Transcript Slide 1

The Electrochemical Double Layer
Lecture 9.1
Gouy – Chapman Model
 (0)
M  -
+ electrostatics
k T (Boltzman) Thermal
Randomization
The number of carriers in a given energy plane (distance away
from electrode) is found to be:
electrostatic
thermal
ni  n  e
0
i
 zi e / kT
charge on e-
Bulk carrier # concentration
The potential profile is:
2kT
 d 

 
 o
 dx 
2
  zkieT

i n  e  1


0
i
For a 1:1 electrolyte (~e.g. NaF, CaSO4)
 8kTn
d
 
dx
  o
 0 is potential
at electrode
1/ 2



 z e 

sinh

2
k
T


 dx
tanh z e x  / 4kT 
tanh z e (0) / 4kT 
cm K  3.3x10 z C 
1
0
7
 1/ 2
mol/L
inverse thickness of diffuse layer
 e x
For aqueous at 25C
The Electrochemical Double Layer
Lecture 9.2
If  (0) is  50 / z mV at 25C
 ( x)   (0)e x
Use unsimplified
equation
−1 is often called
the Debye Length
Wow! Precipitous drop in  ( x) at high  (0);
does not followexp function. Why?
1.0
effectively linear
 ( x)
 (o )
exponential
0

~ 3 4 A
 H 2O
~ 3  5 vs. 78 in bulk!

x, A

10 A
Ah, the outer Helmholtz Plane!
The Electrochemical Double Layer
Lecture 9.3
 
Cd  228 z C
 1/ 2
cosh19.5 z  (0) 
25 C
H 2O
1:1 electrolyte
***Figure 12.3.5***
Chem. Rev. 1947 41, 441
***Figure 12.3.1***
Too large a Cd and too fast a change!
Why does Gouy – Chapman Fail?
The model assumes that the ions are point
charges. As  (0) increases, the separation between
the metal and charged electrolyte decreases to 0.
Not Realistic!
Stern’s Modification Accounts for
1. Finite ionic size
2. Additional radial increase due to
solvation of ions
The Electrochemical Double Layer
Lecture 9.4
Thus, must have plane of closest approach!
-
tanh z e x  / 4kT 
+
tanh z e ( x 2 ) / 4kT 
 e K ( x x2 )
+
For diffuse layer only!!!
+
“x2”
OHP
1.0
This is the compact layer.
Get linear drop of  (x) .
 ( x)
 (o )

x, A
compact
OHP
1
x
 2 
CDL   o  2  z 2 e 2 n 0
o

kT

1
1/ 2



 z e ( x2 ) 

cosh

 z kT 
Exactly what we saw from Helmholtz.
The Electrochemical Double Layer
Lecture 9.5
Effects of Double Layer on ET Reactions
O z  ne  R z ; for z  0
attracts
repels
-
O+
C+ O+
O+ O+ O +
C+ O+
O+ O+ O +
C+
O+ O+
C+ O+ O+
O+ O+
vs.
-
C+
OC+
OC+
OC+
O-
OOOOO-
x2
OHP
Thus apparent concentration of Oz is “similar” to that of the
electrolyte. That is to say we have an electrostatic driving
force attracting the cationic O or repelling anionic O.
If
 M is
+ , then cationic O repelled and anionic O attracted.
So,
Co ( x2 , t )  Coe zF ( x2 ) / RT
So, we will see changes in
i0 and k0 at different [SE]
and [Oz]*, which is what
prompted this study/theory.
Note NO absolute
value of charge.
z is the signed
charge on O.
The Electrochemical Double Layer
Lecture 9.6
Linear Decay of  (x) , à là Helmholtz
OHP
140
 (x)
120
Diffuse Layer,
exponential decay of  (x)
mV 100
80
60
40
20
0
10
20
30
40
50
60
70
80

x, A
OHP
So, Oz does not experience
Eapplied , but Edriving  Eapp   ( x2 ).
force
So, must correct for:
1. electrostatic effects on Co ( x2 , t )
2. electrostatic effects on E in rate equations from Chp. 3.
Recalling:
Totally irreversible reaction of O
kf >>> kb
i  nFAk0Co 0, t e
i  nFAkt0Coe  zF ( x2 ) / RT  e
k 0  kt0  e
R
   nF 
o 
 
( E  E ) 
  RT 

corrections
so
+ne
   nF
 
  RT

o 
 ( E  ( x2 )  E ) 


  n  z F ( x2 ) 


RT


This is the apparent rate constant.
The Electrochemical Double Layer
Lecture 9.7
Examples:
Zn 2  2e   Zn0
[ SE], M
 mA 
j0  2 
 cm 
 ( x2 ), m V
j0,t
0.025
-63.0
12
0.40
0.25
-41.1
2.7
0.38
means
delocalized
and
 e
 An

“An”
 ( x2 ), mV k (cm / s)
 76
5.0
0
kt0 (cm / s)
26
The [ An] at x2 is being depleted due
to - interactions.
Of course, x2, and thus,  ( x2 ) , vary with electrolyte size/type.
Also, we have assumed NO specific adsorption of SE anions,
O, or R.
Thus, the Frumkin Correction is limited, but it works well in
most cases.