Test condition
- No stirring;
- Fast kinetics
- Step to an arbitrary potential suddenly
- Cottrell-like experiment
- Planar plate electrode
Half reaction:
(1)
O
+
n
e
⇌
R
O+ne{\rightleftharpoons} R \tag{1}
O+ne⇌R(1)
In the case of Nerstian equation, the potential can be expressed as
(2)
E
=
E
0
′
−
R
T
n
F
ln
C
O
(
0
,
t
)
C
R
(
0
,
t
)
E=E^{0'}-\dfrac {RT}{nF} \ln \dfrac {C_{O}(0,t)}{C_{R}(0,t)} \tag {2}
E=E0′−nFRTlnCR(0,t)CO(0,t)(2)
Here we define
(3)
θ
=
C
O
(
0
,
t
)
C
R
(
0
,
t
)
=
exp
[
n
F
R
T
(
E
−
E
0
′
)
]
\theta=\dfrac {C_{O}(0,t)}{C_{R}(0,t)}=\exp \left [\dfrac{nF} {RT} (E-E^{0'}) \right] \tag{3}
θ=CR(0,t)CO(0,t)=exp[RTnF(E−E0′)](3)
The governing equations of the reductant and oxidant will be
(4)
∂
C
O
(
0
,
t
)
∂
t
=
D
O
∂
2
C
O
(
0
,
t
)
∂
x
2
\dfrac{\partial C_O(0,t)}{\partial t}=D_O \dfrac{\partial ^2 C_O(0,t)}{\partial x^2} \tag{4}
∂t∂CO(0,t)=DO∂x2∂2CO(0,t)(4)
(5)
∂
C
R
(
0
,
t
)
∂
t
=
D
R
∂
2
C
R
(
0
,
t
)
∂
x
2
\dfrac{\partial C_R(0,t)}{\partial t}=D_R \dfrac{\partial ^2 C_R(0,t)}{\partial x^2} \tag{5}
∂t∂CR(0,t)=DR∂x2∂2CR(0,t)(5)
Initial condition:(Concentration is homogeneous)
(6)
C
O
(
x
,
0
)
=
C
O
∗
C_O(x,0)=C_O ^* \tag{6}
CO(x,0)=CO∗(6)
(7)
C
R
(
x
,
0
)
=
0
C_R(x,0)=0 \tag{7}
CR(x,0)=0(7)
Boundary conditions:
(8)
C
O
(
∞
,
t
)
=
C
O
∗
(
t
>
0
)
C_O(\infty,t)=C_O ^* \quad (t>0) \tag{8}
CO(∞,t)=CO∗(t>0)(8)
(9)
C
R
(
∞
,
t
)
=
0
(
t
>
0
)
C_R(\infty,t)=0 \quad (t>0) \tag{9}
CR(∞,t)=0(t>0)(9)
(10)
D
O
∂
C
O
(
0
,
t
)
∂
x
+
D
R
∂
C
R
(
0
,
t
)
∂
x
=
0
D_O \dfrac{\partial C_O(0,t)}{\partial x} +D_R \dfrac{\partial C_R(0,t)}{\partial x} =0 \tag{10}
DO∂x∂CO(0,t)+DR∂x∂CR(0,t)=0(10)
Apply Laplace transform on Equation(4)-(10) to solve the equations
Laplace transform function
(11)
F
(
s
)
=
∫
0
∞
f
(
t
)
e
−
s
t
d
t
F (s)=\int_0^\infty f(t)e^{-st}dt \tag{11}
F(s)=∫0∞f(t)e−stdt(11)
Apply the transformation on these equations:
(12)
C
‾
O
(
∞
,
s
)
=
C
O
∗
s
\overline C_O(\infty,s)=\dfrac{C_O^*}{s} \tag{12}
CO(∞,s)=sCO∗(12)
(13)
C
‾
O
(
0
,
s
)
=
0
\overline C_O(0,s)=0 \tag{13}
CO(0,s)=0(13)
(14)
D
O
∂
C
‾
O
(
0
,
t
)
∂
x
+
D
R
∂
C
‾
R
(
0
,
t
)
∂
x
=
0
D_O \dfrac{\partial \overline C_O(0,t)}{\partial x} +D_R \dfrac{\partial \overline C_R(0,t)}{\partial x} =0 \tag{14}
DO∂x∂CO(0,t)+DR∂x∂CR(0,t)=0(14)
Using (12) and (13) we can get the solution
(15)
C
‾
O
(
x
,
s
)
=
A
(
s
)
exp
(
−
s
D
O
x
)
+
C
O
∗
s
\overline C_O(x,s)=A(s) \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{15}
