A solution of Fe2(SO4)3 is electrolyzed for 'x' min with a current of 1.5 A to deposit 0.3482 g of Fe. The value of x is ___________. [nearest integer]
Given : 1 F = 96500 C mol$-$1
Atomic mass of Fe = 56 g mol$-$1
Answer (integer)
20
Solution
$\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$
<br/><br/>
$\text {Moles of Fe deposited }=\frac{0.3482}{56}=6.2 \times 10^{-3}$
<br/><br/>
For 1 mole $\mathrm{Fe}$, charge required is $3 \mathrm{~F}$
<br/><br/>
For $6.2 \times 10^{-3}$ mole $\mathrm{Fe}$, charge required is $3 \times 6.2 \times 10^{-3} \mathrm{~F}$
<br/><br/>
Since, charge required $=18.6 \times 10^{-3} \times 96500 \mathrm{C}$
<br/><br/>
$=1794.9 \mathrm{C}$
<br/><br/>
And,
<br/><br/>
$1.5 \times \mathrm{t}=1794.9$
<br/><br/>
$t=\frac{1794.9}{1.5 \times 60} \min$
<br/><br/>
$\mathrm{t} \simeq 20 \mathrm{~min}$
About this question
Subject: Chemistry · Chapter: Electrochemistry · Topic: Electrochemical Cells
This question is part of PrepWiser's free JEE Main question bank. 127 more solved questions on Electrochemistry are available — start with the harder ones if your accuracy is >70%.