1.
The Electrical Double Layer
(Ref.:Delahay: Double Layer and Electrode Kinetics; Mohilner in Bard, Ed.: Electroanalytical Chemistry, A Series of Advances, Vol. 1; Sparney: The Electrical Double
Layer; Bockris and Reddy: Modern Electrochemistry, Vol. 2)
1.1 The electrified interface
Interface between metal electrode and electrolyte solution:
−
+
−
−
−
−
+
+
+
+
Electrical double layer : the arrangement of charges and oriented dipoles constituting the
interphase region at the boundary of an electrolyte
Electrochemical Potential
Outer (Volta) potential) of a given phase, Ψ : the work done to transport a unit test
charge from infinity to a point just outside a charged but dipole-free phase (just outside
the reach of the image forces from the phase)
Surface potential of a given phase, χ : the work done to carry the unit test charge across
the dipole layer at the surface of an uncharged phase
Inner (Galvani) potential, φ : work done to transport the unit test charge from infinity
across the charged surface covered with a dipole layer to a point inside the phase
φ=Ψ + χ
(1)
Chemical potential, μi : work done to bring a mole of i particles from infinity into the
bulk of an uncharged, dipole-layer- free material phase.
1
−
Electrochemical potential, μ : total work done to take a mole of charges from infinity in
vacuum into the bulk of the material phase (it is the sum of potentials, one for material
phase without either charges or dipole layer on the surface, and another involving only
ψ
→
+
+
→
→
χ
φ
→
+
→
→
the charges and the dipole layer). Thus,
−
μ = μ + zFφ = μ + zF (ψ + χ )
(2)
Note: electrical work to bring a unit charge = φ
electrical work to transport one particle bearing a charge zieo = zieoφ
electrical work to bring an Avogadro number (NA) of particles inside a material
phase = NAzieoφ = ziFφ
Total driving force for the flow of a particular species is:
For an interface to be at equilibrium,
−
d μ i / dx
−
( d μ i / dx ) = 0 ,
2
i.e. the electrochemical potential of a species i must be the same on both sides of the
interface, or,
−
−
μ i (metal ) = μ i ( solution )
−
The electrochemical potential μ i of a species i in a particular phase is the change in the
(electrochemical) Gibbs energy of the system resulting from an introduction of a mole of
i particles into the phase (while keeping other conditions constant), i.e.
⎛ −⎞
∂G ⎟
μ i = ⎜⎜
⎜ ∂ni ⎟⎟
⎝
⎠ T , P ,n j
−
(3)
Thus, equality of electrochemical potentials on either side of phase boundary implies that
the change in Gibbs energy of the system resulting from the transfer of particles from one
phase to the other should be the same as that due to transfer in the other direction; i.e.
−
d G = 0 . This implies a free flow of species across the interface. An interface which
maintains an “open border” is a non-polarizable interface. Thus, thermodynamic
equilibrium exists at a non-polarizable interface.
For a non-polarizable interface:
or,
−
S
ΔM μ i =
S
ΔM μ i =
−
S
ΔM ( μ i + z i Fφ ) = 0
(4)
S
ΔM μ i + z i F S ΔM φ = 0
(4’)
where i is the species exchanged across the non-polarizable interface. Rearranging,
S
or,
ΔM φ = −
1 S M
Δ μi
zi F
d ( S ΔM φ ) = −
1
RT
dμi = −
d ln ai
zi F
zi F
(5)
(5’)
1.2 Thermodynamics of the electrified interface
The combined statement of the first and second law of thermodynamics (where only PdV
type of wok is considered) for an open system is:
dG = VdP – SdT + Σμidni
(6)
If other types of work are included, e.g. work of expansion of the interface, of interfacial
tension γ , by an area dA, which is given by γdA, or work involved in altering the charge
3
on the metal by an amount dqM upon the application of a potential φ, which is given by
φ dqM , then the complete expression for the change in Gibbs energy becomes:
dG = VdP – SdT + Σμidni + φ dqM + γdA
(7)
dG = Σμidni + φ dqM + γdA
(8)
G = Σμi ni + φ qM + γ A
(9)
At constant T and P,
Integrating:
Taking the total differential,
dG = Σμidni + φ dqM + γdA + Σni dμi + qM dφ + A dγ
(10)
Comparing with Eq. (8),
Σni dμi + qM dφ + A dγ = 0
(11)
Rearranging, (and taking unit area ,A = 1)
dγ = - Σni dμi - qM dφ
(12)
Surface excess, Γi , is defined as the amount of material over and above that which would
have existed had there been no double layer.
