---
title: "Why a blocking electrode can still charge"
description: "Connect ion profiles to electrode charge, transient current and charging time, then check the Debye–Hückel and thin-layer limits."
canonical_url: "https://sandboxphysics.com/stories/electrochemistry/blocking-double-layer/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/electrochemistry/blocking-double-layer/"
last_modified: "2026-10-28"
---

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

L2 · Chapter 3

# Why a blocking electrode can still charge

Sandbox Physics · 2026-10-28

Ions cannot cross a blocking electrode, but they can rearrange beside it. Electronic charge in the metal faces opposite ionic charge in solution across a short distance: an electric double layer. Current initially transports this charge. As charging proceeds, current decays and most of the voltage falls across the two surface layers.

## Before you begin

Use chapter two’s PNP equations and blocking boundaries. Exponential decay means that the remaining difference shrinks by the same factor over equal intervals. We derive it from a resistance and capacitance.

[Review the preceding chapter](https://sandboxphysics.com/stories/electrochemistry/pnp-boundaries/)

- Compute electrode charge from potential slope and explain external current.
- Derive the charging time for thin, weak-potential double layers.
- Use conservation and limiting behaviour to assess a numerical charging result.

## Neither end allows ions through

Two plates have full separation $L$, with position increasing rightward and initially uniform ion densities. Apply electrode potentials $-V V_T/2$ and $+V V_T/2$. Here $V$ is the total voltage divided by thermal voltage. Both ionic fluxes vanish at both ends.

Cations first move left and anions right. Cations accumulate and anions deplete near the left plate, with the reverse near the right plate. Diffusion increasingly opposes migration and accumulated charge screens the interior field. Each species keeps its total particle number throughout this closed-cell process.

Reading: Bazant

## Calculate charge and current from the profiles

Gauss’s law gives metal charge per area from the electric field normal to the surface and directed into solution. This normal points rightward at the left plate and leftward at the right plate, giving the opposite signs below. Surface charge density is in $\mathrm{C\,m^{-2}}$.

The external circuit changes electronic charge in the metal, so external current is the electrode-charge derivative with respect to dimensional time $\tau$. Ionic current within solution can vary with position as space charge changes. Adding displacement current gives a spatially uniform total current in one dimension.

$$
q_L=-\varepsilon\partial_X\Phi(0),\qquad q_R=\varepsilon\partial_X\Phi(L),\qquad j_{{\rm ext},R}=\frac{dq_R}{d\tau}
$$

External current is positive here when it increases positive charge on the right electrode. The two electrode charging currents have opposite signs.

$$
\partial_X\!\left[j_{\rm ion}+\varepsilon\partial_\tau E\right]=0
$$

Combine charge continuity with the time derivative of Gauss’s law. At a blocking surface, ionic current vanishes and displacement current carries the external current.

Reading: Bazant · MIT 10.626

## A small surface potential decays exponentially

Setting the equilibrium fluxes to zero gives Boltzmann distributions. If local potential relative to the bulk is small compared with thermal voltage, retain the first-order term in each exponential. Charge density becomes proportional to potential. Poisson’s equation then reduces to a linear equation, the Debye–Hückel approximation.

Near a single surface far from the other plate, potential decays from surface value $\zeta$ over a Debye length. Its surface slope gives a charge proportional to voltage. The differential capacitance per area is permittivity divided by Debye length. A short charge separation produces a large capacitance.

$$
n_\pm\simeq n_0\left(1\mp\frac{\Phi}{V_T}\right),\qquad\partial_X^2\Phi=\frac{\Phi}{\lambda_D^2}
$$

Linearization requires local potential relative to the bulk to be much smaller than thermal voltage. In a finite closed cell, each species’ total number sets its concentration normalization.

$$
\Phi(y)=\zeta e^{-y/\lambda_D},\qquad C_D=\frac{dq}{d\zeta}=\frac{\varepsilon}{\lambda_D}
$$

Distance $y$ increases outward from one surface. Here $q$ denotes metal charge; integrated diffuse charge has the opposite sign.

Reading: MIT 10.626 · Bazant

## The charging time combines two lengths

The bulk resistance supplies ions to thin surface layers. A uniform monovalent electrolyte has conductivity $\kappa=2e^2Dn_0/(k_BT)=\varepsilon D/\lambda_D^2$. Area-specific bulk resistance, meaning resistance multiplied by cross-sectional area, is full gap divided by conductivity and has units of ohm square metres. The two equal double-layer capacitors are in series, halving the capacitance.

Multiply resistance by the series capacitance to obtain charging time. It lies between diffusion across the double layer and diffusion across the whole cell. The factor two comes from using a full gap and two equal layers; a half-gap convention expresses the same result differently.

$$
\tau_c=\frac{L}{\kappa}\frac{C_D}{2}=\frac{L\lambda_D}{2D},\qquad t_c=\frac{\delta}{2}
$$

This limit assumes thin layers, small total voltage, equal diffusivities and no compact layer. A separate short charge-relaxation transient precedes the slower circuit response.

