Test condition

1. No stirring;
2. Fast kinetics
3. Step to an arbitrary potential suddenly
4. Cottrell-like experiment Half reaction:

$O+ne{\rightleftharpoons} R \tag{1}$ In the case of Nerstian equation, the potential can be expressed as $E=E^{0'}-\dfrac {RT}{nF} \ln \dfrac {C_{O}(0,t)}{C_{R}(0,t)} \tag {2}$

Initial condition:(Concentration is homogeneous) $\begin{array}{}\text{(6)}\end{array}$

$D_O \dfrac{\partial \overline C_O(0,t)}{\partial x} +D_R \dfrac{\partial \overline C_R(0,t)}{\partial x} =0 \tag{14}$ Using (12) and (13) we can get the solution $\overline C_O(x,s)=A(s) \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{15}$ $\overline C_R(x,s)=B(s) \exp{\left (-\sqrt{\dfrac{s}{D_R}}x \right)} \tag{16}$ Apply (14) the relation between $A(s)$ and $B(s)$ $B(s)=-A(s)\xi \tag{17}$, where $\xi = \sqrt{\dfrac{D_O}{D_R} } \tag{18}$ Hence the results are: $\overline C_O(x,s)=A(s) \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{15}$ $\overline C_R(x,s)=-A(s)\xi \exp{\left (-\sqrt{\dfrac{s}{D_R}}x \right)} \tag{16}$