Question

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

Answer: 0.0069L

Explanation:

2H2O(l) ---->O2(g) + 4H+(aq) + 4e-

no of moles= it/eF

NO of moles of O2 produced = (Current in Ampere x Time in second)/ (Faraday constant x Number of electrons required)

Moles of O2 produced = (0.02x (60 x 60X1.5 s)/(96485 x 4)

= 0.0002798 moles= 2.798x 10 ^-4moles

Using  ideal gas equation,

P V = n R T

Where, P is the pressure,

V is the volume,

n is the number of moles,

R is the gas constant, and T is the temperature

We have, 1 bar = 0.986923 atm

Substituting the values,

V = nRT/P = (2.798 x 10-4moles x 0.08205 L atm mol K x 298 K)/ 0.986923 atm = 0.0069L

Volume of O2 produced = 0.0069L