First slide

Faraday's Laws

Question

Electrolysis of aq. $K_{2} {SO}_{4}$ using inert electrodes is carried out by passing 19300 coulombs. Weight of the substance obtained at anode is

Easy

A

3.2 g
B

1.6 g
C

0.8 g
D

6.4 g

Solution

Sulphate ion can never get oxidized further (Sulphur is in its highest O.S).
Thus at anode O₂ gas is liberated;
$2 H_{2} O \to O_{2} + 4 H^{+} + 4 e;$
$∴ W_{O_{2}} = \frac{Eit}{F} = \frac{Eq}{F} W_{O_{2}} = \frac{8 \times 19300}{96500} W_{O_{2}} = 1.6 g$

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