Question

While charging the lead storage battery,___________.
(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 $Pb$

Answer 1

We need to know the exact working, construction, chemistry involved in it. Knowing this may help us to come to a conclusion within no time. It is used in battery cars.

Complete step by step answer:
The battery which uses sponge lead and lead peroxide for the conversion of chemical energy into electrical power, such type of battery is known as lead battery.
The key usage of lead storage batteries is that it can store lots of charge and will provide high current but only for a short period of time. It can be recharged. The lead storage battery is also known as lead acid batteries. The process of discharging of the stored ion entirely relies on the positive and negative plates and the electrolyte. The positive plate that is cathode is made up of lead peroxide which will be in chocolate brown colour and the negatively charged plate that is anode, a lead sponge which will be in grey in colour. The electrolyte is the mixture of sulphuric acid ($38\%$) and water ($62\%$).
A load is used to connect the two electrodes.
Some of the sulphuric acid splits into ${{H}^{+}}$ and $S{{O}_{4}}^{-}$.
The reaction at cathode:
The ${{H}^{+}}$ moves to the cathode on passing.
$Pb{{O}_{2}}+2H\to PbO+{{H}_{2}}O$
Then, reacts with lead peroxide reacts with the sulphuric acid:
$PbO+{{H}_{2}}S{{O}_{4}}\to PbS{{O}_{4}}+{{H}_{2}}O$
The reaction at anode:
The $S{{O}_{4}}^{-}$ moves towards anode and it gives out 2 electrons which become radical sulphate. The radical sulphate cannot exist alone so, it attacks the lead that is anode:
$Pb+S{{O}_{4}}\to PbS{{O}_{4}}$
As the positive ion takes electrode from electrode and negative ion gives electron to electrode due to this there will flow of current. There is a difference between the potential attributes to the flow of current.
Chemical reaction during charging the lead storage battery.
-The anode and cathode are connected to positive and the negative terminal of DC supply. The molecules of sulphuric acid break down to ${{H}^{+}}$ and $S{{O}_{4}}^{-}$.
The ${{H}^{+}}$ being positively charged moved towards the electrode that is the negative terminal of DC supply. It receives electrons from and forms hydrogen atoms. The hydrogen atom reacts with $PbS{{O}_{4}}$ of cathode to give Pb and sulphuric acid:
$PbS{{O}_{4}}+2H\to Pb+{{H}_{2}}S{{O}_{4}}$
The $S{{O}_{4}}^{-}$ moves towards the electrode connected to the positive terminal of the DC supply. It form radical sulphate which reacts with $PbS{{O}_{4}}$ of anode to give Pb:
$PbS{{O}_{4}}+2{{H}_{2}}+S{{O}_{4}}\to Pb{{O}_{2}}+2{{H}_{2}}S{{O}_{4}}$
From the above question we can conclude that $PbS{{O}_{4}}$ on cathode has been oxidized to $Pb$.

Thus, option C is the correct answer.

Note: The lead acid battery is commonly used in the power stations and substations because it has higher cell voltage and lower cost. It is used in battery cars. Be thorough with the construction, working, the chemistry involved and use it in order to answer questions from this topic very easily and within no time.