Question

What will be the volume of $O_2$ liberated if 2.5 litres of $H_2$ is liberated at cathode during the electrolysis of water?

• A. 1 L
• B. 1.25 L
• C. 1.5 L
• D. 2 L

Answer 1

The correct option is B 1.25 L

We know that v/v of gas obtained at anode to that of cathode is 1 : 2

So from above we get, $\frac{1}{2}=\frac{x}{2.5}$

$\Rightarrow x=\frac{2.5}{2}$

$\Rightarrow x=1.25$

Thus,1.25 L volumes of $O_2$ will be liberated at anode.