If ni is the actual number of moles of species i in the interfacial region and nio is the
number of moles that would have been there if there had been no double layer, then (per
unit area)
Γi = ni - nio
and therefore
(13)
ni = Γi + nio
ni dμi = Γi dμi + nio dμI
(14)
Σ ni dμi = Σ Γi dμi + Σ nio dμi
(15)
Summing over all i :
From the Gibbs-Duhem equation (for bulk material phase, without surface effects):
Σ nio dμi = 0
(16)
4
Thus,
Σ ni dμi = Σ Γi dμi
(17)
Substituting Eq.(17) in Eq. (12):
dγ = - Σ Γi dμi - qM dφ
Note that dφ represents the change in the inner or Galvani potential difference across the
interface under study. It may be written as d ( M 1 ΔS φ ) , where M1 is the metal electrode
under study and S is the solution. Thus,
dγ = - Σ Γi dμi - qM d ( M 1 ΔS φ )
(17’)
Although the inner potential is not measurable, the change in inner potential can be
measured provided that the M1/S interface is polarizable and that it is connected to a nonpolarizable interface M2/S to form an electrochemical cell. If such a cell is connected to
an external source of electricity,
V =
M1
ΔSφ + S ΔM 2 φ +
M2
'
ΔM1 φ
(18)
where M1’ is the connecting metal wire that completes the circuit. Since the last term in
Eq. (18) does not depend upon the potential V supplied from the external source, or the
solution composition,
dV = d(
or
d(
M1
M1
ΔSφ ) + d( S ΔM 2 φ )
ΔSφ ) = dV - d( S ΔM 2 φ )
(19)
(20)
Substituting in Eq. (17’):
dγ = - Σ Γi dμi - qM [dV - d( S ΔM 2 φ )]
(21)
Since there is thermodynamic equilibrium at the non-polarizable interface, (see Eq. (5)),
d ( S ΔM 2 φ ) = −
1
dμj
zjF
where the non-polarizable interface (reference electrode) is reversible with respect to ion
j. Eq. (21) then becomes:
dγ = −q M dV −
qM
zjF
dμ j − ∑ Γi dμ i
(22)
i
5
[Note: The reference electrode is reversible with respect to ion j. However, M2/S refers
to an electrode M2 dipped into the same solution S as M1; i.e. the cell is represented as:
M1/S/M2. Thus, when potential V is read, this value refers to a potential with respect to a
reference electrode consisting of metal M2 immersed in solution of the same composition
as that of the surrounding test electrode. Potentials referred to those reversible to ions of a
given varying concentration are denoted by V+ or V-]
For a solution of fixed composition (and where i = j ), dμi = 0 so that
dγ = −qM dV
⎛ ∂γ ⎞
⎜
⎟ = −q M
⎝ ∂V ⎠ μ
or,
(22)
Eq.(22) is referred to as the Lippmann equation. The potential at which the derivative in
Eq. (22) becomes zero (i.e. qM = 0 ) is known as the potential of zero charge (p.z.c.). The
plot of interfacial tension γ versus potential is known as the electrocapillary curve.
γ
p.z.c
V
The electrocapillary curve
The differential capacitance of the interface is defined as:
⎛ ∂ 2γ
⎛ ∂q ⎞
C = ⎜ M ⎟ = −⎜⎜
2
⎝ ∂V ⎠ μ
⎝ ∂V
⎞
⎟⎟
⎠μ
(23)
Consider the cell Hg⎜HCl⎜AgCl⎜Ag with varying composition using a reversible silversilver chloride electrode. This reference electrode is reversible with respect to the anion
Cl-. At constant potential, Eq. (22) becomes:
6
dγ = −
qM
zjF
dμ j − ∑ Γi dμ i
(24)
i
For i = j = -1, expanding the summation over all i :
dγ =
qM
dμ − − Γ+ dμ + − Γ− dμ −
F
(25)
Since μ = μ+ + μ- , then dμ = dμ+ + dμ-, or dμ+ = dμ - dμ- , so that
dγ =
qM
dμ − − Γ+ dμ + Γ+ dμ − − Γ− dμ −
F
(26)
or
⎛ q + FΓ+ − FΓ− ⎞
dγ = −Γ+ dμ + ⎜ M
⎟dμ −
F
⎝
⎠
(27)
Due to electrical neutrality across the interface,
Fn+ - Fn- = - qM
(28)
whereas in the absence of the double layer
Fn+o - Fn-o = - qM
(28’)
Combining these two equations:
o
o
q M + F (n+ − n+ ) − F (n− − n− ) = 0
(29)
Introducing the definition of surface excess, (Eq.(13),
qM + FΓ+ - FΓ- = 0
(30)
The numerator in the quantity in brackets in Eq. (27) is thus zero, and therefore
or,
dγ = - Γ+ dμ
(31)
⎛ ∂γ ⎞
− Γ+ = ⎜⎜ ⎟⎟
⎝ ∂μ ⎠V−
(32)
Since
7
μ = μ + + μ − = ( μ + o + RT ln a + ) + ( μ − o + RT ln a − ) = ( μ + o + μ − o ) + RT ln a + a −
and
2
a± = a+ a−
then
μ = ( μ + o + μ − o ) + 2RT ln a±
(33)
⎛
∂γ
− Γ+ = ⎜⎜
⎝ 2 RT∂ ln a ±
(34)
so that
⎞
⎟⎟
⎠V−
Excess charge densities in solution side of interface (due to surface excesses of positive
and negative ions) are given by
q+ = z+FΓ+
and
q- = z-FΓ-
qS = q+ + q- = -qM
(36)
-qM = z+FΓ+ + z-FΓ-
(37)
positively ch arg ed electrode
q− = − FΓ−
0
−
q+ = FΓ+
qS = − qM
+
q
(35)
p.z.c.
potential difference
1.3 The Structure of Electrified Interfaces
1.3.1 The Helmholtz Model: electrified interface consists of two sheets of charge, one on
the solution side, and another on the electrode side – similar to a parallel plate condenser.
