Chapter 18, Problem 18.49QP

Interpretation Introduction

Interpretation:

Need to calculate the Faraday of electricity utilized to produce 0.076L of O₂ with 755mmHg pressure, through electrolysis of water.

Concept introduction:

Electrolysis of water will produce oxygen, hydrogen ion and electron, the reaction was given below

${\text{2H}}_2{\text{O}}_{\text{(l)}}\stackrel{}{\to }{\text{O}}_{2\text{(g)}}{\text{+4H}}^{\text{+}}{}_{\text{(aq)}}{\text{+4e}}^{\text{-}}$

Since volume and pressure of oxygen produced was given, by applying it into ideal gas equation the number of mole of oxygen produced can be calculated. Ideal gas equation was given as follows.

$\begin{array}{l}\text{PV=nRT}\\ \text{n=}\frac{\text{PV}}{\text{RT}}\\ \text{n=number}\text{of}\text{moles}\text{of}\text{gas}\\ \text{P=pressure}\text{of}\text{gas}\\ \text{V=Volume}\\ \text{R=universal}\text{gas}\text{constant (0}\text{.08206L}\text{.atm/K}\text{mol)}\\ \text{T=Temperature}\end{array}$

On applying the number of moles of oxygen produced into stoichiometry of the reaction, the number of moles of electron involved in the reaction can be calculated. In case of the given reaction 4 mole of electron was liberated during the production of one mole of oxygen. Since one Faraday is equal to one mole of electron, so 4 Faraday of electricity will be needed to produce one mole of oxygen. Therefore Faraday of electricity utilized to produce required mole of oxygen can be calculated according the formula

$\text{Faraday=}\text{n×}\frac{\text{4F}}{\text{1}\text{mole}{\text{O}}_{\text{2}}}$

To find: Faraday of electricity need to produce 0.076L of O₂ with pressure 755mmHg, at 298K, through electrolysis of water.