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=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=FAk^0 \left [c_O(0,t) \exp \dfrac{-\beta F(E-E^{0\prime})}{RT}- c_R(0,t) \exp \dfrac{(1-\beta)F(E-E^{0\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=i0[cO(0,t)cO∗exp⁡−βF(E−Eeq)RT−cR(0,t)cR∗exp⁡(1−β)F(E−Eeq)RT]

i=i_0 \left [\dfrac{c_O(0,t)}{c_O^*} \exp \dfrac{-\beta F(E-E_{eq})}{RT}- \dfrac {c_R(0,t)}{c_R^*} \exp \dfrac{(1-\beta)F(E-E_{eq})}{RT}\right]

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)(4)k1=k10exp⁡−β1Fη1RT(5)k−1=k−10exp⁡(1−β1)Fη1RT(6)k2=k20exp⁡−β2Fη2RT(7)k−2=k−20exp⁡(1−β2)Fη2RT

\begin{align*} k_1&=k_1^0 \exp \dfrac{-\beta_1 F \eta_1}{R T} \tag {4}\\ \\ k_{-1}&=k_{-1}^0 \exp \dfrac{(1-\beta_1) F\eta_1}{RT} \tag {5}\\ \\ k_2&=k_2^0 \exp \dfrac{-\beta_2 F \eta_2}{R T} \tag {6}\\ \\ k_{-2}&=k_{-2}^0 \exp \dfrac{(1-\beta_2)F \eta_2}{R T} \tag {7} \end{align*}
其中 βjβj\beta_j为各个基元反应 jjj的symmetry factor，近似取值0.5; kj0" role="presentation" style="position: relative;">k0jkj0k_j^0为基元反应 jjj的标准速率常数, kj0" role="presentation" style="position: relative;">k0jkj0k_j^0为反应 jjj的标准速率常数

(8)η1=E−Eeq,1" role="presentation" style="position: relative;">η1=E−Eeq,1(8)(8)η1=E−Eeq,1\eta_1=E-E_{eq,1} \tag{8}
η2=E−Eeq,2(9)(9)η2=E−Eeq,2\eta_2=E-E_{eq,2} \tag{9}

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)(10)r1=k1aH2O(1−θ)=k10aH2O(1−θ)exp⁡−β1Fη1RT(11)r−1=k−1θaOH−=k−10aOH−θexp⁡(1−β1)Fη1RT(12)r2=k2aH2Oθ=k20aH2Oθexp⁡−β2Fη2RT(13)r−2=k−2pH2aOH−(1−θ)=k−20pH2aOH−(1−θ)exp⁡(1−β2)Fη2RT(14)r3=k3θ2(15)r−3=k−3pH2(1−θ)2

\begin{align*} r_1&= k_1 a_{H_2O}(1-\theta)=k_1^0 a_{H_2O}(1-\theta) \exp \dfrac{-\beta_1 F \eta_1}{R T} \tag {10}\\ \\ r_{-1}&= k_{-1} \theta a_{OH^-}=k_{-1}^0 a_{OH^-} \theta \exp \dfrac{(1-\beta_1) F\eta_1}{RT} \tag {11}\\ \\ r_2&= k_2 a_{H_2O} \theta=k_2^0 a_{H_2O} \theta \exp \dfrac{-\beta_2 F \eta_2}{R T} \tag {12}\\ \\ r_{-2}&= k_{-2} p_{H_2} a_{OH^-}(1-\theta)= k_{-2}^0 p_{H_2} a_{OH^-}(1-\theta) \exp \dfrac{(1-\beta_2)F \eta_2}{R T} \tag {13}\\ \\ r_3&= k_3 \theta^2 \tag {14}\\ \\ r_{-3}&= k_{-3} p_{H_2} (1-\theta)^2 \tag {15}\\ \end{align*}
分情况讨论在不同讨论在不同条件下得HER反应速率