Reading: Bazant · MIT 10.626

## Work a charging example

For $L=1.00\;\mathrm{\mu m}$, $\lambda_D=10.0\;\mathrm{nm}$ and $D=1.00\times10^{-9}\;\mathrm{m^2\,s^{-1}}$, charging takes $\tau_c=5.00\;\mathrm{\mu s}$. The layer diffusion time is $0.100\;\mathrm{\mu s}$ and whole-cell diffusion time is $1.00\;\mathrm{ms}$. Each has units of time but describes a different transport distance.

With $\varepsilon=6.95\times10^{-10}\;\mathrm{F\,m^{-1}}$ and total voltage $5.00\;\mathrm{mV}$, each final layer takes about half the voltage. The right electrode approaches $1.74\times10^{-4}\;\mathrm{C\,m^{-2}}$. The table is an analytic thin-layer, linear-circuit reference. A PNP solution should approach it in that limit.

$$
\frac{q_R(\tau)}{q_R(\infty)}=1-e^{-\tau/\tau_c},\qquad \frac{j_{{\rm ext},R}(\tau)}{j_{{\rm ext},R}(0^+)}=e^{-\tau/\tau_c}
$$

These are ideal reference values, not measurements. The instantaneous geometric-capacitance response to the initial voltage step is distinguished from finite-time numerical samples.

Analytic thin-layer, small-potential reference

| $\tau/\tau_c$ | $q/q_\infty$ | $j/j_{0^+}$ |
| --- | --- | --- |
| 0.5 | 0.393 | 0.607 |
| 1.0 | 0.632 | 0.368 |
| 2.0 | 0.865 | 0.135 |
| 3.0 | 0.950 | 0.0498 |

## Larger voltages require the ion distributions

A larger surface potential adsorbs more neutral salt, reducing bulk concentration and changing differential capacitance. Charging then has several time scales. A numerical calculation must resolve both thin charge layers and slower salt diffusion, conserve ions, keep concentrations positive and approach a stable result under spatial and temporal refinement.

At high surface concentrations, a point-ion model also neglects finite ion size, activity and solvent structure. Increasing voltage calls for a fresh assessment of these effects before extending the model.

Reading: Bazant · Kilic

## Try the prediction

### [Read the current and ion profiles together](https://sandboxphysics.com/labs/double-layer-charging/)

1. Start with a small voltage and thin layers. Predict the charged fraction when current has fallen to about one-third of its initial value.
2. Compare profiles, potential and current at the same time. Locate the two concentration layers and the remaining interior field.
3. Increase voltage and identify where the single-exponential reference first departs from the solution. Explain the role of salt adsorption.

## Exercises and explanations

1. Double plate separation in the same solution. What happens to thin-layer linear charging time? What if Debye length alone is halved?

   Read the explanation

   Doubling separation doubles charging time. At fixed diffusivity and separation, halving Debye length halves it. Increasing salt raises conductivity enough to outweigh the accompanying increase in double-layer capacitance.

2. Late in charging, current is nearly zero. Is electrode charge also nearly zero?

   Read the explanation

   Current measures the rate at which charge changes. Near the end, the electrode holds nearly its final charge and only its rate of change approaches zero. A nearly full tank can likewise retain water after its inflow becomes small.

[Previous: How ions reshape the field that moves them](https://sandboxphysics.com/stories/electrochemistry/pnp-boundaries/) [Next: How ion depletion creates a transport limit](https://sandboxphysics.com/stories/electrochemistry/concentration-polarization/)

## 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, Double layer structure, 2011 Lecture 25](https://ocw.mit.edu/courses/10-626-electrochemical-energy-systems-spring-2014/b74a3f2744e2c150142cc1eab0d25fdc_MIT10_626S14_S11lec25.pdf)

   Planar diffuse-layer structure and differential capacitance. The chapter derives its small-potential limit and separately states the ideal point-ion approximation.

3. [Bazant, Thornton & Ajdari, Diffuse-Charge Dynamics in Electrochemical Systems, arXiv:cond-mat/0401118v1 (2004)](https://arxiv.org/abs/cond-mat/0401118v1)

   Sections III–V and the thin-layer analysis define the blocking-plane PNP problem, conservation and charging scales; published as Physical Review E 70, 021506. The paper uses plates at plus and minus a half-gap. Our full-gap convention gives a factor one-half in the two-layer RC time. Our browser omits the compact Stern layer.

4. [MIT 10.626, Impedance of electrodes, 2011 Lecture 6](https://ocw.mit.edu/courses/10-626-electrochemical-energy-systems-spring-2014/6aacd39911619ff5b669d80a0a0c7bbf_MIT10_626S14_S11lec06.pdf)

   Reaction resistance and double-layer capacitance in a lumped small-signal electrode. Our spectrum uses a series solution resistor and one parallel reaction-resistor/capacitor pair, with no diffusion element.

5. [Kilic, Bazant & Ajdari, Steric effects in the dynamics of electrolytes at large applied voltages I, arXiv:physics/0611030v1 (2006)](https://arxiv.org/abs/physics/0611030v1)

   Sections I–II explain why point-ion dilute theory can overpredict near-surface concentrations and introduce finite-size alternatives. These modifications are reading material, not active equations in the ideal PNP browser solver.