CO(x,s)=A(s)exp(−DOs
(16)
C
‾
R
(
x
,
s
)
=
B
(
s
)
exp
(
−
s
D
R
x
)
\overline C_R(x,s)=B(s) \exp{\left (-\sqrt{\dfrac{s}{D_R}}x \right)} \tag{16}
CR(x,s)=B(s)exp(−DRs
Apply (14) the relation between
A
(
s
)
A(s)
A(s) and
B
(
s
)
B(s)
B(s)
(17)
B
(
s
)
=
−
A
(
s
)
ξ
B(s)=-A(s)\xi \tag{17}
B(s)=−A(s)ξ(17),
where
(18)
ξ
=
D
O
D
R
\xi = \sqrt{\dfrac{D_O}{D_R} } \tag{18}
ξ=DRDO
Hence the results are:
(15)
C
‾
O
(
x
,
s
)
=
A
(
s
)
exp
(
−
s
D
O
x
)
+
C
O
∗
s
\overline C_O(x,s)=A(s) \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{15}
CO(x,s)=A(s)exp(−DOs
(16)
C
‾
R
(
x
,
s
)
=
−
A
(
s
)
ξ
exp
(
−
s
D
R
x
)
\overline C_R(x,s)=-A(s)\xi \exp{\left (-\sqrt{\dfrac{s}{D_R}}x \right)} \tag{16}
CR(x,s)=−A(s)ξexp(−DRs
Here we make the assumption: reversible electrochemical reaction at the surface of the electrode
(17)
C
O
(
0
,
s
)
=
θ
C
R
(
0
,
s
)
C_{O}(0,s)=\theta C_{R}(0,s) \tag{17}
CO(0,s)=θCR(0,s)(17)
Thus
(18)
C
O
∗
s
+
A
(
s
)
=
−
ξ
θ
A
(
s
)
\dfrac{C_O^*}{s} +A(s)=-\xi \theta A(s) \tag{18}
sCO∗+A(s)=−ξθA(s)(18)
We can get
A
(
s
)
A(s)
A(s)
(19)
A
(
s
)
=
−
C
‾
O
∗
s
(
1
+
ξ
θ
)
A(s)=-\dfrac{\overline C_O^*}{s(1+\xi \theta)} \tag {19}
A(s)=−s(1+ξθ)CO∗(19)
finally
(20)
C
‾
O
(
x
,
s
)
=
−
C
‾
O
∗
s
(
1
+
ξ
θ
)
exp
(
−
s
D
O
x
)
+
C
O
∗
s
\overline C_O(x,s)=-\dfrac{\overline C_O^*}{s(1+\xi \theta)} \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{20}
CO(x,s)=−s(1+ξθ)CO∗exp(−DOs
(21)
C
‾
R
(
x
,
s
)
=
ξ
C
‾
O
∗
s
(
1
+
ξ
θ
)
exp
(
−
s
D
R
x
)
\overline C_R(x,s)=\dfrac{\xi \overline C_O^*}{s(1+\xi \theta)} \exp{\left (-\sqrt{\dfrac{s}{D_R}}x \right)} \tag{21}
CR(x,s)=s(1+ξθ)ξCO∗exp(−DRs
Fick’s first law states that the flux is proportional to the concentration gradient
(22)
−
J
O
(
x
,
t
)
=
D
O
∂
C
O
(
x
,
t
)
∂
x
-J_O(x,t)=D_O \dfrac {\partial C_O(x,t)}{\partial x} \tag{22}
−JO(x,t)=DO∂x∂CO(x,t)(22)
Laplace tansformation gives
(23)
−
J
O
(
x
,
s
)
=
D
O
∂
C
O
(
x
,
s
)
∂
x
-J_O(x,s)=D_O \dfrac {\partial C_O(x,s)}{\partial x} \tag{23}
−JO(x,s)=DO∂x∂CO(x,s)(23)
Assumption: Cottrell experiment
(24)
i
(
t
)
=
n
F
A
D
O
1
/
2
C
O
∗
π
1
/
2
t
1
/
2
(
1
+
ξ
θ
)
i(t)= \dfrac {nFAD_O^{1/2} C_O^*}{\pi ^{1/2} t^{1/2} (1+\xi \theta)}\tag{24}
i(t)=π1/2t1/2(1+ξθ)nFADO1/2CO∗(24)
then
(25)
i
(
t
)
=
i
d
(
t
)
1
+
ξ
θ
i(t)= \dfrac{i_d(t)}{1+\xi\theta} \tag{25}
i(t)=1+ξθid(t)(25)
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