8
d
q+
−
+
−
−
+
ε
V =
4π d
ε
q
(38)
V = potential difference across the condenser, q = charge on the parallel plates, d =
distance between plates, ε = dielectric constant of material between the plates.
Assuming ε is a constant, and that the charge on the metal is qM
dV =
4π d
ε
From the Lippmann equation,
dγ = -qM dV
Substituting for dV,
dγ = - qM
γ
Integrating,
∫ dγ = −
γ max
γ − γ max
4π d
ε
(39)
dq M
4π d
ε
dq M
(40)
q
∫q
M
dq M
0
4π d q M
=−
ε
2
2
(41)
or, in terms of the potential,
γ − γ max = −
ε 1 2
V
4π d 2
(42)
This is an eqation of a parabola symmetrical about γmax. Experimental electrocapillary
curve, however, is not symmetrical. Also, according to the parallel plate model, the
capacitance of the condenser is given by:
C=
dq
ε
=
= cons tan t
dV 4π d
(43)
(for constant ε and d). The differential capacitance of the interface, however, is not a
constant but depends on potential in a complex manner.
9
1.3.2 Gouy-Chapmann Theory
Consider a lamina in the electrolyte parallel to electrode and at a distance x from the
electrode.
The charge density in the lamina can be expressed as follows (Poisson’s equation):
+
ρx
+
ψ
+
+
+
+
ρx = −
ε d 2ψ x
4π dx 2
(44)
where ψ x is the outer potential difference between the lamina and the bulk of the solution
for which ψ x →∞ = 0 . Note that since Δχ for two points in the same phase is zero,
ψ x = φx .
From the Boltzmann distribution law:
ρ x = ∑ ni ( z i eo ) = ∑ ni o ( z i eo ) exp[−( z i eo )ψ x / kT ]
i
(45)
i
where ni and nio are the concentrations of the ith species in he lamina (at a distance x
from the electrode) and in the bulk of the solution, respectively; zi is the valence of
species i ; eo is the electronic charge.
Combining the Poisson and Boltzmann equations:
d 2ψ
4π
=−
2
ε
dx
∑n
o
i
( z i eo ) exp[−( z i eo )ψ x / kT ]
(46)
i
Note that:
10
2
d ⎛ dψ ⎞
d ⎛ dψ ⎞⎛ dψ ⎞
⎛ d ⎞⎛ dx
⎟ = 2⎜ ⎟⎜⎜
⎟⎜
⎟ =2
⎜
⎜
dψ ⎝ dx ⎠
dψ ⎝ dx ⎠⎝ dx ⎠
⎝ dx ⎠⎝ dψ
⎛ dx
= 2⎜⎜
⎝ dψ
⎞⎛ d 2ψ
⎟⎟⎜⎜ 2
⎠⎝ dx
⎞⎛ dψ ⎞⎛ dψ ⎞
⎟⎟⎜
⎟
⎟⎜
⎠⎝ dx ⎠⎝ dx ⎠
⎛ d 2ψ ⎞
⎞⎛ dψ ⎞
⎟⎟⎜
⎟ = 2⎜⎜ 2 ⎟⎟
dx
⎠
⎝
⎝ dx ⎠
⎠
2
1 d ⎛ dψ ⎞
d 2ψ
or,
⎜
⎟ =
2 dψ ⎝ dx ⎠
dx 2
Using this in the Poisson-Boltzmann equation:
2
d ⎛ dψ ⎞
8π
⎜
⎟ =−
dψ ⎝ dx ⎠
ε
2
Rearranging,
8π
⎛ dψ ⎞
d⎜
⎟ =−
ε
⎝ dx ⎠
8π
⎛ dψ ⎞
⎟ =−
⎜
ε
⎝ dx ⎠
o
i
z i e o exp[ − z i eoψ x / kT ]
(48)
i
∑n
o
i
z i e o e[ − zi eoψ x / kT ] dψ
(49)
i
Integrating,
2
∑n
(47)
∑
∑n
o
i
z i e o exp[ − z i eoψ x / kT ]
i
( − z i eo / kT )
i
+ const
(50)
Since, for x → ∞ , ψ = 0 and ( dψ / dx ) = 0 , then
const = −
2
Therefore,
8π kT
ε
∑n
8π kT
