Solutions to exercises
145
Assuming that the Na
+
activity is constant in the vitreous electrolyte
and that the Na activity in the mixture is equal to its molar fraction, we
obtain
E
F
RT
ln x
x
Na
(1)
Na
(2)
Δ =
b. Numerical evaluation gives
E
96 480
8.314 473
ln 2 10
10
–3
–5
#
Δ =
#
ΔE = − 216 mV
5. When the potential difference between the cell terminals goes to zero, the
sodium activity in the mixture is the same in compartments 1 and 2. Under
these conditions we can write
Δx Na = x
(1)
Na − x
f
Na = x
(2)
Na + x
f
Na
where x
f
Na denotes the molar fraction of sodium in each mixture when ΔE
is zero.
From this we deduce
x
x
x
2
( )
( )
Na
f
Na
Na
1
2
=
−
.
x
2
2 10
10
9 95 10
Na
f
3
5
4
#
#
=
−
=
−
−
−
and
Δx Na = x
(1)
Na − x
f
Na
Δx Na = 2 # 10
−3 − 9.95 # 10
−4
x
1.005 10
Na
3
#
Δ
=
−
We now show that the number Δn Na of Na moles transfered from compartment 1 to compartment 2 is given by
Δn Na = 10
−2
Δx Na
Using
Q = I # t = Δn Na # F
to denote the electric charge that passes through the cell, we obtain
I
t
10
x F
2
Na
Δ
=
−
or
.
I
60
10
1 005 10
96 480
2
3
#
=
−
−
#
#
.
I
mA
16 2
=