect to potential
2. Cations shield surface and
reduce the effective surface
potential
G
N*
 exp  kT
No
Yo
+
Y**o
Charged Surface
Y*o
+
+
+
+
+
+
Y***o
+
X=0
dx *
+
dx
**
dx
***
Surface Potentials
 zF x
n
 e RT
no
Cation distribution has
to account for all species,
i
Poisson-Boltzman equation
Simeon-Denis Poisson
1781-1840
 d 
 o  2  
 dx 
2
Dielectric constant of solution
Permitivity of free space
zi x F
*  RT
i
 z FC e
i
i
Charge near electrode depends
upon potential and is integrated
over distance from surface - affects
the effective surface potential
 d 
  o  2  
 dx 
2
 z FC e
i
*
i
 zi x F
RT
i
Solution to the Poisson-Boltzman equation can be simple if the
initial surface potential is small:
ze o
 1
kT
 o  50mV
 x   o e  x
Potential decays from the surface potential exponentially with distance
 d 
 2 
 dx 
2
 z FC
i
i
  o
*
i
e
zi i F
RT
 z FC
*
i
i

i
  o
2


zi i F 1   zi i F 
 
  ......
1  
RT
2  RT 


Largest term
 d 2 
 2 
 dx 
Let
2
 1
   
 xa 
2
F 2  zi Ci*
i
RT o
x
F 2  zi Ci*
i
RT o
Then:
 d 2   x
 2  2
 dx  xa
General Solution of:
Y x  Ae

x
xa
 d 2   x
 2  2
 dx  xa
 Be
x
xa
Because Y goes to zero as x goes to infinity
B must be zero
Y x  Ae

x
xa
 Ae  x
Because Y goes to Y0 as x goes to zero (e0 =1)
A must be Y0
thus
 x   o e  x
Potential decays from the surface potential exponentially with distance
When =1/x or x=1/ then
 x   o e 1   o (0.367)
The DEBYE LENGTH
x=1/
What is ?
Yo
Charged Surface
+
+
+
+
+
+
Y=0.36 Yo
+
+
+
X=0
X=1/
Petrus Josephus
Wilhelmus Debye
1844-1966
Debye Length
 2n z e 


  o kT 
* 2 2
 2Conc N A z e 


 o kT 

1
2
2 2
1
2
Does not belong
2


moles   6.02 x10 23   1L   100cm 
.
x10 19 C 
2  160218

 2


 ch arg e 
 3 3







L  mole  10 cm
m
ch arg e






 12
2
 23




8
.
85419
x
10
C
m
Nm
138065
.
x
10
J




 78.49 o unitless 



 298 K 
2

25 C
K
N m

  100cm   J  



2


=1/cm
1
2


 # 
 19 2
2
160218
.
x
10





 
2


mole
cm

 




 
23
 23
 78.49 o 138065
   6.022 x10 ions    
.
x
10
298

25 C






  z(3.29 x10 7 )(C * ) 1/ 2
Units are 1/cm
1
2
 2n * z 2 e 2 

  zF 
  o kT 
Debye Length
1
2
  z(3.29 x10 )(C )
7
* 1/ 2
Units are 1/cm
Table 2: Extent of the Debye length as a function of electrolyte
1/κ ( )
C(M)
1
3
0.1
9.6
0.01
30.4
0.001
96.2
0.0001
304
In the event we can not use a series approximation to solve the
Poisson-Boltzman equation we get the following:
ze
ze 0



 exp 2 kT  1  exp 2 kT  1
Check as


Compared to tanh 






x
By Bard exp

ze
ze 0



 exp 2 kT  1  exp 2 kT  1






Ludwig Boltzman
1844-1904
Simeon-Denis Poisson
1781-1840
Set up excel sheet ot have them calc effect
Of kappa on the decay
Example Problem
A 10 mV perturbation is applied to an electrode surface bathed in
0.01 M NaCl. What potential does the outer edge of a Ru(bpy)33+
molecule feel?
Debye length, x?
  z(3.29 x10 7 )(C * ) 1/ 2
Units are 1/cm
108 A
X  1/ 

30
.
4
A
1(3.29 x10 7 )(0.01) 1/ 2
Since the potential applied (10 mV) is less than 50 can use
the simplified equation.
Radius of Ru
 x   o e  x   o e
x

xz
 10e
9

30.4
 7.43
The potential the Ru(bpy)33+ compound experiences
is less than the 10 mV applied.
This will affect the rate of the electron transfer event
from the electrode to the molecule.
Surface Charge Density
The surface charge distance is the integration over all the charge
lined up at the surface of the electrode


a
a
 o    dx   0 
d 2
d
dx


0
2
dx
dx
The full solution to this equation is:
ze o
 o  (8kT o n o ) sinh(
)
2 kT
1
2
1
2
 o  117
. (C*) sinh(19.5z o )
C is in mol/L

a
Yo
Charged Surface
+
+
+
+
+
+
Y=0.36 Yo
+
+
+
X=0
X=1/
Can be modeled as a capacitor:

