Impedance technique (Impedance spectroscopy)
Impedance techniques is conceptually totally different from other transient techniques where rather large perturbations are
applied to the electrode. In impedance technique, only a very small periodic perturbation is applied either at equilibrium or
steady-state, and the frequency of the signal is scanned typically in the range of 0.1Hz…100kH; the amplitude of the signal is
ca. 10 mV. As shown in the picture below, the system causes a phase shift δ between the input and output signals. All the
information of the system is included in the phase shift and the amplitude Î as the function of the angular frequency ω.
input
E = Ê sin(ωt)
system
output
I = Î sin(ωt + δ)
Z=E/I
Z = Z(ω)
Z ≠ Z(t)
Î = Î(ω)
δ = δ(ω)
ω = 2πf
If the input and output signals are plotted as the function of time a
picture like one aside is obtained. Presenting data in this form is,
however, rather pointless since periodical signals repeat themselves
and all the information can be extracted from a single period (cycle).
Therefore, data is presented in a polar coordinate presentation (next
page).
Polar coordinate (complex) presentation:
Impedance Z =
imag
ωt + δ
Eˆe jωt
Iˆe j(ωt +δ)
= Zˆe − jδ ; j = − 1 = imaginary unit
Y = Z −1 = admittance
Î
ωt + δ
Ê
real
 Note that the time dependence is explicitly lost!
 Impedance (and admittance) is a complex quantity.
 Consequence: the analysis must be carried out in the frequency domain. Laplace
(Fourier) transformation converts a function from the time domain to the
frequency domain.
Primer of complex numbers:
z = x + jy = re jφ
x = Re(z) = r cos(φ) = real part
y = Im(z) = r sin(φ) = imaginary part
r = z = zz * = x 2 + y 2 = modulus
y
φ = atan   = phase
x
z* = x − jy = re − jφ = complex conjugate
e ± jx = cos(x) ± j sin( x) Euler' s equation
 Consequence: non-linear equations need to be linearized → ac signal amplitude
must be small (≤ 20 mV peak to peak).
Problem: What is the polar coordinate expression of
1
?
jω
Basic impedance elements
• Resistor, R: E = RI ⇒ Z = R
• Capacitor, C:
I =C
• Inductor, L:
I=
dE
1
= jωCE ⇒ Z =
dt
jωC
Combination of elements
Z1
Z2
ZN
In series: ZTOT = Z1 + Z2 + ... + ZN = Σk Zk
1
1
Edt
=
E ⇒ Z = jωL
∫
L
jωL
• In electrochemistry, resistors represent the solution
(ohmic) resistance and the reaction (kinetic) resistance.
• All electrodes have a double layer capacitance (Lecture
VIII); the cell itself may include a parasitic (stray)
capacitance.
• An inductor is found in corrosion problems, but also as a
parasitic element coming from the wires connecting the
cell and the measurement apparatus.
Z1
Z2
In parallel: YTOT = Y1 + Y2 + ... + YN = Sk Yk
ZN
Why are electrodes (interfaces) capacitive?
Total electric current = Kirchoff’s first law:
Jk = transfer of species k (mol/cm2s)
2

∂ φ 
IT = IF + IC = A  F ∑ zk Jk − ε0εr
∂x∂t 
 k
zk = charge number of species k
A = electrode area
IC
IT
φ = Galvani potential
C
IF
ZF
Electrode reaction = Faradays’s law:
rk = rate of reaction of species k (mol/s)
IF = nF ∑ rk
n = number of electrons in the reaction
k
Electrode reactions and transport are of different tensorial degree and coupled only through the mass balance. Example, one
reactant:
nr
I
zk ( Jk )x =0 = k = F
Here, zk = n
A AF
Formal derivation of the electrode capacitance
 δ


IT δ = A F ∫ ∑ zk Jk dx − jωε 0εr [φ(δ) − φ(0)]
 k

 0

 δ


= A F ∫ ∑ zk Jk dx + jωε0εr E 

 k

 0
IT
ε ε A
= YT = YF + jω 0 r = YF + jωC
E
δ
Faradaic admittance (impedance) is derived considering the kinetics of the electrode reaction and utilizing
the mass balance.
Any impedance can be presented as a serial combination of a pseudo-resistor Rp and pseudo-capacitor Cp:
Z = Rp – j/ωCp , hence Re(Z) = Rp and –Im(Z) = 1/ωCp
An ac bridge was the first experimental technique to measure
impedance. It can still be used and it is very accurate but slow
to operate.
I1R1 = I2 (Rp − j / ωC p )

 I1R2 = I2 Z
R1 Rp − j / ωC p
=
Z
R2
An ac bridge.
R1 = R2 ⇒ Z = Rp – j/ωCp
Kirchoff’s 2nd law
Impedance presentations
Nyquist plot:
Bode plot:
Electrochemists prefer a Nyquist (or
an impedance) plot because several
issues can immediately be seen
from it. Angular frequency ω is not
explicitly seen in the plot.
Process control people prefer a
Bode plot. Main concern is put to
the amplitude of the impedance
vector, phase angle has a smaller
significance for them.
Desibel dB = 10·log|Z|
Recipe for the derivation of impedance:
1. Linearize the current-voltage equation.
2. Solve the problem in Laplace domain.
3. Remove non-periodical components of the solution with the relation
(
)
fac = lim s2 + ω2 F (s)
s → jω
4.
Z (ω) =
In practice, replace s with jω and leave terms not including E (η) or I (I) out.
E (ω)
. Note! No inverse transformations are needed!
I(ω)
This very simple method becomes clear after the following examples. This method is not presented in any
electrochemical textbook.
(
)
Linearization of a multivariable function at point x1* , x2* ,..., xN* :
g(x1 , x2 ,..., xN ) ≈ g(x1* , x2* ,..., xN* ) +
N
∑
 ∂g

