JEE MAIN - Chemistry (2022 - 26th July Morning Shift - No. 21)
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
3
Explanation
For $$\mathrm{Fe}_{3} \mathrm{O}_{4}$$,
$$ x=\frac{+8}{3} $$
where x is oxidation state of Fe.
$$ \mathrm{Fe}_{3} \mathrm{O}_{4}+8 \mathrm{H}^{+}+8 \mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{H}_{2} \mathrm{O} $$
Charge required $$=\frac{8}{3} \times \mathrm{F}=\frac{8 \mathrm{~F}}{3} \simeq 3 \mathrm{~F}$$
$$ x=\frac{+8}{3} $$
where x is oxidation state of Fe.
$$ \mathrm{Fe}_{3} \mathrm{O}_{4}+8 \mathrm{H}^{+}+8 \mathrm{e}^{-} \longrightarrow 3 \mathrm{Fe}+4 \mathrm{H}_{2} \mathrm{O} $$
Charge required $$=\frac{8}{3} \times \mathrm{F}=\frac{8 \mathrm{~F}}{3} \simeq 3 \mathrm{~F}$$
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