Solutions to exercises
89
Taking the average values, we obtain
J
F
1 t
t
M
2
t
M
M
M
(2)
M
(1)
,
σ
μ
μ
=
−
−
#
+
+
+
`
`
j
j
Because E
F
1
1 2
a
M
(2)
M
(1)
μ
μ
Δ
=
−
−
`
j, we obtain
J
F
(1 t )t
E
M
t
M
M
1 2
a
,
σ
Δ
=
−
−
+
+
+
Solution 2.8 – Determination of transport number
by electrochemical semipermeability
1. The unit of the flux J S,O 2
The dimensional equation applied to equation (1) gives
J
cm mol cm
cm s cm
,
S O
3
1
2
2 1
2
/
−
−
or
J
mol s cm
,
S O
1
2
2
/
−
−
2. Numerically evaluating equation (1) gives
#
#
.
.
.
.
.
J
22 4
4 25 10
10
1 932 1 13
0 2
,
S O
3
5
2
=
−
#
#
.
J
m ol s cm
6 49 10
,
S O
10
1
1
2
=
−
−
−
3. The expression for the potential difference is
E
4F
RT
ln P
P
CE
T
Δ =
l
with
P ′
T = P 1 + ∆P
P ′
T = 5 # 10
−2 + 1.932
P ′
T = 1.982 Pa
and
E
4F
RT
ln P
P
CE
T
Δ =
l
#
E
4 96 480
8.314 823
ln 0.21 10
1.982
5
Δ =
#
#
E
0.164 V
Δ = −
4. The total conductivity σ t of the solid solution is obtained from the relation
R
k
t
t
σ =
89
Taking the average values, we obtain
J
F
1 t
t
M
2
t
M
M
M
(2)
M
(1)
,
σ
μ
μ
=
−
−
#
+
+
+
`
`
j
j
Because E
F
1
1 2
a
M
(2)
M
(1)
μ
μ
Δ
=
−
−
`
j, we obtain
J
F
(1 t )t
E
M
t
M
M
1 2
a
,
σ
Δ
=
−
−
+
+
+
Solution 2.8 – Determination of transport number
by electrochemical semipermeability
1. The unit of the flux J S,O 2
The dimensional equation applied to equation (1) gives
J
cm mol cm
cm s cm
,
S O
3
1
2
2 1
2
/
−
−
or
J
mol s cm
,
S O
1
2
2
/
−
−
2. Numerically evaluating equation (1) gives
#
#
.
.
.
.
.
J
22 4
4 25 10
10
1 932 1 13
0 2
,
S O
3
5
2
=
−
#
#
.
J
m ol s cm
6 49 10
,
S O
10
1
1
2
=
−
−
−
3. The expression for the potential difference is
E
4F
RT
ln P
P
CE
T
Δ =
l
with
P ′
T = P 1 + ∆P
P ′
T = 5 # 10
−2 + 1.932
P ′
T = 1.982 Pa
and
E
4F
RT
ln P
P
CE
T
Δ =
l
#
E
4 96 480
8.314 823
ln 0.21 10
1.982
5
Δ =
#
#
E
0.164 V
Δ = −
4. The total conductivity σ t of the solid solution is obtained from the relation
R
k
t
t
σ =