Question

Four moles of electrons were transferred from anode to cathode in an experiment on electrolysis of water. The total volume. of the two gases (dry and at STP) produced will be approximately: (in litres)

• A. $22.4$
• B. $44.8$
• C. $67.2$
• D. $89.4$

Answer 1

The correct option is C $67.2$

According to first law of faraday,

$\frac{Wt}{Ewt}=\frac{Q}{F}$

$\frac{Q}{F}=$ farady No.

$\therefore$ charge of 1 mol of electrons is equal to 1 farady.

$\frac{Wt}{Ewt}=\frac{ne}{F}\therefore Q=ne$ and $n=4$

$mole\times V.f.=\frac{4F}{F}$

$mole\times 2=4$

mols of $H_{2}O=2$

Hence the two moles of $H_{2}O$ will decomposed into it's gaseous components

$2H_{2}O⟶2H_2+O_2$

$2moles$ of $H_2$(at cathode) and $1mol$ of $O_2$(at anode) will produce.

$moles=\frac{v}{22.4}$

So volume of $H_2$ and $O_2$ will be = 44.8+22.4 = 67.2