Solutions to exercises
195
In yttria-stabilized zirconia, the reaction that introduces the dopant Y 2 O 3 is
Y 2 O 3 $ 2Y ′
Zr + 3O
#
O + V O
::
This doping leads to an increase in the rate of oxygen vacancies and, consequently, to a decrease in the rate of zirconium vacancies. Because of the
small charge relative to the oxide ions and of the high concentration of
oxygen vacancies, the electric conduction is due entirely to the oxide ions
via a vacancy mechanism.
2. a. Expression for current I required to purify the gas inside the tube
Gas purification corresponds to a decrease Δn /s of the number of moles
of oxygen per second of
/s
n
4F
I
RT
(P P )V
/s
e
ex
Δ
=
=
−
which gives
(
)
I
RT
F P P V
4
/
e
ex
s
=
−
#
b. Inserting the numerical values gives the current
.
(
)
I
A
8 314 913 3 600
4 96 480 10
10 10 5 10
7 10
3
5
5
3
3
#
#
=
−
=
−
−
−
−
#
#
#
#
#
I
mA
7
=
3. The average pressure is
P
b ar
2
10
10
5 10
3
5
4
#
=
+
=
−
−
−
with
.
log P
3 33
= −
As per figure 72, this average pressure corresponds to a normalized polarization resistance equal to log R η = 1.85
R η ≈ 71 Ω cm
2
or a polarization resistance R p of
R
S
R
p =
η
R
2
20
71
p
π
= # #
R
0.565
p
Ω
=
The expression relating cathodic overpotential η c to the current,
195
In yttria-stabilized zirconia, the reaction that introduces the dopant Y 2 O 3 is
Y 2 O 3 $ 2Y ′
Zr + 3O
#
O + V O
::
This doping leads to an increase in the rate of oxygen vacancies and, consequently, to a decrease in the rate of zirconium vacancies. Because of the
small charge relative to the oxide ions and of the high concentration of
oxygen vacancies, the electric conduction is due entirely to the oxide ions
via a vacancy mechanism.
2. a. Expression for current I required to purify the gas inside the tube
Gas purification corresponds to a decrease Δn /s of the number of moles
of oxygen per second of
/s
n
4F
I
RT
(P P )V
/s
e
ex
Δ
=
=
−
which gives
(
)
I
RT
F P P V
4
/
e
ex
s
=
−
#
b. Inserting the numerical values gives the current
.
(
)
I
A
8 314 913 3 600
4 96 480 10
10 10 5 10
7 10
3
5
5
3
3
#
#
=
−
=
−
−
−
−
#
#
#
#
#
I
mA
7
=
3. The average pressure is
P
b ar
2
10
10
5 10
3
5
4
#
=
+
=
−
−
−
with
.
log P
3 33
= −
As per figure 72, this average pressure corresponds to a normalized polarization resistance equal to log R η = 1.85
R η ≈ 71 Ω cm
2
or a polarization resistance R p of
R
S
R
p =
η
R
2
20
71
p
π
= # #
R
0.565
p
Ω
=
The expression relating cathodic overpotential η c to the current,
