Diffusion, Migration and Einstein Equation

In the textbook: Electrochemical Systems by Newman and Alyea, 3rd edition, chapter 11.9: Moderately Dilute Solutions, equation for the mole flux of the component $i$ is given by: $$ N\_i = - \frac {u\_i c\_i} {z\_i F} \nabla \bar\mu\_i\ + c\_i v \tag {1}$$


where $u\_i$ is the ionic mobility, $c\_i$ is the concentration, $\bar \mu\_i$ is the electrochemical potential and $v$ is the velocity of the streaming fluid (assuming a low concentration of the component $i$ as situation becomes more complicated at higher concentrations).


One thing to note is that I wrote ionic mobility $u\_i$ a bit differently compared to the textbook. I defined it as a terminal velocity of the ion in the unitary electric field, so its measuring unit is $[\frac {m^2}{Vs}]$ in my case. In the textbook they defined it also as a terminal velocity in the unitary electric field, but divided by $z\_i F$. This is something to keep in mind to avoid confusion.


Basically, equation (1) states that the mole flux of the component has three contributions: diffusion, migration and convection. Diffusion and migration are accounted for in the first term (electrochemical potential gradient) and convection is accounted for in the second term.


My problem is with the first term $-\frac {u\_i c\_i} {z\_i F} \nabla \mu\_i$ which I know how to derive and I will do this here as I don't understand something about the equation (1).


To start, we define diffusion mole flux from as the 1st Fick's law: $$N\_{i,dif} = -\frac {D\_i c\_i} {RT} \nabla \mu\_i \tag {2}$$


where $D\_i$ is the diffusion coefficient and $\mu\_i$ is the chemical potential of the component $i$


Migration mole flux is given by: $$N\_{i,mig} = -u\_i c\_i \nabla \phi \tag {3}$$


where $\phi$ is the electric potential.


If there is no convection, we can write total mole flux of the component as: $$N\_i = -(\frac {D\_i c\_i} {RT} \nabla \mu\_i + u\_i c\_i \nabla \phi) \tag {4}$$


We recall the definition of electrochemical potential $ \bar \mu\_i$: $$ \bar \mu\_i = \mu\_i + z\_i F \phi \tag {5} $$


If we assume that system is in thermodynamic equilibrium than electrochemical potential gradient and net mole flux vanish: $$ \nabla \bar \mu\_i = 0 \tag {6} $$ $$ N\_i = 0 \tag {7}$$


From equations (5) and (6), we can write: $$\frac {\nabla \mu\_i}{\nabla \phi} = -z\_i F \tag {8}$$


From equations (4) and (7), we can write: $$ \frac {\nabla \mu\_i}{\nabla \phi} = - \frac {u\_i RT}{D\_i} \tag {9}$$


By equating (8) and (9), we get the Einstein equation, which relates diffusion coefficient $D\_i$ and ionic mobility $u\_i$ in the thermodynamic equilibrium: $$D\_i = \frac {u\_i RT}{z\_i F} \tag {10}$$


If we know substitute the equation (10) into equation (4), we can write: $$N\_i = -u\_i c\_i(\frac {1}{z\_i F} \nabla \mu\_i + \nabla \phi) = - \frac {u\_i c\_i} {z\_i F} \nabla \bar\mu\_i\ \tag {11}$$


After the derivation of the first term in the equation (1), I can get to my point. Namely, that the first term in the equation must be equal to zero since it is impossible to write this term in that form (via gradient of the electrochemical potential) without using Einstein equation. We saw earlier that Einstein equation is only valid in the thermodynamic equilibrium and therefore equation (6) applies.




Before writing, I understand your derivation, but I do not agree with Eq. (8). The gradient of a scalar field is a vector, so it is not correct to divide a vector quantity like $\underline{\nabla}\mu\_i$ with $\underline{\nabla}\phi$.


Indeed the flux of species $j$, in a $1$-dimensional system, can be written as
$$ N\_j = -D\_j\dfrac{dC\_j}{dx} - u\_j C\_j \dfrac{d\phi}{dx} \tag{1} $$


Here is a somewhat informal derivation of your Eq. (10). Imagine the charged species $ j $ being subjected to two forces, the drag force and the electrical force, but both being balanced. By Newton's 2nd law we have
\begin{align}
 F\_D &= F\_E \\
 6 \pi \mu r\_j v\_D &= z\_j e E \\
 \dfrac{v\_D}{E} &= \dfrac{z\_j e}{6 \pi \mu r\_j} \\
 u\_j &= \dfrac{k\_BT}{6 \pi \mu r\_j}\dfrac{z\_j e}{k\_BT} \hspace{1 cm}
 \left(D\_j = \dfrac{k\_BT}{6 \pi \mu r\_j} \hspace{0.5 cm} \text{(Einstein-Stokes law)}\right) \\
 u\_j &= D\_j\dfrac{z\_j e N\_A}{N\_Ak\_B T} \\
 u\_j &= D\_j\dfrac{z\_j F}{RT} \tag{2} \\
\end{align}
This is the Einstein-Smoluchowski equation. Combining Eqs. (1) and (2) we have
$$ N\_j = -D\_j\dfrac{dC\_j}{dx} - \dfrac{F}{RT} z\_jD\_jC\_j \dfrac{d\phi}{dx} \tag{3} $$
If you couple Eq. (3) with the convection part, i.e. $ C\_j\mathbf{v} $, the result is the famous **Nernst-Planck equation**.


Eq. (2) was derived by considering that the charged species $ j $ is in an equilibrium between its forces. Although I didn't, you can derive Einstein-Stokes law by doing the same, but considering: (1) the balance between the chemical force $d\mu\_j/dx $ and the drag force, and (2) comparing the derived expression with Fick's law.
Thus, **how come this relation, obtained by an equilibrium between forces, is combined with the expression to describe the motion of species in $N\_j$?**. The easiest answer is the following: **so we can reduce the number of phenomenological and constant parameters, to describe the motion, from two to one**.


Nernst-Planck equation is a good approximation to the molar flux, if the concentrations are sufficiently low. This means that you can neglect ion-ion interactions in the medium. This is the reason why you are able to change the activity by the concentration in the chemical part of $\tilde{\mu}\_j$. It also assumes that the flux can be written as a linear combination of driving forces, and the phenomenological coefficients, are typically considered as constants. But indeed, coupling it with Eq. (2) goes much further. If you want to be more precise, Eq. (1) should be written
$$ N\_j = -D\_j^\infty\dfrac{dC\_j}{dx} - u\_j^\infty C\_j \dfrac{d\phi}{dx} \tag{4} $$
So you need two phenomenological parameters to evaluate the flux. However, sometimes it is hard to obtain both coefficients, $ D\_j^\infty $ and $ u\_j^\infty $, so we go around with this by applying Eq. (2). I will give you an example of my own. I am modeling lithium-sulfur batteries, and there are publications where $\ce{Li+}$ (and other sulfur species) are as high as $4-5$ mol/dm$^3$, and other authors and myself still use Eq. (3).


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