Question

Molten $AlC{l}_3$ is electrolysed with the current of $0.5$ empere to produced 27 g $Al$. Number of Farady used are :

• A. $3$
• B. $5$
• C. $6$
• D. None of these

Answer 1

The correct option is A $3$

$A{l}^{3+}+3e^{-}\to Al$
From Faraday's laws of electrolysis, we know that
$W=\frac{EQ}{96500}$.
$\frac{Q}{96500}=\frac{W}{E}$.
$E=27/3=9$.
$\frac{Q}{96500}=27/9=3F$.
Hence option A is the right answer.