⎛ dψ ⎞
⎜
⎟ =
ε
⎝ dx ⎠
o
(51)
i
i
∑n
o
i
{exp[ − z i e oψ x / kT ] − 1}
(52)
For a z-z valent electrolyte, where ⎜z+⎜ = ⎜z-⎜ = z and n+o = n-o = no
2
8π kTn o ze oψ x / kT
⎛ dψ ⎞
e
− 1 + e − ze oψ x / kT − 1
⎜
⎟ =−
ε
dx
⎝
⎠
2
or,
(
8π kTn o ze oψ x / kT
⎛ dψ ⎞
e
+ e − ze oψ x / kT − 2
⎜
⎟ =
ε
dx
⎝
⎠
(
)
)
(53)
This may be rewritten as:
11
2
(
2
(
8π kTn o ze oψ x / kT
⎛ dψ ⎞
e
+ e − ze oψ x / kT − 2 ( e ze oψ x / kT )( e − ze oψ x / kT )
⎜
⎟ =
ε
⎝ dx ⎠
8π kTn o ze oψ x / 2 kT
⎛ dψ ⎞
e
− e − ze oψ x / 2 kT
⎟ =
⎜
ε
⎝ dx ⎠
)
2
(54)
Since ex – e-x = 2 sinh x
2
ze ψ
32π kTn o
⎛ dψ ⎞
=
sinh 2 o x
⎜
⎟
ε
2 kT
⎝ dx ⎠
⎛ 32π kTn o
⎛ dψ ⎞
⎜
⎟ = ± ⎜⎜
ε
⎝ dx ⎠
⎝
⎞
⎟⎟
⎠
1/ 2
⎛ 32π kTn o
⎛ dψ ⎞
⎜⎜
=
±
⎜
⎟
ε
⎝ dx ⎠
⎝
⎞
⎟⎟
⎠
1/ 2
sinh
ze oψ x
2 kT
(55)
Assume that sinh x ≈ x; then
or,
⎛ dψ ⎞
⎜
⎟ = ± bψ x
⎝ dx ⎠
⎛ 8 π n o ( ze o ) 2
⎛ ze oψ x ⎞
⎜⎜
=
±
⎜
⎟
ε kT
⎝ 2 kT ⎠
⎝
⎞
⎟⎟
⎠
1/ 2
ψx
(56)
1/ 2
where
⎛ 8 π n o ( zeo ) 2 ⎞
⎟⎟
b = ⎜⎜
kT
ε
⎝
⎠
(57)
At a positively charged electrode, ψ > 0, but (dψ / dx ) < 0, and at a negatively charged
electrode, ψ < 0, while (dψ / dx ) > 0. [Note also Gauss’s theorem where
4π q M = −ε (dψ / dx ) .] Thus, only the negative root of Eq. (56) corresponds to the
physical situation:
⎛ dψ ⎞
⎜
⎟ = − bψ x
⎝ dx ⎠
The integrated form is :
as shown below.
ψ x = ψ o exp(−bx)
(58)
and the potential - distance profile is
12
)
ψ
x
From Gauss’s law,
⎛ 32π kTno ⎞
⎛ dψ ⎞
−ε ⎜
⎟
⎟=ε ⎜
ε
⎝ dx ⎠
⎝
⎠
and
sinh
ε ⎛ 32π kTno ⎞
qM =
⎜
⎟
ε
4π ⎝
⎠
qM
where
1/ 2
1/ 2
⎛ 2kTno ε ⎞
= −q s = ⎜
⎟
⎝ π ⎠
⎛ kTno ε ⎞
A=⎜
⎟
⎝ 2π ⎠
zeoψ x
= 4π q M
2kT
sinh
(59)
zeoψ x
= −q s
2kT
1/ 2
sinh
zeoψ x
ze ψ
= 2 A sinh o x
2kT
2kT
(60)
1/ 2
(61)
For x = 0, ψ x = ψ o = potential at x = 0 relative to the bulk of the solution where the
potential is zero, or ψ ∞ = 0 . The diffuse charge density (qd = qs) is therefore
⎛ 2kTno ε ⎞
q d = −⎜
⎟
⎝ π ⎠
1/ 2
sinh
zeoψ x
2kT
(62)
The differential capacity may be evaluated from
⎛ ∂q
C = ⎜⎜ M
⎝ ∂ψ M
⎛ ∂q
⎞
⎟⎟ = −⎜⎜ d
⎠
⎝ ∂ψ o
⎞ ⎛ z 2 eo 2 n o ε ⎞
⎟
⎟⎟ = ⎜
⎜ 2πkT ⎟
⎠ ⎝
⎠
1/ 2
cosh
zeoψ o
2kT
(63)
The cosh function gives an inverted parabola, and the Gouy-Chapmann theory therefore
predicts an inverted parabola dependence of C on potential across the interface. This is
approximated experimentally only at very dilute solutions.
13
C
G − C Theory
C
Exp' t (dil. solution )
Exp' t (conc. solution)
Potential
Potential
1.3.3 Stern’s modification of the Gouy-Chapmann theory
Stern: ions cannot reach electrode beyond “plane of closest approach” – which is same
for cations and anions.
