13.8 IONIC CONDUCTIVITY AND DIFFUSION
13.8 Ionic Conductivity and Diffusion 459
σNa+ =
(
×) (
×) (
×)
(
×) ( )
− −
−
1 602 10 3 5 10 1 33 10 1 38 10 300
19 2 27 9
23
. . .
.
σNa+ =28 8. S m/ =0 28. S cm/ and that for the chlorine ions
σCl− =
(
×) (
×) (
×)
(
×) ( )
− −
−
1 602 10 3 5 10 2 032 10 1 38 10 300
19 2 27 9
23
. . .
.
σCl− =44 1. S m/ =0 441. S cm/ to give a total conductivity of
σ σ= Na++σCl− =0 288 0 441 0 729. + . = . S cm/
which makes it a pretty good “semiconductor” at room temperature because its electrical conductiv- ity is between a good conductor like a metal (σ≅ 106 S/cm) and a good insulator such as SiO2
(σ ≅ 10–13 S/cm). On the other hand, the conduction of electrons by water is essentially zero so it has a very low electronic conductivity: an electron conductivity of a good insulator. The combina- tion of high ionic conductivity and low electronic conductivity is necessary for a material to be an electrolyte in a battery or fuel cell: ions are easily conducted through the electrolyte between the electrodes, while the electrons must go through an external circuit. This also highlights the confus- ing word “semiconductor,” discussed in Section 9.5.3. The term is commonly interpreted as refer- ring to materials, such as, silicon, germanium, or gallium arsenide, that are used to make electronic devices. However, the term really is comparative and refers to how good an electrical conductor a material happens to be: a semiconductor has conductivity somewhere between that of an insulator and a good conductor like a metal. In fact, it would be better to separate materials into just two gen- eral classes. (1) Metals, materials that have free electrons to conduct at T = 0 K, that is, they do not require energy to remove electrons from chemical bonds: all the free electrons are the chemical bond in a metal. (2) Nonmetals, materials that have no free electrons to conduct at T = 0 K because it takes energy to create electron–hole pairs from chemical bonds as described in Chapter 9. Then, the only difference between a semiconductor and an insulator is the amount of energy to create electron–
hole pairs: PbS, Eg ≅ 0.25 eV, a semiconductor; MgO, Eg ≅ 5.5 eV, an insulator. Furthermore, as shown above, a sodium chloride solution is a good semiconductor as an ionic conductor but is an insulator to electron conduction. Without such ionic semiconductors, batteries and fuel cells would not be possible (Xu 2004).
13.8.3 cao-D
oPeDZ
iRconiuMo
xiDe As As
oliDe
lectRolyteAs was discussed in Chapters 7 and 9, the fluorite crystal structure of ZrO2 forms extensive solid solutions with other oxides that form oxygen vacancies in ZrO2, including MgO, CaO, Y2O3, etc.
A classic example is a CaO-ZrO2 solid solution having the composition Ca0.15Zr0.85O1.85; that is, a 15 m/o (mole percent) solid solution of CaO and ZrO2. As with Y2O3, CaO goes into a solid solution of zirconia with the formation of oxygen vacancies:
CaO ZrO CaZr OOx V
O
2
′′ + + •• (13.25)
in the Kröger–Vink notation. Solving the electrical neutrality condition
2 2 2
2 2
e Ca e V
Ca ZrO
V ZrO
Ca Zr
V
Zr O
Zr O Zr O
η η η
η
η η
η η
η
′′ = ••⇒ ′′ = ⇒ ′′ =
•• •••
ηO
because ηzr =ηZrO2 andηO=2ηZrO2and gives
V Ca
O V
O O Zr
•• −
=η •• = ′′ = = ×
η 2
0 15
2 7 5 102
. .
where ηVO•• are the number of vacant oxygen sites/m3 and ηO are the number of oxygen ion sites/m3 in the doped ZrO2. Data in the literature for the diffusion of oxygen ions in this solid solution are (Kingery et al. 1976)
D O D e e cm s
Q RT
kJ mole 2 RT
0 3
81
10 2
− − − −
( )
= = / /and give a diffusion coefficient at 1000°C of
D O2 3e cm s m
81 000
8 314 1273 7 2 11 2
10 4 7 10 4 7 10
− − −
− −
( )
= ( . ,)( ) = . × / = . × // .sTo calculate the ionic conductivity of this solid solution at 1000°C it is the number of oxygen ions/m3, ηO, that must be used for η in Equation 13.24, in this case, because the oxygen ions are doing the conducting. From the literature (Haynes 2013), ρ(ZrO2) = 5.6 g/cm3 and M(ZrO2) = 123.22 g/mol, neglecting the differences between densities of the solid solution and pure ZrO2, which is a valid first approximation. (Is it? See exercises.)
η η ρ
O ZrO ZrO
ZrO
M NA kg m
kg mole
= = =
(
×)
× −
2 2 2 5 6 10
123 22 10
2
2 2
3 3
3
. /
( . / )) .
. .
×
(
×)
= × −
6 022 10 5 47 10
23
28 3
ηO m
Substituting these values of D and η into Equation 13.24
σO 1000oC 2 1 602 10 5 47 10 4 7 10 1 38 10
19 2 28 11
( )
=(
× ×) (
2 ×) (
×)
×
− −
−
. . .
. 33 1273
( ) ( )
σO
(
1000οC)
=15 0. S m/ =0 15. S cm/which is a significant ionic conductivity for oxygen in calcia-doped zirconia. Because of its high ionic conductivity, doped zirconia is used as an oxygen-conducting electrolyte in solid oxide fuel cells (SOFCs), in automobile and combustion oxygen sensors, and in oxygen pumps for medical applications. However, the temperature of 1000°C is rather high for many of these applications, and there is a strong desire to reduce the temperature by several hundred degrees to below 600°C while maintaining high ionic conductivity, to simplify the engineering design of the devices. Over the last few decades, considerable progress has been made with various dopant additions to modify the mobility and vacancy concentra- tions to achieve high conductivities at lower temperatures. Recent materials now demonstrate excellent properties at 600°C or lower, an enabling technology for solid state energy storage devices. Even better would be a good solid oxygen ion conductor like doped zirconia with a similar conductivity below 100°C!
A point worth noting with regard to ionic conductivity in something like doped zirconia is that the diffusion coefficient for oxygen can be written as (see Chapter 9)
D O f a e V
Q
RT O
2 1 2 m
6
− − ••
( )
= .13.9 Coupled Diffusion in Ionic Systems 461 This equation could be written as D O
( )
2− =DVO•• =VO•• DVO••(
ηVO•• ηO)
, where DVO•• is the vacancydiffusion coefficient for oxygen vacancies. Replacing D O
(
2−)
in Equation 13.24 with this expression, the Nernst–Einstein equation becomesσ η
O V V
B
ze D
k T
O O
=
( )
2 •• ••(13.26) implying that the electrical conductivity can be considered to be a result of either the oxygen ions moving or the oxygen vacancies moving. As long as the correct values for concentrations and dif- fusion coefficients are chosen, the results are the same. If the zirconia were indeed operating as the electrolyte in a cell with an oxygen chemical potential gradient, then the negatively charged oxygen ions would move one way in the cell—from high to low oxygen chemical potential—and the oxygen vacancies would move in the opposite direction, because they are positively charged relative to the crystal lattice.