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
