How many hours does it take to form 15.0 L of O₂ measured at 750 torr and 30°C from water by passing 3.55 A of current through an electrolytic cell?

The correct answer is 17.845 hours.

Explanation:

To solve the question, that is, to determine the hours required there is a need to combine the Faraday's law of electrolysis with the Ideal gas law.  

Based on Faraday's law, m = Mit/nF

Here m is the mass in grams, M is the molecular mass, i is the current in amperes, t is time, n is the number of moles of electron per mole of oxygen formed and F is the Faraday's constant (the value of F is 96487 coulombs/mole).  

From the above mentioned equation,  

t = mnF/Mi (i)

Now based on ideal gas law's, PV = nRT or PV = m/M RT, here n = mass/molecular mass.  

So, from the above gas law's equation, m = PVM/RT

Now putting the values of m in the equation (i) we get,  

t = PVMnF/MiRT = PVnF/iRT

Based on the given information, the value of P is 750 torr or 750/760 atm = 0.98 atm, the value of v is 15.0 L, T is 30 degree C or 273 + 30 K = 303 K, i is 3.55 Amperes, and the value of R is 0.0821 atm L/mol K.  

1 mole of oxygen gives 2 moles of electrons, therefore, 2 moles of oxygen will give 4 moles of electrons.  

Now putting the values we get,  

t = PVnF/iRT

= 0.98 atm × 15.0 L × 4 moles of electron × 96487 coulombs per mole / 3.55 coulomb per sec × 0.0821 atm L per mole-K × 303 K

= 64243.81 secs or 64243.81/3600 hr  

= 17.845 hours