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Sandbox Physics
Electrochemistry and ion transport

L3 · Chapter 7

What can one impedance arc tell us?

Sandbox Physics ·

A complete impedance arc can determine solution resistance, effective reaction resistance and capacitance. It may still leave microscopic reaction pathways unresolved. First ask what each frequency measures, then fit a saved spectrum to distinguish uncertain parameters from parameters these data cannot determine at all.

Rock the voltage slowly, then quickly

Apply a small sinusoidal voltage to a stable electrode at frequency ff, completing ff cycles per second. Angular frequency is ω=2πf\omega=2\pi f in radians per second. If response is linear at this amplitude, current oscillates at the same frequency with a different amplitude and possibly a shifted peak time.

Impedance records both the voltage-to-current amplitude ratio and their phase difference. Amplitude alone misses essential information: resistor current peaks with voltage, whereas capacitor current leads by a quarter cycle. At high frequency, charging the same capacitor with a given current requires a smaller voltage excursion.

Reading: MIT 10.626

Use a complex number to keep magnitude and phase

Real and imaginary parts record two components a quarter cycle apart. The imaginary unit i\mathrm i obeys i2=−1\mathrm i^2=-1; physical current remains II. Factor out the shared time dependence eiωτe^{\mathrm i\omega\tau}. The remaining complex amplitude is a phasor. Actual voltage or current is the real part of the complete expression.

A resistor has in-phase voltage and current, giving real impedance. Capacitor current is capacitance multiplied by the voltage derivative. Differentiating the shared exponential multiplies by iω\mathrm i\omega, giving negative imaginary capacitive impedance. Phasors turn the time derivative into algebra.

Z=V^I^,ZR=R,ZC=1iωC=−iωCZ=\frac{\widehat V}{\widehat I},\qquad Z_R=R,\qquad Z_C=\frac{1}{\mathrm i\omega C}=-\frac{\mathrm i}{\omega C}
Hats denote complex amplitudes and impedance is in ohms. We use positive-exponential time dependence. Nyquist plots show negative imaginary part upward for the capacitive arc.

Reading: MIT 10.626

Combine reaction and charging in parallel

Reaction current and double-layer charging feel the same small interfacial overpotential, so they form parallel branches. Solution resistance RsR_s carries their summed current and is in series. Add the parallel admittances, or reciprocal impedances, then invert to obtain this simplified Randles-type model.

Capacitance CdlC_{\rm dl} and charge-transfer resistance RctR_{\rm ct} describe the whole modelled electrode. Their product τr=RctCdl\tau_r=R_{\rm ct}C_{\rm dl} sets the time scale for current partition. The model holds temperature and surface concentrations fixed and includes neither diffusion impedance nor distributed capacitance.

Z(ω)=Rs+(Rct−1+iωCdl)−1Z(\omega)=R_s+\left(R_{\rm ct}^{-1}+\mathrm i\omega C_{\rm dl}\right)^{-1}
At low frequency the capacitor is effectively open, so impedance approaches the sum of resistances. At high frequency it effectively shorts the reaction branch, leaving solution resistance.
Z′=Rs+Rct1+(ωτr)2,−Z′′=Rctωτr1+(ωτr)2Z'=R_s+\frac{R_{\rm ct}}{1+(\omega\tau_r)^2},\qquad -Z''=\frac{R_{\rm ct}\omega\tau_r}{1+(\omega\tau_r)^2}
Primes Z′Z\prime and Z′′Z^{\prime\prime} denote real and imaginary parts. Eliminating frequency gives a semicircle, whose top occurs at ωτr=1\omega\tau_r=1.

Reading: MIT 10.626

Read three parameters from three features

Suppose the high-frequency intercept is 35.0  Ω35.0\;\Omega, the low-frequency intercept 215  Ω215\;\Omega, and the top of the arc occurs at 49.1  Hz49.1\;\mathrm{Hz}. The first gives solution resistance. Their difference gives Rct=180  ΩR_{\rm ct}=180\;\Omega. The peak condition then gives Cdl≈18.0  μFC_{\rm dl}\approx18.0\;\mathrm{\mu F}.

The relaxation time is 3.24  ms3.24\;\mathrm{ms}. The table gives noiseless circuit truth with frequency scaled by the peak frequency fcf_c. The default synthetic record covers 0.1  Hz0.1\;\mathrm{Hz} to 105  Hz10^5\;\mathrm{Hz}. Reproducible seeded Gaussian noise has standard deviation 0.8  Ω0.8\;\Omega independently in each real and imaginary component.

