Biomedical Engineering Reference
In-Depth Information
hence, smaller, more polarizable electrodes are used. The requirements for the coun-
terelectrode of the two-electrode system include a high exchange current (fast elec-
tron transfer kinetics), a very large surface area (to lower the current density) and
a high concentration of the species involved in the redox reaction, such that the
concentrations are not significantly changed by the passage of a current.
Two approaches have been developed in biosensor applications as the current
and voltage change while using polarizable electrodes. Electrochemistry techniques
are based on the current (
i
) measurement as a function of voltage (
ΔΦ
appl
). The key
electronic component for electrochemical measurement is a potentiostat, which
can be configured for potentiometric (apply fixed current and measure output volt-
age) or amperometric (apply fixed voltage and measure output current) readout.
A potentiostat is used to apply a constant potential to the working electrode with
respect to a counterelectrode (a reference electrode). A potentiostat is a simple elec-
tronic circuit that can be constructed using a battery, two operational amplifiers,
and several resistors.
A problem with potentiometric sensors in sensor applications is the require-
ment of rapid electrode kinetics for faster response. The standard redox potential
of two coupled half-cells generating a voltage
ΔΦ
0
is related to the free energy
change of the reaction, by the following equation:
Δ=
G E
(9.5)
0
0
where
F
is Faraday's constant and
z
is the number of electrons transferred. Since
Δ
G
can be related to the
K
D
value,
E
0
is a function of the kinetics of the reaction.
However, many biological reactions are slow interactions.
EXAMPLE 9.3
Using yeast alcohol dehydrogenase, which contains NAD, Mr. Dickerson tested the con-
version of alcohol to acetaldehyde. At the end of the reaction, NAD (the oxidized form)
is reduced to NADH. At 25
°
C, the dissociation constant was determined to be 50 mM.
Calculate the electrode potential.
Solution: From (7.3),
G TK
ln
D
Δ=
Equating to (9.4),
K
zFE
RT
ln
D
−=
0
Hence,
RT
E
ln
D
K
=
0
zF
−
E
=−
25.70[mV]ln0.05
=
77 mV
0
