Half-wave potential（半波电位）推导之一
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(−DOsx)+sCO∗(15)
(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(−DRsx)(16)
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(18)
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(−DOsx)+sCO∗(15)
(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(−DRsx)(16)
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(−DOsx)+sCO∗(20)
(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(−DRsx)(21)
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)