Question

Write the anode and the cathode reactions and the overall reaction occurring in a lead storage battery.

Answer 1

Hint: In a lead storage battery, the cathode is lead dioxide and the anode is lead, while the electrolyte is an acid, usually sulphuric acid. Oxidation happens at the anode and reduction happens at the cathode. So, we can start off by writing the reactions of oxidation of lead with the help of the electrolyte and then balance it to get our equation. We can adopt a similar strategy to find the reaction at cathode (where lead oxide would get reduced).

Complete step by step answer:
The electrolyte in a lead storage battery is sulphuric acid in aqueous form:
${H_2}S{O_4} \rightleftharpoons 2{H^ + } + SO_4^{2 - }$
In a lead storage battery, the anode is solid lead, and oxidation reaction happens here. Therefore, the lead gets oxidised to lead ion with the sulphate ion from the sulphuric acid acting as the oxidising agent, with the liberation of two electrons. The oxidation state of lead changes from $0$ in solid lead to $ + 2$, as oxidation means removal of electrons, resulting in an increase of oxidation number. The anode reaction is as follows:
\[Pb + SO_4^{2 - } \to PbS{O_4} + 2{e^ - }\]
The cathode is lead dioxide, and reduction happens here. The lead in lead dioxide has an oxidation state of $ + 4$. Thus, the lead ion, which is in $ + 4$ oxidation state gets reduced by accepting 2 electrons. Thus, the oxidation state of lead changes from $ + 4$ to $4 - 2 = + 2$, as reduction means gaining electrons. This lead ion, which is now in $ + 2$ oxidation state, combines with a sulphate ion from the electrolyte to form lead sulphate. The reaction is as follows:
\[Pb{O_2} + SO_4^{2 - } + 4{H^ + } + 2{e^ - } \to PbS{O_4} + 2{H_2}O\]
Therefore, as we know, the overall reaction is the sum of the anode and cathode reactions:
\[Pb + SO_4^{2 - } + Pb{O_2} + SO_4^{2 - } + 4{H^ + } + 2{e^ - } \to PbS{O_4} + 2{e^ - } + PbS{O_4} + 2{H_2}O\]
Combining the similar terms and cancelling the two electrons on both sides of the reaction, we get the overall reaction as:
\[Pb + Pb{O_2} + 2SO_4^{2 - } + 4{H^ + } \to 2PbS{O_4} + 2{H_2}O\]

Note: The above reactions occur during discharge of the lead storage battery, that is, when we are drawing current from it. For charging the battery, the exact reverse of these reactions happens at the anode and cathode. Also note that during oxidation, electrons are removed, so the oxidation number increases, and during reduction, electrons are added to the species, resulting in a decrease in the oxidation number. Note that oxidation always happens at the anode and reduction always occurs at the cathode.