Question

How many liters of O2 at 298 K and 1.00 bar are produced in 3.25 hr in an electrolytic cell operating at a current of 0.0200 A?

Answer 1

Sol: mass of any substance deposited or formed at any electrode is given by faradah first law.

Mass = M × I × t / nF

Where M is the molar mass, I is the current, t is the time in seconds, n is the no. Of e- change and F is the faraday constant.(96485 C)

Mass/ M = no. Of moles = I × t/ nF

2 O2- ---> O2 + 4e- ( n = 4 )

No. Of moles = 0.02 A ×( 3.25 × 60×60)/ ( 4 × 96485 C) =

0.000606312 moles if oxygen.

Now, PV = nRT

P = 1bar , V = ? , n = 0.000606312 moles, R = 0.083 Lbar/mol.K , T = 298K

V = 0.000606312 × 0.083 × 298 / 1 = 0.015 L of oxygen gas = 15 ml