Question

When lead storage battery is charged
A. Lead dioxide dissolves
B. Sulphuric acid is regenerated
C. The lead electrode becomes coated with lead sulphate
D. The amount of sulphuric acid decreases

Answer 1

Hint: As we know although the batteries are reliable, they have a limited life. Because they contain toxic materials that require specific removal methods at the end of their useful life.

Complete step by step answer:
A lead storage battery also termed as a lead-acid battery, is the oldest rechargeable battery. We know that it is one of the most common energy storage devices. We also know that lead-acid batteries have moderate power density and good response time. It is affected by temperature and that is why it must be maintained to achieve maximum life expectancy.
Now let us discuss the charge chemistry of lead-acid batteries in brief: as we know this type of battery can be recharged.

1.) Therefore, during the charged state, each cell contains negative plates of lead (Pb) and positive plates of lead (IV) oxide \[Pb{{O}_{2}}\] in an electrolyte.
2.) And thus, the charging process is driven by the removal of electrons from the positive plate and the force shifts them to the negative plate by the help of a charging source.

Now let us see the chemical reaction for the charged battery:
Reaction on negative plate:
\[PbS{{O}_{4}}\left( s \right)+{{H}^{+}}\left( aq \right)+2{{e}^{-}}\to Pb\left( s \right)+HS{{O}_{4}}{{^{-}}_{{}}}\]
Reaction on positive plate:
\[2PbS{{O}_{4}}\left( s \right)+2{{H}_{2}}O\left( l \right)\to Pb{{O}_{2}}\left( s \right)+HS{{O}_{4}}^{-}{{\left( aq \right)}_{{}}}+3{{H}^{+}}\left( aq \right)+2{{e}^{-}}\]
Now combining these two reactions we will get,
\[2PbS{{O}_{4}}\left( s \right)+2{{H}_{2}}O\left( l \right)\to Pb\left( s \right)+Pb{{O}_{2}}\left( s \right)+2{{H}^{+}}\left( aq \right)+HS{{O}_{4}}^{-}{{\left( aq \right)}_{{}}}\]
Now we can conclude from the above reaction is that the concentration of \[{{H}_{2}}S{{O}_{4}}\] increases as it is generated during the chemical reaction.
Therefore, the correct option is (B).

Note: Do not confuse here option B with option D, because option D is in discharge state where the concentration of \[{{H}_{2}}S{{O}_{4}}\] is consumed and hence it has degenerated.