Double layer divided into two regions: compact double layer (between electrode and
plane of closest approach, and a diffuse double layer, from plane of closest approach to
bulk of solution.
Due to charge separation, a potential drop results:
Differentiating the above:
or,
ψ M = ψ M − ψ S = (ψ M − ψ 2 ) + (ψ 2 − ψ s )
(64)
∂ψ M ∂ (ψ M − ψ 2 ) ∂ψ 2
=
+
∂q M
∂q M
∂q d
(65)
1
1
1
=
+
C C M − 2 C 2− s
(66)
Model of double layer: two capacitors in series
⎯⎜⎜⎯⎯⎜⎜⎯
CM-2 C2-S
14
plane of closest approach ( 2)
−
−
−
−
−
Electrode
compact double
−
scattered ions
−
−
−
−
−
diffuse double layer
layer
Note: for large concentrations, no is large and from the Gouy-Chapman expression, for
capacitance, C2-S becomes large and hence 1/C2-S becomes small compared to 1/CM-2 .
Therefore C ≈ CM-2. Hence the disagreement of the G-C theory with experiment for
large electrolyte concentrations. But for low concentrations, C ≈ C2-S in agreement with
experiment.
CM-2 is not experimentally accessible. However, C can be measured, and q can be
calculated from the C versus Potential curve by integration. From magnitude of q, ψ can
be calculated and hence also C2-S .
Grahame assumed that CM-2 depends only on charge on electrode and not on electrolyte
concentration. Thus, CM-2 can be calculated from C and C2-S. For concentrated electrolyte
solutions, C2-S >> CM-2 . (Electrolyte to be chosen: one that is not specifically adsorbed.)
CM − 2
C
Calc ' d
C
1 M NaF
Expt ' l
0.01 M NaF
Expt ' l
Potential
Potential
15
1.3.4 Grahame Model of double layer – Classical picture
OHP
IHP
adsorbed anion
electrode
solvated cation
<
>
-
+
>
<
<
+
<
<
-
>
-
+
>
<
+
<
>
+
-
solvent dipole
diffuse layer
1.4 Ionic concentration in plane of closest approach and ionic contribution of diffuse
double layer to total charge
Ionic concentration ni(2) in plane of closest approach (x = x2) may be obtained from the
Boltzmann distribution law with ψ = ψ2 = φ2:
n1( 2 ) = ni e − zi eoφ 2 / kT
o
(67)
[Note that since Δχ for two points in the same phase is zero, ψ x = φ x .]
ni(2) > nio for ziφ2 < 0 and ni(2) < nio for ziφ2 > 0 . Thus, since φ2 > 0 when E >Epzc
and φ2 < 0 when E < Epzc , cations are attracted [ attraction: ni(2) > nio ] for E < Epzc
and are repelled [repulsion: ni(2) < nio ] for E >Epzc (in absence of specific adsorption).
The excess charge due to ionic species i in a lamina of thickness dx at a distance x from
the electrode is (assuming ψo = ψs = 0):
zeo (ni − ni )dx = zeo ni [e − zeoψ x / kT − 1]dx
x
o
o
(68)
The total excess of cations in the diffuse layer is then:
qi
2− s
∞
∞
= ∫ zeo (ni − ni )dx = ∫ zeo ni [e − zeoφ x / kT − 1]dx
x2
x
o
o
(69)
x2
16
Let zeoφx / kT = Φ. In view of : (e − Φ − 1) = [(e − Φ − 1) 2 ]1 / 2 , then
qi
2− s
= zeo ni
∞
∫ (e
o
−2Φ
− 2e
−Φ
+ 1) dx = zeo ni
1/ 2
∞
∫ [(e
o
x2
= zeo ni
∞
o
−Φ
− 2 + e Φ )e −Φ ]1 / 2 dx
x2
∫ [ (e
−Φ / 2
−e
Φ/2 2
) e
−Φ 1 / 2
]
dx = zeo ni
o
x2
(70)
∞
∫ [(e
−Φ / 2
−e
Φ/2
)e
−Φ / 2
dx
x2
From the previously derived result (eq. (54):
2
8π kTn o ze oψ x / 2 kT
⎛ dψ ⎞
e
− e − ze oψ x / 2 kT
⎜
⎟ =
dx
ε
⎝
⎠
(
o
⎛ d ψ ⎞ ⎛ 8π kTn ⎞
⎟⎟
⎜
=
⎜
⎟ ⎜
ε
⎝ dx ⎠ ⎝
⎠
⎛ 8π kTn ⎞
⎟⎟
= ⎜⎜
ε
⎠
⎝
o
Therefore,
(e
Φ/2
−e
−Φ / 2
)
⎛
ε
= ⎜⎜
o
⎝ 8π kTn
⎞
⎟⎟
⎠
1/ 2
1/ 2
1/ 2
(e
(e
ze oψ x / 2 kT
Φ/2
)
2
− e − zeoψ x / 2 kT
− e −Φ / 2
)
(71)
)
⎛ dφ ⎞
⎜
⎟
⎝ dx ⎠
(72)
Substituting this, and changing variables from dx to dφ ,
qi
2− s
⎛
ε
= zeo ni ⎜⎜
o
⎝ 8π kTn
o
⎛ kTn ε ⎞
⎟⎟
= ⎜⎜
⎝ 2π ⎠
o
⎞
⎟⎟
⎠
1/ 2 0
∫e
φ
−Φ / 2
dφ
(73)
2
1/ 2
[e − zeoφ 2 / 2 kT − 1]
i.e.
