恒电势电流曲线和浓度分布（i

电势电流曲线和浓度分布（i"/>

恒电势电流曲线和浓度分布（i

Test condition

1. No stirring;
2. The concentration of reactant at the electrode surface is always 0.
   Half reaction:
   O+ne→R(a) (a) O + n e → R $$O+ne\rightarrow R \tag{a}$$
   The equation of the limiting current versus concentration will be
   ∂CO∂t=DO∂2CO∂x2(1) (1) ∂ C O ∂ t = D O ∂ 2 C O ∂ x 2 $$\dfrac{\partial C_O}{\partial t}=D_O \dfrac{\partial ^2 C_O}{\partial x^2} \tag{1}$$
   Initial condition:(Concentration is homogeneous)
   CO(x,0)=C∗O(2) (2) C O ( x , 0 ) = C O ∗ $$C_O(x,0)=C_O ^* \tag{2}$$
   Boundary conditions:
   CO(∞,t)=C∗O(3) (3) C O ( ∞ , t ) = C O ∗ $$C_O(\infty,t)=C_O ^* \tag{3}$$
   CO(0,t)=0(t>0)(4) (4) C O ( 0 , t ) = 0 ( t > 0 ) $$C_O(0,t)=0 \quad (t>0) \tag{4}$$

Apply Laplace transform to solve Eq.(1)

Laplace transform function

F(s)=∫∞0f(t)e−stdt(5) (5) F ( s ) = ∫ 0 ∞ f ( t ) e − s t d t $$F (s)=\int_0^\infty f(t)e^{-st}dt \tag{5}$$
Apply Laplace transform on the left part of Eq(1)
L{∂CO∂t}=∫∞0∂CO∂te−stdt=COe−st∣∣∣∞0+s∫∞0COe−stdt=CO(x,∞)e−s∞−CO(x,0)e0+s∫∞0COe−stdt=−C∗O+s∫∞0COe−stdt(6) L { ∂ C O ∂ t } = ∫ 0 ∞ ∂ C O ∂ t e − s t d t = C O e − s t | 0 ∞ + s ∫ 0 ∞ C O e − s t d t = C O ( x , ∞ ) e − s ∞ − C O ( x , 0 ) e 0 + s ∫ 0 ∞ C O e − s t d t (6) = − C O ∗ + s ∫ 0 ∞ C O e − s t d t $$\begin{align*} \mathcal L \left \{\dfrac{\partial C_O}{\partial t} \right\} &=\int_0^\infty \dfrac{\partial C_O}{\partial t} e^{-st}dt \\&= C_O e^{-st} \bigg |_0^\infty+s\int_0^\infty C_O e^{-st}dt\\&=C_O(x,\infty) e^{-s\infty}-C_O(x,0) e^0+s\int_0^\infty C_O e^{-st}dt\\&=-C_O^*+s\int_0^\infty C_O e^{-st}dt \tag{6}\\ \end{align*}$$
Define
C¯¯¯¯O(x,s)=∫∞0COe−stdt(7) (7) C ¯ O ( x , s ) = ∫ 0 ∞ C O e − s t d t $$\overline C_O(x,s) =\int_0^\infty C_O e^{-st}dt \tag{7}$$
L{DO∂2CO∂x2}=∫∞0DO∂2CO∂x2e−stdt=DO∂2∂x2∫∞0COe−stdt=DO∂2C¯¯¯¯O(x,s)∂x2(8) L { D O ∂ 2 C O ∂ x 2 } = ∫ 0 ∞ D O ∂ 2 C O ∂ x 2 e − s t d t = D O ∂ 2 ∂ x 2 ∫ 0 ∞ C O e − s t d t (8) = D O ∂ 2 C ¯ O ( x , s ) ∂ x 2 $$\begin{align*} \mathcal L \left \{D_O\dfrac{\partial^2 C_O}{\partial x^2} \right\} &=\int_0^\infty D_O \dfrac{\partial^2 C_O}{\partial x^2} e^{-st}dt \\&= D_O \dfrac{\partial^2}{\partial x^2}\int_0^\infty C_O e^{-st}dt\\&=D_O \dfrac{\partial^2 \overline C_O(x,s) }{\partial x^2} \tag{8}\\ \end{align*}$$
After Laplace transform, Eq(1) becomes
sC¯¯¯¯O(x,s)=DO∂2C¯¯¯¯O(x,s)∂x2+C∗O(9) (9) s C ¯ O ( x , s ) = D O ∂ 2 C ¯ O ( x , s ) ∂ x 2 + C O ∗ $$s \overline C_O(x,s) =D_O \dfrac{\partial^2 \overline C_O(x,s) }{\partial x^2} +C_O^* \tag{9}$$
The solution of the ODE is:
C¯¯¯¯O(x,s)=A1expsDO−−−−√x+A2exp(−sDO−−−−√x)+A3(10) (10) C ¯ O ( x , s ) = A 1 exp ⁡ s D O x + A 2 exp ⁡ ( − s D O x ) + A 3 $$\overline C_O(x,s)=A_1\exp{\sqrt{\dfrac{s}{D_O}}x}+A_2\exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+A_3 \tag{10}$$
where A1 A 1 $A_1$, A2 A 2 $A_2$ and A3 A 3 $A_3$are constant.
Apply Laplace transform on boundary conditions, the we can get

C¯¯¯¯O(∞,s)=C∗Os(11) (11) C ¯ O ( ∞ , s ) = C O ∗ s $$\overline C_O(\infty,s)=\dfrac{C_O^*}{s} \tag{11}$$
C¯¯¯¯O(0,s)=0(12) (12) C ¯ O ( 0 , s ) = 0 $$\overline C_O(0,s)=0 \tag{12}$$
After substitution, we can solve
A1=0(13) (13) A 1 = 0 $$A_1=0 \tag{13}$$
A2=−C∗Os(14) (14) A 2 = − C O ∗ s $$A_2=-\dfrac{C_O^*}{s} \tag{14}$$
A3=C∗Os(15) (15) A 3 = C O ∗ s $$A_3=\dfrac{C_O^*}{s} \tag{15}$$
Hence the solution should be

