Question

1 g equivalent of sodium metal is formed from electrolysis of fused ${\text{NaCl}}$. Moles of ${\text{Al}}$ formed from fused ${\text{N}}{{\text{a}}_{\text{3}}}{\text{Al}}{{\text{F}}_{\text{6}}}$ will be __________. When same charge used in electrolysis of molten ${\text{NaCl}}$ is crossed by ${\text{N}}{{\text{a}}_{\text{3}}}{\text{Al}}{{\text{F}}_{\text{6}}}$.
A.$\dfrac{1}{2}$
B.$\dfrac{1}{3}$
C.2
D.3

Answer 1

Hint:The Faraday’s laws of electrolysis state that the amount of chemical change produced by current at an electrode-electrolyte boundary is directly proportional to the current supplied.
The amount of the chemical changes produced by the same quantity of electricity in different substances is proportional to their equivalent weights.

Complete step by step answer:
The quantity of electricity that will cause a chemical change of one equivalent weight has been designated by one Faraday. It is equal to 96500 coulombs of electricity. The electrolysis of fused sodium chloride deposits 23 grams of sodium cations at the cathode and $35.5$ grams of chlorine gas at the anode.
By the Faraday’s law of electrolysis,
1 equivalent of sodium = 1 equivalent of aluminium.
$\left( {\dfrac{{\text{W}}}{{\dfrac{{{\text{at}}{\text{. wt}}{\text{.}}}}{{\text{n}}}}}} \right)$ of sodium = $\left( {\dfrac{{\text{W}}}{{\dfrac{{{\text{at}}{\text{. wt}}{\text{.}}}}{{\text{n}}}}}} \right)$ of aluminium
The oxidation state of aluminium is${\text{N}}{{\text{a}}_{\text{3}}}{\text{Al}}{{\text{F}}_{\text{6}}}$, is $\left( { + 3} \right)$.
Hence, at the electrode, one faraday of electricity will deposit $\dfrac{1}{3}$ grams of aluminium.

So, the correct answer is option B.

Note:
 The word “electrolysis” was introduced by Michael Faraday in the 19th century. Each electrode has an individual charge. The positive one being the anode that takes up the negative ions or the anions while the negative electrode being the cathode that takes up the positive ions or the cations.