Answer :
Time taken to form O₂ at 99.8kPa and 28°C from water if a current of 1.3 A passes through the electrolytic cell is 1969.283 minutes.
How do you calculate the time?
In order to calculate the time
We have:
- volume of O₂(V) = 10.0 L
- pressure P= 99.8 kPa = 0.984 atm
- temperature T = 28°C = 301 K
- current (I) = 1.3 A
Now The Values That We Have To Remember For This Question Is :
- Molar Mass of O₂(M) = 32 g/mol
- Value Of Universal Gas Constant (R) = 0.0821 L.atm/mol.K
- Charge On One Electron (Qₐ) = 96485 C/mol
First of all we have to calculate mass of Oxygen Gas
Mass of O₂ = n × molar mass of O₂
= [tex]\frac{PV}{RT}[/tex] × molar mass of O₂
where
n = number of moles of O₂
Mass of O₂ = [tex]\frac{PV}{RT}[/tex] × molar mass of O₂
= [tex]\frac{(0.984atm) (10L) (32g/mol)}{(0.0821 L.atm/mol.K)(301K)}[/tex]
= 12.74 g
Now,
Reaction that takes place in electrolytic cell
2H₂O (l) → O₂ (g) + 2H₂ (g)
Hence , No. of moles of electron = number of moles of O₂ × 4 mol e⁻¹
= 1.592 mol
Now ,For total charge Q,
Q = no. of moles of electron × Qₐ
= (1.592 mol) × (96485 C/mol)
= 153604.12 C
Now, Total time taken for production of 0₂ (t)
t = [tex]\frac{Q}{I}[/tex]
= [tex]\frac{ 153604.12 C}{1.3 A}[/tex]
=- 118157.015 sec
= 1969.283 min
Thus from above conclusion we can say that the time taken for the production of 0₂ is 1969.283 min.
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