Acidified water was electrolysed using an inert electrode. The volume of gases liberated at STP was 168 ml. The amount of charge passed through the acidified water was: (in coloumb)

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Solution

$2 H_{2} O \to 2 H_{2} (g) 2 x + O_{2} (g) x$
$3 x = 168 or, x = 56 ml V_{H_{2}} = 2 x = 112 ml; V_{O_{2}} = x = 56 ml$
11200 ml of $H_{2}$ at STP = 1 F
112 ml of $H_{2}$ at STP = 0.01 F
5600 ml of $O_{2}$ at STP = 1 F
56 ml of $O_{2}$ at STP = 0.01 F
$Amount of electricity passed = 0.01 F = 0.01 \times 96500 = 965 C$
Note: We get identical results whether we consider $H_{2}$ or $O_{2}$ since the number of equivalents are the same throughout the reaction.

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