离子液体格气模型

参考文献:A mean-field theory on the differential capacitance of asymmetric ionic liquid electrolytes

体系如上图所示,蓝、红圆分别为正、负离子。

体系自由能

\begin{equation} F=e\varphi(N_+-N_-)-kT\ln\Omega \label{freeenergy} \end{equation}

其中 $e$ 为基元电荷电量,$\varphi$为电势,$N_+$和$N_-$ 分别为正负离子数目,$\Omega$ 为离子可能的分布的数目。正负离子大小不一样,半径分别为$r_+$和$r_-$,体积之比为:

\begin{equation} \xi=\frac{r_-^3}{r_+^3} \label{volratio} \end{equation}

对于常见离子液体,$\xi \lt 1$。

体系总格子数为 $N$,阳离子占据格子数为 $N_+$,留给阴离子的位置有$[(N-N_+)/\xi]$个,这里 $[]$ 表示取整。因此,有

\begin{equation} \Omega=\frac{N!}{N_+!(N-N_+)!}\frac{[(N-N_+)/\xi]}{N_-!\{[(N-N_+)/\xi]-N_-\}!} \label{distnum} \end{equation}

应用斯特灵公式,$\ln N!=N\ln N-N, N\gg 1$,有

\begin{equation} \begin{split} \ln\Omega=&N\ln N-N_+\ln N_+-(N-N_+)\ln(N-N_+)\\ &+\frac{N-N_+}{\xi}\ln\frac{N-N_+}{\xi}-N_-\ln N_-\\ &-\left (\frac{N-N_+}{\xi}-N_- \right )\ln \left (\frac{N-N_+}{\xi}-N_- \right ) \end{split} \label{lndistnum} \end{equation}

正负离子化学势分别为

\begin{equation} \begin{split} \frac{\mu_+}{kT}=&\frac{1}{kT}\frac{\partial F}{\partial N_+}\\ =&\frac{e\varphi}{kT}-\ln(N-N_+)+\ln N_++\frac{1}{\xi}\ln\left(\frac{N-N_+}{\xi}\right)-\frac{1}{\xi}\ln\left(\frac{N-N_+}{\xi}-N_- \right)\\ =&-\ln(N-N_0)+\ln N_0+\frac{1}{\xi}\ln\left(\frac{N-N_0}{\xi}\right)-\frac{1}{\xi}\ln\left(\frac{N-N_0}{\xi}-N_0 \right) \end{split} \label{mu+} \end{equation}

\begin{equation} \begin{split} \frac{\mu_-}{kT}=&\frac{1}{kT}\frac{\partial F}{\partial N_-}\\ =&-\frac{e\varphi}{kT}+\ln N_--\ln\left(\frac{N-N_+}{\xi}-N_- \right)\\ =&\ln N_0-\ln\left(\frac{N-N_0}{\xi}-N_0 \right) \end{split} \label{mu-} \end{equation}

上两式中最后一个等号为本体化学势。以上两式可分别重新整理为

\begin{equation} \frac{e\varphi}{kT}-\ln\frac{N-N_+}{N-N_0}+\ln \frac{N_+}{N_0}+\frac{1}{\xi}\ln\frac{N-N_+}{N-N_0}-\frac{1}{\xi}\ln\frac{N-N_+-\xi N_-}{N-N_0-\xi N_0}=0 \label{mu+eq} \end{equation}

\begin{equation} -\frac{e\varphi}{kT}+\ln \frac{N_-}{N_0}-\ln\frac{N-N_+-\xi N_-}{N-N_0-\xi N_0}=0 \label{mu-eq} \end{equation}

令 $u=e\varphi/(kT)$, $2/\gamma=N/N_0$,$\eta=2/\gamma-1-\xi$,$\gamma$ 和 $\eta$ 分别叫做压缩率和孔隙率。令 $N_{\pm}/N_0=c_{\pm}/c_0$,$c_0$ 为离子液体本体浓度,$c_+$ 和 $c_-$ 分别为正负离子浓度。方程\eqref{mu+eq} 和 \eqref{mu-eq}分别可重新写为:

\begin{equation} -u=\ln\frac{c_+}{c_0}-\frac{\xi-1}{\xi}\ln\frac{\frac{2}{\gamma}-\frac{c_+}{c_0}}{\xi+\eta}-\frac{1}{\xi}\ln\frac{\frac{2}{\gamma}-\frac{c_+}{c_0}-\xi\frac{c_-}{c_0}}{\eta} \label{c+eq} \end{equation}

\begin{equation} u=\ln\frac{c_-}{c_0}-\ln\frac{\frac{2}{\gamma}-\frac{c_+}{c_0}-\xi\frac{c_-}{c_0}}{\eta} \label{c-eq} \end{equation}

两边取指数,有:

\begin{equation} e^{-u}=\frac{c_+}{c_0} \left(\frac{\frac{2}{\gamma}-\frac{c_+}{c_0}}{\xi+\eta}\right)^{-\frac{\xi-1}{\xi}}\left(\frac{\frac{2}{\gamma}-\frac{c_+}{c_0}-\xi\frac{c_-}{c_0}}{\eta}\right)^{-\frac{1}{\xi}} \label{ec+eq} \end{equation}

\begin{equation} e^u=\frac{\eta \frac{c_-}{c_0}}{\frac{2}{\gamma}-\frac{c_+}{c_0}-\xi\frac{c_-}{c_0}} \label{ec-eq} \end{equation}

