246
5 – Applications
or
F
F
e
Pt
7
e
Cu
8
μ
ϕ
μ
ϕ
−
=
−
Combining the preceding relations allows us to write
E
F
1
2
1
4
1
8
1
Ag
int.ref.
O
salt
O
2
2
ϕ
ϕ
μ
μ
μ
Δ =
−
=
−
−
+
+
−
8
B
Because
RT ln a
Ag
int.ref.
Ag
Ag
int.ref.
μ
μ
=
+
+
+
+
°
RT ln a
O
salt
O
O
salt
2
2–
2
μ
μ
=
+
−
−
°
RT ln P
O
O
O
2
2
2
μ
μ
=
+
°
given that
P O 2 = 0.2 bar
By inserting this into the expression for the emf ΔE, we obtain
E
F
1
RT ln a
2
1
2
RT
ln a
4
RT
ln P
Ag
Ag
int.ref.
O
O
salt
O
2
2
2
μ
μ
Δ = −
+
+
+
−
+
+
−
−
°
°
8
B
By joining activity and molality in molten salts, we obtain
RT ln m
Ag
int.ref.
Ag
Ag
int.ref.
μ
μ
=
+
+
+
+
°
RT ln m
O
salt
O
O
salt
2
2
2
μ
μ
=
+
−
−
−
°
or
2.3 RT pO
O
salt
O
2
2
2
μ
μ
=
−
−
−
−
°
and E
R Tlnm
4
RT
ln0.2
2F
2.3RT
logm
F
1
Ag
Ag
int.ref. 2
1 O
O
salt
2
2
μ
μ
Δ =−
+
+
−
−
+
+
−
−
°
°
8
B
By writing
E
F
1
RT ln m
2
1
4
RT
ln 0.2
Ag
Ag
int.ref.
O 2
μ
μ
Δ
= −
+
+
−
+
+
−
°
°
°
8
B
we have
E
E
2F
2.3RT
log m O
salt
2
Δ
Δ
=
−
−
°
and
E
E
2F
2.3RT
pO
2
Δ
Δ
=
+
−
°
3. a. For high pO
2−
, the solution is not saturated, so the emf varies with pO
2−
.
The horizontal part corresponds to a saturated solution. The intersection
between the two line segments indicates the onset of oxide precipitation
and allows the solubility to be evaluated.
b. Determination of experimental slope
s
pO
( E)
expt
2
Δ
Δ Δ
=
−