Solutions to exercises
269
The molar mass of Ag 2 SO 4 is
M Ag 2 SO 4 = (2 # 107.9) + 32.1 + (4 # 16) = 311.9 g mol
−1
We deduce the number of moles of Ag 2 SO 4
n
311.9
1.078 10
3.46 10 mol
Ag SO
4
7
2
4
=
=
−
−
#
#
The time t required for all the Ag 2 SO 4 to disappear is
t
I
Q
I
2F n Ag SO
2
4
=
=
#
t
1.133 10
2 96 480 3.46 10
6
7
=
−
−
#
#
#
#
t 58 927 s 16.36 h
=
=
We repeat the same calculation for Nb 2 O 4 . The volume of Nb 2 O 4 is
V
4
0.5
0.1 4
1
4.91 10 cm
Nb O
2
3
3
2 4
π
=
=
−
#
#
#
#
The molar mass of Nb 2 O 4 is
M Nb 2 O 4 = (2 # 92.9) + (4 # 16) = 249.8 g mol
−1
The number of moles of Nb 2 O 4 is
n
249.8
4.91 10
5, 9 1.16 10 mol
Nb O
3
4
2 4
=
=
−
−
#
#
#
The time t ′ required for all the Nb 2 O 4 to disappear is
t
I
2F n Nb O
2 4
=
#
l
.
.
t
1 133 10
2 96 480 1 16 10
6
4
=
−
−
#
#
#
#
l
t
1.97 10 s 5488 h
7
=
=
#
l
c. The results show that the amount of Ag 2 SO 4 is what limits the lifetime
of the sensor.
d. The use of a voltage shifter allows us to decrease the current I. For example, if we shift the voltage by 1 V, we would have
ΔE ′ = − 1.133 + 1 = − 0.133 V
2 The current is approximately divided by 10.
2 We work at a more sensitive scale.