Question

What is the ratio of a volume of the gases liberated at STP at the cathode and the anode when water acidulated with $H_2{SO}_4$ is electrolyzed?

• A. $2:1$
• B. $1:2$
• C. $1:1$
• D. $1:8$

Answer 1

The correct option is A $2:1$

$H_{2}S{O}_4⟶2H^{+}+S{O}_4^{2-}$

$1$ volume of $H_{2}S{O}_4\equiv 2$ volume of $H^{+}\equiv 1$ volume of $S{O}_4^{2-}$

$H^{+}$ will be deposited on cathode

and $S{O}_4^{2-}$ will be deposited on anode

$\therefore \frac{Volumeofgasliberatedatcathode}{Volumeofgasliberatedatanode}=\frac{V_H^{+}}{V_{S{O}_4^{2-}}}$

$=\frac{2V}{V}$

$2:1$