Question

If $9$ gm $H_{2}O$ is electrolysed completely with the current of $50\%$ efficiency then?

• A. $96500$ charge is required
• B. $2\times 96500$ C charge is required
• C. $5.6$ L of $O_2$ at STP will be formed
• D. $11.2$ L of $O_2$ at STP will be formed

Answer 1

The correct option is C $5.6$ L of $O_2$ at STP will be formed

Number of moles of water= $\frac{9}{18}=0.5$

Balanced equation for electrolysis of water

$2H_{2}O⟶2H_2+O_2$

$\therefore 2$ moles of $H_{2}O$ dissociates to produce $1$ mole of $O_2$ at STP.

$\therefore 0.5$ moles of $H_{2}O$ produces $0.25$ moles of $O_2$ at STP.

$\therefore 0.25$ moles of $O_2$ will occupy= $0.25\times 22.4=5.6litres$ .