How ion depletion creates a transport limit

Join a reservoir to a selective interface

Let $x\in[0,1]$ run from the left reservoir to the right cation-selective interface. Scale concentration by reservoir density and potential by thermal voltage. At the reservoir, $c_+=c_-=1$ and $\phi=0$ . At the selective boundary, $c_+=1$ , $J_-=0$ and $\phi=-V$ .

The fixed boundary cation concentration represents an ideal prescribed cation activity. Together with the maintained reservoir, it allows steady transport. This half-cell does not resolve a real membrane’s thickness and fixed-charge profile. Its voltage includes the interfacial charge layer at the right.

Reading: MIT 10.626 · Rubinstein & Zaltzman

Two flux conditions determine the neutral bulk

Outside a thin interfacial layer, first approximate $c_+=c_-=c$ . Anions cannot cross the interface, so their steady flux vanishes at every section. Setting their Nernst–Planck flux to zero gives a potential gradient equal to the logarithmic concentration gradient.

Substitute this gradient into the cation flux. Diffusion and migration contribute equally, giving $J_+=-2c\prime$ . Call the constant cation flux $j$ and integrate from reservoir concentration one. The concentration decreases linearly with position.

\phi=\ln c,\qquad j=-2\partial_xc,\qquad c(x)=1-\frac j2x

Dimensionless current

j=J_+-J_-

equals cation flux because

J_-=0

. Its dimensional current-density scale is

eDn_0/L

Reading: MIT 10.626

Match the interfacial layer to get total voltage

Let $c_s$ denote bulk concentration just outside the interfacial layer, where $\phi_s=\ln c_s$ . This bulk concentration generally differs from the imposed boundary cation value one. The thin layer connects them.

Under local equilibrium across that layer, cation electrochemical potential is continuous. On its bulk side it is $\ln c_s+\phi_s$ ; on its boundary side it is $\ln1-V$ . Equating these gives $2\ln c_s=-V$ . One logarithmic voltage drop lies in the depleted bulk and another in the Donnan partition layer.

c_s=e^{-V/2},\qquad j_{\rm EN}(V)=2\left(1-e^{-V/2}\right),\qquad j_{\rm lim}=2

This electroneutral outer solution belongs to the stated boundaries. Ignoring the interfacial drop and applying the boundary concentration directly to the bulk changes the problem.

Reading: MIT 10.626 · Rubinstein & Zaltzman

Calculate a current plateau

At $V=2$ , the room-temperature total voltage is about $51.4\;\mathrm{mV}$ . The interfacial bulk concentration is $e^{-1}=0.368$ of the reservoir value. Current is $1.264$ , about sixty-three percent of the diffusion limit.

At $V=6$ , concentration falls to $e^{-3}=0.0498$ and current reaches $1.900$ . Tripling voltage has increased current by only about half. The table uses the analytic electroneutral solution and shows why its current approaches a plateau.

Analytic electroneutral, thin-interface reference
$V$	$c_s$	$j_{\rm EN}$	$j_{\rm EN}/j_{\rm lim}$
1	0.607	0.787	0.393
2	0.368	1.264	0.632
4	0.135	1.729	0.865
6	0.0498	1.900	0.950

Depletion challenges electroneutrality

If the straight concentration profile carried $j>2$ , it would become negative near the right end. That exposes the approximation’s limit. Fewer ions mean weaker screening, so potential curvature and the difference between ion densities become important. Retaining Poisson’s equation allows extended space charge to support voltage beyond the neutral description.

A finite-Debye-length one-dimensional PNP solution can therefore exceed the electroneutral diffusion limit. It still contains no transverse flow. In real devices, surface conduction, side-wall electro-osmosis and electroconvection can also carry overlimiting current. Geometry and observed flow help distinguish them.

Compare the two models first at moderate driving with well-separated thin layers, then increase voltage. Under strong driving, resolving the charge layer under mesh refinement is essential before assigning a discrepancy to physics.

Reading: Rubinstein & Zaltzman · Dydek et al.

Try the prediction

Follow depletion along the current plateau

At a small voltage, compare PNP with the electroneutral reference and predict where concentration first falls.
Increase voltage gradually and record minimum salt concentration, current and space charge. Distinguish the outer bulk from the thin interface layer in the profiles.
Reduce the Debye-length ratio. Check agreement at weak driving, then explain why charge must be retained as depletion grows.

Exercises and explanations

At ninety percent of the electroneutral limiting current, what is the interfacial bulk concentration and approximate room-temperature total voltage?

Read the explanation

The linear profile gives $c_s=0.1$ . Matching the interface gives $V=-2\ln0.1=4.605$ , or about $118\;\mathrm{mV}$ . This model reference omits additional voltage drops through a real membrane, contacts and the external circuit.

A one-dimensional calculation produces overlimiting current. Can it locate the onset of electroconvection?

Read the explanation

No. Electroconvection needs transverse electric fields, charge and velocity perturbations to amplify one another. A stationary one-dimensional model has no such degrees of freedom. It supplies a conduction base state; stability analysis requires a transverse direction and fluid equations.

Sources and calculation scope

MIT 10.626, Ion concentration polarization, 2011 Lecture 23

Selective transport, concentration gradients and diffusion-limited current. Our reservoir-to-interface problem states its own cation-activity boundary and explicitly includes the Donnan voltage in the total applied voltage.

Rubinstein & Zaltzman, The driving factors of electro-convective instability in concentration polarization, arXiv:1602.00137v1 (2016)

Sections I–II, equations (3)–(9) and Figures 1–3 distinguish selective conduction, fluid coupling and a transverse stability calculation. The course reads this published research; its one-dimensional PNP experiment does not reproduce the neutral-stability curves.

Dydek et al., Overlimiting current in a microchannel, arXiv:1108.2325v1 (2011)

Figures 1–3 and the separate surface-conduction, side-wall-flow and instability mechanisms. Their channel geometry and voltage convention differ from the course half-cell; our PNP curve is not a reproduction of those figures.