Solutions to exercises
197
b. In the domain of high oxygen partial pressure (P O 2 > 7.4 # 10
−3
bar), for
example, in air, we can equally well write
log R η = − α ′ log P O 2
where α ′ is a constant. An approach identical to that following question 5(a) leads to
I
I P
.
c
c O
0 46
2
=
,
,
−
°
where I c
, ° is a constant.
6. a. Resistance of the electrolyte
We start by determining the value of A
1.78 10
1000
A
e
2
1.38 10
1000
0.85 1.6 10
23
19
#
σ =
=
−
−
#
#
#
#
−
−
or
A = 339 175 S cm
−1
K
We deduce the conductivity σ at 913 K
913
339175
e 1.38 10
0.85 1.6 10
913
23
19
σ =
−
#
#
#
#
−
−
σ = 7.62 # 10
−3
S cm
−1
and the resistance R of the electrolyte
R
1 S
,
σ
=
#
R
7.62 10
1
2
20
0.2
3
#
π
=
− # # #
R 0.21Ω
=
b. Value of potential difference
The expression for the potential difference ΔU to apply to the terminals
of the pump is
ΔU = ΔE th + RI + η a − η c
where ΔE th is the thermodynamic potential (at I = 0), η a is the anodic
overpotential, η c is the cathodic overpotential, and I is the current in the
circuit.
The thermodynamic potential ΔE th is given by the expression
E
4F
RT
ln P (air)
P
th
ext
int
Δ
=
E
4 96 480
8.314 913
ln 0.21
5 10
th
4
#
Δ
=
−
#
#
.
E
V
0 119
th =