Chapter 18, Problem 18.60QP

Interpretation Introduction

Interpretation:

Need to calculate the faraday of electricity consumed during electrolysis of water to H₂gas.

Concept introduction:

The half-cell reaction at the cathode during the electrolysis of water was given below

${\text{Cathode (reduction) H}}^{\text{+}}{}_{\text{(aq)}}{\text{+ 2e}}^{\text{-}}\stackrel{}{\to }{\text{H}}_{\text{2}}{}_{\text{(g)}}$

Since volume, pressure and temperature of hydrogen was given, the number of moles can be attained from ideal gas equation

$\begin{array}{l}\text{PV}\text{=}\text{nRT}\\ \text{n}\text{=}\frac{\mathrm{PV}}{\text{RT}}\end{array}$

From the stoichiometry of the reaction, it was known that two moles of electrons will be needed to produce one mole of hydrogen.

So,

$\begin{array}{l}\text{Number}\text{of}\text{moles of electrons = Number of moles of hydrogen }\times \text{ 2}\\ \text{then}\\ \text{1 Farday of electricity }\text{=}\text{one moles of electrons}\\ \text{So }\\ \text{Farday of electricity = number of moles of hydrogen }\times \text{ 2}.\end{array}$