Question

If $9gm{H}_{2}O$ is electrolysed completely with the current of $50$% efficiency then :

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

Answer 1

The correct option is B $2\times 96500C$ charge is required

$2\times 96500C$ charge is required

18 g of $H_{2}O=1\text{mole}$

9 g of $H_{2}O=\frac{1}{18}\times 9=0.5\text{mole}$

$\underset{0.5\text{mole}}{H_{2}O}⟶\underset{0.5\text{mole}}{H_2}+\underset{0.25\text{mole}}{\frac{1}{2}{O}_2}$

for $\frac{1}{4}$ mole of $O_2$,

volume $=22.4\times \frac{1}{4}=5.6L$ at STP

Now for the given equation,

No. of moles of electron invoved for 1 mole of $H_{2}O=2\text{mole}$

$\therefore$ No. of moles of electron invoved for $\frac{1}{2}$ mole of $H_{2}O=1\text{mole}=1F=96500C$

$\because$ current efficiency $=50\left(Given\right)$

$\therefore$ charge required $=2\times 96500C$

Hence, $2\times 96500C$ charge is required.