Question

When 4.5 A current is pass through 1 litre of 0.6 M CuCl

2

solution for 1.15 hour, calculate mass

of Cu and change in concentration of solution. [Cu = 63.5u]​

Answer 1

Answer:

releted expalinatuon

Explanation:

The electrode reaction are:

1 mole

Cu

2+

+

2×96500 C

2e

−

→Cu(Cathode)

Cu×Cu

2+

+2e

−

(Anode)

Thus, cathode increases in mass as copper is deposited on it and the anode decreases in mass as copper from it dissolves.

Charge passed through cell =2.68×60×60 coulomb

Copper deposited or dissolved =

2×96500

63.5

×2.68×60×60=3.174 g

Increase in mass of cathode = Decrease in mass of anode =3.174 g.