Solutions to exercises
247
for the case of ZnO, we obtain
s
95 mV u
expt
pO
1 2
.
− −
where u pO
1 2
− − denotes the units of pO
2−
.
The result of question 2 allows us to calculate the theoretical slope s th .
.
s
F
RT
2
2 3
th =
.
.
s
2 96 480
2 3 8 314 723
th =
#
#
#
.
s
m V u
71 7
th
pO
1 2
=
− −
We observe that s expt ≈ s th , which validates the expression for the emf
as a function of pO
2−
established in question 2.
c. The solubility equilibria and the corresponding constants
2 for ZnO are
ZnO (s) m Zn
2+ + O
2−
with
K s (ZnO) = [Zn
2+
] [O
2−
]
s(ZnO) = [O
2−
]
where s denotes the solubility.
2 for Li 2 O are
Li 2 O (s) m 2Li
+ + O
2−
with
K s (Li 2 O) = [Li
+
]
2
[O
2−
]
and
s (Li 2 O) = [O
2−
]
d. From figure 83 the solubilities of ZnO and Li 2 O may be read from the
respective intersections marking the onset of precipitation.
2 For ZnO
log s(ZnO) = pO
2−
sat = − 1.8
s 1.58 10 m l kg
2
1
#
ο
=
−
−
2 For Li 2 O
log s(Li 2 O) = pO
2−
sat = − 1.95
s 1.12 10 m l kg
2
1
#
ο
=
−
−
Solution 5.4 – Calculation of equilibrium constants for defect
formation in Cu 2 O
1. 2 Formation of singly ionized copper vacancies
4
1
O 2(g) m 2
1
O
#
O + V ′
Cu + h
•