Question

Nuclear submarines can stay under water nearly indefinitely because they can produce their own oxygen by the electrolysis of water.
How many liters of O2 at 298 K and 1.00 bar are produced in 2.00 hr in an electrolytic cell operating at a current of 0.0500 A?

Answer 1

Using the formula

m/E = it/F

m = itE/F

Where , m = mass of oxygen

E = equvalent mass of oxygen = 16/2 = 8g/mol ( because oxygen have always 2 valency)

i = current = 0.0500A

t = time = 2.00hr = 2×60×60s = 7200s

F = 96485.33charge/mole

m = 0.0500A×7200s×8g/mol/96485.33charge/mol

m = 0.0298490g

Number of moles of oxyy gas = mass of oxygen gas/molar mass of oxygen gas

Number of moles = 0.0298490g/32g/mol = 0.0009328mole

We know that the volume of one mole gas at STP 22.4liter

STP mean 1 atm or 1.01325bar pressure and 273.15K temperature

Using the ideal gas equation

PV = nRT

PV/nT = R

So

(P1)(V1)/(n1)(T1) = (P2)(V2)/(n2)(T2)

(V2)= (P1)(V1)(n2)(T2)/(P2)(n1)(T1)

Where at STP

P1 = 1.01325bar

T1 = 273.15K

n1 = 1mol

V1 = 22.4 L

in submarine

P2 = 1.0bar

V2 = volume of oxygen gas

n2 = number of moles of oxygen gas = 0.0009328mol

T2 = 298K

Putting all value in formula

V2 = (1.01325bar)(22.4L)(0.0009328mol)(298K)/(1.0bar)(1.0mol)(273.15K)

V2 = 0.023097L = 23.097ml oxygen