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

The quantity of charge (in Faraday) required to liberate 33.6 litre Cl2C{l_2}from molten NaClNaCl.

Explanation

Solution

To solve such questions we must understand that If n electrons are involved in electrode reaction, the passage of n faraday's (n×96500C)\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 Cl2C{l_2} from the molten NaClNaCl. So, the reaction becomes
2ClCl2(g)+2e2C{l^ - } \to C{l_2}_{\left( g \right)} + 2{e^ - }
We can see that we have a balanced reaction where Cl2C{l_2} is liberated. Now we can apply stoichiometry in this.
From the balanced reaction we obtain information such as, 2 moles of ClC{l^ - } is required to generate 1 mole of Cl2C{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 Cl2C{l_2} from 2 moles of ClC{l^ - } and 2 moles e{e^ - }, In other words we require 2×22.4L2 \times 22.4L volume of ClC{l^ - } to liberate 22.4L of Cl2C{l_2}.
If we want to liberate 33.6L of Cl2C{l_2} i.e., 1.5 moles of Cl2C{l_2} (33.622.4=1.5moles\dfrac{{33.6}}{{22.4}} = 1.5 moles) we need twice the number of moles of Cl2C{l_2} as per the stoichiometry discussed above.
number  of  moles  of  e  required  =  2×1.5=3{\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 Cl2C{l_2} from NaClNaCl.

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.