JEE Main 2022 · 25 Jun · Shift 2 · Chemistry Q85

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ChemistryInteger+4 / -1Question 85 of 90

A solution of $\mathrm{Fe_2(SO_4)_3}$ 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 $^{-1}$

Your answer

Answer and worked solution

Official answer: 20

Charge. $\mathrm{Fe^{3+}+3e^-\rightarrow Fe}$ : $\dfrac{0.3482}{56}\times3\times96500\approx1800$ C.

Time. $t=\dfrac{1800}{1.5}=1200$ s $=20$ min.

Answer: 20

Topics:Electrolysis and Faraday's laws

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JEE Main 2022 · 25 Jun · Shift 2, Chemistry question 85