C
d
differential
d
For the full equation
1
2
 2 z e  0 n 
 ze o 
 cosh
C 

d
 2 kT 
kT


2
2
Cd 228zC
o
1
* 2

Differential capacitance
Ends with units of uF/cm2
cosh19.5z o 
At 25oC, water
Conc. Is in mol/L
y  cosh x
12000
10000
8000
6000
4000
2000
0
-15
-10
-5
0
5
10
15
Can be simplified if
(o ~ 25 mV),
 o   o o
Specific Capacitance is the differential
space charge per unit area/potential
Cspecific
dq
d


Ad d
A
C
  o
A
Specific Capacitance
Independent of potential
For small potentials
1
2
 2 z e  0 n 
 ze o 
 cosh
C 

 2 kT 
kT


2
2
o
120
100
Capacitance
80
60
 o
40
20
0
-500
-400
-300
-200
-100
0
E-Ezeta
Gouy-Chapman Model
100
200
300
400
500
Flat in this region
Henrik Jensen , David J. Fermn and Hubert H. Girault*
Received 16th February 2001 , Accepted 3rd April 2001
Published on the Web 17th May 2001
Physical Chemistry Chemical Physics
DOI: 10.1039/b101512p
Paper
Photoinduced electron transfer at liquid/liquid
interfaces. Part V. Organisation of water-soluble
chlorophyll at the water/1,2-dichloroethane interface�
Laboratoire d'Electrochimie, D�partement de Chimie, Ecole Polytechnique F�d�rale de Lausanne, CH-1015,
Switzerland
Real differential capacitance plots appear to roll off instead of
Steadily increasing with increased potential
Yo
+
Exponential
in the thermally
equilibrated or
diffuse layer
CHelmholtz or Stern
Charged Surface
Linear drop
in potential
first in the
Helmholtz or
Stern specifically
adsorbed layer
+
+
+
Hermann Ludwig
Ferdinand von Helmholtz
1821-1894
+
+
+
+
+
X=0
x2
Cdiffuse
O. Stern
Noble prize 1943
Capacitors in series
1
2
 2 z e  0 n 
 ze o 
 cosh
CDiffuse  

 2 kT 
kT


2
2
o
 C
  o
 
 A  Helmholtz or Stern
Cseries 
1
Cseries
Wrong should be x distance of stern layer
1
1
1
1

 ......
C1 C2
CN
Cseries
1
1
1


 ......
C1 C2
CN
For large applied potentials and/or for large salt concentrations
1.
ions become compressed near the electrode surface to
create a “Helmholtz” layer.
2.
Need to consider the diffuse layer as beginning at the
Helmholtz edge
x2
1


C  0
1
1
2
 2 z 2 e 2  0 n o 
 ze o 

 cosh

 2 kT 
kT


Capacitance
Due to Helmholtz
layer
Capacitance due to diffuse
layer
0.71
0.7
0.69
Capacitance
0.68
0.67
0.66
0.65
Deviation
Is dependent upon
The salt conc.
0.64
0.63
-500
-400
-300
-200
-100
0
E-Ezeta
100
200
300
400
500
The larger the “dip”
For the lower
The salt conc.
Create an excel problem
And ask students to determine the smallest
Amount of effect of an adsorbed layer
Experimental data does not
Correspond that well to the
Diffuse double layer double capacitor
model
(Bard and Faulkner 2nd Ed)
Physical Chemistry Chemical
Physics
DOI: 10.1039/b101279g
Complex formation between halogens and
sulfoxides on metal surfaces
Paper
*
and
Andrew
A.
Gewirth
pacitance�potential curve for the Au(111)/25 mM KI in DMSO interface w
Department of Chemistry, and Frederick Seitz Materials Research
Laboratory, Uni rsity of Illinois at Urbana-Champaign, Urbana, IL,
61801, USA
Received 8th February 2001 , Accepted 20th April 2001
Published on the Web 1st June 2001
Siv K. Si
e
Model needs to be altered to account
For the drop with large potentials
This curve is pretty similar to predictions except where specific
Adsorption effects are noted
Graphs of these types were (and are) strong evidence of the
Adsorption of ions at the surface of electrodes.
Get a refernce or two of
deLevie here
Charged Surface
Yo
Introducing the Zeta Potential
Imagine a flowing solution
along this charged surface.
Some of the charge will be carried
+
away with the flowing solution.
+
+
+
+
+
+
+
+
Yzeta
Introducing the Zeta Potential, given the symbo lz
Yo
Shear Plane
Flowing solution
Charged Surface
+
+
Sometimes
assumed
zeta
corresponds
to Debye
Length, but
Not
necessarily
true
+
+
+
+
+
+
+
z  C
1

