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: $2 H ^{+}+2 e ^{-} \rightarrow H _{2}$

It $2 F$ charge flows $=2 F \rightarrow 2 gH _{2}$

$\Rightarrow 2=x \times 16$

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

At Anode: $4 OH ^{-} \rightarrow 2 H _{2} O + O _{2}+4 e ^{-}$

$4 F$ deposits $32\, g\, O _{2}$

$2F\, $ deposits $ = \frac{32}{4} \times2O_{2} =16 gm$

$\Rightarrow 22.4L =y \times11.2L$

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

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