Chapter 21, Problem 49A

Interpretation Introduction

Interpretation: The given electrolysis diagram needs to be modified for the electrolysis of brine to represent the formation of hydrogen and chlorine gas.

Concept Introduction: In an electrochemical cell, there are two electrodes; anode and cathode. At the anode, oxidation takes place and at the cathode reduction takes place.

Explanation of Solution

The given diagram is as follows:

  

In the electrolysis of brine, electrolyte other than water is taken which can be more easily oxidized or reduced is taken. This will result in the formation of products of electrolysis other than hydrogen and oxygen. In the electrolysis of brine, the aqueous solution of sodium chloride is taken.

This results in the formation of three products that is chlorine gas, hydrogen gas, and sodium hydroxide. In the electrolysis process, oxidation of chloride ions takes place at the anode to form chlorine gas. The reduction reaction of water takes place to form hydrogen gas at the cathode. Here, water molecules get easily reduced to hydrogen gas as compared to sodium ions; thus, sodium metal is not formed.

Here, the electrolyte solution becomes sodium hydroxide as the reduction of water also produces hydroxide ions.

The reactions at the cathode and anode are represented as follows:

Anode-Oxidation: $2{\text{Cl}}^{-}\left(aq\right)\to {\text{Cl}}_{\text{2}}\left(g\right)+2e$

Cathode-Reduction: $2{\text{H}}_{\text{2}}\text{O}\left(l\right)+2e\to {\text{H}}_2\left(g\right)+2{\text{OH}}^{-}\left(aq\right)$

Thus, the overall reaction will be:

  $\begin{array}{l}2{\text{Cl}}^{-}\left(aq\right)\to {\text{Cl}}_{\text{2}}\left(g\right)+2e\\ 2{\text{H}}_{\text{2}}\text{O}\left(l\right)+2e\to {\text{H}}_2\left(g\right)+2{\text{OH}}^{-}\left(aq\right)\\ \overline{\underset{¯}{2{\text{Cl}}^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cl}}_{\text{2}}\left(g\right)+{\text{H}}_2\left(g\right)+2{\text{OH}}^{-}\left(aq\right)}}\end{array}$

Here, a spectator sodium ion is also introduced in the reaction. Thus, the overall reaction becomes:

  $2\text{NaCl}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{Cl}}_{\text{2}}\left(g\right)+{\text{H}}_2\left(g\right)+2\text{NaOH}\left(aq\right)$

Thus, the given diagram can be modified to show the electrolysis of brine as follows:

  

It shows the same amount of hydrogen and chlorine gas produced in the reaction as the number of moles are same.