Question

While charging the lead storage battery _________________.
This question has multiple correct options
A.$PbS{O_4}$on anode is reduced to $Pb$
B.$PbS{O_4}$on cathode is reduced to $Pb$
C.$PbS{O_4}$on cathode is oxidized to $Pb$
D.$PbS{O_4}$on anode is oxidized to $Pb{O_2}$

Answer 1

Hint : At anode (positive) attract anions (negative charge) due to the electric potential is forced to release an electron (oxidation) happens, opposite happens at cathode.

Complete step by step solution :
Let’s start with understanding what actually happens at the cathode and anode end in a battery. Anode (positive) anions (negative charge) due to the electric potential is forced to release an electron (oxidation) happens. The cathode accepts this electron and reduces itself. But, we have to note one thing that this happens when the battery is being charged. In case the battery is being used all the reactions are reversed.
Looking above we have a basic understanding of what happens at anode and cathode during charging. Now, moving back to the question in$PbS{O_4}$, $Pb$is having 2+ charge on it so, to convert itself to $Pb$it is necessary that $Pb$ accepts 2 electron therefore reduction of $PbS{O_4}$ to $Pb$ will happen. There are two options in which $PbS{O_4}$ is being reduced but on anode side oxidation happens during charging. So, the correct answer will be $PbS{O_4}$ on cathode is reduced to$Pb$. On the anode side $Pb$is oxidized to$Pb{O_2}$. The reactions are given below.
At Cathode ${\text{PbS}}{{\text{O}}_{\text{4}}}\left( {\text{s}} \right){\text{ + 2}}{{\text{e}}^{{\text{ - }}}}{\text{ > Pb}}\left( {\text{s}} \right){\text{ + S}}{{\text{O}}^{{\text{2 - }}}}_{\text{4}}{\text{}}\left( {{\text{aq}}} \right){\text{ }}\left( {{\text{Reduction}}} \right)$
At anode ${\text{PbS}}{{\text{O}}_{\text{4}}}\left( {\text{s}} \right){\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{}}^{{\text{}}}}{\text{ > Pb}}{{\text{O}}_{\text{2}}}\left( {\text{s}} \right){\text{ S}}{{\text{O}}^{{\text{2 - }}}}_{\text{4}}{\text{ + 4}}{{\text{H}}^{\text{ + }}}{\text{ + 2}}{{\text{e}}^{{\text{ - }}}}\left( {{\text{Oxidation}}} \right)$
Overall reaction ${\text{2PbS}}{{\text{O}}_{\text{4}}}\left( {\text{s}} \right){\text{ + 2}}{{\text{H}}_{\text{2}}}{\text{O}}{{\text{}}^{{\text{}}}}{\text{ > Pb}}\left( {\text{s}} \right){\text{ + Pb}}{{\text{O}}_{\text{2}}}\left( {\text{s}} \right){\text{ + 4}}{{\text{H}}^{\text{ + }}}\left( {{\text{aq}}{\text{.}}} \right){\text{ + 2S}}{{\text{O}}^{{\text{2 - }}}}_{\text{4}}{\text{(aq)}}$
So, the answer to this question is B. $PbS{O_4}$ on cathode is reduced to $Pb$and D. $PbS{O_4}$on anode is oxidized to$PbO2$.

Note : We must know that lead acid batteries can store a high amount of charge and can provide high current for a short amount of time. Lead-acid batteries are capable of being recharged and due to this property it is being used in automobiles. In charged state, each cell contains $Pb\left( s \right)$and$Pb{O_2}\left( s \right)$, during their usage they react with ${H_2}S{O_4}$and become$PbS{O_4}$, $Pb$is present at anode and $Pb{O_2}$is present at cathode.