Which way do ions move in salt water?

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.

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.

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.

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.

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.