Question

A current of 0.75A is passed through an acidic solution of $Cu{SO}_4$ for 10 minutes. The volume of oxygen liberated at anode (at STP) will be:

Answer 1

Given:-

Time $\left(t\right)=10min=10\times 60=600s$

Current passed $\left(i\right)=0.75A$

From Faraday's law of electrolysis,

$Q=nF$

Whereas, $n$ is the no. of moles of electrons used

$\Rightarrow i\times t=nF(\because q=I\times t)$

$\Rightarrow n=\frac{i\times t}{F}$

$\Rightarrow n=\frac{0.75\times 600}{96500}=\frac{4.5}{965}\text{mol}$

Now,

$H_{2}O⟶4H^{+}+O_2+4e^{-}$

Amount of $O_2$ released at STP when $4$ mole of electrons are used $=22400mL$

Amount of $O_2$ released at STP when $\frac{4.5}{965}$ mole of electrons are used $=\frac{22400}{4}\times \left(\frac{4.5}{965}\right)=26.11mL$

Hence te volume of oxygen liberated at anode (at STP) will be $26.11mL$.