Solutions to exercises
279
Consider the following equilibria:
Na m Na
+ + e
F
Na
Na
NASICON
e
Na
1
Na
μ
μ
μ
ϕ
=
+
−
+
u
At the NASICON (Na
+
) / NaNO 3 interface, we have
Na
+
NASICON m Na
+
NaNO 3
Na
NASICON
Na
NaNO 3
μ
μ
=
+
+
u
u
At the NaNO 3 (porous layer) / Au,NO 2 ,O 2 interface, we have
Na
+ + NO 2 + 2
1
O 2 + e m NaNO 3
2
1
F
NaNO
Na
NaNO
NO
O
e
Au
2
Au
3
3
2
2
μ
μ
μ
μ
μ
ϕ
=
+
+
+
−
+
u
As a first approximation, assume that the chemical potential of electrons
is the same in the sodium and in the gold. We obtain the following expression for the emf ∆E Ia of the cell (Ia):
E Ia
2
Au
1
Na
ϕ
ϕ
Δ
=
−
E
F
1
2
1
Ia
Na
NO
O
N aNO
2
2
3
μ
μ
μ
μ
Δ
=
+
+
−
`
j
The factor in parentheses is the opposite of the free enthalpy of the following reaction:
Na + NO 2 + 2
1
O 2 m NaNO 3
(1)
which gives
E
F
G
Ia
r 1
Δ
Δ
= −
E
F
1
RT ln P
2
RT
ln P
G
Ia
r 1
NO
O
2
2
Δ
Δ
=
−
+
+
°
`
j
where ∆ r G 1 ° is the change in standard free enthalpy of reaction (1) and
P NO 2 and P O 2 are the respective NO 2 and O 2 partial pressures expressed
in bars.
b. The expression for emf as a function of the molar fractions x NO 2 and x O 2 ,
for the case in which the total pressure is 1 bar, is
E
F
1
RT ln x
2
RT
ln x
G
Ia
r 1
NO
O
2
2
Δ
Δ
=
−
+
+
°
`
j
2. To verify that the expression for emf is valid, we must compare the theoretical slope s th to the experimental slope s expt .