Solutions to exercises
291
RT ln K
RT ln P (P )
P
G
f T
T
H
O
an
H O
2
2
2
1 2
Δ
= −
=−
°
from which we obtain
ln P
2 RT
2ln P
P
G
O
an
f T
H
H O
2
2
2
Δ
=
+
°
The emf is expressed as
E
4 F
RT
ln P
2F
RT
ln P
P
2F
G
O
c
H
H O
f T
2
2
2
Δ
Δ
=
−
−
°
or
E
4 F
RT
ln P
2F
RT
ln P
P
2F
H
T S
O
c
H
H O
f T
f T
2
2
2
Δ
Δ
Δ
=
−
−
−
°
°
2. Numerical evaluation at 1 000 K gives
E
4 96 480
8.314 1000
ln 0.21
2 96 480
8.314 1000
ln 0.98
0.02
2 96 480
247 900 (1000 55.3)
Δ =
−
+
+
#
#
#
#
#
#
E 1.132 V
Δ =
3. The change in emf as a function of temperature depends on the sign of the
change in the entropy of the overall operating reaction of the cell, itself given
by the change Δn (g) of the number of moles of gas. The results are given in
table 52.
Table 52 – Change in emf ΔE T with temperature for cells supplied with various fuels.
Overall reaction
Δn (g) Sign of Δ f S ΔE = f (T)
H 2(g) + 2
1
O 2(g) m H 2 O (g)
− 0.5
< 0
decreasing
CH 4(g) + 2O 2(g) m CO 2(g) + 2H 2 O (g)
0
0
≈ constant
CH 3 OH (g) + 2
3
O 2(g) m CO 2(g) + 2H 2 O (g)
0.5
> 0
increasing
4. The thermodynamic efficiency of the cell as a function of thermodynamic
quantities is expressed as
R
H
G
H
H
T S
th
r T
r T
r T
r T
r T
Δ
Δ
Δ
Δ
Δ
=
=
−