Question

Which among the following equations represents the reduction reaction taking place in lead accumulator at positive electrode, while it is being used as a source of electrical energy?

• A. $PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)\to Pb{O}_2\left(s\right)+S{O}_4^{2-}+4H^{+}+2e^{-}$
• B. $PbS{O}_4\left(s\right)+2e^{-}\to Pb\left(s\right)+S{O}_4^{2-}\left(aq\right)$
• C. $Pb\left(s\right)+S{O}_4^{2-}\left(aq\right)\to PbS{O}_4\left(s\right)+2e^{-}$
• D. $Pb{O}_2\left(s\right)+S{O}_4^{2-}\left(aq\right)+4H^{+}\left(aq\right)+2e^{-}\to PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)$

Answer 1

The correct option is D $Pb{O}_2\left(s\right)+S{O}_4^{2-}\left(aq\right)+4H^{+}\left(aq\right)+2e^{-}\to PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)$

In lead storage battery, lead grids filled with spongy lead will act as anode (negative plate) and lead grids filled with $Pb{O}_2$ will act as cathode (positive plate). 38% solution of sulphuric acid will act as the electrolyte for the cell.

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:
$Pb\left(s\right)+Pb{O}_2\left(s\right)+2H_{2}S{O}_4\left(aq\right)\to 2PbS{O}_4\left(s\right)+2H_{2}O\left(l\right)$
$E_{cell}=2.05V$