Question

A current of 10 A is passed for 80 min. and 27 seconds through a cell containing dilute sulphuric acid.

1. How many moles of oxygen gas will be liberated at the anode?
2. Calculate the amount of zinc deposited at the cathode when another cell containing ZnSO₄ solution is connected in series (Zn = 65).

Answer 1

Given: Current (I) = 10 A

Time (t) = 80 min + 27 sec = 80 × 60 + 27 = 4827 sec

Zn atomic mass = 65 g/mol

Faraday’s constant (F) = 96500 C mol⁻¹

Q = I × t

= 10 × 4827

= 48270 C

i. In the electrolysis of dilute H₂SO₄, oxygen is liberated at the anode via:

\[\ce{4OH- -> O2 + 2H2O + 4e-}\]

This shows that 4 moles of electrons are needed to produce 1 mole of O₂.

So,

Moles of O₂ = `Q/(4 F)`

= `48270/(4 xx 96500)`

= `48270/386000`

= 0.125 mol

ii. In the second cell:

\[\ce{Zn^{2+} + 2e- −> Zn}\]

So, 2 Faradays deposit 65 g of Zn, i.e.,

Mass of Zn = `(65 xx Q)/(2 xx F)`

= `(65 xx 48270)/(2 xx 96500)`

= `3137550/193000`

= 16.25 g