Solutions to exercises
87
2 for cationic conductivity
.
E
e V
1 81
,
a c =
We observe a significant difference between the two activation energies, which is shown by the difference between the anionic and cationic
conductivities.
d. We find behavior similar to that of CaF 2 in terms of ionic-conducting
substance (i.e., a dominant anionic conduction) such as ionic crystals
with the same structure (e.g., BaF 2 , SrCl 2 , PbF 2 , ThO 2 , CeO 2 , UO 2 ,
ZrO 2 , Bi 2 O 3 , …).
Solution 2.7 – Electrochemical semipermeability
1. a. We have a mixed conductor containing cations M
+
and electrons that are
subjected to a difference in chemical potential between the two extremities of a metal M. The general expression for the potential difference
∆E between terminals is
E
F
1
t d
ion
M
1
2
μ
Δ =
#
where t ion is the ionic transport number of the crystal, which here equals
the cationic transport number t M +.
For a purely cationic conductor (t M + = 1), we obtain
E
F
1
1 2
a
M
(2)
M
(1)
μ
μ
Δ
=
−
−
`
j
b. If we consider the average ionic transport number t M + in the domain
M
(1)
M
(2)
μ
μ
−
  , the preceding relation becomes
E
F
t
d
M
M
1
2 μ
Δ =
+ #
E
F
t M
M
(2)
M
(1)
μ
μ
Δ =
−
+
`
j
and the electromotive force ∆E
b
1−2 is
E
F
t
t
E
1 2
b
M
M
(2)
M
(1)
M
1 2
a
μ
μ
Δ
Δ
=
=
−
−
−
+
+
`
j
2. The cationic and electric current densities are given by the following
expressions
2 cationic current i
F
F
M
M
M
d
d
σ
μ
ϕ
= −
+
+
+
+
^
h