How many liters of o2 at 298 k and 1.00 bar are produced in 1.25 hr in an electrolytic cell operating at a current of 0.0500 a?

0.0144 L

Explanation:

Step 1:

Data obtained from the question. This includes:

Temperature (T) = 298k

Pressure (P) = 1 bar

Time (t) = 1.25 hr

Current (I) = 0.0500 A

Step 2:

Determination of the quantity of electricity (Q) used. This is illustrated below:

Q = it

Time (t) = 1.25 hr = 1.25 x 3600 = 4500 secs

Current (I) = 0.0500 A

Quantity of electricity (Q) =?

Q = it

Q = 0.05 x 4500

Q = 225C

Step 3:

Determination of the number of mole of O2 liberated in the process.

In the electrolytic process, O2 will be liberated according to the equation:

2O^2- + 4e- —> O2

From the above illustration, 4 faraday are needed to liberate 1 mole of O2.

1 faraday = 96500C

Therefore of 4 faraday = 4x96500C = 386000C

From the above equation,

386000C of electricity liberated 1 mole of O2.

Therefore, 225C will liberate = 225/386000 = 5.83x10^-4 mole of O2.

Step 4:

Determination of the volume of the O2 liberated.

Number of mole (n) = 5.83x10^-4 mole

Temperature (T) = 298k

Pressure (P) = 1 bar = 0.987 atm

Gas constant (R) = 0.082atm.L/Kmol

Volume (V) =?

Applying the ideal gas equation:

PV = nRT

The volume of O2 can be obtained as follow:

PV = nRT

0.987 x V = 5.83x10^-4 x298x0.082

Divide both side by 0.987

V = (5.83x10^-4 x298x0.082)/0.987

V = 0.0144 L

c. electrical charges of ionic bonds.