Question

A solution of $F{e}_2{\left(S{O}_4\right)}_3$ is electrolyzed for ' $x$ ' min with a current of $1.5$ A to deposit $0.3482g$ of $Fe$. The value of $x$ is____ [nearest integer]
Given : $1F=96500Cmo{l}^{-1}$
Atomic mass of $Fe=56gmo{l}^{-1}$

Answer 1

Answer: 20

$F{e}^{3+}+3e^{-}⟶Fe$
$3F⟶1$ mole $Fe$ is deposited
For $56g⟶3\times 96500$ (required charge)
For $1g⟶\frac{3\times 96500}{56}$ (required charge)
For $0.3482g⟶\frac{3\times 96500}{56}\times 0.3482$
$=1800.06$
$Q=$ it
$1800.06=1.5t$
$t=20\min$