The electrochemical potential of an ion i in an electrolytic solvent is given by:

\begin{align}\mu_i(\vec r) &= \mu_i^0 + RT \ln(a_i(\vec r))\\& = \mu_i^0 + RT \ln(\gamma_i(\vec r) c_i(\vec r)/c_0) \\&= \mu_i^0 + RT \ln(c_i(\vec r)/c_0) + RT \ln(\gamma_i(\vec r)) \\&= \mu_i^0 + RT \ln(c_i(\vec r)/c_0)+z_i e \phi(\vec r) \end{align}

where $\mu_i^0$ is the chemical potential of the ion in an ideal solution with ionic concentration $c_0$, $\gamma$ is the activity coefficient, $\phi$ the electrostatic potential of the ions in solution (given by the Poisson equation).

Is this expression correct? The authors in this paper (I. Rubinstein, Electro-Diffusion of Ions (1990) (SIAM link)) claim the following:

... the entire system, including the boundary layers. This implies in turn the establishment of a certain degree of smoothness on the microscopic scale $r_d$ ($\sqrt{\varepsilon}$ dimensionless terms) of some function of concentration and the electric potential, termed the electrochemical potential and defined as

$$\tilde{\mu}_i = RT \ln{\tilde{C}_i} + z_iF\tilde{\varphi} \tag{1.19a}$$

or in dimensionless terms

$$\mu_i = \ln{C_i} + z_i\varphi . \label{eq:1.19b} \tag{1.19b}$$

Indeed, by $\eqref{eq:1.19b}$

$$j_i = -\alpha_iC_i\nabla\mu_i, \label{eq:1.19c} \tag{1.19c}$$

so that $j_i = O(1)$ implies that the variation $(\delta \mu)|_{\varepsilon^{1/2}}$ on the length $\sqrt{\varepsilon}$ is of the order

$$(\delta \mu)|_{\varepsilon^{1/2}} = O\left(\varepsilon^{1/2}\right).$$


To illustrate some of the notions introduced so far, let us consider the $C_i$ and $\varphi$ fields in an electrodialysis cell at equilibrium. For simplicity, let us limit our consideration to a $1,1$ valent electrolyte at bulk (feed) concentration $C_0$. Assume a constant fixed charge density $\tilde{N}(-\tilde{N})$ for the an- (cat-) ion membrane.

Recall that equilibrium is a steady state without macroscopic fluxes. By $\eqref{eq:1.19c}$ this implies constancy of ionic electrochemical potentials in the system. Thus,

$$\mu_i = \ln{C_i} + z_i\varphi = 0 \tag{1.23a}$$


$$C_i = e^{-z_i\varphi} \label{eq:1.23b} \tag{1.23b}$$

(For the univalent electrolyte under consideration $M = 2$, $z_1 = 1$, $z_2 = -1$ and the dimensionless bulk concentration at $\varphi = 0$ is normalized to unity.)

Equation $\eqref{eq:1.23b}$ is the equilibrium Boltzmann distribution in a potential field. Substitution of $\eqref{eq:1.23b}$ into (1.9c) yields the Poisson-Boltzmann ...

When you assume equilibrium, the ionic flux $j_i \propto \nabla \mu_i$ should be zero. But what people do in this paper, is setting $$ RT \ln(c_i(\vec r)/c_0)+z_i e \phi(\vec r)=0 $$ and therefore we get:

$$c_i(\vec r) = c_0 \exp(-z_i e/(RT) \phi(\vec r)),$$

which is the Boltzmann distribution of the ions according o the electrostatic potential. Does anyone understand this logic? Why can we set the additional term to zero in equilibrium? And why do we neglect $\mu_i^0$?


1 Answer 1


You state

Then you can assume equilibrium, thus the ionic flux $j_i\propto\nabla\mu$ should be zero

So it seems to me that the authors you read are not defining $$ RT \ln(c_i(\mathbf r)/c_0)+z_i e \phi(\mathbf r)\equiv0 $$ but instead defining $$ \mu_i(\mathbf r)=\mu_i^0. $$ That is, the chemical potential doesn't change.

  • $\begingroup$ Thanks for the reply! I have edited my question and posted the important part of the original paper. Please take another look at it. $\endgroup$
    – Guiste
    Commented Feb 5, 2015 at 7:47
  • $\begingroup$ @Guiste: As far as I can tell, my answer has an explanation for your update already. $\endgroup$
    – Kyle Kanos
    Commented Feb 5, 2015 at 13:22
  • $\begingroup$ I accepted this answer, because I think this is what the authors are claiming. However, I believe that really $\mu_i-\mu_i^0$ has to be a constant and not zero. This is the constant shift of the electrostatic potential which is unknown. $\endgroup$
    – Guiste
    Commented Jan 11, 2018 at 16:24

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