碱性溶液中OER动力学分析

Bulter-Volmer 方程：

i = i_{c} - i_{a} = F A k^{0} [c_{O} (0, t) \exp \frac{- β F (E - E^{0'})}{R T} - c_{R} (0, t) \exp \frac{(1 - β) F (E - E^{0'})}{R T}]

或者

i = i_{0} [\frac{c_{O} (0, t)}{c_{O}^{*}} \exp \frac{- β F (E - E_{e q})}{R T} - \frac{c_{R} (0, t)}{c_{R}^{*}} \exp \frac{(1 - β) F (E - E_{e q})}{R T}]

在碱性溶液中HER可由如下三步基元反应组成
碱性溶液中OER动力学分析
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，

\begin{aligned} (6) & k_{1} & = k_{1}^{0} \exp \frac{(1 - β_{1}) F η_{1}}{R T} \\ (7) & k_{- 1} & = k_{- 1}^{0} \exp \frac{- β_{1} F η_{1}}{R T} \\ (8) & k_{2} & = k_{2}^{0} \exp \frac{(1 - β_{2}) F η_{2}}{R T} \\ (9) & k_{- 2} & = k_{- 2}^{0} \exp \frac{- β_{2} F η_{2}}{R T} \\ (10) & k_{3} & = k_{3}^{0} \exp \frac{(1 - β_{3}) F η_{3}}{R T} \\ (12) & k_{- 3} & = k_{- 3}^{0} \exp \frac{- β_{3} F η_{3}}{R T} \\ (13) & k_{5} & = k_{5}^{0} \exp \frac{(1 - β_{5}) F η_{5}}{R T} \\ (14) & k_{- 5} & = k_{- 5}^{0} \exp \frac{- β_{5} F η_{5}}{R T} \end{aligned}

其中

β_{j}

为各个基元反应

j

的symmetry factor，近似取值0.5;

k_{j}^{0}

为基元反应

j

的标准速率常数,

k_{j}^{0}

为反应

j

的标准速率常数

$\begin{matrix} (15) & η_{1} = E - E_{e q, 1} \end{matrix}$
$\begin{matrix} (16) & η_{2} = E - E_{e q, 2} \end{matrix}$
$\begin{matrix} (17) & η_{3} = E - E_{e q, 3} \end{matrix}$
$\begin{matrix} (18) & η_{5} = E - E_{e q, 5} \end{matrix}$

则每一个反应的反应速率为：

\begin{aligned} (10) & r_{1} & = k_{1} a_{O H^{-}} θ_{V} = k_{1}^{0} a_{O H^{-}} θ_{V} \exp \frac{(1 - β_{1}) F η_{1}}{R T} \\ (11) & r_{- 1} & = k_{- 1} θ_{O H} = k_{- 1}^{0} θ_{O H} \exp \frac{- β_{1} F η_{1}}{R T} \\ (12) & r_{2} & = k_{2} a_{O H^{-}} θ_{O H^{-}} = k_{2}^{0} a_{O H^{-}} θ_{O H^{-}} \exp \frac{(1 - β_{2}) F η_{2}}{R T} \\ (13) & r_{- 2} & = k_{- 2} a_{H_{2} O} θ_{O} = k_{- 2}^{0} a_{H_{2} O} θ_{O} \exp \frac{- β_{2} F η_{2}}{R T} \\ (14) & r_{3} & = k_{3} a_{O H^{-}} θ_{O} = k_{3}^{0} a_{O H^{-}} θ_{O} \exp \frac{(1 - β_{3}) F η_{3}}{R T} \\ (15) & r_{- 3} & = k_{- 3} θ_{O O H} = k_{- 3}^{0} θ_{O O H} \exp \frac{- β_{3} F η_{3}}{R T} \\ (16) & r_{4} & = k_{4} a_{O H^{-}} θ_{O O H} \\ (17) & r_{- 4} & = k_{4} a_{H_{2} O} θ_{O O^{-}} \\ (18) & r_{5} & = k_{5} θ_{O O^{-}} = k_{5}^{0} θ_{O O^{-}} \exp \frac{(1 - β_{5}) F η_{5}}{R T} \\ (19) & r_{- 5} & = k_{- 5} p_{O_{2}} θ_{V} = k_{- 5}^{0} p_{O_{2}} θ_{V} \exp \frac{- β_{5} F η_{5}}{R T} \end{aligned}

分情况讨论在不同讨论在不同条件下得OER反应速率

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

即表面覆盖率接近于零 $θ \approx 0$ ，那么 $θ_{V} = 1 - θ \approx 1$ 则整体反应速率 $r$

\begin{matrix} (20) & r = r_{1} \end{matrix}

\begin{matrix} (21) & r = k_{1}^{0} a_{O H^{-}} θ_{V} \exp \frac{(1 - β_{1}) F η_{1}}{R T} = k_{1}^{0} a_{O H^{-}} \exp \frac{(1 - β_{1}) F η_{1}}{R T} \end{matrix}

则电流

\begin{matrix} (22) & i = n F A r = n F A k_{1}^{0} a_{O H^{-}} \exp \frac{(1 - β_{1}) F η_{1}}{R T} \end{matrix}

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

\begin{matrix} (15) & r_{1} = r_{- 1} \end{matrix}

即

\begin{matrix} (16) & k_{1}^{0} a_{O H^{-}} θ_{V} \exp \frac{(1 - β_{1}) F η_{1}}{R T} = k_{- 1}^{0} θ_{O H} \exp \frac{- β_{1} F η_{1}}{R T} \end{matrix}

解出

θ_{O H}

\begin{matrix} (17) & θ = \frac{k_{1}^{0} a_{H_{2} O} \exp \frac{- β_{1} F η_{1}}{R T}}{k_{1}^{0} a_{H_{2} O} \exp \frac{- β_{1} F η_{1}}{R T} + k_{- 1}^{0} a_{O H^{-}} \exp \frac{(1 - β_{1}) F η_{1}}{R T}} = \frac{k_{1}^{0} a_{H_{2} O}}{k_{1}^{0} a_{H_{2} O} + k_{- 1}^{0} a_{O H^{-}} \exp \frac{F η_{1}}{R T}} \end{matrix}

因此总反应速率

\begin{matrix} (18) & r = r_{2} \end{matrix}

把(19)带入(11)则得到总反应速率

\begin{matrix} (19) & r = k_{2}^{0} a_{H_{2} O} θ \exp \frac{- β_{2} F η_{2}}{R T} \end{matrix}

则总电流为

\begin{matrix} (20) & i = n F A \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 \frac{F η_{1}}{R T}} \exp \frac{- β_{2} F η_{2}}{R T} \end{matrix}

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 $r = r_{3}$ ：
$\begin{matrix} (21) & r = k_{3} θ^{2} = k_{3} {[\frac{k_{1}^{0} a_{H_{2} O}}{k_{1}^{0} a_{H_{2} O} + k_{- 1}^{0} a_{O H^{-}} \exp \frac{F η_{1}}{R T}}]}^{2} \end{matrix}$

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