How many hours does it take to form 15.0 L of Oz measured at 750 torr and 30°C from water by passing 1.15 A of current through an
electrolytic cell?

55.7 hrs

Explanation:

We'll begin by calculating the number of mole of O2 produced during the process. This can be obtained as follow:

Volume (V) = 15L

Pressure (P) = 750torr = 750/760 = 0.99atm

Temperature (T) = 30°C = 30°C + 273 = 303K

Gas constant (R) = 0.0821 atm.L/Kmol

Number of mole (n) =?

PV = nRT

0.99 x 15 = n x 0.0821 x 303

Divide both side by 0.0821 x 303

n = (0.99 x 15 ) /(0.0821 x 303)

n = 0.597 mole

Next, we shall determine the quantity of electricity required to liberate 1 mole of O2. This is illustrated below:

2O^2- + 4e —> O2

Thus, 4 moles of electron (e) is needed to produce 1 mole of O2.

Recall:

1 electron (e) = 1 Faraday = 96500C

1e = 96500C

Therefore, 4e = 4 x 96500C = 386000C.

Therefore, 386000C of electricity is required to liberate 1 mole of O2.

Next, we shall determine the quantity of electricity required to liberate 0.597 mole of O2.

This is illustrated below:

From the balanced equation above,

386000C of electricity is required to liberate 1 mole of O2.

Therefore, XC of electricity will be required to liberate 0.597 mole of O2 i.e

XC of electricity = 386000 x 0.597

XC of electricity = 230442C

Therefore, 230442C of electricity is needed to liberate 0.597 mole of O2.

Now, we can obtain the time taken to produce 0.597 mole of O2 as follow:

Current (I) = 1.15A

Quantity of electricity (Q) = 230442C

Time (t) =.?

Q = It

230442 = 1.15 x t

Divide both side by 1.15

t = 230442/1.15

t = 200384.35 secs

Finally, we shall convert 200384.35 secs to hours. This is illustrated below:

3600 secs = 1 hr

Therefore, 200384.35 secs = 200384.35/3600 = 55.7 hrs

Therefore, it will take 55.7 hrs to produce 15L of O2.