Half-wave potential（半波电位）推导之一_orr里半波电位怎么计算-优快云博客

本文探讨了在无搅拌条件下，快速动力学电化学实验中，平面板电极上半反应的动力学特性。通过Nerst方程描述电位变化，并使用Laplace变换解析求解氧化剂和还原剂浓度随时间变化的偏微分方程，得到电流密度与时间的关系。

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Test condition

No stirring;
Fast kinetics
Step to an arbitrary potential suddenly
Cottrell-like experiment
Planar plate electrode
Half reaction:

$O+ne⇌R(1)O+ne{\rightleftharpoons} R \tag{1}$
In the case of Nerstian equation, the potential can be expressed as
$E=E0′−RTnFln⁡CO(0,t)CR(0,t)(2)E=E^{0'}-\dfrac {RT}{nF} \ln \dfrac {C_{O}(0,t)}{C_{R}(0,t)} \tag {2}$
Here we define
$θ=CO(0,t)CR(0,t)=exp⁡[nFRT(E−E0′)](3)\theta=\dfrac {C_{O}(0,t)}{C_{R}(0,t)}=\exp \left [\dfrac{nF} {RT} (E-E^{0'}) \right] \tag{3}$
The governing equations of the reductant and oxidant will be
$∂CO(0,t)∂t=DO∂2CO(0,t)∂x2(4)\dfrac{\partial C_O(0,t)}{\partial t}=D_O \dfrac{\partial ^2 C_O(0,t)}{\partial x^2} \tag{4}$
$∂CR(0,t)∂t=DR∂2CR(0,t)∂x2(5)\dfrac{\partial C_R(0,t)}{\partial t}=D_R \dfrac{\partial ^2 C_R(0,t)}{\partial x^2} \tag{5}$

Initial condition:(Concentration is homogeneous)
$C_O(x,0)=C_O ^* \tag{6}$
$C_R(x,0)=0 \tag{7}$
Boundary conditions:
$CO(∞,t)=CO∗(t>0)(8)C_O(\infty,t)=C_O ^* \quad (t>0) \tag{8}$
$CR(∞,t)=0(t>0)(9)C_R(\infty,t)=0 \quad (t>0) \tag{9}$
$DO∂CO(0,t)∂x+DR∂CR(0,t)∂x=0(10)D_O \dfrac{\partial C_O(0,t)}{\partial x} +D_R \dfrac{\partial C_R(0,t)}{\partial x} =0 \tag{10}$
Apply Laplace transform on Equation(4)-(10) to solve the equations
Laplace transform function
$(s)=\int_0^\infty f(t)e^{-st}dt \tag{11}$
Apply the transformation on these equations:
$C‾O(∞,s)=CO∗s(12)\overline C_O(\infty,s)=\dfrac{C_O^*}{s} \tag{12}$
$C‾O(0,s)=0(13)\overline C_O(0,s)=0 \tag{13}$

$DO∂C‾O(0,t)∂x+DR∂C‾R(0,t)∂x=0(14)D_O \dfrac{\partial \overline C_O(0,t)}{\partial x} +D_R \dfrac{\partial \overline C_R(0,t)}{\partial x} =0 \tag{14}$
Using (12) and (13) we can get the solution
$C‾O(x,s)=A(s)exp⁡(−sDOx)+CO∗s(15)\overline C_O(x,s)=A(s) \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{15}$
$C‾R(x,s)=B(s)exp⁡(−sDRx)(16)\overline C_R(x,s)=B(s) \exp{\left (-\sqrt{\dfrac{s}{D_R}}x \right)} \tag{16}$
Apply (14) the relation between $A (s)$ and $B (s)$
$B(s)=−A(s)ξ(17)B(s)=-A(s)\xi \tag{17}$ ,
where
$ξ=DODR(18)\xi = \sqrt{\dfrac{D_O}{D_R} } \tag{18}$
Hence the results are:
$C‾O(x,s)=A(s)exp⁡(−sDOx)+CO∗s(15)\overline C_O(x,s)=A(s) \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{15}$
$C‾R(x,s)=−A(s)ξexp⁡(−sDRx)(16)\overline C_R(x,s)=-A(s)\xi \exp{\left (-\sqrt{\dfrac{s}{D_R}}x \right)} \tag{16}$

Here we make the assumption: reversible electrochemical reaction at the surface of the electrode
$CO(0,s)=θCR(0,s)(17)C_{O}(0,s)=\theta C_{R}(0,s) \tag{17}$
Thus
$CO∗s+A(s)=−ξθA(s)(18)\dfrac{C_O^*}{s} +A(s)=-\xi \theta A(s) \tag{18}$
We can get $A (s)$
$A(s)=−C‾O∗s(1+ξθ)(19)A(s)=-\dfrac{\overline C_O^*}{s(1+\xi \theta)} \tag {19}$
finally
$C‾O(x,s)=−C‾O∗s(1+ξθ)exp⁡(−sDOx)+CO∗s(20)\overline C_O(x,s)=-\dfrac{\overline C_O^*}{s(1+\xi \theta)} \exp{\left (-\sqrt{\dfrac{s}{D_O}}x \right)}+\dfrac{C_O^*}{s} \tag{20}$
$C‾R(x,s)=ξC‾O∗s(1+ξθ)exp⁡(−sDRx)(21)\overline C_R(x,s)=\dfrac{\xi \overline C_O^*}{s(1+\xi \theta)} \exp{\left (-\sqrt{\dfrac{s}{D_R}}x \right)} \tag{21}$
Fick’s first law states that the flux is proportional to the concentration gradient
$−JO(x,t)=DO∂CO(x,t)∂x(22)-J_O(x,t)=D_O \dfrac {\partial C_O(x,t)}{\partial x} \tag{22}$
Laplace tansformation gives
$−JO(x,s)=DO∂CO(x,s)∂x(23)-J_O(x,s)=D_O \dfrac {\partial C_O(x,s)}{\partial x} \tag{23}$
Assumption: Cottrell experiment
$\dfrac {nFAD_O^{1/2} C_O^*}{\pi ^{1/2} t^{1/2} (1+\xi \theta)}\tag{24}$
then
$\dfrac{i_d(t)}{1+\xi\theta} \tag{25}$