Question

During charging of lead storage battery:

• A. Sulphuric acid is regenerated
• B. $Pb{O}_2$ dissolves
• C. Lead sulphate gets deposited over the lead electrode
• D. Concentration of $H_{2}S{O}_4$ decreases

Answer 1

The correct option is A Sulphuric acid is regenerated

Lead Storage Battery:-
When the battery discharge,
At anode: $Pb\left(s\right)+S{O}_4^{2-}\left(aq\right)\to PbS{O}_4\left(s\right)+2e^{-}$
Most of the lead sulphate precipitates formed gets sticks to the lead rod. Thus, it is not consumed during discharging.

At cathode: $Pb{O}_2\left(s\right)+S{O}_4^{2-}\left(aq\right)+4H^{+}\left(aq\right)+2e^{-}\to PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)$
Here also formed lead sulphate precipitate gets sticks
to cathode rod.

Overall reaction during discharge:
$Pb\left(s\right)+Pb{O}_2\left(s\right)+2H_{2}S{O}_4\left(aq\right)\to 2PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)$
$E_{cell}=2.05V$
$H_{2}O$ is produced in the cell reaction.
$H_{2}S{O}_4$ is consumed in the cell reaction during discharge. So, its concentration in the solution decreases and results into decrease in the density of the solution.

On charging the lead storage battery, the reactions are reversed
and produce $H_{2}S{O}_4$

At anode: $PbS{O}_4$ gets converted
into $Pb$

At cathode: $PbS{O}_4$ gets converted into $Pb{O}_2$

Thus, option (c) is correct.