Cdl=12πfcRct,fc=12πτrC_{\rm dl}=\frac{1}{2\pi f_cR_{\rm ct}},\qquad f_c=\frac{1}{2\pi\tau_r}
Use hertz and ohms to obtain farads. The horizontal position of the arc top is halfway between its intercepts.
Noiseless default circuit; synthetic recorded points separately include labelled noise
f/fcf/f_cZ′Z\prime / Ω\Omega−Z′′-Z^{\prime\prime} / Ω\Omega
0.1213.217.82
1125.090.00
1036.7817.82

Save a spectrum before estimating its parameters

Acquisition saves frequency, real part, imaginary part and noise settings. Fitting reads that record and varies positive resistances and capacitance to compare predictions with the saved points. Changing resistance, capacitance or noise acquires a new record. Changing the pathway fraction or fitted frequency band repeatedly analyses the same saved points.

For equal real and imaginary noise scales, add the squared residuals of both components and minimize their sum. Dividing by a positive common noise variance gives the expression below with the same minimizing parameters. A noiseless reference uses unweighted residuals. Residuals should fluctuate around zero. A coherent frequency-dependent trend can indicate missing physics. A small residual also needs adequate coverage of both intercepts and the arc top.

At high frequencies alone, much of the reaction information lies in small corrections. Many reaction resistances can produce nearly coincident spectra within the noise. Broader frequency coverage can improve this precision problem.

χ2=∑k(Zk,rec′−Zk,model′)2+(Zk,rec′′−Zk,model′′)2σ2\chi^2=\sum_k\frac{(Z'_{k,{\rm rec}}-Z'_{k,{\rm model}})^2+(Z''_{k,{\rm rec}}-Z''_{k,{\rm model}})^2}{\sigma^2}
Index kk labels frequency and σ\sigma is the known noise standard deviation of each component. Correlated errors require a covariance instead of this independent equal-variance assumption.

Some ambiguity survives perfect measurements

Split the effective reaction branch into two parallel pathways. For any 0<p<10<p<1, choose R1=Rct/pR_1=R_{\rm ct}/p and R2=Rct/(1−p)R_2=R_{\rm ct}/(1-p). Their admittances always add to the original reaction admittance. Keep the same capacitor in parallel with both.

For example, p=0.5p=0.5 gives two 360  Ω360\;\Omega resistors. Taking p=0.2p=0.2 gives 900  Ω900\;\Omega and 225  Ω225\;\Omega. Both pairs equal 180  Ω180\;\Omega in parallel. Their complete impedance spectra are identical, so even noiseless measurements cannot determine pp.

Separating microscopic pathways requires an observation or intervention that changes their relative responses, such as reactant, temperature or steady-potential variation with a corresponding reaction model. Effective parameters may be precise even when a microscopic interpretation needs another experiment.

1R1+1R2=pRct+1−pRct=1Rct\frac1{R_1}+\frac1{R_2}=\frac{p}{R_{\rm ct}}+\frac{1-p}{R_{\rm ct}}=\frac1{R_{\rm ct}}
This exactly equivalent family of linear circuits demonstrates structural non-identifiability. Assigning microscopic pathways to a real electrode requires separate physical evidence.

Try the prediction

Give the same record another interpretation

  1. Fit the default synthetic record. Estimate resistance from the intercepts first and compare with the fit.
  2. Change the fitted frequency band and check that the recorded points stay fixed. Compare the information in full and high-frequency fits. Finally change resistance or capacitance to acquire a new spectrum.
  3. Move the two-pathway fraction. Predict whether impedance or residuals change, then explain which quantities the record can determine.

Exercises and explanations

  1. Why does a small residual on a short high-frequency arc segment fail to establish precise values for all three parameters?

    Read the explanation

    The high-frequency end primarily determines solution resistance. Reaction resistance and capacitance enter smaller corrections and can compensate for one another. Many parameter combinations may overlap within noise. Coverage of the peak and low-frequency end, or independent information, is needed.

  2. How low must noise become to distinguish a half-and-half reaction split from a one-tenth split when effective resistance stays fixed?

    Read the explanation

    No noise level can do it in this model. Total parallel admittance is exactly unchanged, giving identical predictions at every frequency. This differs from noise hiding a small difference and requires an experiment that changes the pathways’ relative response.

Sources and calculation scope

  1. MIT 10.626, Impedance spectroscopy, 2011 Lecture 5

    Sinusoidal response, complex amplitudes and circuit impedances. The course introduces positive-exponential phasors before using them and plots negative imaginary impedance upward.

  2. 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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