How many hours does it take to form 15.0 L of O₂ measured at 750 torr and 30°C from water by passing 4.55 A of current through an electrolytic cell?

Answer 1

Find the moles of O₂

PV = nRT

n = PV/RT = (750 torr)(15.0 L) / (62.36 Ltorr/Kmol)(303K)

n = 0.595 moles O₂

Electrolysis of H₂O

2H₂O ==> O₂ + 4H⁺ + 4e- ... oxidation at anode

2H₂O + 2e- ==> 2H₂ + 4OH^- ... reduction at cathode

overall rxn: 2H₂O ==> 2H₂ + O₂ with the transfer of FOUR electrons

Find moles of electrons transferred

0.595 mols O₂ x 4 mol e- / mol O₂ = 2.38 mol e-

Find time needed

2.38 mol e- x 96,500 C / mol e- = 229,670 C (C = coulombs)

1 amp = 1 Coulomb/sec

4.55 amp = 4.55 C / sec

229,670 C x 1 sec / 4.55 C = 50,477 seconds

50,477 seconds x 1 hour / 3600 seconds = 14.0 hours