Question

Question 14:
The weight of substance deposited is
0.450 g,calculate current in amperes when
the process of electrolysis is carried for 20
minutes. Given ECE=0.000306
7.352 Ampere
73.52 Ampere
1.22 Ampere
0.122 Ampere​

Answer 1

Answer:

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.

Answer 2

Answer:

W=Zit= 0.000304×2.5×20×60=0.000304×300

= 0.0912g