∂x
k =1 k



*
(xk − xk* )
N
*
⇒ ∆g ≈ ∑  ∂g  ∆xk
∂x
k =1 k 
Linearization of the current-overpotential equation at the equilibrium: i = 0, η = 0, cs = cb.
i cRs αfη cOs (α −1) fη
= e − be
⇒
i0 cRb
cO
∂  i 
∂  i 
∂  i 
1
1
  = αf − (α − 1) f = f




−
=
=
;
;
b
s i 
b
s i 
∂
η
cO
∂cR  0 eq. cR
∂cO  0 eq.
 i0 eq.
 i cRs cOs 
i cRs − cRb cOs − cOb
cRs cOs
RT
 − + 
≈
−
+ fη = b − b + fη ⇔ η =
b
b
i0
cR
cO
cR cO
nF  i 0 cRb cOb 
Now this linearized equation in time domain is transformed to Laplace domain, and the surface concentrations are inserted:

RT  i cRs cOs  RT  i
i
i
 ; i = I ; A = electrode surface area
η=
−
+
=
+
+
A
nF  i 0 cRb cOb  nF  i 0 nFcRb sDR nFcOb sDO 


1
1
RT
RT
RT
η
 + 1  1
 + 1  1 ⇒ Z (ω) = RT +
=
+ 2 2
nFAi0 n2F 2 A DR  cRb ξcOb  jω
I nFAi0 n F A DR  cRb ξcOb  s
The impedance thus is Z = Rct + W = Rct +
Rct =
σ
(1 − j) where Rct = charge transfer resistance, W = Warburg impedance
ω
RT
1
1 
RT

σ= 2 2
+ b
b

n F A DR 2  cR ξcO 
( )α (cb )1−α
n2F 2 Ak 0 cOb
R
The total impedance with the capacitance and the solution resistance is given by Randles’ equivalent circuit:
Cd
100
Rs
Rct
W
-Zimag / Ω
80
In a Nyquist plot aside, characteristics of an impedance
of a simple electrode reaction is shown. At high
frequencies, the plot intercepts the real (x) axis at Rs;
the diameter of the semicircle is Rct; the slope rising
with 45 degrees angle is characteristic to diffusion.
Depending on the element values, the diffusion slope
overlaps with the semicircle.
60
40
20
0
Rct
Rs
0
50
100
150
200
Zreal / Ω
250
At an arbitrary potential the faradaic impedance is obtained as follows:
[
]
I
∆I (s)
= kox cRs − kred cOs ⇒
= kox ∆cRs − kred ∆cOs + α kox cRs + (1 − α)kred cOs f ∆E (s)
nFA
nFA
∆cRs (s) = −
∆I (s)
nFA sDR
and ∆cOs (s) =
[
∆I (s)
nFA sDO
]
2 2


n
k
k
s
s
ox
red
 = αkox cR + (1 − α) kred cO F A ∆E (s)
∆I (s) 1 +
+

RT
sDR
sDO 

Z=

1 + λ / jω
λ 
RT


⋅
=
R
+
1
ct 
2 2
s
s
jω 
n F A αkox cR + (1 − α)kred cO

λ=
kox
k
+ red .
DR
DO
The average surface concentrations cRs and cOs due to the dc current can be solved as follows:
(
cOb
cRb
DR cR (s) + DO cO (s) = DR
+ DO
⇒ cRs − cRb = ξ cOb − cOs
s
s
s
s
)
(
Realizing that the rate constants kox and kred can be expressed as kox = k 0θα and kred = k 0θα −1 where θ = exp  nF E − E 0'
 RT
Idc
s α
s α −1
c
c
=
θ
−
θ
Butler-Volmer equation becomes in the following form:
R
O
0
nFAk
Combining this with the expression in previous page, it is obtained:
(
)
b
b
1−α
ξcOb + cRb + ξIdc θ1−α / nFAk 0
/ nFAk 0
s θ ξcO + cR − Idc θ
and cO =
cR =
1 + ξθ
1 + ξθ
s
Provided that the kinetics is not very slow and the measurement is not done at high dc currents, the Idc term in the
numerator can be neglected, and the surface concentrations become
(
b
b
ξcOb + cRb
s θ ξcO + cR
and cO ≈
cR ≈
1 + ξθ
1 + ξθ
s
)
The charge transfer resistance finally becomes as
Rct =
1 + ξθ
RT
⋅
θα
n2F 2 Ak 0 ξcOb + cRb
(
)
),
Accordingly, the Warburg impedance can be derived into the form
(
1 + ξθ  k 0θα k 0θα −1  1 − j
1 + ξθ)2
RT
RT
+
= 2 2
⋅
(1 − j)
W= 2 2 0 b b ⋅ α ⋅
b
b


θ
DO  2ω n F A ξcO + cR DO 2ω
θ
n F Ak ξcO + cR
 DR
(
)
(
50
2500
40
2000
1500
'W'
'R'
30
1 + ξθ
'R' = α
θ
20
1000
10
500
0
-200
0
E - E° / mV
200
0
-200
)
The plot aside tells why measuring impedance
at potentials far from equilibrium does not
pay off: the increase of the Warburg
impedance is much stronger than that of the
charge transfer resistance and we won’t see
anything else than the effect of diffusion –
again.
(
1 + ξθ)2
'W ' =
θ
0
E - E° / mV
200
Plots of the dimensionless charge transfer resistance and the Warburg impedance.