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Page 20 of 57
Accepted Manuscript
20
1
)
exp(
1
)
/
exp(
0
0
k
k
k
L
k
k
k
P
L
x
P
C
C
C
C
(56)
where
RT
FL
z
D
L
U
P
k
k
k
k
(57)
P
k
is the analog of the Peclet number (from the theory of convective mass transfer) for
k
-th
defect. We see that
P
k
is the ratio of the intensity of migration to that of diffusion.
As follows from Equation (57), for
P
k
>> 1, migration dominates in the positive direction
and practically in all of the volume of the barrier layer, and hence we have with great accuracy
0
k
k
C
C
with the exception of the very thin transient area near
x = 0
(bl/ol interface). In the
opposite case, for
P
k
<< -1, migration dominates in negative direction and practically in all
volume of the barrier layer we have with great accuracy
L
k
k
C
C
with the exception of the very
thin transient region near
x
=
L
(m/bl interface).
Let us consider the case of metal interstitials (
k
= 1). If, for example, χ = 3,
2.3
10
6
V/cm,
T
= 295 K,
L
10
-5
cm Equation (57) yields
P
1
≈
- 1810, i.e.
P
1
has a large, negative value
and with great accuracy
L
C
C
1
1
, with the exception of the very thin transient area near the bl/ol
interface (with the thickness ~ 10
-3
L
), where the concentration changes sharply with distance.
Accordingly, for the partial cation interstitials flux density, we have:
L
L
C
RT
FD
C
U
k
const
J
1
1
1
1
2
1
(58)
and, for the concentration of cation interstitials inside the barrier layer we have:
Page 21 of 57
Accepted Manuscript
21
i
i
L
i
i
FD
RT
k
FD
RT
k
C
C
2
2
(59)
Please, note that, in the coordinate system used here (increasing x from right to left), with the
origin at the barrier layer/solution interface,
> 0.
By analogy, for cation vacancies (z
2
= -
χ,
J
2
= k
4
) we have:
FD
RT
k
C
C
4
0
(60)
and for oxygen vacancies (z
3
= 2,
J
3
=(
χ/2)
k
3
) we find:
o
L
o
o
FD
RT
k
C
C
4
3
(61)
If, for example, Equation (52) holds, electronic conductivity of the film is:
e
e
e
FC
(58)
and, in accordance with Bojinov [42-44], the electronic conductivity of the barrier layer is:
o
e
i
e
e
e
e
D
D
k
F
D
D
Fk
FC
2
3
2
(59)
It was shown above that, with great accuracy, the electronic conductance does not depend
on the position in the film. Accordingly, Equation (50) can be simplified to yield
ˆ
ˆ
0
j
L
Z
e
e
(60)
Equations (59) and (60) allow us to estimate the electronic impedance.
Page 22 of 57
Accepted Manuscript
22
It must be noted that electronic impedance can be measured, but only in conjunction with
the reaction impedance represented by Randles circuit (see Figure 3).
Figure 3.
(Randles circuit)
It is evident that Randles Impedance can be described by the following equation.
b
R
R
b
O
O
C
D
C
D
A
F
n
Rt
Z
2
1
2
1
2
2
1
1
|
|
(61)
C
j
Z
Z
1
C
1
R
(62)
where
2
1
2
1
s
s
ct
C
j
R
Z
,
s
is the Warburg coefficient for semi-infinite diffusion in
solution, and
R
ct
is the charge transfer resistance of the redox reaction. The Warburg coefficient,
s
is given by:
R
R
O
O
s
C
D
C
D
F
n
RT
2
/
1
2
/
1
2
2
1
1
2
(63)
where
C
O
and
C
R
are the bulk concentrations of oxidized and reduced components, respectively,
of the redox couple and
D
O
and
D
R
are the corresponding diffusion coefficients.
Our calculations show that, due to the extremely low concentration of the oxygen in the
system (1 ppb), the leading cathodic reaction in the system is water reduction, i.e.
OH
H
e
O
H
2
2
2
1
(64)
Page 23 of 57
Accepted Manuscript
23
It is evident that this reaction, in the written direction, has no diffusion limitation, and
accordingly we can neglect the Warburg Impedance (
Z
w
≈ 0).
On the other hand, the current density corresponding to Reaction (64) is:
RT
FE
B
i
O
H
2
exp
(65)
where
O
H
2
is transfer coefficient of Reaction (64) and coefficient
B
depends on pH and the
temperature (but not on potential
E
). Accordingly, we have:
Fi
RT
RT
FE
exp
/
FB
RT
E
i
R
O
H
O
H
O
H
1
ct
2
2
2
(66)
For example, calculations performed by using OLI commercial software [46] yields
B
= 3.92×10
-
11
A/cm
2
(
E
is measured relative to the
SHE
electrode) and
5
.
0
2
O
H
at pH = 8.15 and
T
= 21
o
C and calculations yield
R
ct
= 0.308×10
10
Ω.cm
2
at
E
= 0.044 (
SHE
) and
R
ct
= 0.5938×10
14
Ω.cm
2
at
E
= 0.544 (
SHE
).
For the case of
Z
w
≈ 0, the Randles impedance has the form:
1
1
C
j
R
Z
ct
R
(67)
and the total (parallel) impedance of our system has the form:
''
'
1
0
1
ˆ
ˆ
jZ
Z
C
j
R
j
L
Z
Z
Z
ct
e
R
e
(68)
Page 24 of 57
Accepted Manuscript
24
2
2
1
2
0
2
'
)
(
)
ˆ
ˆ
(
C
R
R
L
Z
ct
ct
e
e
(69)
and
2
2
ct
1
ct
2
0
2
e
0
''
)
C
(
R
C
R
)
ˆ
ˆ
(
ˆ
ˆ
L
Z
(70)
By using Equations (68) to (70) and Equation (66) for
R
ct
, we can easily calculate the modulus:
2
2
r
e
Z
Z
Z
(71)
and the phase angle:
'
''
Z
Z
arctg
(72)
As noted above, the electronic impedance in parallel with the barrier layer is a complex
number [Equation (68)], which yields real [Equation (69)] and imaginary [Equation (70)]
components, as well as a modulus [Equation (71)] and phase angle [Equation (72)], all as a
function of frequency. Because the real and imaginary components are frequency-dependent,
they must be incorporated into the optimization procedure.
Figures 4 to 6 show typical Nyquist and Bode plots for the parallel impedance for the
case of iron in borate buffer solution [0.3 M
H
3
BO
3
/ 0.075 M
Na
2
B
4
O
7
, as appropriate] + 0.001
M
EDTA
[Ethylenediaminetetraacetic acid,
EDTA
, disodium salt], pH = 8.15,
T
= 21
o
C and
E
=
0.044 V (
SHE
). The parameter values used in the calculations are given in Table 2.
Table 2
(Parameter values).