Question

What amount of $C{{l}_{2}}$ gas liberated at anode if 1 ampere current is passed for 30 minutes from $NaCl$ solution?
A. 0.66mole
B. 0.33mole
C. 0.66 g
D. 0.33g

Answer 1

Hint This question is based on Faraday’s first law of electrolysis.
Hint: 1 faraday charge always deposits 1 equivalent of an element on an electrode and 1 faraday charge is equal to the charge present on 1 mole electron.

Complete Step by step solution:
Faraday’s first law of electrolysis states that
Mass of element deposited on an electrode will be directly proportional to the charge pass through the electrolyte solution.
\[W\propto Q\]
\[W=ZQ\]
\[W=ZIt\]
Here: proportionality constant Z is the electrochemical equivalent of the element, from the experiment its value comes as follows
\[Z=\dfrac{E}{96500}\]
E is the equivalent weight of the element.
\[W=\dfrac{E}{96500}Q\]
96500 C is the faraday constant and it is called as 1 faraday charge.
If 96500 C charge is passed through an electrolyte solution then
\[W=\dfrac{E}{96500}\times 96500\]
\[W=E\]
It means when we pass 96500 C charges or 1 faraday charge from an electrolyte solution, the deposited weight of the element will be equal to its equivalent weight.
Let’s calculate the charge on 1 mole electrons
Charge on 1 electron=$1.6\times {{10}^{-19}}$
Charge on 1 mole electrons=$6.022\times {{10}^{23}}\times 1.6\times {{10}^{-19}}$
Charge on 1 mole electrons=$9.6352\times {{10}^{4}}$
Charge on 1 mole electrons= 96352 C
This 96352 C is approximately equal to 96500 C, that’s why we say charge on 1 mole electrons is equal to 1 faraday.
Magnitude of charge passed through the solution
\[Q=I\times t\]
Here current is in ampere and time is in second.
\[Q=1\times 30\times 60\]
\[Q=1800C\]
Now let’s convert this charge into faraday
1 faraday=96500 C
Charge passed through solution=$\dfrac{1800}{96500}F$
Charge passed through solution= 0.01865 faraday
As we know that 1 faraday charge discharges 1 equivalent of substance, therefore 0.01865faraday will discharge 0.01865 equivalent of chloride ion.
Equivalents of chloride ion discharged= 0.01865
     \[\] \[\]
This is the reaction of discharging chloride ions at anode.
\[C{{l}^{-}}\to \dfrac{1}{2}C{{l}_{2}}+{{e}^{-}}\] \[\]
According to this reaction valency factor of chloride ion will be 1 as charge on chloride ion is -1
Number of moles=number of equivalents
Number of moles of chloride ions= 0.01865
According to reaction stoichiometry 1 mole chloride ion discharge will liberate half mole chlorine gas, so
Number of moles of chlorine gas liberated=$\dfrac{0.01865}{2}$
Number of moles of Chlorine gas liberated=$9.325\times {{10}^{-3}}$
Gram molecular weight of chlorine gas= 71g
Weight of chlorine liberated=$n\times M$
Weight of chlorine liberated=$9.325\times {{10}^{-3}}\times 71$ g
Weight of chlorine liberated= 0.66g

So, correct answer is Option(C)

Additional information:
Electrolysis is a very important process in metallurgy. This process is often being used to get ultra pure metals. For purification of different metals different processes are available. For copper we do electrolytic refining and for Aluminum we have hall-heroult process and serpek process and for some highly reactive metals like sodium we directly get them by electrolysis of their ore like rock salt.

Note: Equivalents of the elements get deposited on both the electrodes are always equal and this is actually the faraday’s second law of electrolysis. So if we have the information about the quantity of one element deposited we can actually calculate the other element's quantity deposited or discharged on the other electrode.