由方程\eqref{ec-eq}得,

\begin{equation} \frac{c_-}{c_0}=\frac{e^u}{\xi e^u+\eta}\left(\frac{2}{\gamma}-\frac{c_+}{c_0}\right) \label{cme} \end{equation}

将此式代入\eqref{ec+eq},可得

\begin{equation} \frac{c_+}{c_0}=\frac{2}{\gamma}\frac{e^{-u}}{e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi}} \label{pion} \end{equation}

将此式代入\eqref{cme},可得

\begin{equation} \frac{c_-}{c_0}=\frac{2}{\gamma}\frac{e^{u}\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi-1}}{e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi}} \label{nion} \end{equation}

体系还满足泊松方程:

\begin{equation} \frac{d^2\varphi}{dx^2}=-\frac{e(c_+-c_-)}{\epsilon_0\epsilon} \label{Poisson} \end{equation}

\begin{equation} \begin{split} \frac{d^2u}{dx^2}=&-\frac{e^2(c_+-c_-)}{kT\epsilon_0\epsilon}\\ =&-\frac{e^2c_0}{kT\epsilon_0\epsilon}\frac{2}{\gamma}\frac{e^{-u}-e^{u}\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi-1}}{e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi}}\\ =&-\frac{1}{2L_D^2}\frac{2}{\gamma}\frac{e^{-u}-e^{u}\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi-1}}{e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi}} \end{split} \label{uPoisson} \end{equation}

其中$L_D=\sqrt{\frac{kT\epsilon_0\epsilon}{2e^2c_0}}$ 为德拜长度。令$x=L_DX$,上式可化为:

\begin{equation*} \begin{split} \frac{d^2u}{dx^2}=&\frac{1}{L_D^2}\frac{d^2u}{dX^2}=\frac{1}{L_D^2}\frac{du}{dX}\frac{d}{du}\left(\frac{du}{dX}\right)\\ =&\frac{1}{2L_D^2}\frac{d}{du}\left(\frac{du}{dX}\right)^2\\ =&-\frac{1}{2L_D^2}\frac{2}{\gamma}\frac{e^{-u}-e^{u}\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi-1}}{e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi}} \end{split} \end{equation*}

\begin{equation} \frac{d}{du}\left(\frac{du}{dX}\right)^2=-\frac{2}{\gamma}\frac{e^{-u}-e^{u}\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi-1}}{e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi}} \label{nonDuPoisson} \end{equation}

将上式积分一次,注意到分母对$u$的导数恰是分子,得

\begin{equation} \begin{split} \left(\frac{du}{dX}\right)^2=&-\frac{2}{\gamma}\int \frac{e^{-u}-e^{u}\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi-1}}{e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi}} du\\ =&\frac{2}{\gamma}\int \frac{d\left\{e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi} \right\}}{e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi}} \\ =&\frac{2}{\gamma}\ln\left [e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi} \right ]+C\\ =&\frac{2}{\gamma}\ln\left [e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi} \right ]-\frac{2}{\gamma}\ln\frac{2}{\gamma} \end{split} \label{dudX2} \end{equation}

上式中最后一个等号用到边界条件,$X\rightarrow\infty$时,$u=0$,$du/dX=0$

$X=0$处,$u(0)=u_0$。$u_0>0$时,$du/dX$ 为$X$的减函数,$u_

\begin{equation} \frac{du}{dX}=-\mathrm{sgn}(u_0)\sqrt{\frac{2}{\gamma}}\sqrt{\ln\left [e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi} \right ]+\ln\frac{\gamma}{2}} \label{dudX} \end{equation}

当$\xi=1$时得离子大小没有差异时的结果,即

\begin{equation} \frac{du}{dX}=-\mathrm{sgn}(u_0)\sqrt{\frac{2}{\gamma}}\sqrt{\ln\left [1+2\gamma \sinh^2\frac{u}{2} \right ]} \label{dudXKKBA} \end{equation}

取极限$\gamma\rightarrow 0$,得Gouy-Chapmann双电层结果:

\begin{equation} \frac{du}{dX}=-2\sinh\frac{u}{2} \label{dudXGC} \end{equation}

将\eqref{dudX}、\eqref{dudXKKBA}和\eqref{dudXGC}积分得

\begin{equation} X=\mathrm{sgn}(u_0)\sqrt{\frac{\gamma}{2}}\int_u^{u_0}\frac{du}{\sqrt{\ln\left [e^{-u}+(\xi+\eta)\left[\frac{\xi e^u+\eta}{\xi+\eta}\right]^{1/\xi} \right ]+\ln\frac{\gamma}{2}}} \label{Xu} \end{equation}

\begin{equation} X=\mathrm{sgn}(u_0)\sqrt{\frac{\gamma}{2}}\int_u^{u_0}\frac{du}{\sqrt{\ln\left [1+2\gamma \sinh^2\frac{u}{2} \right ]}} \label{XuKKBA} \end{equation}

\begin{equation} X=-\ln\frac{\tanh(u/4)}{\tanh(u_0/4)} \label{XuGC} \end{equation}

标签: 离子液体

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