Question

Oxygen Supply in Submarines Nuclear submarines can stay under water nearly indefinitely because they can produce their own oxygen by the electrolysis of water.

How many liters of O₂ at 298 K and 1.00 bar are produced in 1.50 hr in an electrolytic cell operating at a current of 0.0200 A?

Answer 1

Electrolysis of water is given below:

At cathode: 4H+(aq) + 4e- -----------> 2H2(g)

At Anode: 2H2O(l) -------> O2(g) + 4H+ + 4e-

1.5 hrs = 5400 sec

Total charge = Q = current * time (in sec)

Q = 0.02*5400 = 108 coulombs

We know that,

96500 Coulombs charge is carried by 1 mole electrons

So, 108 Coulomb charge is carried by 102/96500 moles of electrons.

So, moles of e- = 108/96500 = 0.00112 moles

From reaction at Anode, it is clear that 4 moles of e- produces 1 mole of O2

So, moles of O2 (n)= 0.00112/4 = 0.00028 moles

P = 1 bar = 0.987 atm , T = 298K , R = 0.0821 L atm/mol K

Let volume of O2 is V liters

Also,

PV = nRT then V = nRT/P

V = 0.00028*0.0821*298/0.987 = 0.00695 liters

OR

V = 0.007 Liters ...... Answer