Question

If 90 g of water is electrolyzed completely with 50% current efficiency.

$1)$ 10 Faraday of electricity will be consumed

$2$) 20 faraday of electricity will be consumed

$3)$ 168 L (STP) of gases will be produced

$4)$ 84 L (STP) of gases will be produced.

Select the correct options.

• A. $2,3$
• B. $1,2,3$
• C. $1,3,4$
• D. $1,2,3,4$

Answer 1

The correct option is A $2,3$

Moles of water $=\frac{90}{18}=5$
$H^{+}+e^{-}\to \frac{1}{2}{H}_2$

$2H^{+}+2e^{-}\to H_2$

$H_{2}O\to H_2+\frac{1}{2}{O}_2$
$\therefore$ 1 mole of $H_{2}O$ requires 2 mole electronic charge

Charge required $=2\times 5=10F$

Since, efficiency is 50%.
Charge required $=\frac{10}{\frac{1}{2}}=20F$
5 mole $H_{2}O$ produce $(\frac{1}{2}+1)\times 5=\frac{15}{2}molesgases$
$\therefore \text{Volume of gas at STP}=\frac{15}{2}\times 22.4=168L$