Chapter 19, Problem 19.47QP

Interpretation Introduction

Interpretation:

Need to identify the cheaper metal for production through electrolysis.

Concept introduction:

In order to identify the cheaper metal among sodium and aluminum, which can be produced by electrolysis?

First the number of moles of metal produced need to be identified. From the number of moles of the metal produced we can calculate the number of electron involved in the reaction. The multiplication of number of moles of electron and faraday constant coulombs of charges involved in the process can be determined. Since number of coulombs of charges is the multiple of current and time. The metal which consumes fewer amounts of charges will obviously need less amount of electricity and time, therefore such metal can be assigned as cheaper one correspondingly.

The cell reaction for the production of sodium and aluminum can be written as

   $\begin{array}{l}{\text{Na}}^{\text{+}}{}_{\text{(aq)}}{\text{+ e}}^{\text{-}}\stackrel{}{\to }{\text{Na}}_{\begin{array}{l}\text{(s)}\\ \end{array}}\\ {\text{Al}}^{3+}{}_{\text{(aq)}}{\text{+ 3e}}^{\text{-}}\stackrel{}{\to }{\text{Al}}_{\text{(s)}}\end{array}$

Let us take the production of sodium as the representative example and the model calculation steps are given below.

  $\begin{array}{l}\text{Mass}\text{}\text{of}\text{Na=}\text{Number}\text{of}\text{mole}\text{of}\text{Na×Atomic}\text{mass}\text{of}\text{Na}\\ \text{Number}\text{of}\text{mole}\text{of}\text{Na= }\frac{\text{Mass}\text{}\text{of}\text{Na}}{\text{Atomic}\text{Na}}\\ \text{Number}\text{of}\text{mole}\text{of}\text{Na}\text{=Number of moles of electron ×}\frac{\text{1}\text{mole}\text{of}\text{Na}}{\text{1mole}\text{of}{\text{e}}^{\text{-}}}\\ \frac{\text{1}\text{mole}\text{of}{\text{e}}^{\text{-}}}{\text{1mole}\text{of}\text{Na}}\text{×Number}\text{of}\text{mole}\text{of}\text{Na=Number of moles of electron}\\ \text{Number}\text{of}\text{moles}\text{ofelectrons=charges/96500C/mole}\text{of}{\text{e}}^{\text{-}}\\ \text{Charges=Number}\text{}\text{of}\text{moles}\text{of}\text{electrons×96500}\text{C/mole}\text{of}{\text{e}}^{\text{-}}\\ \text{Charges=Current×time}\end{array}$

To find: Among sodium and aluminum, need to identify which will be cheaper for production through electrolysis.