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

Question

In the electrolysis of water, one faraday of electrical energy would evolve (A) one mole of oxygen (B) one g atom of oxygen (C) 8 g of oxygen (D) 22.4 L of oxygen

Tardigrade Admin · Accepted Answer

Hydrolysis of water can be represented by the following equations 2 H 2 O leftharpoons O 2+4 H ++4 e- ∵ 4 faraday of charge liberate O 2=32 g ∴ 1 faraday of charge liberate O 2=(32/4)=8 g

Q. In the electrolysis of water, one faraday of electrical energy would evolve

one mole of oxygen

one g atom of oxygen

8 g of oxygen

22.4 L of oxygen

Hydrolysis of water can be represented by the following equations
$2 H_{2} O ⇌ O_{2} + 4 H^{+} + 4 e^{-}$
$∵ 4$ faraday of charge liberate $O_{2} = 32 g$
$∴ 1$ faraday of charge liberate $O_{2} = \frac{32}{4} = 8 g$

Q. In the electrolysis of water, one faraday of electrical energy would evolve

one mole of oxygen

one g atom of oxygen

8 g of oxygen

22.4 L of oxygen

Hydrolysis of water can be represented by the following equations 2H2​O⇌O2​+4H++4e− ∵4 faraday of charge liberate O2​=32g ∴1 faraday of charge liberate O2​=432​=8g

Hydrolysis of water can be represented by the following equations
$2 H_{2} O ⇌ O_{2} + 4 H^{+} + 4 e^{-}$
$∵ 4$ faraday of charge liberate $O_{2} = 32 g$
$∴ 1$ faraday of charge liberate $O_{2} = \frac{32}{4} = 8 g$