2.
(a) A current of 1 ampere flows in an electrolytic cell. In how many seconds
will Avogadro’s number of electrons pass through the cell?
(b) A plate of area 1 m2 is to be covered with a 1 ¹m layer of Zn by electrolytic
deposition, using a current of 1 ampere. Estimate how long the process will take.
Find in tables the density (or the lattice constant) of metallic Zinc, and assume
that the Zn atoms are deposited as Zn++ ions.
A one ampere current corresponds to the following number of electrons per
sec.
1 amp =
Ã
1 coul
sec
!
=
Ã
!
6:242 £ 1018 e
:
sec
So then the number of second it takes for Avogadro’s number of electrons to
pass through a cross sectional area when there is one ampere of current is
6:022 £ 10
23
µ
¶
sec
e
= 9:648 £ 104 sec:
6:242 £ 1018 e
ZnCl
2
++
_
+
2 Cl
source of zinc
Zn
plate
To cover the plate requires 10¡6 m3 of Zn, which at a density of 6:57 £
103 kg=m3 corresponds to 6:57 £ 10¡3 kg of Zn. Since Zn has an atomic mass of
69:723, this corresponds to 9:423 £ 10¡2 moles of Zn or 5:675 £ 1022 atoms of Zn.
If the Zn ions are doubly charged, this corresponds to 1:135 £ 1023 elementary
charges that have to be moved. Depositing that many charges using an ampere
of current should take
1:135 £ 1023 e
µ
¶
sec
= 1:818 £ 104 sec = 5:051 hours:
6:242 £ 1018 e
(But if the negative ions are moving, don’t we need 2:27 £1023 elementary charges
above? No, we need both the positive and negative ions to complete the circuit,
but the current is the same throughout the circuit and near the plate the current
carried entirely by Zn++ .)
2