Solutions to exercises
297
E
E
(2n m)F
RT
ln
P
P
P
P
C H /CO
CO
n
H O
m/2
C H
O
(n/2) (m/4)
n m
2
2
2
n m
2
Δ
Δ
=
+
+
+
#
#
#
°
2 For an alkane C n H 2n+2 m = 2n + 2
The preceding expressions become
• total electrochemical oxidation
E
E
(6n 2)F
RT
ln
P
P
P
P
C H
/CO
CO
n
H O
n 1
C H
O
(3n 1)/2
n 2n 2
2
2
2
n 2n 2
2
Δ
Δ
=
+
+
+
+
#
#
#
+
+
°
• partial electrochemical oxidation
E
E
(4n 2)F
RT
ln
P
P
P
P
C H
/CO
CO
n
H O
n 1
C H
O
(3n/2) 1
n 2n 2
2
2
2
n 2n 2
2
Δ
Δ
=
+
+
+
+
#
#
#
+
+
°
Solution 5.18 – Thermodynamic study of methane reforming in SOFC
1. The reactions to consider to calculate the open-circuit voltage of the cell
are the electrochemical reactions
2 at the anode CH 4 + 4O
2−
m CO 2 + 2H 2 O + 8e
2 at the cathode
2O 2 + 8e m 4O
2−
The balance equation corresponds to the total combustion reaction of methane
and is
CH 4 + 2O 2 m CO 2 + 2H 2 O
The electromotive force ΔE that appears at the terminals of the cell is due
to the electric potential difference φ e between the two electrodes
E
e
c
e
a
ϕ ϕ
Δ =
−
where φ e
c
and φ e
a
are the electrostatic potentials of the electrons at the positive
and negative electrodes, respectively. Moreover, thermodynamic equilibrium
leads to the following equalities:
2
8
4
O
c
e
c
O
c
2
2
μ
μ
μ
+
=
−
u
u
and
4
2
8
CH
a
O
a
CO
a
H O
a
e
a
4
2
2
2
μ
μ
μ
μ
μ
+
=
+
+
−
u
u
for the cathode and anode, respectively.
By combining these relations and given that
F
e
e
e
μ
μ
ϕ
=
−
u