The amount of charge in $\mathrm{F}$ (Faraday) required to obtain one mole of iron from $\mathrm{Fe}_{3} \mathrm{O}_{4}$ is ___________. (Nearest Integer)
Answer (integer)
3
Solution
For $\mathrm{Fe}_{3} \mathrm{O}_{4}$,
<br/><br/>
$x=\frac{+8}{3}$
<br/><br/>
where x is oxidation state of Fe.
<br/><br/>
$$
\mathrm{Fe}_{3} \mathrm{O}_{4}+8 \mathrm{H}^{+}+8 \mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{H}_{2} \mathrm{O}
$$
<br/><br/>
Charge required $=\frac{8}{3} \times \mathrm{F}=\frac{8 \mathrm{~F}}{3} \simeq 3 \mathrm{~F}$
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%.