Question

Calculate the total volume (in L) of $O_2$ produced at STP, when a current of $30A$ is passed through a $K_{2}S{O}_4\left(aq\right)$ solution for 193 minutes using platinum electrodes?

• A. 20.16
• B. 40.3
• C. 60.48
• D. 80.64

Answer 1

The correct option is A 20.16

Electrolyis of aqueous $K_{2}S{O}_4$ with platinum electrodes will produce $H_2$ at cathode due to higher reduction potential of $H^{+}$ compare to $K^{+}$and $O_2$ at anode due to higher oxidation potential.
When 30 amperes is passed for 193 minutes, the quantity of current passed is,
$=193\times 60\times 30=347400$ coulombs.
$1\text{coloumb}=\frac{1}{96500}F$
$\therefore$
$\Rightarrow \frac{347400}{96500}=3.6\text{Faraday}$

$2H_{2}O\underset{K_{2}S{O}_4\left(aq\right)}{\overset{\text{Electrolysis}}{\to }}{O}_2+2H_2$

At cathode:
$2H_{2}O\left(l\right)+2e^{-}\rightleftharpoons H_2\left(g\right)+2O{H}^{-}\left(aq\right)$

So, on passing 2 Faraday of electricity, 1 mole of Hydrogen is produced.

At anode:
$2H_{2}O\left(l\right)\rightleftharpoons 4H^{+}\left(aq\right)+O_2\left(g\right)+4e^{-}$

On passing 4 Faraday of electricity, 1 mole of Oxygen is produced.
$\therefore$
On passing 3.6 Faradays of electricity, 1.8 moles of Hydrogen will be formed and $0.9$ moles of oxygen will be formed.
$4\text{Faradays gives}=1\text{mole of}{O}_2$
$3.6\text{Faradays gives}=\frac{3.6}{4}=0.9\text{mole of}{O}_2$

$\text{Volume of oxygen produced at STP}=0.9\times 22.4=20.16L$

$\text{Volume of hydrogen produced at STP}=1.8\times 22.4=40.32L$