Cottrell方程推导

平板电极电势阶跃扩散电流的严格推导

1. 控制方程建立

1.1 基本假设

  • 一维半无限扩散体系( x ≥ 0 x \geq 0 x0
  • 氧化还原电对: O + n e − ⇌ R O + ne^- \rightleftharpoons R O+neR
  • 初始条件:
    c O ( x , 0 ) = c ∗ , c R ( x , 0 ) = 0 c_O(x,0) = c^*,\quad c_R(x,0) = 0 cO(x,0)=c,cR(x,0)=0
  • 边界条件:
    c O ( ∞ , t ) = c ∗ , c R ( ∞ , t ) = 0 c_O(\infty,t) = c^*,\quad c_R(\infty,t) = 0 cO(,t)=c,cR(,t)=0

1.2 Fick第二定律

∂ c O ∂ t = D O ∂ 2 c O ∂ x 2 ( 1 ) ∂ c R ∂ t = D R ∂ 2 c R ∂ x 2 ( 2 ) \frac{\partial c_O}{\partial t} = D_O \frac{\partial^2 c_O}{\partial x^2} \quad (1) \\ \frac{\partial c_R}{\partial t} = D_R \frac{\partial^2 c_R}{\partial x^2} \quad (2) tcO=DOx22cO(1)tcR=DRx22cR(2)

2. Laplace变换求解

2.1 变换后方程

定义变换:
L { c ( x , t ) } = c ^ ( x , s ) \mathcal{L}\{c(x,t)\} = \hat{c}(x,s) L{c(x,t)}=c^(x,s)

得到:
s c ^ O − c ∗ = D O d 2 c ^ O d x 2 ( 3 ) s c ^ R = D R d 2 c ^ R d x 2 ( 4 ) s\hat{c}_O - c^* = D_O \frac{d^2\hat{c}_O}{dx^2} \quad (3) \\ s\hat{c}_R = D_R \frac{d^2\hat{c}_R}{dx^2} \quad (4) sc^Oc=DOdx2d2c^O(3)sc^R=DRdx2d2c^R(4)

2.2 通解形式

c ^ O ( x , s ) = c ∗ s + A ( s ) e − x s / D O ( 5 ) c ^ R ( x , s ) = B ( s ) e − x s / D R ( 6 ) \hat{c}_O(x,s) = \frac{c^*}{s} + A(s)e^{-x\sqrt{s/D_O}} \quad (5) \\ \hat{c}_R(x,s) = B(s)e^{-x\sqrt{s/D_R}} \quad (6) c^O(x,s)=sc+A(s)exs/DO (5)c^R(x,s)=B(s)exs/DR (6)

2.3 边界条件处理

(1) 流量平衡:

D O d c ^ O d x ∣ x = 0 + D R d c ^ R d x ∣ x = 0 = 0 D_O \left.\frac{d\hat{c}_O}{dx}\right|_{x=0} + D_R \left.\frac{d\hat{c}_R}{dx}\right|_{x=0} = 0 DOdxdc^O x=0+DRdxdc^R x=0=0

(2) Nernst条件:

c ^ O ( 0 , s ) c ^ R ( 0 , s ) = θ = exp ⁡ [ n F R T ( E f − E ∘ ) ] \frac{\hat{c}_O(0,s)}{\hat{c}_R(0,s)} = \theta = \exp\left[\frac{nF}{RT}(E_f - E^\circ)\right] c^R(0,s)c^O(0,s)=θ=exp[RTnF(EfE)]

3. 电流密度推导

3.1 表面流量计算

i ^ ( s ) = n F D O d c ^ O d x ∣ x = 0 = − n F D O A ( s ) s D O \hat{i}(s) = nFD_O \left.\frac{d\hat{c}_O}{dx}\right|_{x=0} = -nFD_O A(s)\sqrt{\frac{s}{D_O}} i^(s)=nFDOdxdc^O x=0=nFDOA(s)DOs

3.2 系数求解

联立方程解得:
A ( s ) = − c ∗ s ( 1 + 1 θ D O D R ) − 1 A(s) = -\frac{c^*}{s}\left(1 + \frac{1}{\theta}\sqrt{\frac{D_O}{D_R}}\right)^{-1} A(s)=sc(1+θ1DRDO )1

3.3 Laplace逆变换

利用变换对:
L − 1 { 1 s } = 1 π t \mathcal{L}^{-1}\left\{\frac{1}{\sqrt{s}}\right\} = \frac{1}{\sqrt{\pi t}} L1{s 1}=πt 1

得到Cottrell方程:
i ( t ) = n F c ∗ D O π t ( 1 + 1 θ D O D R ) − 1 i(t) = \frac{nFc^*\sqrt{D_O}}{\sqrt{\pi t}} \left(1 + \frac{1}{\theta}\sqrt{\frac{D_O}{D_R}}\right)^{-1} i(t)=πt nFcDO (1+θ1DRDO )1

4. 特例分析

4.1 完全还原( θ → ∞ \theta \to \infty θ

i ( t ) = n F c ∗ D O π t i(t) = nFc^*\sqrt{\frac{D_O}{\pi t}} i(t)=nFcπtDO

4.2 等扩散系数( D O = D R D_O = D_R DO=DR

i ( t ) = n F c ∗ D O π t ⋅ θ 1 + θ i(t) = \frac{nFc^*\sqrt{D_O}}{\sqrt{\pi t}} \cdot \frac{\theta}{1+\theta} i(t)=πt nFcDO 1+θθ

5. 数学验证

5.1 量纲分析

[ i ] = C ⋅ m o l − 1 ⋅ m o l ⋅ c m − 3 ⋅ c m 2 ⋅ s − 1 / 2 s 1 / 2 = A ⋅ c m − 2 [i] = \frac{C\cdot mol^{-1} \cdot mol\cdot cm^{-3} \cdot cm^2\cdot s^{-1/2}}{s^{1/2}} = A\cdot cm^{-2} [i]=s1/2Cmol1molcm3cm2s1/2=Acm2

5.2 极限行为

  • t → 0 + t\to 0^+ t0+ i ( t ) → ∞ i(t) \to \infty i(t)
  • t → ∞ t\to \infty t i ( t ) → 0 i(t) \to 0 i(t)0

6.参数影响分析

参数变化对电流的影响对浓度分布的影响物理机制说明
c*线性增大
i ∝ c*
各位置浓度幅值同比增加本体浓度增加直接提升电流
D_O ↑平方根增大
i ∝ √D_O
浓度梯度区域扩展加快传质速率提升
θ ↑
(电势负移)
渐近趋近极限值电极表面[O]降低,[R]升高Nernst平衡移动导致

附录:关键公式

公式说明
erfc ( x ) = 1 − erf ( x ) \text{erfc}(x) = 1-\text{erf}(x) erfc(x)=1erf(x)互补误差函数
L { t − 1 / 2 } = π / s \mathcal{L}\{t^{-1/2}\} = \sqrt{\pi/s} L{t1/2}=π/s Laplace变换对
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