Q. When 1 F of electricity is passed through acidulated water, $O_{2}$ evolved is
A) $1.0\ {dm}^{3}$
B) $5.6\ {dm}^{3}$
C) $11.2\ {dm}^{3}$
D) $22.4\ {dm}^{3}$
Explanation
(b)
Reaction for electrolysis of water is
$2H_{2}O\ \rightleftharpoons 4H^{+} + 2O^{2 -}$
$2O^{2 -} \rightarrow O_{2} + 4e^{-}$
$4e^{-} + 4H^{+} \rightarrow 2H_{2}$
$\therefore n = 4,$
So, 4 F charge liberates = 1 mol = 22.4 ${dm}^{3}$ oxygen
$\therefore\$1 F charge will liberate$\ = \frac{22.4}{4} = 5.6\ {dm}^{3}\ oxygen$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.