g Calculate the time (in min.) required to collect 0.0760 L of oxygen gas at 298 K and 1.00 atm if 2.60 A of current flows through water. (Hint: Ideal gas law)

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.

your answer would be “dry”