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

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

r=r1" role="presentation">r=r1r=r1

r=r_1

r=k01aH2O(1−θ)exp−β1Fη1RT(13)(13)r=k10aH2O(1−θ)exp⁡−β1Fη1RT

r=k_1^0 a_{H_2O}(1-\theta) \exp \dfrac{-\beta_1 F \eta_1}{R T} \tag{13}
则电流

i=nFAr=nFAk01aH2O(1−θ)exp−β1Fη1RT(14)(14)i=nFAr=nFAk10aH2O(1−θ)exp⁡−β1Fη1RT

i=nFAr=nFAk_1^0 a_{H_2O}(1-\theta) \exp \dfrac{-\beta_1 F \eta_1}{R T} \tag{14}

2.\quad 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)(15)r1=r−1

r_1=r_{-1} \tag{15}
即

k01aH2O(1−θ)exp−β1Fη1RT=k0−1aOH−θexp(1−β1)Fη1RT(16)(16)k10aH2O(1−θ)exp⁡−β1Fη1RT=k−10aOH−θexp⁡(1−β1)Fη1RT

k_1^0 a_{H_2O}(1-\theta) \exp \dfrac{-\beta_1 F \eta_1}{R T}=k_{-1}^0 a_{OH^-} \theta \exp \dfrac{(1-\beta_1) F\eta_1}{RT} \tag{16}
解出 θθ\theta

θ=k01aH2Oexp−β1Fη1RTk01aH2Oexp−β1Fη1RT+k0−1aOH−exp(1−β1)Fη1RT=k01aH2Ok01aH2O+k0−1aOH−expFη1RT(17)(17)θ=k10aH2Oexp⁡−β1Fη1RTk10aH2Oexp⁡−β1Fη1RT+k−10aOH−exp⁡(1−β1)Fη1RT=k10aH2Ok10aH2O+k−10aOH−exp⁡Fη1RT

\theta=\dfrac {k_1^0 a_{H_2O} \exp \dfrac{-\beta_1 F \eta_1}{R T}}{k_1^0 a_{H_2O} \exp \dfrac{-\beta_1 F \eta_1}{R T}+k_{-1}^0 a_{OH^-}\exp \dfrac{(1-\beta_1) F\eta_1}{RT}} =\dfrac {k_1^0 a_{H_2O}}{k_1^0 a_{H_2O} +k_{-1}^0 a_{OH^-} \exp \dfrac{ F\eta_1}{RT}} \tag{17}
因此总反应速率

r=r2(18)(18)r=r2

r=r_2 \tag {18}
把(19)带入(11)则得到总反应速率

r=k02aH2Oθexp−β2Fη2RT(19)(19)r=k20aH2Oθexp⁡−β2Fη2RT

r= k_2^0 a_{H_2O} \theta \exp \dfrac{-\beta_2 F \eta_2}{R T} \tag{19}
则总电流为

i=nFAk01k02a2H2Ok01aH2O+k0−1aOH−expFη1RTexp−β2Fη2RT(20)(20)i=nFAk10k20aH2O2k10aH2O+k−10aOH−exp⁡Fη1RTexp⁡−β2Fη2RT

i= nFA \dfrac {k_1^0 k_2^0 a_{H_2O}^2}{k_1^0 a_{H_2O} +k_{-1}^0 a_{OH^-} \exp \dfrac{ F\eta_1}{RT}} \exp \dfrac{-\beta_2 F \eta_2}{R T} \tag{20}

3.\quad 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=r3r=r3r=r_3：
r=k3θ2=k3⎡⎣⎢⎢k01aH2Ok01aH2O+k0−1aOH−expFη1RT⎤⎦⎥⎥2(21)(21)r=k3θ2=k3[k10aH2Ok10aH2O+k−10aOH−exp⁡Fη1RT]2r= k_3 \theta^2=k_3 \left [\dfrac {k_1^0 a_{H_2O}}{k_1^0 a_{H_2O} +k_{-1}^0 a_{OH^-} \exp \dfrac{ F\eta_1}{RT}}\right]^2 \tag {21}

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