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Q.

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

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Detailed Solution

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${Fe}^{3 +} + 3 e^{-} ⟶ Fe$

$3 F ⟶ 1 mole Fe is deposited$

$For 1 g ⟶ \frac{3 \times 96500}{56} (required charge)$

$For 0 .3482 g ⟶ \frac{3 \times 96500}{56} \times 0 .3482 = 1800 .06$

$Q = it 1800.06 = 1.5 t$

$t = 20 \min$

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