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

Answer: (C)

According to first law of faraday,
$<br/>\frac{Wt}{\text{ Ewt }}=\frac{Q}{F}$
$\frac{Q}{F}=$ farady No.
$\therefore$ charge of 1 mol of electrons is equal to 1 farady.
$\frac{Wt}{\text{ Ewt }}=\frac{\text{ ne }}{F}\therefore Q=$ 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 $=44.8+22.4=67.2$