Solutions to exercises
73
2 or by choosing two pairs (σ,T), which leads to the following expression for the activation energy:
E
T T
RT T ln T
T
a
1
2
1 2
2 2
1 1
σ
σ
=
−
The pairs
(T 1 = 375 K,σ 1 = 70.15 S cm
−1
)
and
(T 2 = 671 K,σ 1 = 153.2 S cm
−1
)
lead to
E a = 9 631 J mol
−1
E a = 0.1 eV
Note – This result is low compared with those associated with the classic ionic conductors (NaCl, CaF 2 , YSZ).
5. a. The resistance measured between points A and B include the polarization
resistance R p of the electrode and the ohmic resistance R 2 of the material
itself
R A-B = R p + R 2
If we assume a constant cross section for the sample, R 2 may be calculated from the following relation:
R
R
1
2
1
2
,
,
=
which gives
R
R
2
1 1
2
,
,
=
.
.
.
R
00421 1 496
0 5
2 =
#
R
0.0141
2
Ω
=
This result shows that the polarization resistance R p is negligible (R p ≈ 0).
b. Taken together, the preceding results (low activation energy, negligible polarization resistance, role of the electrode material) indicate that
La 0.8 Sr 0.2 MnO 3−δ is primarily an electronic conductor in air.
Solution 2.2 – Measurement of electric quantities
by complex impedance spectroscopy
1. The equation for dipole devices associated in parallel is
Z
Z
1
1
tot
k
k
= /
which gives for two dipole devices 1 and 2 Z
Z Z
Z Z
tot
1
2
1 2
=
+ .