Chapter 21.1, Problem 5LC

Interpretation Introduction

Interpretation: The electrolyte in a lead storage battery needs to be identified. Half-reactions for such a battery need to be determined.

Concept Introduction: In an electrochemical process, a redox reaction takes place. Here, two half-reactions occur where in one reaction oxidation takes place and in the other reaction, reduction takes place. The sum of these two-half reactions gives the overall reaction for the process.

In oxidation, the loss of electrons takes place by an element and its oxidation state increases Similarly in reduction, the gain of electrons takes place by an element. Here, the oxidation state of the element decreases.

Explanation of Solution

An electrolyte is defined as a liquid containing ions that can be decomposed by an electrolysis reaction in the battery.

In a lead storage battery, oxidation of lead at the anode takes place. Here, the electrolyte is dilute sulphuric acid. The lead ions formed at the anode in the presence of sulfate ions (from sulphuric acid). Thus, the insoluble lead sulfate is formed. Here, the flow of electrons takes place from the anode to reduce lead dioxide at the cathode. Here, also lead ions formed react with aqueous sulfate ions to form insoluble lead sulfate.

The two half-reactions for lead storage battery are as follows:

Anode-Oxidation takes place:

  $\text{Pb}\left(s\right)\to {\text{Pb}}^{2+}\left(aq\right)+2e$

Since sulfuric acid is an electrolyte the overall anode reaction will be:

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

Cathode-Reduction takes place:

  ${\text{PbO}}_{\text{2}}\left(s\right)+4{\text{H}}^{+}\left(aq\right)+2e\to {\text{Pb}}^{2+}+2H_{2}O$

Since sulfuric acid is an electrolyte the overall cathode reaction will be:

  ${\text{PbO}}_{\text{2}}\left(s\right)+4{\text{H}}^{+}\left(aq\right)+2e+{\text{SO}}_4^{2-}\left(aq\right)\to {\text{PbSO}}_{\text{4}}\left(aq\right)+2H_{2}O$