Question

Enter your answer in the provided box. One of the half-reactions for the electrolysis of water is 2H2O(l) → O2(g) + 4H+(aq) + 4e− If 3.696 L of O2 is collected at 25°C and 755 mmHg, how many faradays of electricity had to pass through the solution?

Answer 1

Answer : The number of faradays of electricity had to pass through the solution will be, 0.596 F

Explanation :

First we have to calculate the moles of oxygen gas.

using ideal gas equation:

`PV=nRT`

where,

P = pressure of gas = 755 mm Hg = 0.99 atm

conversion used : (1 atm = 760 mmHg)

V = volume of gas = 3.696 L

T = temperature of gas =
`25^oC=273+25=298K`

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get the number of moles of oxygen gas.

`(0.99atm)* (3.696L)=n* (0.0821L.atm/mole.K)* (298K)`

`n=0.149mole`

Now we have to calculate the number of faradays of electricity had to pass through the solution.

The balanced half-reactions for the electrolysis of water is,

`2H_2O(l)\rightarrow O_2(g)+4H^+(aq)+4e^-`

From this we conclude that,

As, 1 mole of oxygen gas require 4 mole of electrons that means 4 F (faraday) of electricity.

So, 0.149 mole of oxygen gas require
`0.149* 4=0.596F` of electricity.

Therefore, the number of faradays of electricity had to pass through the solution will be, 0.596 F