Question

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 1

Correct Answer: 20

Fe³⁺ + 3e^– → Fe

Moles of Fe deposited

\(=\frac{0.3482}{56}=6.2×10^{-3}\)

For 1 mole Fe, charge required is 3 F

For 6.2 × 10^–3 mole Fe, charge required is

3 × 6.2 × 10^–3 F

Since, charge required = 18.6 × 10^–3 × 96500 C

= 1794.9 C

And,

1.5 × t = 1794.9

\(t=\frac{1794.9}{1.5×60}\) min

t ≃20 min