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(a) How many minutes does it take to form 10.0 L of [tex]$O_2$[/tex] measured at 99.8 kPa and 28°C from water if a current of 1.3 A passes through the electrolytic cell?

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:

  1. volume of O₂(V) = 10.0 L
  2. pressure P= 99.8 kPa = 0.984 atm
  3. temperature T = 28°C = 301 K
  4. 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.

Learn more about the Electrolytic Cell here; https://brainly.com/question/27548707

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