Question

During the electrolysis of Acidulated water, the mass of hydrogen obtained is $x$ times that of $O_2\&$ the volume of $H_2$ is y times that of $O_2$ at STP. Calculate the ratio y: $x$.

• A. 16: 1
• B. 1: 16
• C. 12: 1
• D. 1: 12

Answer 1

Answer: (A)

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

It $2F$ charge flows $=2F\to 2g{H}_2$

$\Rightarrow 2=x\times 16$

$x=\frac{1}{8}$

At Anode: $4O{H}^{-}\to 2H_{2}O+O_2+4e^{-}$

$4F$ deposits $32g{O}_2$

$2F$ deposits $=\frac{32}{4}\times 2O_2=16gm$

$\Rightarrow 22.4L=y\times 11.2L$

$y=\frac{2}{1}$

Ratio $y:x=\frac{\frac{2}{1}}{\frac{1}{8}}=16:1$