2
exp
1
 C 2
The zeta potential is dependent upon how the electrolyte
concentration compresses the double layer. ,  are constants
and sigma is the surface charge density.
Shear Plane can be talked about in
two contexts
Charged Surface
Yo
Shear Plane
+
+
+
+
++
+
+
+
+
+
+
+
+
+
+
+
+
In either case if we “push” the solution along
Shear
a plane we end up with charge separation which Plane
leads to potential
+
+
+
Particle in motion
Streaming Potentials
From the picture on preceding slide, if we shove the solution
Away from the charged surface a charge separation develops
= potential
Y oz

P

  solution resis tan ce   m
z  zeta potential
kg
  vis cos ity
m s
Sample problem here
Reiger- streaming potential
apparatus.
Can also make measurements on blood capillaries
In the same way, we can apply a potential and move ions and
solution
Anode
Vapp
Charged Surface
Yo
Jm
+
+
+
+
+
+
+
+
X=0
+
Jo
Jm
Cathode
+
Movement of a charged ion in an electric field
Electrophoretic mobility


f i v  zi e 
i
  applied electric field
f  frictional drag  6r
v  electrophoretic velocity
The force from friction is equal to the electric driving force
The frictional drag comes
about because the migrating
ion’s atmosphere is moving
in the opposite direction, dragging
solvent with it, the drag is related to the ion atmosphere
f frictional  6r
  vis cos ity
r  ionic radius
  ion velocity
Drag Force
f electrical  zi e
Electric Force
Direction of Movement
Ion accelerates in electric field until the electric force
is equal and opposite to the drag force = terminal velocity
At terminal velocity
f frictional  f electric
6r  zi e

zi e
6r
The mobility is the velocity normalized for the electric field:
zi e

ui  
 6r

vi
zi e
zi e


 uep

f
6r
Stokes-Einstein
equation

Frictional drag  6r
(Stokes Law)
r = hydrodynamic
radius
Typical values of the electrophoretic mobility are
small ions
5x10-8 m2V-1s-1
proteins
0.1-1x10-8 m2V-1s-1
Sir George Gabriel
Stokes 1819-1903
Reiger p. 97
Insert a sample calculation
When particles are smaller than the Debye length you get
The following limit:
2 oz
uep 
3
Remember: velocity is mobility x electric field
Reiger p. 98
What controls the hydrodynamic radius?
- the shear plane and ions around it
Compare the two equations for electrophoretic mobility
2 oz
f oz
uep 

3

foz


zi e
6r
zi e

uep  
 6r
r
zi e
f oz 6
Where f is a shape term which is 2/3 for spherical
particles
Relation of electrophoretic mobility to diffusion
Thermal “force”
kT
D
f
Frictional drag  6r
kT
kT
D

f
6r
zi e

ui  
 6r
kT kT
D

uelectrophoretic migration
f
ze
Measuring Mobilities (and therefore Diffusion)
from Conductance Cells
+
-
+
+
+
-
-
-
+
-
+
+
+
-
To make measurement need to worry about all the processes
Which lead to current measured
-
Ac Voltage +
Solution
Charge
Motion = resistance
+
Charging + +
RO
RO
Electron
Transfer
+
- +
+
+
Ct
Ct
Zf1
Rs
Zf2
An aside


1

Z f   Rct 
1

 
Cs 2

2

  1
 
1

  Cs 2





2
1
2





diffusion
Related to ket
Electron transfer at electrode surface can be modeled as the
Faradaic impedance, Z2
Ct
Ct
Zf1
Rs
Zf2
Solving this circuit leads to
Zf1
Zf 2
RT 
 Rs 
Zf1
Zf 2
1
1
Ct
Ct
1
1
RT 
 Rs 
1
1
1
1


Z f 1( ) Ct
Z f 2( ) Ct
Applying a high frequency, w, drops out capacitance and Faradaic
Impedance so that RT=Rs
What frequency would you have to use
To measure the solution resistance between
Two 0.5 cm2 in 0.1 M NaCl?
Cspecific
A
d d ( o )


  o
d
d
  z(329
. x10 )(C )
7
* 1/2
1
 104
. x10
m
C  Cspecific A  Ao
7
Check
Calculation
To show that
It is cm converted to
C  Cspecific A  Ao
2
2 


