Answer:
7.67 mins.
Explanation:
Data obtained from the question include the following:
Volume (V) = 0.0760 L
Temperature (T) = 298 K
Pressure (P) = 1 atm
Current (I) = 2.60 A
Time (t) =?
Next, we shall determine the number of mole (n) of O2 contained in 0.0760 L.
This can be obtained by using the ideal gas equation as follow:
Note:
Gas constant (R) = 0.0821 atm.L/Kmol
PV = nRT
1 x 0.0760 = n x 0.0821 x 298
Divide both side by 0.0821 x 298
n = 0.0760 / (0.0821 x 298)
n = 0.0031 mole
Next, we shall determine the quantity of electricity needed to liberate 0.0031 mole of O2.
This is illustrated below:
2O²¯ + 4e —> O2
Recall:
1 faraday = 1e = 96500 C
4e = 4 x 96500 C
4e = 386000 C
From the balanced equation above,
386000 C of electricity liberated 1 mole of O2.
Therefore, X C of electricity will liberate 0.0031 mole of O2 i.e
X C = 386000 X 0.0031
X C = 1196.6 C
Therefore, 1196.6 C of electricity is needed to liberate 0.0031 mole of O2
Next, we shall determine the time taken for the process. This can be obtained as follow:
Current (I) = 2.60 A
Quantity of electricity (Q) = 1196.6 C
Time (t) =?
Q = It
1196.6 = 2.6 x t
Divide both side by 2.6
t = 1196.6/2.6
t = 460.23 secs.
Finally, we shall convert 460.23 secs to minute. This can be achieved by doing the following:
60 secs = 1 min
Therefore,
460.23 secs = 460.23/60 = 7.67 mins
Therefore, the process took 7.67 mins.
