JEE MAIN - Chemistry (2022 - 25th June Evening Shift - No. 19)
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
20
Explanation
$\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}$
$$ \text {Moles of Fe deposited }=\frac{0.3482}{56}=6.2 \times 10^{-3} $$
For 1 mole $\mathrm{Fe}$, charge required is $3 \mathrm{~F}$
For $6.2 \times 10^{-3}$ mole $\mathrm{Fe}$, charge required is $3 \times 6.2 \times 10^{-3} \mathrm{~F}$
Since, charge required $=18.6 \times 10^{-3} \times 96500 \mathrm{C}$
$$ =1794.9 \mathrm{C} $$
And,
$$ 1.5 \times \mathrm{t}=1794.9 $$
$t=\frac{1794.9}{1.5 \times 60} \min$
$$ \mathrm{t} \simeq 20 \mathrm{~min} $$
$$ \text {Moles of Fe deposited }=\frac{0.3482}{56}=6.2 \times 10^{-3} $$
For 1 mole $\mathrm{Fe}$, charge required is $3 \mathrm{~F}$
For $6.2 \times 10^{-3}$ mole $\mathrm{Fe}$, charge required is $3 \times 6.2 \times 10^{-3} \mathrm{~F}$
Since, charge required $=18.6 \times 10^{-3} \times 96500 \mathrm{C}$
$$ =1794.9 \mathrm{C} $$
And,
$$ 1.5 \times \mathrm{t}=1794.9 $$
$t=\frac{1794.9}{1.5 \times 60} \min$
$$ \mathrm{t} \simeq 20 \mathrm{~min} $$
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