Chapter 21, Problem 47A

Interpretation Introduction

Interpretation: The equation for the decomposition of water by electrolysis needs to be written.

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

Explanation of Solution

In the electrolysis of water, two electrodes are immersed in pure water. Here, a small amount of electrolytes such as sulphuric acid or potassium nitrate is added to pure water. The solution conducts electric current and electrolysis of water takes place. The products formed in the process are hydrogen gas and oxygen gas. Here, the oxidation of water occurs at the anode and its reduction occurs at the cathode.

Thus, the oxidation and reduction reactions are represented as follows:

Oxidation: $2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{O}}_{\text{2}}\left(g\right)+4{\text{H}}^{+}\left(aq\right)+4e$

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

In this process, the anode region turns acidic due to an increase in hydrogen ions. On the other hand, the cathode region turns basic as the number of hydroxide ions increases. The above two reactions can be added to get the overall cell reaction:

  $\begin{array}{l}2{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{O}}_{\text{2}}\left(g\right)+4{\text{H}}^{+}\left(aq\right)+4e\\ 2\left(2{\text{H}}_{\text{2}}\text{O}\left(l\right)+2e\to {\text{H}}_{\text{2}}\left(g\right)+2{\text{OH}}^{-}\left(aq\right)\right)\\ \overline{\underset{¯}{{\text{6H}}_{\text{2}}\text{O}\left(l\right)\to {\text{O}}_{\text{2}}\left(g\right)+4{\text{H}}^{+}\left(aq\right)+2{\text{H}}_{\text{2}}\left(g\right)+4{\text{OH}}^{-}\left(aq\right)}}\\ \text{Or,}\\ {\text{2H}}_{\text{2}}\text{O}\left(l\right)\to {\text{O}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\left(g\right)\end{array}$

Therefore, the equation for the decomposition of water by electrolysis is as follows:

  ${\text{2H}}_{\text{2}}\text{O}\left(l\right)\to {\text{O}}_{\text{2}}\left(g\right)+2{\text{H}}_{\text{2}}\left(g\right)$