q+
q−
2− s
2− s
⎛ kTn o ε
= ⎜⎜
⎝ 2π
⎞
⎟⎟
⎠
1/ 2
⎛ kTn ε ⎞
⎟⎟
= − ⎜⎜
⎝ 2π ⎠
o
[ e − ze oφ 2 / 2 kT − 1]
(74)
1/ 2
[ e − ze oφ 2 / 2 kT − 1]
For φ2 = 0, (i.e. at the potential of zero charge), q +
2− s
= q−
2− s
=0 .
In the absence of specific adsorption,
q+
2− s
= z + FΓ+
q−
2− s
= z − FΓ−
(75)
17
Generally,
q±
2− s
(
)
= ± A e − zeoφ 2 / 2 kT − 1
⎛ kTn o ε
A = ⎜⎜
⎝ 2π
where
⎞
⎟⎟
⎠
(75’)
1/ 2
(76)
The total charge is then:
(
q M = −q s = − q +
2− s
+ q−
2− s
) = 2 A sinh ze2kTφ
o
2
(77)
1.5 Criteria for specific adsorption
Since dγ is a perfect differential,
it follows that:
dγ = -qM dE± - Γm dμ
(78)
⎛ ∂E ±
⎜⎜
⎝ ∂μ
(79)
⎞
⎛ ∂Γ ⎞
⎟⎟ = −⎜⎜ m ⎟⎟ Esin – Markov equation
⎠q
⎝ ∂q ⎠ μ
Expressing the charge in terms of q m , the charge in solution due to anions or cations,
respectively, (i.e. qi 2-s = ziFΓi ),
⎛ ∂E ±
⎜⎜
⎝ ∂μ
⎛ ∂E ±
⎞
⎟⎟ = ⎜⎜
⎠ q ⎝ RT∂ ln a s
⎞
1 ⎛ ∂q m
⎟⎟ = −
⎜
z m F ⎜⎝ ∂q
⎠q
⎞
⎟⎟
⎠μ
(80)
2
where as = activity of the electrolyte = a± . Note that E ± refers to potential measured
against a reference electrode which is reversible to cation (E+) or anion (E-) in a cell with
no liquid junction. If reference electrode of constant electrolyte activity is used, then,
upon variation of the activity of electrolyte in contact with ideal polarized electrode,
E ref = E ± ±
RT
ln a ± + const
zF
(81)
provided the liquid junction potential is constant. [For z-z valent electrolyte, |z| = z+ = -z-]
Substituting in the above,
⎛ ∂E ref ⎞
RT ⎡⎛ ∂q m
⎟
⎜
⎢⎜⎜
=
±
⎜ ∂ ln a 2 ⎟
|
z
|
F
⎢⎣⎝ ∂q M
± ⎠q
⎝
⎞
1⎤
⎟⎟ ± ⎥
⎠ as 2 ⎥⎦
(82)
18
i.e. a shift in Eref with ln a (or ln
2
a± ) at constant q depends on ⎛⎜⎜ ∂qm ⎞⎟⎟ . In absence of
⎝ ∂qM ⎠ as
specific adsorption, q m can be computed from the Gouy-Chapman theory and one can
compare experimental and predicted shifts of potential.
In absence of specific adsorption,
and
so that
q± = q±
2− s
q± = q±
2− s
⎛ kTn o ε ⎞
⎟⎟
= ±⎜⎜
π
2
⎝
⎠
1/ 2
⎛ RTc s ε ⎞
= ±⎜
⎟
⎝ 2π ⎠
1/ 2
(e
− z ± eoφ 2 / 2 kT
−1
)
(e
− z ± Fφ 2 / 2 RT
− 1 = ± A e − z± Fφ2 / 2 RT − 1
)
(83)
(
)
⎛ ze φ ⎞
q M = −(q +2− s + q −2− s ) = −q 2− s = 2 A sinh ⎜ o 2 ⎟
⎝ 2kT ⎠
⎛ ∂q 2− s
− ⎜⎜ −
⎝ ∂q M
⎛ ∂q 2− s
⎞
⎟⎟ = ⎜⎜ −2− s
⎠ a ⎝ ∂q
⎞
zF
⎟⎟ = −
RT
⎠
⎛ ∂E +
⎜⎜
2
⎝ ∂ ln a ±
⎡
⎛ q 2− s
⎛1⎞
= ⎜ ⎟ exp ⎢sinh −1 ⎜⎜ −
⎝2⎠
⎝ 2A
⎣
(85)
⎞
⎟⎟
⎠q
⎞⎤ ⎡ ⎛ q 2− s
⎟⎟⎥ ⎢1 + ⎜⎜
⎠⎦ ⎢⎣ ⎝ 2 A
(84)
⎞
⎟⎟
⎠
2
⎤
⎥
⎥⎦
−1 / 2
(86)
⎛ ∂q ⎞
Note: for q2-s → ∞ or q → -∞, ⎜⎜ − ⎟⎟ → 0
⎝ ∂q ⎠ a
⎛ ∂q ⎞
For q2-s → -∞ , or q → ∞, ⎜⎜ − ⎟⎟ → −1
⎝ ∂q ⎠ a
⎛ ∂q
Also, ⎜⎜ −
⎝ ∂q
⎞
1
⎟⎟ = − for q2-s = 0 ( q = 0). Therefore, potential of zero charge is
2
⎠a
independent of electrolyte activity when it is measured vs a reference electrode of
constant electrolyte activity in a cell with liquid junction.
