碱性溶液中HER动力学分析

Bulter-Volmer 方程：

i=ic−ia=FAk0[cO(0,t)exp−βF(E−E0′)RT−cR(0,t)exp(1−β)F(E−E0′)RT]$$i=i_c-i_a=FA{k}^0\left[c_O(0,t)\exp {\displaystyle \frac{-\beta F(E-E^{0\mathrm{\prime }})}{RT}}-c_R(0,t)\exp {\displaystyle \frac{(1-\beta )F(E-E^{0\mathrm{\prime }})}{RT}}\right]$$

或者

i=i0[cO(0,t)c∗Oexp−βF(E−Eeq)RT−cR(0,t)c∗Rexp(1−β)F(E−Eeq)RT]$$i=i_0\left[{\displaystyle \frac{c_O(0,t)}{c_O^{\ast }}}\exp {\displaystyle \frac{-\beta F(E-E_{eq})}{RT}}-{\displaystyle \frac{c_R(0,t)}{c_R^{\ast }}}\exp {\displaystyle \frac{(1-\beta )F(E-E_{eq})}{RT}}\right]$$

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在碱性溶液中HER可由如下三步基元反应组成

其中*代表吸附活性位，按照Bulter-Volmer方程可得各个反应的反应速率常数，

k1k−1k2k−2=k01exp−β1Fη1RT=k0−1exp(1−β1)Fη1RT=k02exp−β2Fη2RT=k0−2exp(1−β2)Fη2RT(4)(5)(6)(7)$$\begin{array}{r}\text{(4)}& k_1& =k_1^0\exp {\displaystyle \frac{-{\beta }_{1}F{\eta }_1}{RT}}\\ \text{(5)}& k_{-1}& =k_{-1}^0\exp {\displaystyle \frac{(1-{\beta }_1)F{\eta }_1}{RT}}\\ \text{(6)}& k_2& =k_2^0\exp {\displaystyle \frac{-{\beta }_{2}F{\eta }_2}{RT}}\\ \text{(7)}& k_{-2}& =k_{-2}^0\exp {\displaystyle \frac{(1-{\beta }_2)F{\eta }_2}{RT}}\end{array}$$

其中βj${\beta }_j$为各个基元反应j$j$的symmetry factor，近似取值0.5; k0j$k_j^0$为基元反应j$j$的标准速率常数, k0j$k_j^0$为反应j$j$的标准速率常数

η1=E−Eeq,1(8)$\begin{array}{}\text{(8)}& {\eta }_1=E-E_{eq,1}\end{array}$
η2=E−Eeq,2(9)$\begin{array}{}\text{(9)}& {\eta }_2=E-E_{eq,2}\end{array}$

则每一个反应的反应速率为：

r1r−1r2r−2r3r−3=k1aH2O(1−θ)=k01aH2O(1−θ)exp−β1Fη1RT=k−1θaOH−=k0−1aOH−θexp(1−β1)Fη1RT=k2aH2Oθ=k02aH2Oθexp−β2Fη2RT=k−2pH2aOH−(1−θ)=k0−2pH2aOH−(1−θ)exp(1−β2)Fη2RT=k3θ2=k−3pH2(1−θ)2(10)(11)(12)(13)(14)(15)$$\begin{array}{r}\text{(10)}& r_1& =k_1{a}_{H_{2}O}(1-\theta )=k_1^0{a}_{H_{2}O}(1-\theta )\exp {\displaystyle \frac{-{\beta }_{1}F{\eta }_1}{RT}}\\ \text{(11)}& r_{-1}& =k_{-1}\theta a_{O{H}^{-}}=k_{-1}^0{a}_{O{H}^{-}}\theta \exp {\displaystyle \frac{(1-{\beta }_1)F{\eta }_1}{RT}}\\ \text{(12)}& r_2& =k_2{a}_{H_{2}O}\theta =k_2^0{a}_{H_{2}O}\theta \exp {\displaystyle \frac{-{\beta }_{2}F{\eta }_2}{RT}}\\ \text{(13)}& r_{-2}& =k_{-2}{p}_{H_2}{a}_{O{H}^{-}}(1-\theta )=k_{-2}^0{p}_{H_2}{a}_{O{H}^{-}}(1-\theta )\exp {\displaystyle \frac{(1-{\beta }_2)F{\eta }_2}{RT}}\\ \text{(14)}& r_3& =k_3{\theta }^2\\ \text{(15)}& r_{-3}& =k_{-3}{p}_{H_2}(1-\theta {)}^2\end{array}$$

分情况讨论在不同讨论在不同条件下得HER反应速率

1.$$ 当反应(1)为控制步骤（Rate-Determining Step, RDS)

