Solutions to exercises
153
Solution 3.12 – Electrical conductivity of solid vitreous solution
(SiO 2 ) 1−x (Na 2 O) x
1. The junction potential between vitreous phases G1H and G2H is given by the
general expression
F
1
z
t d
1
2
i
i
i
1
2
i
ϕ
ϕ
μ
−
=
/
#
Applying this to the given problem leads to
F
1
d
F
1
1
2
M
1
2
M
(2)
M
(1)
ϕ
ϕ
μ
μ
μ
−
=
=
−
+
+
+
`
j
#
2. Expression of emf ∆E at the terminals of the electrochemical chain
Each electrode is at the following equilibrium:
2
1
O 2(gas) + 2e GPtH m O
2−
GglassH
which allows us to write
2 for electrode 1
2
1
RT ln P
2
F
2 F
O
O
(1)
e
O
1
1
2
2
2
μ
μ
ϕ
μ
ϕ
+
+
−
=
−
α
−
°
°
8
8
B
B
2 for electrode 2
2
1
RT ln P
2
F
2 F
O
O
(2)
e
O
2
2
2
2
2
μ
μ
ϕ
μ
ϕ
+
+
−
=
−
α
−
°
°
l
8
8
B
B
Given that the oxygen partial pressures P
(1)
O 2 and P
(2)
O 2 are equal and that we
have the same electrode metal (Pt), the emf ∆E reduces to
∆E = φ (α) − φ (α′)
E
2F
1
O
(2)
O
(1)
1
2
2
2
μ
μ
ϕ
ϕ
Δ =
−
+
−
−
−
8
8
B
B
By using the result of question 1, we obtain
E
2F
1
2F
1
2
2
O
(2)
O
(1)
M
(2)
M
(1)
2
2 –
μ
μ
μ
μ
Δ =
−
+
−
−
+
+
8
8
B
B
By considering that the equilibrium
M 2 O m 2M
+ + O
2−
is realized in the vitreous phases G1H and G2H, we obtain
E
2F
RT
ln a
a
M O
1
M O
2
2
2
Δ =