Question

The quantity of charge (in Faraday) required to liberate 33.6 litre $C{l_2}$from molten $NaCl$.

Answer 1

Hint: To solve such questions we must understand that If n electrons are involved in electrode reaction, the passage of n faraday's $\left( {n \times 96500C} \right)$ of electricity will deposit or liberate 1 mole of the substance.

Complete step by step solution:
We know that one faraday means 96500 coulombs of charge or the charge carried by one mole of electrons. As we said in the hint, if one electron is involved in electrode reaction, then if we pass an electricity of 96500 C of charge we can deposit or liberate 1 mole of substance in that.
Now, coming to our question we can see that we have to liberate 33.6 litre $C{l_2}$ from the molten $NaCl$. So, the reaction becomes
$2C{l^ - } \to C{l_2}_{\left( g \right)} + 2{e^ - }$
We can see that we have a balanced reaction where $C{l_2}$ is liberated. Now we can apply stoichiometry in this.
From the balanced reaction we obtain information such as, 2 moles of $C{l^ - }$ is required to generate 1 mole of $C{l_2}$. Also there are 2 mole electrons involved in the reaction.
And we know that 1 mole of substance contain a volume of 22.4L. Here we have liberated 1 mole $C{l_2}$ from 2 moles of $C{l^ - }$ and 2 moles ${e^ - }$, In other words we require $2 \times 22.4L$ volume of $C{l^ - }$ to liberate 22.4L of $C{l_2}$.
If we want to liberate 33.6L of $C{l_2}$ i.e., 1.5 moles of $C{l_2}$ ($\dfrac{{33.6}}{{22.4}} = 1.5 moles$) we need twice the number of moles of $C{l_2}$ as per the stoichiometry discussed above.
${\text{number}}\;{\text{of}}\;{\text{moles}}\;{\text{of}}\;{{\text{e}}^ - }\;{\text{required}}\; = \;2 \times 1.5 = 3$
From Faraday's first law of electrolysis we required 3 faradays of charge to liberate 33.6 litre of $C{l_2}$ from $NaCl$.

Note: If we pass the same quantity of electricity through solutions of different electrolytes connected in series, the amount of different substances deposited at the electrodes is directly proportional to their equivalent weights.