How many liters of O 2 at 298 K and 1.00 bar are produced in 3.50 hr in an electrolytic cell operating at a current of 0.0300 A? Solution According to Faraday’s law ,W = (ECt) / 96500 Where W = mass of metal deposited = ? E = Equivalent weight of O = 16 / 2 = 8 C = current = 0.0300 A t = time taken = 3.50 hr x(60min/hr)x(60s/hr) = 12600 s Plug the values we get W = 0.0313 g --------------------------------------------------------- We know that ideal gas equation is PV = nRT Where T = Temperature = 298 K P = pressure = 1.00 barx(0.987 atm/bar) = 0.987 atm n = No . of moles = mass/molar mass = 0.0313 g/16(g/mol) = 1.96x10 -3 mol R = gas constant = 0.0821 L atm / mol - K V= Volume of the gas = ? Plug the values we get V = (nRT) / P = 0.048 L .

1. How many liters of O 2 at 298 K and 1.00 bar are produced in 3.50 hr in an electrolytic cell
operating at a current of 0.0300 A?
Solution
According to Faraday’s law ,W = (ECt) / 96500
Where W = mass of metal deposited = ?
E = Equivalent weight of O = 16 / 2 = 8
C = current = 0.0300 A
t = time taken = 3.50 hr x(60min/hr)x(60s/hr) = 12600 s
Plug the values we get W = 0.0313 g
---------------------------------------------------------
We know that ideal gas equation is PV = nRT
Where
T = Temperature = 298 K
P = pressure = 1.00 barx(0.987 atm/bar) = 0.987 atm
n = No . of moles = mass/molar mass = 0.0313 g/16(g/mol) = 1.96x10 -3
mol
R = gas constant = 0.0821 L atm / mol - K
V= Volume of the gas = ?
Plug the values we get V = (nRT) / P = 0.048 L