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 litres of oxygen.

Tardigrade Admin · Accepted Answer

Hydrolysis of water: 2 H 2 O leftharpoons 4 H ++4 e-+ O 2 4 F charge will produce =1 mole of O 2=32 g O 2 1 F charge will produce =(32/4)=8 g of O 2

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

$22.4$ litres of oxygen.

Hydrolysis of water :

$2 H_{2} O ⇌ 4 H^{+} + 4 e^{-} + O_{2}$

$4 F$ charge will produce $= 1 m o l e$ of $O_{2} = 32 g O_{2}$

$1 F$ charge will produce $= \frac{32}{4} = 8 g$ of $O_{2}$

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

one mole of oxygen

one g atom of oxygen

8g of oxygen

22.4 litres of oxygen.

Hydrolysis of water : 2H2​O⇌4H++4e−+O2​ 4F charge will produce =1mole of O2​=32gO2​ 1F charge will produce =432​=8g of O2​

one $g$ atom of oxygen

$8 g$ of oxygen

$22.4$ litres of oxygen.

Hydrolysis of water :

$2 H_{2} O ⇌ 4 H^{+} + 4 e^{-} + O_{2}$

$4 F$ charge will produce $= 1 m o l e$ of $O_{2} = 32 g O_{2}$

$1 F$ charge will produce $= \frac{32}{4} = 8 g$ of $O_{2}$