Question

The number of mole of electrons required to deposit 1 g equivalent aluminum (atomic wt = 27) from a solution of aluminum chloride will be:
A. 3
B. 1
C. 4
D. 2

Answer 1

Hint:. Equivalent weight of chemical can be calculated by using the following formula.
\[Equivalent Weight=\dfrac{MW}{acidity\text{ }or\text{ }basicity}\]
Here MW = Molecular weight of the chemical
Acidity or basicity means capacity to lose or gain electrons

Complete step by step answer:
- In the given it is given to calculate the number of moles of electrons required to deposit 1 g of equivalent aluminum from a solution of aluminum chloride.
- In the question it is given that the molecular weight of aluminum is 27.
- By using the below formula we can calculate the equivalent weight of aluminum.
\[Equivalent Weight=\dfrac{MW}{acidity\text{ }or\text{ }basicity}\]
Here MW = Molecular weight of the aluminum = 27
Acidity of the aluminum is 3 (means aluminum can lose three electrons easily)

- Substitute all the known values in the above formula to get the equivalent weight of aluminum.
\[\begin{align}
 & Equivalent Weight=\dfrac{MW}{acidity\text{ }or\text{ }basicity} \\
 & =\dfrac{27}{3} \\
 & =9 \\
\end{align}\]
- The aluminum chloride dissociates in the following way in a solution.
\[AlC{{l}_{3}}\to A{{l}^{3+}}+3C{{l}^{-}}\]

- Means aluminum loses three electrons and those electrons are gained by three chlorine atoms.
- Now to deposit pure aluminum form the solution we should provide 3 moles of electrons to aluminum ions produced in the solution.
- Therefore
\[A{{l}^{3+}}+3{{e}^{-}}\to Al\]

- Means three faraday of charges required to deposit one mole of aluminum.
- Three faraday of charge required to deposit 27 g of aluminum.
- One faraday of charge required to deposit 9 g of aluminum.
- Therefore one faraday of charge required to deposit one gram equivalent of aluminum.
- Thus, we need one mole of electron to deposit one gram equivalent of aluminum.
So, the correct answer is “Option B”.

Note: We are not supposed to take total molecular weight while calculating the number of moles of electrons required depositing 1 g equivalent of aluminum. We are supposed to consider the equivalent weight of aluminum to find the number of moles of electrons required to deposit 1 g equivalent of aluminum.