Question

A well stirred (1 L) solution of 0.1 M CuSO, is
electrolysed at 25°C using copper electrodes with
a current of 25 mA for 6 hours. If current efficiency
is 50%. At the end of the duration what would be
the concentration of copper ions in the solution?
(1) 0.0856 M.
(2) 0.092 M
(3) 0.0986 M.
(4) 0.1 M​

Answer 1

Using Faraday's law of electrolysis,

W=Zit

where W is the weight deposited and Z is the electrochemical equivalent and i is the current.

W=

2×96500

63.5

×25×10

−3

×6×3600×

2

1

W=0.089 gm

Moles deposited =1.398 milimoles

Initial moles of CuSO

4

=M×V=1×0.1=0.1 moles=100 milimoles

∴ Remaining moles of Cu

2+

in the solution =100−1.398=98.6 milimoles

Concentration of Cu

2+

in the solution (Molarity) =

1×1000

98.6

=0.0986 M

Answer 2

Answer: At the end of the duration 0.0986 M be the concentration of copper ions in the solution