⎛ ∂E ref ⎞
RT ⎡⎛ ∂q m
⎟
⎜
⎢⎜
=
±
⎜ ∂ ln a 2 ⎟
| z | F ⎢⎜⎝ ∂q M
± ⎠q
⎝
⎣
⎞
1⎤
⎟⎟ ± ⎥ = 0
⎠ as 2 ⎥⎦
(87)
for q2-s = 0, or qM = 0.
(See Note below.)
Note that, for the system:
Hg/HCl/AgCl/Ag,
19
E- = EAg/AgCl/Cl - EHg = [EoAg/AgCl/Cl – (RT/F)ln aCl-] - EHg
constant
whereas for the system:
Hg/HCl/KClsat’d/AgCl/Ag
Eref = EAg/AgCl/Cl - EHg = [EoAg/AgCl/Cl – (RT/F)ln aCl-(sat’d)] - EHg
Constant
The difference between the two is:
Eref – E- = (RT/zF) ln aCl- + constant = (RT/zF) ln a± + constant
If it is assumed that single ion activities may be replaced by mean ionic activities.
qM = −12
E+
q M = −8
E+
qM
qM
qM
qM
qM
qM
qM = −4
qM = 0
qM = 4
qM = 8
=0
=2
=4
=6
=8
= 10
qM = 12
RT ln a±2
RT ln a±2
Hg / NaF
Hg / KCl
Eref
qM = −12
q M = −8
qM = −4
qM = 0
qM = 4
qM = 8
qM = 12
RT ln as
20
Note:
zeoφ 2
2kT
2− s
2− s
⎛ ∂q ⎞
⎛ ∂q ⎞
zF ⎛ ∂E + ⎞
⎜
⎟
− ⎜⎜ − ⎟⎟ = ⎜⎜ −2− s ⎟⎟ = −
RT ⎜⎝ ∂ ln a ±2 ⎟⎠
⎝ ∂q M ⎠ a ⎝ ∂q ⎠
ze φ
zFφ 2
⎞
⎛
⎛
⎞
q −2− s = − A⎜ exp
− 1⎟ = − A⎜ exp o 2 − 1⎟
2 RT
2kT
⎝
⎠
⎝
⎠
zFφ 2
⎛q ⎞
but
= sinh ⎜ M ⎟
2 RT
⎝ 2A ⎠
q M = −q 2− s = −q d = 2 A sinh
⎡ ⎛
⎛ q ⎞⎞ ⎤
and ∴ q −2− s = − A⎢exp⎜⎜ sinh ⎜ M ⎟ ⎟⎟ − 1⎥
⎝ 2A ⎠⎠ ⎦
⎣ ⎝
and
⎛ ∂q −2− s
⎜⎜
⎝ ∂q M
−1
⎞
⎛
⎛ q ⎞ ⎞⎛ ∂ sinh (q M / 2 A) ⎞
⎟⎟ = − A exp⎜⎜ sinh ⎜ M ⎟ ⎟⎟⎜⎜
⎟⎟
∂q M
⎝ 2 A ⎠ ⎠⎝
⎝
⎠
⎠
Since
1
d sinh −1 u
du
= 2
1/ 2
dx
(u + 1) dx
then
⎛ ∂q −2− s
⎜⎜
⎝ ∂q M
Since
sinh −1 x = ln x + ( x 2 + 1)1 / 2
2
⎞
⎛
1
⎛ q ⎞ ⎞⎛ ⎛ q ⎞ ⎞
⎟⎟ = − exp⎜⎜ sinh⎜ M ⎟ ⎟⎟⎜1 + ⎜ M ⎟ ⎟
2
⎝ 2 A ⎠ ⎠⎜⎝ ⎝ 2 A ⎠ ⎟⎠
⎝
⎠
[
⎛ ∂q −2− s
⎜⎜
⎝ ∂q M
]
⎞
(q M / 2 A)
1 1⎛
⎟⎟ = − − ⎜
2 2 ⎜⎝ 1 + (q M / 2 A) 2
⎠
[
]
1/ 2
or, since
∂E ±
RT ⎛ ∂q m2− s
⎜
=
−
z m F ⎜⎝ ∂q M
∂ ln a ±2
then
(q M / 2 A)
∂E ±
RT ⎛⎜
1+
=
−
2
2 zF ⎜⎝
∂ ln a ±
1 + (q M / 2 A) 2
For qM >>1,
∂E±
= - (RT/zF);
∂ ln a±2
For qM = 0,
∂E±
= - 1/2(RT/zF);
∂ ln a±2
⎞
⎟
⎟
⎠
⎞
⎟⎟
⎠
[
For qM/2A << - 1, or – (qM/2A) >> 1,
−1 / 2
]
1/ 2
∂E ref
∂ ln a ±2
∂E ref
∂ ln a ±2
∂E±
= 0;
∂ ln a±2
⎞
⎟
⎟
⎠
= − 1/2(RT/zF)
=0
∂E ref
∂ ln a ±2
= 1/2(RT/zF)
21
1.6 Amount of specifically adsorbed ions
Cations are assumed to be not specifically adsorbed. Thus,
q+ = q+
2− s
(
= z + FΓ+ = A e − z+ Fφ 2 / 2 RT − 1)
)
(88)
Γ+ can be determined experimentally. Therefore, φ2 can be computed and hence also
q-2-s which is given by:
q−
2− s
= − A[e − z −eoφ 2 / 2 kT − 1]
Setting q − = z − FΓ− = q − ' + q −
latter may easily be computed .
