302
5 – Applications
n
2F
Q
Cl 2 =
We thus deduce E
2F
RT
ln 2F
Q
V
RT
F
G
f T
Δ
Δ
=
−
#
°
c
m
b. Numerical evaluation
The charge Q that goes through the cell is
Q = I # t
Q = 10
−2 # 5 # 60
Q
C
3
=
and
E
96 480
109.8 10
2 96 480
8.314 298
ln 2 96 480
3
2 96 480
8.314 298
3
Δ =
+
#
#
#
#
#
#
#
`
j
The emf is
E 1.274 V
Δ =
c. For this integrator to operate continuously simply by reading the emf, it
requires
2 constant temperature. This requires a thermal management to counter
any Joule heating,
2 that the thermodynamic equilibria be in place.
d. The maximum charge Q max that should pass through the cell is obtained
from the relations established in 2(a)
P
n
V
RT
and n
F
Q
2
( )
Cl
Cl
Cl
2
2
2
2
=
=
or
Q
RT
FVP
2
( )
max
Cl
2
2
=
.
Q
8 314 298
2 96 480 10
2 10
max
6
5
=
−
#
#
#
#
#
.
Q
C
15 58
max =
5 – Applications
n
2F
Q
Cl 2 =
We thus deduce E
2F
RT
ln 2F
Q
V
RT
F
G
f T
Δ
Δ
=
−
#
°
c
m
b. Numerical evaluation
The charge Q that goes through the cell is
Q = I # t
Q = 10
−2 # 5 # 60
Q
C
3
=
and
E
96 480
109.8 10
2 96 480
8.314 298
ln 2 96 480
3
2 96 480
8.314 298
3
Δ =
+
#
#
#
#
#
#
#
`
j
The emf is
E 1.274 V
Δ =
c. For this integrator to operate continuously simply by reading the emf, it
requires
2 constant temperature. This requires a thermal management to counter
any Joule heating,
2 that the thermodynamic equilibria be in place.
d. The maximum charge Q max that should pass through the cell is obtained
from the relations established in 2(a)
P
n
V
RT
and n
F
Q
2
( )
Cl
Cl
Cl
2
2
2
2
=
=
or
Q
RT
FVP
2
( )
max
Cl
2
2
=
.
Q
8 314 298
2 96 480 10
2 10
max
6
5
=
−
#
#
#
#
#
.
Q
C
15 58
max =