Download Electrodes

Biopotential electrodes
A complex interface
Basics of Instrumentation, Measurement and Analysis 2011, 2012
the interface problem
metal
e-
electrolyte
I
?
2
M+
-
A
To sense a signal
a current I must flow !
But no electron e- is
passing the interface!
metal cation
leaving into the electrolyte
No current
One atom M out of the metal
is oxidized to form
one cation M+ and giving off
one free electron eto the metal.
4
metal cation
joining the metal
No current
One cation M+
out of the electrolyte
becomes one neutral atom M
taking off one free electron
from the metal.
6
half-cell voltage
No current, 1M salt concentration, T = 25ºC
metal:
Li
Vh / Volt -3,0
8
Al
Fe
negativ
Pb
H
Ag/AgCl
0
0,223
Cu
Ag
positiv
Pt
Au
1,68
Nernst equation
For arbitrary concentration and temperature
E = RT/(zF)·ln(c/K)
E – electrode potential
R = 8.314 J /(mol*K) – molar gas constant
T – absolute temperature
z – valence
F = 96485 C/mol – Faraday’s constant
c – concentration of metal ion in solution
K – “metal solution pressure”,
or tendency to dissolve
9
electrode double layer
No current
13
current influence



with current flowing
the half-cell voltage changes
this voltage change is called
overpotential or polarization:
Vp = Vr + Vc + Va
activation, depends on direction of reaction
concentration (change in double layer)
ohmic (voltage drop)
16
polarizable electrode


18
“perfectly” polarizable electrode:
- only displacement current,
electrode behave like a capacitor
example: noble metals like platinum Pt
nonpolarizable electrode


20
“perfectly” nonpolarizable electrode:
- current passes freely across
interface,
- no overpotential
examples:
- silver/silver chloride (Ag/AgCl),
- mercury/mercurous chloride
(Hg/Hg2Cl2) (calomel)
chemical reactions
silver / silver chloride
22
electrical behaviour
equivalent circuit
26
equivalent circuit
electrode-electrolyte
27
more precise approximation of
double layer – Randles circuit
electrode-electrolyte
Rct – active charge transfer resistance
W – Warburg element reflecting diffusion
with impedance ZW = AW/(jω)0.5
AW – Warburg coefficient
28