1
m
C




 78.54 8.854 x1012

C  Ao   104
. x107   2 x0.5cm2 x





 100cm  
m 
J  m

2
2 


1
m
C




7
2

12

C  Ao  104
. x10
. x10
  2 x0.5cm x
  78.54 8854

 100cm  
m 
J  m

C  7..2 x10
7
2
C2
C
7 C
 7..2 x10
 7..2 x107  7..2 x107 F
J
CV
V
The predicted capacitance of both electrodes in 0.1 M NaCl would
Be 0.72 microfarads
For the capacitive term to drop out of the electrical circuit
We need:
1
 1
Ct
 
1
1
6


14
.
x
10
Ct
7.2 x10 7
The frequency will have to be very large.
Solution Resistance Depends upon
Cell configuration
A
length
Resistivity of soln.
R

A
Sample calculation in a thin layer cell
Resistance also depends upon the shape
Of an electrode
Disk Electrode
R

4a
a is the radius
Spherical electrode

R
4a
Hemispherical
electrode

R
2a
Scan rate 1000 V/s at two different size electrodes for
Thioglycole at Hg electrode
From Baranski, U. Saskatchewan
Conductivity is the inverse of Resistance

k 
 RA
1
Resistivity and conductivity both depend upon
Concentration. To get rid of conc. Term divide

k
C
1



C RCA
  molar conductivity
A plot of the molar conductivity vs Concentration has a slope
Related to the measurement device, and an intercept related to
The molar conductivity at infinite dilution
o  s tan dard molar conductivity
This standard molar conductivity depends upon the solution
Resistance imparted by the motion of both anions and cations
Moving in the measurement cell.
   t 
o
   t 
o
Where t is a transference number which accounts for the
Proportion of charge moving
Transference
Numbers can be
Measured by capturing
The number of ions
Moving.
Once last number needs
To be introduced:
The number of moles of ion
Per mole of salt
o  v    v  
Compute the resistance of a disk electrode
Of 0.2 cm radius in a 0.1 M CaCl2 solution
  v    v  
o

  2Ca  1Cl  2
o
2
m
0.00763   mol
  1
2
m
0.0119   mol

2
m
 0.02716   mol
m2
1
1
  0.02716


3
  mol C
 mol   L   100cm 
  01
.
 3 3 


L   10 cm   m 

1
m2  mol   L   100cm 
0.02716
.
 01
 3 3 






  mol
L
m 
10 cm
3
 0.368  m
The resistance is computed from

0.368  m
R

 4.6
01
. m
4a

4 0.2cmx


cm 
Remember – we were trying to get to mobility
From a conductance measurement!!!!
i
ui 
zi F
o
Also remember that mobility and diffusion coefficients are
related
o
kT
kT oi kToi

 7 i J  mol
D
u
 2
 2.66x10
ze
ze zF z eF
z 2 C2
D  2.66x107
oi J  mol
z2
C2
We can use this expression to calculate
Diffusion coefficients
D  2.66x107
D3  2.66 x107
oi J  mol
z2
C2
2
m
302.7 x104
2
J

mol
m
J
10
  mol
 8.92 x10
2
2
( 3)
C
C 2
 
m2 J     VC  m2
 

2 Vs 
s
C    J 
 C
D4   2.66 x107
2
m
442 x104
2
J

mol
m
10
  mol

7
.
34
x
10
s
( 4) 2
C2
Fe(CN)63- diffusion coefficient is
9.92x10-10 m2/s
Fe(CN)64- diffusion coefficient is
7.34x10-10 m2/s
The more highly charged ion has more solution solutes around
It which slows it down.
How does this effect the rate of electron transfer?
Activation energy
*
ket   el
Probability factor
 kT 
Z ~

 2m 
 G
Z exp kT
Collisional factor
1
2
Where m is the reduced mass.
Z is typically, at room temperature,
104 cm/s
  G 


o
G *
Free energy change
2
4
  in  out
work required to change bonds
And bring molecules together
a D  donor radii
a A  acceptor radii
op  optical dielectric cons tan t
 s  regular dielectric cons tan t
e  electron ch arg e
out
 e 2 
 1





1
1
1
1




 



 4o   2a D 2a A rDA   op  s 




Formal potential
G  eE  ( w  w )
o
o
p
r
Work of bringing ions together
a A 
 za z pe2   ea D
e
rDA

( w p  wr )  U r  

e

 40   1  a D 1  a A 




The larger kappa the smaller the activation energy, the closer
Ions can approach each other without work
When one ion is very large with respect to other (like an electrode)
Then the work term can be simplified to:
( w p  wr )  Ur  Yze