280
5 – Applications
2 Theoretical slope s
ln x
E
F
RT
th
Δ
Δ
=
=
.
.
s
m V per decade
96 480
8 314 423
36 4
th =
=
#
2 Experimental slope s
2.3 log x
E
expt
Δ
Δ
=
. ( .
. )
(
)
.
s
m V per decade
2 3 1 65 0 77
150
50
35 9
expt =
+
− −
=
The slopes s th and s expt are essentially equal, which validates the expression
for emf. The number of electrons exchanged is 1.
3. The expression for the change in free entropy of the reaction is deduced
from that for the emf
F E RT ln x
2
RT
ln x
G
r 1
NO
O
2
2
Δ
Δ
= −
+
+
°
This is evaluated numerically with x NO 2 = 10
−5
, x O 2 = 0.21, and T = 423 K.
Based on figure 92, for a NO 2 concentration of 10 ppm, we determine a
potential difference between the sensor terminals of 91.3 # 10
−3
V. We obtain
91.3 10
96 480 8.314 423 ln 10
2
8.314 423
ln 0.21
G
r 1
3
5
Δ
= −
+
+
−
−
#
#
#
#
°
52.04 kJ mol
G
r 1
1
Δ
= −
−
°
4. Expression of emf for cell (Ib)
Na / NASICON (Na
+
) / NaNO 2 (porous layer) / Au,NO,O 2
as a function of NO partial pressure
We apply the same reasoning as for question 1(a).
Consider the following equilibria:
Na m Na
+ + e
F
Na
Na
NASICON
e
Na
1
Na
μ
μ
μ
ϕ
=
+
−
+
l
u
At the NASICON (Na
+
) / NaNO 2 interface, we have
Na
+
NASICON m Na
+
NaNO 2
Na
NASICON
Na
NaNO 2
μ
μ
=
+
+
u
u
At the NaNO 2 (porous layer) / Au,NO,O 2 interface, we have
5 – Applications
2 Theoretical slope s
ln x
E
F
RT
th
Δ
Δ
=
=
.
.
s
m V per decade
96 480
8 314 423
36 4
th =
=
#
2 Experimental slope s
2.3 log x
E
expt
Δ
Δ
=
. ( .
. )
(
)
.
s
m V per decade
2 3 1 65 0 77
150
50
35 9
expt =
+
− −
=
The slopes s th and s expt are essentially equal, which validates the expression
for emf. The number of electrons exchanged is 1.
3. The expression for the change in free entropy of the reaction is deduced
from that for the emf
F E RT ln x
2
RT
ln x
G
r 1
NO
O
2
2
Δ
Δ
= −
+
+
°
This is evaluated numerically with x NO 2 = 10
−5
, x O 2 = 0.21, and T = 423 K.
Based on figure 92, for a NO 2 concentration of 10 ppm, we determine a
potential difference between the sensor terminals of 91.3 # 10
−3
V. We obtain
91.3 10
96 480 8.314 423 ln 10
2
8.314 423
ln 0.21
G
r 1
3
5
Δ
= −
+
+
−
−
#
#
#
#
°
52.04 kJ mol
G
r 1
1
Δ
= −
−
°
4. Expression of emf for cell (Ib)
Na / NASICON (Na
+
) / NaNO 2 (porous layer) / Au,NO,O 2
as a function of NO partial pressure
We apply the same reasoning as for question 1(a).
Consider the following equilibria:
Na m Na
+ + e
F
Na
Na
NASICON
e
Na
1
Na
μ
μ
μ
ϕ
=
+
−
+
l
u
At the NASICON (Na
+
) / NaNO 2 interface, we have
Na
+
NASICON m Na
+
NaNO 2
Na
NASICON
Na
NaNO 2
μ
μ
=
+
+
u
u
At the NaNO 2 (porous layer) / Au,NO,O 2 interface, we have
