Chapter 14, Problem 13E

Interpretation Introduction

Interpretation:

The working of an automobile battery is to be explained and the reactions occurring at cathode and anode are to be discussed.

Concept introduction:

An automobile battery is the type of the battery that converts chemical energy into electrical energy.

Oxidation is the addition of an electronegative element or the removal of an electropositive element in a chemical reaction.

Reduction is the addition of an electropositive element or the removal of an electronegative element in a chemical reaction.

The chemical reaction in which oxidation process and reduction process take place simultaneously, is called a redox reaction.

Electrolysis is the process in which a reaction takes place at electrodes. The positively charged electrode is called anode. The negatively charged electrode is called cathode. During electrolysis, reduction takes place at the cathode and oxidation takes place at the anode.

Answer to Problem 13E

Solution:

The automobile battery is formed when lead oxide and sponge lead are immersed in dilute sulphuric acid.

The sulphuric acid completely dissociates into $\left(2{\text{H}}^{+}\right)$ and $\left({\text{SO}}_4^{2-}\right)$ in an aqueous solution.

The positively charged hydrogen moves towards the electrode that is connected to the negative terminal. The negatively charged sulfate ion moves towards the electrode that is connected to the positive terminal.

The two half reactions are as:

$\begin{array}{l}\text{Anode}-\text{Pb}\left(s\right)+{\text{ SO}}_4^{2-}\left(aq\right)\to 4{\text{PbSO}}_4\left(s\right)+2e^{-}\\ \text{Cathode}-{\text{PbO}}_2\left(s\right)+{\text{ SO}}_4^{2-}\left(aq\right)+2e^{-}+4{\text{H}}^{+}\left(aq\right)\to {\text{PbSO}}_4\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\\ \overline{\text{Net}-\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+{\text{2SO}}_4^{2-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\to 2{\text{PbSO}}_4\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)}\end{array}$.

Explanation of Solution

The automobile battery is a lead-acid storage battery. The lead-acid storage battery is the type of the battery that converts chemical energy into electrical energy. The electrical energy is converted into chemical energy with help of an external source and is called charging of the automobile battery. The chemical energy is converted into electrical energy with the help an external load and is called discharging of the automobile battery. The materials used in the automobile battery are lead oxide, sponge lead and dilute sulphuric acid.

Sulphuric acid completely dissociates into $\left(2{\text{H}}^{+}\right)$ and $\left({\text{SO}}_4^{2-}\right)$ in the aqueous solution. The automobile battery is formed when lead oxide and sponge lead are immersed in the dilute sulphuric acid. The positively charged hydrogen moves towards the electrode that is connected to the negative terminal. The negatively charged sulfate ion moves towards the electrode that is connected to the positive terminal.

${\text{PbO}}_2$ half-cell reaction will occur as reduction at the cathode. ${\text{PbO}}_2$ is reduced in the presence of sulfate ion to form a ${\text{Pb}}^{2+}$ ion.

${\text{PbO}}_2\left(s\right)+{\text{SO}}_4^{2-}\left(aq\right)+2e^{-}+4{\text{H}}^{+}\left(aq\right)\to {\text{PbSO}}_4\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)$.

$\text{Pb}$ half-cell reaction will occur as oxidation at the anode. $\text{Pb}$ is oxidized in the presence of sulfate ion to form ${\text{PbSO}}_4$.

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

Adding two half reaction gives the overall cell reaction as follows:

$\begin{array}{l}\text{Pb}\left(s\right)+{\text{ SO}}_4^{2-}\left(aq\right)\to 4{\text{PbSO}}_4\left(s\right)+2e^{-}\\ {\text{PbO}}_2\left(s\right)+{\text{ SO}}_4^{2-}\left(aq\right)+2e^{-}+4{\text{H}}^{+}\left(aq\right)\to {\text{PbSO}}_4\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)\\ \overline{\text{Pb}\left(s\right)+{\text{PbO}}_2\left(s\right)+{\text{2SO}}_4^{2-}\left(aq\right)+4{\text{H}}^{+}\left(aq\right)\to 2{\text{PbSO}}_4\left(aq\right)+2{\text{H}}_2\text{O}\left(l\right)}\end{array}$

Conclusion

The reaction occurring at the cathode is as follows:

${\text{PbO}}_2\left(s\right)+{\text{SO}}_4^{2-}\left(aq\right)+2e^{-}+4{\text{H}}^{+}\left(aq\right)\to {\text{PbSO}}_4\left(s\right)+2{\text{H}}_2\text{O}\left(l\right)$

The reaction occurring at the anode is as follows:

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