Page 89 - MODERN ASPECTS OF ELECTROCHEMISTRY

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72
from the proton conducting membrane). The cell potential under working
conditions is
–
+
+
E(j) = E c (j)– E a (j) –R e j =E r –(  η + η  R e j ≤ E r (10)
)
a
c
Thus the energy efficiency will be decreased proportionally to the so-
called voltage efficiency:
(11)
ε =E( j) /E r
E
2
For a DMFC working at 200 mA/cm and 0.5 V, this ratio will be
ε E = O.5/1.21 =41.3% (12)
and the overall efficiency of the fuel cell will be
ε cell = ε rev × ε E = 0.967 x 0.413 ≈ 40% (13)
assuming a Coulombic efficiency of 100%, i.e., a complete combustion
of methanol. This is acceptable for an autonomous power source when the
oxidation reaction is complete, i.e., it gives a theoretical number of
faradays n th = 6 F per mole of methanol. However, for a methanol fuel
cell, the complete oxidation of CH OH to CO (involving n = 6 faradays
3
2
per mole) is rarely complete, so that a Coulombic efficiency is introduced:
(14)
ε =n exp /n th
F
where n exp is the number of faradays effectively exchanged in the half-cell
reaction, and n th = 6 is the number of faradays exchanged for a complete
combustion.
The overall energy efficiency becomes
n expFE(j) n th FE r E(j) n exp
We
ε cell = _ = _ = _ × × = ε ×ε ×ε F (15)
E
rev
( ∆H) ( ∆H) ( ∆ H ) E r n th
Therefore the overall efficiency may be dramatically decreased, e.g.,
if the electrooxidation stops at the formaldehyde stage:
+
CH OH → HCHO + 2 H +2 e – (nexp= 2),
3 aq
or at the formic acid stage:
+
CH 3 OH + H 2 O → HCOOH + 4 H aq +4 e – (n = 4),
exp
thus leading, respectively, to a Coulombic efficiency: