Answer:
0.0144 L
Step-by-step 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