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In the electrolysis of water, one faraday of electrical energy would liberate

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one mole of oxygen

one gram atom of oxygen

8 g oxygen

22.4 lit. of oxygen

answer is C.

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Detailed Solution

According to the definition 1 F or 96500 C is the charge carried by 1 mol of electrons when water is electrolysed
$2 H_{2} O \to 4 H^{+} + O_{2} + 4 e^{-}$
So, 4 Faraday of electricity liberate = 32 g of O₂.
Thus 1 Faraday of electricity liberate

$= \frac{32}{4} g o f O_{2} = 8 g o f O_{2}$

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