Question

On electrolysis of dil. $H_{2}S{O}_4$ with platinum electrodes, 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:
In dilute sulphuric acid, we have $H^{+},O{H}^{-}\text{and}S{O}_4^{2-}ions.$

$H^{+}$ will migrate to the cathode and $O{H}^{-},S{O}_4^{2-}$ migrates to the anode.

At cathode:
$2H^{+}+2e^{-}\to H_2.....\left(1\right);E^0=0V$

At Anode:
Possible reactions:
$2S{O}_4^{2-}\left(aq\right)\rightleftharpoons S_2{O}_8^{2-}\left(aq\right)+2e^{-};E_{red}^0=2.0V$

$2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}+O_2\left(g\right)+4e^{-};E_{red}^0=1.23V$

Hence, $H_{2}O$ has lower reduction, it will undergo oxidation at anode and liberates $O_2$.

Actual reaction at anode:
$2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}+O_2\left(g\right)+4e^{-}...\left(2\right)$

Multiplying equation (1) by 2 and adding to equation (2) gives overall reaction.
Overall reaction:
$2H_{2}O\to 2H_2+O_2$

2 moles of $H_{2}O$ electrolyse to give 2 moles of $H_2$and 1 mole of $O_2$

$\therefore$
$\text{No. of moles of}{H}_2\text{liberated at cathode}=2$
$\text{Amount of}{H}_2\text{liberated at cathode}=4g$

$\text{No. of moles of}{O}_2\text{liberated at anode}=1$
$\text{Amount of}{O}_2\text{liberated at anode}=32g$
$\therefore$
$\text{Ratio of the amount of substance liberated at the cathode and anode}=4:32\Rightarrow 1:8$