First slide

Faraday's Laws

Question

If electrolysis of CuSO₄ solution is carried out with 100g impure Cu as anode of 90% purity and pure Cu of mass 1009 as cathode by passing 2 amp. current for 9650 sec. Then mass of cathode and anode will be: (At. mass of Cu = 63)

Easy

A

112.6 g and 112.6 g
B

106.3 g and 93 g
C

112.6 g and 86 g
D

172.6 g and 114 g

Solution

$Q_{F} = \frac{IE}{F} = \frac{2 \times 9650}{96500} = 0 .2$
At cathode Cu²⁺ + 2e^- → Cu 0.1 mole deposited
Weight of cathode = 100 +0.1 x 63
= 106.3 gm
At anode Cu + Cu²⁺ + 2e^-
$Weight of anode = 100 - 0 .1 \times 63 \times \frac{100}{90}$
= 93 gm

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