---
title: "Which way do ions move in salt water?"
description: "Combine diffusion and electric forces into electrochemical potential, derive dilute-ion flux, and work out the sign of an ionic current."
canonical_url: "https://sandboxphysics.com/stories/electrochemistry/chemical-potential-flux/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/electrochemistry/chemical-potential-flux/"
last_modified: "2026-10-28"
---

[Electrochemistry and ion transport](https://sandboxphysics.com/stories/electrochemistry/)

L1 · Chapter 1

# Which way do ions move in salt water?

Sandbox Physics · 2026-10-28

An ion responds to both concentration and electric potential. Random motion spreads ions out of crowded regions; an electric field pulls positive and negative ions in opposite directions. Combining these tendencies lets us predict when salt water carries current and when its ions settle into equilibrium.

## Before you begin

You need distance, time, area and the idea of a flow rate. A derivative is a local slope. A logarithm turns a concentration ratio into an additive energy difference.

- Predict each ion’s flux from concentration and potential gradients.
- Derive the dilute Nernst–Planck flux from electrochemical potential.
- Distinguish particle motion from electric current.

## Count ions through a small window

Take a straight narrow channel with coordinate $X$ increasing to the right, measured in metres. At a small window across the channel, subtract the number of ions crossing leftward each second from the number crossing rightward. Divide by window area to obtain number flux $N_i$, in ions per square metre per second.

Number density $n_i$ counts ions per cubic metre. The species label is $i$, and its signed valence is $z_i$. Each ion carries charge $z_i e$. Multiply each number flux by that charge and add the species to find current density. A negative ion moving left contributes rightward current.

$$
j=e\sum_i z_i N_i
$$

The current density $j$ has units $\mathrm{A\,m^{-2}}$. Keep the charge sign when converting particle motion into electric current.

Reading: MIT 10.626

## Diffusion spreads a concentration difference

Ions in two neighbouring boxes cross their shared boundary at random. The more crowded box sends more ions across. Subtract the crossings in the opposite direction and a net flux remains from high to low concentration. No individual ion needs to know which side is emptier.

A concentration that increases rightward has positive slope $\partial_X n_i$ and negative diffusive flux. Fick’s law expresses this with a diffusion coefficient $D_i$ in $\mathrm{m^2\,s^{-1}}$. Increasing this coefficient increases the flux driven by the same concentration slope.

$$
N_i^{\rm diff}=-D_i\partial_X n_i
$$

The minus sign points the flux down the concentration gradient. We assume a dilute isothermal continuum and a constant diffusion coefficient.

## Put both driving forces into an energy

Electric potential $\Phi$ is measured in volts. The electric field $E=-\partial_X\Phi$ points toward lower potential. It pushes positive ions along the field and negative ions against it. Add an ion’s chemical potential to its electrical potential energy to obtain its electrochemical potential $\widetilde\mu_i$.

For an ideal dilute solution, the concentration term is $k_BT\ln(n_i/n_*)$. The reference density $n_*$ makes the logarithm dimensionless and drops out of the spatial gradient. Temperature $T$ is in kelvin and $k_B$ is Boltzmann’s constant. At equilibrium there is no electrochemical energy difference to drive a net flux between neighbouring positions.

Near equilibrium, flux is proportional to the downward electrochemical-potential gradient. Substituting the ideal-solution expression produces a diffusion term and an electric-migration term. The Einstein relation supplies the link between random diffusion and force-driven drift.

$$
\widetilde\mu_i=\mu_i^*+k_BT\ln\frac{n_i}{n_*}+z_i e\Phi
$$

This is energy per ion in joules. A concentration ratio and a voltage difference become comparable only after conversion into energy.

$$
N_i=-\frac{D_i n_i}{k_BT}\partial_X\widetilde\mu_i=-D_i\left(\partial_Xn_i+\frac{z_i e n_i}{k_BT}\partial_X\Phi\right)
$$

The liquid is stationary in this Nernst–Planck expression. A moving liquid adds an advective flux equal to number density multiplied by liquid velocity.

