Solutions to exercises
147
Numerical evaluation gives
3 10
10
e
Br
K
4
–2
1.38 10
900
1.35 1.6 10
23
19
#
σ
σ =
#
#
−
#
#
−
−
11.9
Br
K
σ
σ =
Assuming that the electronic transport number is negligible (t e ≈ t h ≈ 0), the
expression for the transport number for K
+
is
t
1
K
K
Br
K
Br
K
Br
K
σ
σ
σ
=
+
=
+
σ
σ
σ
σ
.
.
t
12 9
11 9
K =
.
t
0922
K =
3. At the bromine partial pressure P Br 2 = 10
−10
bar, we can read from the
diagram
[V ′
K ] = [V
•
Br ] = 10
18
cm
−3
and [h
•
] = 10
14
cm
−3
We thus deduce
σ h ≈ 10
−4
σ K
and
t h
h
K
Br
h
σ
σ
σ
σ
=
+
+
.
t
1 10
0 084 10
1
h
4
4
#
= +
+
.
t
922 10
h
5
#
=
−
that we round off to10
−4
, which is the order of magnitude of the hole transport number.
4. a. The electrochemical chain in question is
P Br 2 , Me / KBr (s) / Me, P Br 2
(c)
(a)
where Me denotes the electrode material, (c) is the cathode, and (a) is
the anode.
Each interface is at equilibrium:
2
1
Br 2(g) + V
•
Br(SE) + e ′
(Me) m Br
#
Br
where we verify the following relation:
2
RT
ln P
F
const.
2
1 Br
Br
V
e
2
2
Br
•
μ
μ
μ
ϕ
+
+
+ −
=
°
u