即表面∗H$\ast H$浓度接近于零θ≈0$\theta \approx 0$，则整体反应速率r$r$

r=r1$$r=r_1$$

r=k01aH2O(1−θ)exp−β1Fη1RT(13)$$\begin{array}{}\text{(13)}& r=k_1^0{a}_{H_{2}O}(1-\theta )\exp {\displaystyle \frac{-{\beta }_{1}F{\eta }_1}{RT}}\end{array}$$

则电流

i=nFAr=nFAk01aH2O(1−θ)exp−β1Fη1RT(14)$$\begin{array}{}\text{(14)}& i=nFAr=nFA{k}_1^0{a}_{H_{2}O}(1-\theta )\exp {\displaystyle \frac{-{\beta }_{1}F{\eta }_1}{RT}}\end{array}$$

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2.$$ 当反应(2)为控制步骤（Rate-Determining Step, RDS)

即反应(1)到达平衡状态，则

r1=r−1(15)$$\begin{array}{}\text{(15)}& r_1=r_{-1}\end{array}$$

即

k01aH2O(1−θ)exp−β1Fη1RT=k0−1aOH−θexp(1−β1)Fη1RT(16)$$\begin{array}{}\text{(16)}& k_1^0{a}_{H_{2}O}(1-\theta )\exp {\displaystyle \frac{-{\beta }_{1}F{\eta }_1}{RT}}=k_{-1}^0{a}_{O{H}^{-}}\theta \exp {\displaystyle \frac{(1-{\beta }_1)F{\eta }_1}{RT}}\end{array}$$

解出θ$\theta$

θ=k01aH2Oexp−β1Fη1RTk01aH2Oexp−β1Fη1RT+k0−1aOH−exp(1−β1)Fη1RT=k01aH2Ok01aH2O+k0−1aOH−expFη1RT(17)$$\begin{array}{}\text{(17)}& \theta ={\displaystyle \frac{k_1^0{a}_{H_{2}O}\exp {\displaystyle \frac{-{\beta }_{1}F{\eta }_1}{RT}}}{k_1^0{a}_{H_{2}O}\exp {\displaystyle \frac{-{\beta }_{1}F{\eta }_1}{RT}}+k_{-1}^0{a}_{O{H}^{-}}\exp {\displaystyle \frac{(1-{\beta }_1)F{\eta }_1}{RT}}}}={\displaystyle \frac{k_1^0{a}_{H_{2}O}}{k_1^0{a}_{H_{2}O}+k_{-1}^0{a}_{O{H}^{-}}\exp {\displaystyle \frac{F{\eta }_1}{RT}}}}\end{array}$$

因此总反应速率

r=r2(18)$$\begin{array}{}\text{(18)}& r=r_2\end{array}$$

把(19)带入(11)则得到总反应速率

r=k02aH2Oθexp−β2Fη2RT(19)$$\begin{array}{}\text{(19)}& r=k_2^0{a}_{H_{2}O}\theta \exp {\displaystyle \frac{-{\beta }_{2}F{\eta }_2}{RT}}\end{array}$$

则总电流为

i=nFAk01k02a2H2Ok01aH2O+k0−1aOH−expFη1RTexp−β2Fη2RT(20)$$\begin{array}{}\text{(20)}& i=nFA{\displaystyle \frac{k_1^0{k}_2^0{a}_{H_{2}O}^2}{k_1^0{a}_{H_{2}O}+k_{-1}^0{a}_{O{H}^{-}}\exp {\displaystyle \frac{F{\eta }_1}{RT}}}}\exp {\displaystyle \frac{-{\beta }_{2}F{\eta }_2}{RT}}\end{array}$$

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3.$$ 当反应(3)为控制步骤（Rate-Determining Step, RDS)

则反应(1)和反应(2)都达到平衡，那么(19)式依旧有效，在此情况下，总的反应速率r=r3$r=r_3$：
r=k3θ2=k3⎡⎣⎢⎢k01aH2Ok01aH2O+k0−1aOH−expFη1RT⎤⎦⎥⎥2(21)$\begin{array}{}\text{(21)}& r=k_3{\theta }^2=k_3{\left[{\displaystyle \frac{k_1^0{a}_{H_{2}O}}{k_1^0{a}_{H_{2}O}+k_{-1}^0{a}_{O{H}^{-}}\exp {\displaystyle \frac{F{\eta }_1}{RT}}}}\right]}^2\end{array}$

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参考文献：

[1]. Tatsuya Shinagawa, Angel T. Garcia-Esparza, Kazuhiro Takanabe. Insight on Tafel slopes from a microkinetic analysis of aqueous electrocatalysis for energy conversion. Scientific reports. 2015,5,13801