H2 (g) and O2 (g), can be produced by the
electrolysis of water. What total volume (in L) of
O2 and H2 are produced at STP when a current of
30 A is passed through a K2SO4 (aq) solution for
965 second
Answers
When 30 amperes is passed for 193 minutes, the quantity of current passed =193×60×30=347400 coulombs.
=347400/96500=3.6 Faradays.
H
2
O
electrolysis
H
2
+0.5O
2
So on passing 1 Faraday of electricity, 1 mole of Hydrogen and 0.5 mole of Oxygen are produced.
On passing 3.6 Faradays of electricity 3.6 moles of Hydrogen and (3.6×0.5)=1.8 moles of Oxygen will be formed.
So the volume of Hydrogen produced at STP =3.6×22.4=80.64 liters.
Volume of oxygen produced at STP =1.8×22.4=40.32 liters.
I know this not the same answer but u change the digit and.do like this method so u can do easily ok I hope you understand
thanks ◉‿◉｡◕‿◕｡
Given:
Current, I = 30 A
Time, t = 965 s
To Find:
The total volume (in L) of O2 and H2 produced at STP by electrolysis.
Calculation:
- The amount of current:
Q = 30 × 965/96500 = 0.3 F
- Electrolysis reaction is given as:
H2O ⇄ H2 + 1/2 O2
⇒ On passing 1 Faraday electricity, 1 mole of Hydrogen and 0.5 mole of Oxygen is produced.
- The amount of hydrogen evolved on passing 0.3 F electricity = 0.3 mole
- The amount of oxygen evolved on passing 0.3 F electricity = 0.15 mole
- Volume of 0.3 mole of Hydrogen at STP = 0.3 × 22.4 = 6.72 L
- Volume of 0.15 mole of Oxygen at STP = 0.15 × 22.4 = 3.36 L
- Total volume of H2 and O2 = 6.72 + 3.36