During the electrolysis of acidulated water, the mass of hydrogen obtained is ‘x’ times that of O₂ and the volume of H₂ is ‘y’ times that of O₂. The ratio of ‘y’ and ‘x’ is

Electrolysis of acidulated water gives 2 volumes of hydrogen gas (H₂) and 1 volume of oxygen gas (O ₂). It can be shown by the following reaction.

$2{\mathrm{H}}_2\mathrm{O}\hspace{0.17em}\to 2{\mathrm{H}}_2 +\hspace{0.17em}{\mathrm{O}}_2$

Here, mass of hydrogen gas obtained is ‘x’ times that of oxygen gas, which can also be written as, mass of H₂​ = 'x'  mass of  O₂.

$\frac{\mathrm{Mass} \mathrm{of} {\mathrm{O}}_2}{\mathrm{Mass} \mathrm{of} {\mathrm{H}}_2}= \mathrm{x} \times \frac{2\times 2}{1\times 32}=\frac{1}{8}$

Also the volume of H₂ is ‘y’ times of O₂.

$$\begin{gathered} \frac{\mathrm{volume} \mathrm{of} {\mathrm{H}}_2}{\mathrm{volume} \mathrm{of} {\mathrm{O}}_2}= \mathrm{y}\frac{\mathrm{moles} \mathrm{of} {\mathrm{H}}_2}{\mathrm{moles} \mathrm{of} {\mathrm{O}}_2}=\frac{2}{1} \\ \mathrm{y}:\hspace{0.17em}\mathrm{x} = 2:\hspace{0.17em}\frac{1}{8}=16:1 \end{gathered}$$