2− s
, where q-’ is the specifically adsorbed charge, the
2− s
(89)
Degree of specific adsorption varies with electrolyte concentration (see figure below).
Also the potential of zero charge varies with electrolyte concentration.
γ
Anion specific adsorption
incr. concentration
p.z.c.
Potential
q=6
I−
CH 3COO −
−
Br
Cl −
− q−
'
CO3
NO3
4
=
− q−
'
2
0
−
−2
0
Potential vs NCE
log a
22
1.7 Distribution of potential
0.001 M NaF
0.01 M
0.1 M
−
φ2
KCl
−
KBr
KAc
φ2
0
E
+
+
KF
0
+
E − Ez
−
NaF : not specifically adsorbed
With specific adsorption
1.8 Adsorption Isotherms
Abulk = Aads
−
(90)
−
μ A bulk = μ A ads
(91)
μ A o ,bulk + RT ln a A bulk = μ A o ,ads + RT ln a A ads
(92)
At equilibrium:
(μ A
o ,bulk
− μA
o , ads
) = RT ln
− ΔG o = RT ln
−
aA
aA
ads
bulk
f (Γ )
bulk
aA
ΔG o
f (Γ )
= ln bulk
RT
aA
(93)
f (Γ ) = a A
bulk
e −ΔG
o
/ RT
=β a
Γ = Γ( β a )
23
The relationship between the bulk (a )and adsorbed (Γ ) amounts (concentrations) is the
adsorption isotherm.
π = ξo -ξ
(94)
Surface pressure defined as:
where the auxiliary variable ξ is given by:
ξ
Name
Henry’s law
Virial
Langmuir
Frumkin
Volmer
Equation of State
π = RTΓ
π = RTΓ + gΓ2
π = RTΓs ln (1 - θ )
π = RTΓs ln (1 - θ ) + gΓ2
π = RT Γ /(1-bΓ)
= γ + qE
(95)
Isotherm
βa = RTΓ
βa = RTΓ exp(2gΓ /RT)
βa = Γ / (Γs - Γ ) = θ / (1-θ )
βa = [θ / (1-θ )]exp (2gΓ /RT )
βa = RT Γ /(1-bΓ) exp (bΓ/(1-bΓ )
g = interaction parameter, positive for attraction and negative for repulsion
θ = Γ /Γs (subscript s indicates saturation coverage)
Exercise / Group Work
Read and give brief presentations of the following articles:
1. D.C. Grahame, Chem. Rev. 41, 441(1947).
2. R. Parsons in Bockris et al, Eds., Modern Aspects of Electrochemistry, Vol. 1, Butterworth's
Publications Ltd. London, 103 (1954).
3. D.M. Mohilner in Bard et al, Eds., Electroanalytical Chemistry : A Series of Advances,
Vol. 1, 241 (1966).
Exercise
1. The interfacial tension of Hg in 1 mol dm-3 CsCl solution is given below. (a) Using a graphical
method determine the potential of zero charge. (b) Calculate the capacitance of the double layer
as a function of potential from the data below.
-V / SCE
0
0.1
0.2
0.3
0.4
0.5
γ / N m-1
0.345
0.376
0.397
0.411
0.419
0.423
-V / SCE
0.6
0.7
0.8
0.9
1.0
γ / N m-1
0.420
0.414
0.406
0.395
0.383
2. Explain how one determines the presence/absence, and the amount of specifically adsorbed
ions.
3. Give one example where an equation of state is used to derive an adsorption isotherm or viceversa.
(See the end of the Lecture Notes for further exercises.)
24