Question

How long would it take to deposit $50$g of $Al$ from an electrolytic cell containing $A{l}_2{O}_3$ using a current of $105$ ampere?

• A. $1.54$ h
• B. $1.42$ h
• C. $1.32$ h
• D. $2.15$ h

Answer 1

The correct option is B $1.42$ h

Faraday's First Law of Electrolysis states that only, According to this law, the chemical deposition due to flow of current through an electrolyte is directly proportional to the quantity of electricity (coulombs) passed through it.

The reaction of Al is,

$A{l}^{3+}+3e^{-}\to Al$

Now as n=3, Equivalent weight of Al=$\frac{27}{3}=9$

As per Faraday's first law,

$W=Z\times I\times t⟹Z=\frac{Eq.wt}{96500}$

$⟹t=\frac{W\times 96500}{Eq.wt\times 1}=\frac{50\times 96500}{9\times 105}=5102.2$ or $1.42$ $hours$