Question

The number of electrons required to deposit 1g atom of Al (atomic mass=27) from a solution of $AlC{l_3}$ are:
(A) 1N
(B) 3N
(C) 2N
(D) 4N

Answer 1

Hint: Aluminium chloride, $AlC{l_3}$, is an ionic compound, containing both metallic and non-metallic ions. When $AlC{l_3}$ is subjected to electrolysis, it dissociates to form both positive and negative ions.
Electrolysis is the process of decomposition of ionic compounds to obtain its constituent elements when an electric current is passed through that compound in the molten form.
Therefore, on passing an electric current through molten $AlC{l_3}$, it will dissociate into $A{l^{ + 3}}$ and $C{l^ - }$ ions.

Complete step by step solution: Let us first understand what exactly happens during the electrolysis of $AlC{l_3}$. On dissociation of $AlC{l_3}$ during electrolysis, the ions $A{l^{ + 3}}$ and $C{l^ - }$ move independently in solution. The $A{l^{ + 3}}$ then undergoes reduction at cathode, depositing aluminium metal on the cathode.
The following half reaction taking place at the cathode.
$A{l^{ + 3}} + 3{e^ - } \to Al$
The $C{l^ - }$ formed get oxidised at anode, releasing chlorine gas.
The half reaction taking place at anode is shown below.
\[2C{l^ - } \to C{l_2} + 2{e^ - }\]
The chemical equation for overall reaction taking place is:
$AlC{l_3}\xrightarrow{{}}A{l^{3 + }} + 3C{l^ - }$
The electric current is nothing but the flow of electrons. To calculate the number of electrons required for deposition of 1g of Al from the solution of $AlC{l_3}$, we need to understand Faraday's second law of electrolysis.
According to the Faraday's second law of electrolysis, on passing an equal amount of charge or electricity through different electrolytes, the mass which will be deposited will be directly proportional to that of their equivalent weights.
If we consider the following reaction taking place at cathode, we get,
$A{l^{ + 3}} + 3{e^ - } \to Al$
This equation shows that 3 equivalents of aluminium are deposited at cathode and 3 mole of electrons gives 3 mole of aluminium.
To calculate the 1g equivalent of aluminium, we need to divide the atomic weight of aluminium by 3.
We will get,
$1equivalent\;of\;Al = \dfrac{{27g}}{{3g}} = 9g$
As per Faraday’s second law of electrolysis, 3g of charge deposits 27g of aluminium at cathode during electrolysis of $AlC{l_3}$.
Therefore, 1g of charge deposits 9g of aluminium at cathode during electrolysis of $AlC{l_3}$.
As we are aware of, 1F is equal to one mole of electrons.
From the above steps, we concluded that 3 mole electrons are required to deposit 1g atom Al.
We know that 1 mole $ = $ 1N and N represents Avogadro’s number.

Hence, the correct option is option (B).

Note: While solving the question, the most important point is to balance the chemical reaction. The chemical reaction of the cell should be balanced then only we will get the correct number of moles of the electron.
The Faraday’s law gives us following formula,
$F = N \times e$
Here, F represents Faraday’s constant which is equal to charge of 96320 coulombs. The letter N represents Avogadro’s number, equal to that of $6.023 \times {10^{23}}moles$ and e represents the charge on an electron.