Chapter 21, Problem 35A

Interpretation Introduction

Interpretation: The reason for the density of the electrolyte to decrease during discharging of a lead storage battery needs to be explained.

Concept Introduction: Lead-storage battery is a rechargeable battery. When the battery is charged, the chemical energy is stored as the potential difference between pure lead on the negative side and lead oxide on the positive side.

Explanation of Solution

A battery is made up of a group of voltaic cells. In a lead-storage battery, the anode is lead solid and the cathode is lead (IV) oxide. For both the half-cells, sulphuric acid works as an electrolyte.

At the time of discharging of a lead-storage battery, electrical energy is produced. The reactions at the anode and cathode are represented as follows:

Oxidation (anode):

  $\text{Pb}\left(s\right)+{\text{HSO}}_4^{-}\left(aq\right)\to {\text{PbSO}}_{\text{4}}\left(s\right)+{\text{H}}^{+}\left(aq\right)+2e^{-}$

Reduction (cathode):

  ${\text{PbO}}_{\text{2}}\left(s\right)+{\text{HSO}}_4^{-}\left(aq\right)+3{\text{H}}^{+}\left(aq\right)+2e^{-}\to {\text{PbSO}}_4\left(s\right)+{\text{2H}}_{\text{2}}\text{O}\left(l\right)$

Thus, the overall reaction will be the sum of oxidation and reduction reactions as follows:

  $\text{Pb}\left(s\right)+{\text{PbO}}_{\text{2}}\left(s\right)+2{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to 2{\text{PbSO}}_{\text{4}}\left(s\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

In the above equation, it can be seen that lead (II) oxide is formed during discharging. Here, sulfate builds up slowly on the plates as a result of which the concentration of sulphuric acid in the electrolyte decreases. Therefore, this is the reason why the density of the electrolyte decreases at the time of discharging of a lead-storage battery.