What volume of Oxygen measured at s.t.p would be liberated on electrolysis by 9650 coulombs of electricity?
(Molar volume of gas = 22.4 dm\(^3\), 1F = 96,500 C/mol)
- 22.4 dm3
- 11.2 dm3
- 1.12 dm3
-
0.560 dm3
The correct answer is: D
Explanation
For a diatomic divalent element like oxygen, the following reaction takes place:
2H\(_2\)0 → O\(_2\) + 4H\(^+\) + 4e\(^-\)
⇒ 4e\(^-\) are transferred to liberate 1 mole of O\(_2\)
So, to calculate the volume of O\(_2\)
4 Faradays (4 x 96,500) of electricity to produce 22.4 dm\(^3\) of O\(_2\) at S.T.P
4 X 96,500 C = 22.4 dm\(^3\)
9650 C = x dm\(^3\)
x = \(\frac{9650 X 22.4}{ 4 X 96,500}\)
x = 0.56dm\(^3\)