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Sandbox Physics
Electrochemistry and ion transport

L3 · Chapter 6

When ions exchange electrons at an electrode

Sandbox Physics ·

A reacting electrode gives ions another route: exchange electrons at the interface and enter a different chemical state. Net current is the difference between forward and reverse reaction rates. At equilibrium both reactions continue at equal rates, producing zero net current.

Define reaction direction and local potential

Consider R⇌O+nee−\mathrm R\rightleftharpoons\mathrm O+n_e e^-. The forward oxidation reaction releases nen_e electrons, and anodic current is positive. Let Φm\Phi_m be metal potential and Φs\Phi_s the solution potential at the reaction plane. Activation overpotential ηact\eta_{\rm act} is their difference relative to the local equilibrium value.

Equilibrium potential depends on local reactant and product activities: effective concentrations that include nonideal interactions. A dilute ideal approximation uses concentration ratios. If the reaction plane is depleted, reservoir concentration cannot be substituted for surface concentration.

ηact=Φm−Φs−ΔΦeq\eta_{\rm act}=\Phi_m-\Phi_s-\Delta\Phi_{\rm eq}
Overpotential is in volts. It describes driving at the reaction plane. The total terminal voltage may also drop across the electrolyte, other electrodes and interfacial layers.

Reading: MIT 10.626

Subtract the two reaction currents

Butler–Volmer theory represents the potential response of a transition state as changes in forward and reverse activation barriers. Increasing anodic overpotential favours oxidation and suppresses reduction. Multiplying rates by transferred charge and subtracting gives net current.

Exchange current density i0i_0 is the magnitude of either direction’s equilibrium current. Transfer coefficient α\alpha describes how potential shifts the activation barriers, usually between zero and one. It is not universally fixed by invoking symmetry. We use a single-step local mass-action form with concentrations held at the values relevant to the small perturbation.

iF=i0 ⁣[exp⁡ ⁣((1−α)neFηactRT)−exp⁡ ⁣(−αneFηactRT)]i_F=i_0\!\left[\exp\!\left(\frac{(1-\alpha)n_eF\eta_{\rm act}}{RT}\right)-\exp\!\left(-\frac{\alpha n_eF\eta_{\rm act}}{RT}\right)\right]
Faraday constant FF converts moles of electrons to charge; RR is the molar gas constant. Both iFi_F and i0i_0 are in A m−2\mathrm{A\,m^{-2}}. The voltage sign follows our anodic convention.

Reading: MIT 10.626 · MIT 10.626

Match reaction current to ion flux

Suppose each reaction consumes one ion and transfers one electron. Reactions per area per second determine both its solution-side boundary flux and the Faradaic electrode current. For other valences and stoichiometries, apply each coefficient explicitly. Different species’ number fluxes cannot all be identified with the same current.

The electrode also has double-layer capacitance. Some external current changes stored interfacial charge and some drives reaction. A small voltage step can therefore produce substantial initial charging current even when reaction is slow. Impedance exploits the different frequency responses of these processes.

Iext=Cdldηactdτ+IFI_{\rm ext}=C_{\rm dl}\frac{d\eta_{\rm act}}{d\tau}+I_F
This lumped relation holds with fixed local equilibrium potential during the perturbation and an ideal double-layer capacitor. CdlC_{\rm dl} is whole-electrode capacitance, and II is the whole-electrode current in amperes.

Reading: MIT 10.626

Read a resistance from the equilibrium slope

For ∣neFηact∣≪RT|n_eF\eta_{\rm act}|\ll RT, retain the constant and first-order terms of each exponential. Constants cancel and slopes add. The transfer coefficient cancels too, leaving a current proportional to overpotential.

For uniform area AA, exchange current is I0=Ai0I_0=Ai_0. Whole-electrode charge-transfer resistance RctR_{\rm ct} is the inverse current–voltage slope. It belongs in a small-signal equivalent circuit. Larger excursions require the nonlinear reaction law.

IF≃neFI0RTηact,Rct=RTneFI0I_F\simeq\frac{n_eFI_0}{RT}\eta_{\rm act},\qquad R_{\rm ct}=\frac{RT}{n_eFI_0}
This resistance is in ohms. Using current density gives an area-specific resistance in ohm square metres. Convert between these resistances using electrode area.

Reading: MIT 10.626 · MIT 10.626

How small is a small signal?

Take T=298.15  KT=298.15\;\mathrm K, a one-electron reaction, I0=0.100  mAI_0=0.100\;\mathrm{mA} and α=0.5\alpha=0.5. Charge-transfer resistance is 257  Ω257\;\Omega. At 10.0  mV10.0\;\mathrm{mV}, the linear current is 38.9  μA38.9\;\mathrm{\mu A} and the full exponential result is about 39.2  μA39.2\;\mathrm{\mu A}, less than one percent apart.

At 100  mV100\;\mathrm{mV}, the linear estimate is 0.389  mA0.389\;\mathrm{mA} but the full result is about 0.686  mA0.686\;\mathrm{mA}. A tenfold voltage has made linearization substantially less accurate. Frequency sweeps also require a system that responds with small excursions around the same steady state.

The impedance experiment defaults to Rct=180  ΩR_{\rm ct}=180\;\Omega, corresponding to I0≈0.143  mAI_0\approx0.143\;\mathrm{mA} at the same temperature and electron count. Its fixed-concentration branch represents the local linear reaction response. Diffusion depletion, activity changes and membrane voltage require additional physical relations.

Try the prediction

Read reaction speed from the impedance arc

  1. Predict whether increasing exchange current raises or lowers charge-transfer resistance.
  2. Change reaction resistance, acquire a new spectrum and compare Nyquist arc diameters.
  3. Keep capacitance fixed. Predict how the arc’s peak frequency moves, then check it in the record.

Exercises and explanations

  1. Double electrode area at fixed exchange current density and capacitance per area. How do total resistance, capacitance and their product change?

    Read the explanation

    Exchange current doubles, so total charge-transfer resistance halves. Total capacitance doubles. Their product is unchanged, so the ideal reaction–capacitance relaxation frequency stays fixed.

  2. Does a large voltage between two electrodes imply a large overpotential at one reaction plane?

    Read the explanation

    No. Most voltage may fall across the electrolyte, a selective interface or the other electrode. The reaction law requires local activation overpotential. Allocate the voltage before deciding whether linearization applies.

Sources and calculation scope

  1. MIT 10.626, Faradaic reactions in dilute solutions, 2011 Lecture 12

    Local reaction stoichiometry and the transition-state picture behind charge transfer. The course uses anodic current positive and separates local activity from reservoir concentration.

  2. MIT 10.626, Butler–Volmer equation, 2011 Lecture 13

    Forward and reverse activation currents, equilibrium potential and overpotential. Our worked example expands the local fixed-concentration law near equilibrium and uses whole-electrode exchange current in the impedance model.

  3. MIT 10.626, Impedance of electrodes, 2011 Lecture 6

    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.

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