C¯¯¯¯O(x,s)=−C∗Osexp(−sDO−−−−√x)+C∗Os(16) (16) C ¯ O ( x , s ) = − C O ∗ s exp ⁡ ( − s D O x ) + C O ∗ s $$\overline C_O(x,s)=-\dfrac{C_O^*}{s} \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{16}$$

Current-time profile

It is known that the flux at the electrode surface is proportional to the current, we can get the current

−JO(0,t)=i(t)nFA=DO∂CO(x,t)∂x∣∣∣x=0(17) (17) − J O ( 0 , t ) = i ( t ) n F A = D O ∂ C O ( x , t ) ∂ x | x = 0 $$-J_O(0,t)=\dfrac{i(t)}{nFA}=D_O \dfrac{\partial C_O(x,t)}{\partial x} \bigg |_{x=0}\tag{17}$$
Applying the Laplace transform on Eq(17) gives
i¯(t)nFA=DO∂C¯¯¯¯O(x,s)∂x∣∣∣x=0(18) (18) i ¯ ( t ) n F A = D O ∂ C ¯ O ( x , s ) ∂ x | x = 0 $$\dfrac{\overline i(t)}{nFA}=D_O \dfrac{\partial \overline C_O(x,s)}{\partial x} \bigg |_{x=0}\tag{18}$$
Substitution of Eq (16) into Eq (18) gives

i¯(s)nFA=DOC∗OssDO−−−−√exp(−sDO−−−−√)x∣∣∣x=0(19) (19) i ¯ ( s ) n F A = D O C O ∗ s s D O exp ⁡ ( − s D O ) x | x = 0 $$\dfrac{\overline i(s)}{nFA}=D_O \dfrac{C_O^*}{s} \sqrt{\dfrac{s}{D_O}} \exp{\left(-\sqrt{\dfrac{s}{D_O}}\right)x}\bigg |_{x=0}\tag{19}$$
it becomes

i¯(s)nFA=DOC∗OssDO−−−−√=C∗ODOs−−−−√(20) (20) i ¯ ( s ) n F A = D O C O ∗ s s D O = C O ∗ D O s $$\dfrac{\overline i(s)}{nFA}=D_O \dfrac{C_O^*}{s} \sqrt{\dfrac{s}{D_O}} =C_O^* \sqrt{\dfrac{D_O}{s}}\tag{20}$$
It yields
i¯(s)=nFAD1/2OC∗Os1/2(21) (21) i ¯ ( s ) = n F A D O 1 / 2 C O ∗ s 1 / 2 $$\overline i(s)=\dfrac{nFAD_O^{1/2}C_O^*}{s^{1/2}}\tag{21}$$
applying the form
L−1{1s√}=L−1{1π−−√⋅π−−√20s√}=1π−−√L−1{π−−√20s√}(22) L − 1 { 1 s } = L − 1 { 1 π ⋅ π 2 0 s } (22) = 1 π L − 1 { π 2 0 s } $$\begin{align*} \mathcal L^{-1}\left \{\dfrac{1}{\sqrt s}\right \} &=\mathcal L^{-1}\left \{\dfrac{1}{\sqrt \pi} \cdot \dfrac{\sqrt{\pi}}{2^0\sqrt s}\right \} \\&= \dfrac{1}{\sqrt \pi}\mathcal L^{-1}\left \{ \dfrac{\sqrt{\pi}}{2^0\sqrt s}\right \} \tag{22} \\\end{align*}$$
use Inverse Laplace Transform table
L−1{(2n−1)!π−−√2nsn+1/2}=tn−1/2(23) (23) L − 1 { ( 2 n − 1 ) ! π 2 n s n + 1 / 2 } = t n − 1 / 2 $$\mathcal L^{-1}\left \{ \dfrac{(2n-1)!\sqrt{\pi}}{2^n s^{n+1/2}}\right \}=t^{n-1/2} \tag{23}$$
L−1{1s√}=πt−−√(24) (24) L − 1 { 1 s } = π t $$\mathcal L^{-1}\left \{\dfrac{1}{\sqrt s}\right \}=\sqrt{\pi t} \tag{24}$$

so the applying the inverse Laplace transform on Eq(21) gives

i(t)=nFAD1/2OC∗Oπ1/2t1/2(25) (25) i ( t ) = n F A D O 1 / 2 C O ∗ π 1 / 2 t 1 / 2 $$i(t)=\dfrac{nFAD_O^{1/2}C_O^*}{\pi^{1/2} t^{1/2}}\tag{25}$$
which is called Cottrell equation

Concentration-distance profile

Applying Inverse Laplace transform on Eq(16) gives

CO(x,t)=C∗O(1−erfcx2DOt−−−−√)(26) (26) C O ( x , t ) = C O ∗ ( 1 − e r f c x 2 D O t ) $$C_O(x,t)=C_O^*\left (1-erfc \dfrac{x}{2\sqrt{D_Ot}} \right) \tag{26}$$
or
CO(x,t)C∗O=1−erfcx2DOt−−−−√=erfx2DOt−−−−√(27) (27) C O ( x , t ) C O ∗ = 1 − e r f c x 2 D O t = e r f x 2 D O t $$\dfrac{C_O(x,t)}{C_O^*}=1-erfc \dfrac{x}{2\sqrt{D_Ot}} =erf \dfrac{x}{2\sqrt{D_Ot}} \tag{27}$$