Question

On passing electricity through dil.$H_2{SO}_4$ solution, the amount of substance liberated at the cathode and anode are in the ratio :

• A. $1:8$
• B. $8:1$
• C. $16:1$
• D. $1:16$

Answer 1

The correct option is A $1:8$

Electrolysis of dilute sulphuric acid:-

At cathode-

$2H^{+}+2e^{-}⟶H_2$

At anode-

$4{OH}^{-}+⟶2H_{2}O+2O_2+4e^{-}$

Overall reaction-

$H_{2}O⟶2H_2+O_2$

$\therefore$ Amount of substance liberated at cathode = 2 moles of $H_2=4g$

Also, amount of substance liberated at anode = 1 mole of $O_2=32g$

$\therefore$ Ratio of the amount of substance liberated at the cathode and anode = $4:32=1:8$