Question

During electrolysis, 4828.4 mg of iodine requires 3671.3 coulombs of electricity for its deposition at the anode. Calculate the value of Faraday’s constant.

Answer 1

The reaction involved at anode is:

\[\ce{2I- -> \underset{1 mole}{I2} + \underset{2 moles}{2e-}}\]

The equation shows that one mole of I₂ requires 2 moles of electrons to deposit at the anode. Given that one mole of electrons carries one Faraday (1F) charge, the charge necessary to deposit one mole of I₂ is 2F.

In the current case,

Mass of I₂ = 4828.4 mg = 4828.4 × 10⁻³ g

Molecular mass of I₂ = 126.9 × 2 = 253.8

Moles of I₂ deposited = `(4828.4 xx 10^-3)/253.8`

= 0.0190244

Thus, 0.0190244 moles of I₂ require a charge of 3671.3 C.

∴ Charge required to deposit 1 mole of I₂ = `3671.3/0.0190244` = 192978.5 C

This charge must be equal to 2F. Thus,

2F = 192978.5 C

or, F = `192978.5/2`

= 96489.25 C

∴ The value of Faraday’s constant is 96489.25 C.