Question

When a lead storage battery is discharged, then:

• A. $\mathrm{Pb}$ is formed
• B. ${\mathrm{SO}}_2$ is evolved
• C. ${\mathrm{H}}_2{\mathrm{SO}}_4$ is consumed
• D. ${\mathrm{PbSO}}_4$ is consumed

Answer 1

The correct option is C

${\mathrm{H}}_2{\mathrm{SO}}_4$ is consumed

The correct option is (C):

The explanation for the correct answer:

${\mathrm{H}}_2{\mathrm{SO}}_4$ is consumed

1. Lead storage battery is basically a secondary voltaic cell that can be recharged. Here the reactions taking place during the recharging process are reverse of the reactions taking place during the discharge.
2. Anode is actually made from a series of lead plates and the spongy lead. Whereas the cathode is a series of lead plates coated with Lead oxide.
3. And the aqueous Sulfuric acid solution around is basically the electrolyte.
4. Sulfuric acid ionizes to give protons and sulfate ions.
5. At the anode, Lead reacts with the sulfate ions to form Lead sulfate and the two electrons.
6. At the cathode, Lead dioxide reacts with the protons and the sulfate ions to form Lead sulfate and water.
7. The net cell reaction is the reaction between Lead, Lead dioxide, and Sulfuric acid to form Lead sulfate and Water.
8. Reactions for this would be:

$$\begin{gathered} \underset{\mathrm{Hydrosulfuric}\mathrm{acid}}{2{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Hydrogen}\mathrm{ion}}{4{\mathrm{H}}^{+}\left(\mathrm{aq}\right)}+\underset{\mathrm{Sulfate}\mathrm{ion}}{2{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)} \\ \underset{\mathrm{Lead}}{\mathrm{Pb}\left(\mathrm{s}\right)}+\underset{\mathrm{Sulfate}\mathrm{ion}}{{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)}\to \underset{\mathrm{Lead}\mathrm{sulfate}}{{\mathrm{PbSO}}_4\left(\mathrm{aq}\right)}+2{\mathrm{e}}^{-}\left(\mathrm{anode}\right) \\ \underset{\mathrm{Lead}\mathrm{oxide}}{{\mathrm{PbO}}_2(\mathrm{s}})+\underset{\mathrm{Hydrogen}\mathrm{ion}}{4{\mathrm{H}}^{+}(\mathrm{aq}})+\underset{\mathrm{Sulfate}\mathrm{ion}}{{\mathrm{SO}}_2^{-4}}\left(\mathrm{aq}\right)+2{\mathrm{e}}^{-}\to \underset{\mathrm{Lead}\mathrm{sulfate}}{{\mathrm{PbSO}}_4\left(\mathrm{s}\right)}+\underset{\mathrm{Water}}{{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}\left(\mathrm{cathode}\right) \\ \underset{\mathrm{Lead}}{\mathrm{Pb}(\mathrm{s}})+\underset{\mathrm{Lead}\mathrm{peroxide}}{{\mathrm{PbO}}_2\left(\mathrm{s}\right)}+\underset{\mathrm{Hydrosulfuric}\mathrm{acid}}{2{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{aq}\right)}\to \underset{\mathrm{Lead}\mathrm{sulfate}}{2{\mathrm{PbSO}}_4(\mathrm{s}})+\underset{\mathrm{Water}}{2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)}(\mathrm{overall}\mathrm{reaction}) \end{gathered}$$

Therefore, the correct option is (C): When Lead storage is discharged, the Sulfuric acid is consumed.