恒电势电流曲线和浓度分布（i-t curve）

Test condition

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

Apply Laplace transform to solve Eq.(1)

Laplace transform function

$$F (s)=\int_0^\infty f(t)e^{-st}dt \tag{5}$$
Apply Laplace transform on the left part of Eq(1)
$$\begin{align*} \mathcal L \left \{\dfrac{\partial C_O}{\partial t} \right\} &=\int_0^\infty \dfrac{\partial C_O}{\partial t} e^{-st}dt \\ &= C_O e^{-st} \bigg |_0^\infty+s\int_0^\infty C_O e^{-st}dt\\ &=C_O(x,\infty) e^{-s\infty}-C_O(x,0) e^0+s\int_0^\infty C_O e^{-st}dt\\ &=-C_O^*+s\int_0^\infty C_O e^{-st}dt \tag{6}\\ \end{align*}$$
Define
$$\overline C_O(x,s) =\int_0^\infty C_O e^{-st}dt \tag{7}$$
$$\begin{align*} \mathcal L \left \{D_O\dfrac{\partial^2 C_O}{\partial x^2} \right\} &=\int_0^\infty D_O \dfrac{\partial^2 C_O}{\partial x^2} e^{-st}dt \\ &= D_O \dfrac{\partial^2}{\partial x^2}\int_0^\infty C_O e^{-st}dt\\ &=D_O \dfrac{\partial^2 \overline C_O(x,s) }{\partial x^2} \tag{8}\\ \end{align*}$$
After Laplace transform, Eq(1) becomes
$$s \overline C_O(x,s) =D_O \dfrac{\partial^2 \overline C_O(x,s) }{\partial x^2} +C_O^* \tag{9}$$
The solution of the ODE is:
$$\overline C_O(x,s)=A_1\exp{\sqrt{\dfrac{s}{D_O}}x}+A_2\exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+A_3 \tag{10}$$
where $A_1$, $A_2$ and $A_3$are constant.
Apply Laplace transform on boundary conditions, the we can get

$$\overline C_O(\infty,s)=\dfrac{C_O^*}{s} \tag{11}$$
$$\overline C_O(0,s)=0 \tag{12}$$
After substitution, we can solve
$$A_1=0 \tag{13}$$
$$A_2=-\dfrac{C_O^*}{s} \tag{14}$$
$$A_3=\dfrac{C_O^*}{s} \tag{15}$$
Hence the solution should be

$$\overline C_O(x,s)=-\dfrac{C_O^*}{s} \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{16}$$

Current-time profile

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

$$-J_O(0,t)=\dfrac{i(t)}{nFA}=D_O \dfrac{\partial C_O(x,t)}{\partial x} \bigg |_{x=0}\tag{17}$$
Applying the Laplace transform on Eq(17) gives
$$\dfrac{\overline i(t)}{nFA}=D_O \dfrac{\partial \overline C_O(x,s)}{\partial x} \bigg |_{x=0}\tag{18}$$
Substitution of Eq (16) into Eq (18) gives

$$\dfrac{\overline i(s)}{nFA}=D_O \dfrac{C_O^*}{s} \sqrt{\dfrac{s}{D_O}} \exp{\left(-\sqrt{\dfrac{s}{D_O}}\right)x}\bigg |_{x=0}\tag{19}$$
it becomes

$$\dfrac{\overline i(s)}{nFA}=D_O \dfrac{C_O^*}{s} \sqrt{\dfrac{s}{D_O}} =C_O^* \sqrt{\dfrac{D_O}{s}}\tag{20}$$
It yields
$$\overline i(s)=\dfrac{nFAD_O^{1/2}C_O^*}{s^{1/2}}\tag{21}$$
applying the form
$$\begin{align*} \mathcal L^{-1}\left \{\dfrac{1}{\sqrt s}\right \} &=\mathcal L^{-1}\left \{\dfrac{1}{\sqrt \pi} \cdot \dfrac{\sqrt{\pi}}{2^0\sqrt s}\right \} \\ &= \dfrac{1}{\sqrt \pi}\mathcal L^{-1}\left \{ \dfrac{\sqrt{\pi}}{2^0\sqrt s}\right \} \tag{22} \\ \end{align*}$$
use Inverse Laplace Transform table
$$\mathcal L^{-1}\left \{ \dfrac{(2n-1)!\sqrt{\pi}}{2^n s^{n+1/2}}\right \}=t^{n-1/2} \tag{23}$$
$$\mathcal L^{-1}\left \{\dfrac{1}{\sqrt s}\right \}=\sqrt{\pi t} \tag{24}$$

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

$$i(t)=\dfrac{nFAD_O^{1/2}C_O^*}{\pi^{1/2} t^{1/2}}\tag{25}$$
which is called Cottrell equation

Concentration-distance profile

Applying Inverse Laplace transform on Eq(16) gives

$$C_O(x,t)=C_O^*\left (1-erfc \dfrac{x}{2\sqrt{D_Ot}} \right) \tag{26}$$
or
$$\dfrac{C_O(x,t)}{C_O^*}=1-erfc \dfrac{x}{2\sqrt{D_Ot}} =erf \dfrac{x}{2\sqrt{D_Ot}} \tag{27}$$