Verified Solution • 2025–26 Session moderate

QJEEM22Chda51: JEE Main 2022 Chemistry - Class 12 JEEM22Chda51 Solution: A solution of Fe 2 (SO 4 ) 3 is electrolyzed for 'x' min with a …

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

Step-by-Step Verified Solution

Fe3++3e−→Fe\mathrm{Fe}^{3+}+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe}

Moles of Fe deposited =0.348256=6.2×10−3

For 1 mole Fe\mathrm{Fe} , charge required is 3 F3 \mathrm{~F}

For 6.2×10−36.2 \times 10^{-3} mole Fe\mathrm{Fe} , charge required is 3×6.2×10−3 F3 \times 6.2 \times 10^{-3} \mathrm{~F}

Since, charge required =18.6×10−3×96500C=18.6 \times 10^{-3} \times 96500 \mathrm{C}

=1794.9C

And,

1.5×t=1794.9

t=1794.91.5×60min⁡t=\frac{1794.9}{1.5 \times 60} \min

t≃20 min

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring