Question

Consider the electrolysis of an aqueous soltion of calcium nitrate.

If the solution was electrolyzed for 1.00 hours, 10.5L of gaseous product was formed at the anode, at 1.00atm and 298k. What was the current in amps used for the electrolysis? What was the volume of a gas formed at the cathode at the same time?

Answer 1

no of mol of gaseous product liberated = PV/RT

                       = 1*10.5/(0.0821*298)

                       = 0.43 mol

reaction : 4OH-(aq) ---> o2(g) + 2H2O(l) + 4e-

amount of o2 gas liberated = 0.43*32 = 13.76 g

faradays first law

equivalent weight of O2 (E)= atwt / charge = 32/2 = 16 g/equiv

W = Zit

Z= E/F

w = E/Fit

F = faraday = 96500 C

i = current = x

t = time = 1*60*60 = 3600 sec

w = weight of o2 reacted = 13.76 g

13.76 = (16/96500)*x*3600

i = current = 23 A