Reading: MIT 10.626 · MIT 10.626

## Work out the field that balances diffusion

At $T=298.15\;\mathrm K$, let a monovalent positive ion rise from $1.0\;\mathrm{mM}$ at the left end of a $10\;\mathrm{\mu m}$ segment to $1.1\;\mathrm{mM}$ at the right. Ask what end-to-end voltage stops its net flux. Set the flux to zero and integrate the electrochemical-potential gradient across the segment.

The thermal voltage is $V_T=k_BT/e=25.69\;\mathrm{mV}$. The right-hand potential must be lower by $25.69\ln(1.1)=2.45\;\mathrm{mV}$. The average field points rightward with magnitude about $245\;\mathrm{V\,m^{-1}}$. It drives positive ions toward the more concentrated end and balances their leftward diffusion.

For a negative ion with the same concentration ratio, the balancing voltage changes sign. One field generally cannot balance two arbitrarily prescribed concentration gradients. We will need to solve for both ion distributions together with their potential.

$$
\Delta\Phi=-\frac{V_T}{z_i}\ln\frac{n_{i,R}}{n_{i,L}}
$$

The two ends have the same temperature and this species has zero net flux. The result is an equilibrium condition for that particular species.

## A neutral solution can carry current

Positive and negative number densities can remain equal throughout a uniform electrolyte. Under an electric field, the ions travel in opposite directions but carry current in the same direction. Local electroneutrality describes the balance of charge at a place; it does not require every ion to stop.

An electrode introduces a different question. If ions reach its surface and cannot cross, local charge builds up. That charge alters the field and eventually suppresses further accumulation. The next chapter turns this feedback into a solvable set of equations.

## Try the prediction

### [Follow the two ions toward the electrodes](https://sandboxphysics.com/labs/double-layer-charging/)

1. Predict which ion will accumulate at the negative electrode, then play the charging process.
2. Reverse voltage, compare the profiles and infer current direction from ion motion.
3. Explain why a concentration difference near the surface remains after the net ion flux becomes small.

## Exercises and explanations

1. In a uniform monovalent symmetric electrolyte, both ions have the same diffusivity and number density. Do their currents cancel when they move in opposite directions?

   Read the explanation

   No. The negative ion reverses both charge and velocity, so its current points the same way as the positive ion’s. With uniform concentration, the two migration currents have equal magnitude and add.

2. A monovalent positive ion has a twofold end-to-end concentration ratio. What voltage stops it at room temperature? What changes if absolute temperature rises ten percent?

   Read the explanation

   The concentrated end must be lower in potential by $V_T\ln2\approx17.8\;\mathrm{mV}$. At a fixed ratio within the ideal dilute model, the required voltage is proportional to absolute temperature, so it becomes about $19.6\;\mathrm{mV}$.

[Next: How ions reshape the field that moves them](https://sandboxphysics.com/stories/electrochemistry/pnp-boundaries/)

## Sources and calculation scope

1. [MIT 10.626, Diffuse charge in electrolytes, 2011 Lecture 24](https://ocw.mit.edu/courses/10-626-electrochemical-energy-systems-spring-2014/34aaca3a97887695dd295db7cc0fa3c0_MIT10_626S14_S11lec24.pdf)

   Electrochemical potential, Nernst–Planck flux, charge continuity and dilute-electrolyte screening. Our notation uses number density, a full cell gap and a thermal-voltage scale.

2. [MIT 10.626, Electrokinetics, 2011 Lecture 29](https://ocw.mit.edu/courses/10-626-electrochemical-energy-systems-spring-2014/resources/mit10_626s14_s11lec29/)

   Thin-double-layer electro-osmosis and viscous force balance. The chapter derives slip with the wall normal, zeta-potential sign and tangential field defined before use.