Question

A lead storage battery is the most important type of secondary cell having a lead anode a grid of lead packed with $Pb{O_2}$ as a cathode. A $38\% $ solution of sulphuric acid is used as electrolyte. (Density $1.294gm{L^{ - 1}}$) The battery holds $3.5L$ of the acid. During the discharge of the battery, the density of ${H_2}S{O_4}$ falls to $1.139gm{L^{ - 1}}$.
1.Write the reaction taking place at the cathode when the battery is in use.
2.How much electricity in terms of Faraday is required to carry out the reduction of one mole of $Pb{O_2}$ ?
3.What is the molarity of sulphuric acid before discharge?
4.Lead storage battery is considered a secondary cell. Why?
5.Write the products of electrolysis when dilute sulphuric acid is electrolyzed using Platinum electrodes.

Answer 1

Hint:This question gives the knowledge about the secondary cell. A secondary cell is the cell which can be charged or recharged a number of times. Secondary cell is also known as a storage battery or a rechargeable battery.

Formula used: The formula used to determine the number of moles as follows:
$n = \dfrac{m}{{{M_m}}}$
Where $n$ is the number of moles, $m$ is the given mass and ${M_m}$ is the molecular mass.
The formula used to determine the molarity of the solution is as follows:
$M = \dfrac{n}{V}$
Where $M$ is the molarity of the solution, $n$ is the number of moles and $V$ is the volume of the solution.

Complete step-by-step answer:
A secondary cell is the cell which can be charged or recharged a number of times. Secondary cell is also known as a storage battery or a rechargeable battery. Examples of secondary cells are lead storage batteries, lithium ion batteries and so forth.
1.The reaction which is taking place at cathode when the battery is in use is as follows:
$Pb{O_2} + 3{H^ + } + HS{O_4}^ - + 2{e^ - } \to PbS{O_4} + 2{H_2}O$
2.The electricity essential to carry out the reduction of one mole of $Pb{O_2}$ is $2$ Faraday.
3.To determine the molarity of sulphuric acid, firstly we will determine the moles of sulphuric acid as
follows:
$ \Rightarrow n = \dfrac{m}{{{M_m}}}$
Substitute given mass as $38$ and molecular formula of sulphuric acid as $98$in the above equation as follows:
$ \Rightarrow n = \dfrac{{38}}{{98}}$
On simplifying, we get
$ \Rightarrow n = 0.387755$
The number of moles of sulphuric acid are $0.387755$.
Now, we will determine the molarity of sulphuric acid as follows:
$ \Rightarrow M = \dfrac{n}{V}$
Substitute number of moles of sulphuric acid as $0.387755$ and volume determined using density as $0.1$ in the above equation as follows:
$ \Rightarrow M = \dfrac{{0.387}}{{0.1}}$
On simplifying, we get
$ \Rightarrow M = 3.877$
Therefore, the molarity of sulphuric acid is $3.877M$.

4.Lead storage batteries are considered as secondary cells because they can be charged or recharged a number of times.

5.The products of electrolysis when dilute sulphuric acid is electrolyzed using Platinum electrodes are oxygen and hydrogen.
The overall reaction for the reactions occurring at cathode and anode is as follows:
$2{H_2}O \to 2{H_2} + {O_2}$

Note:Lead storage batteries, lithium ion batteries are the types of Secondary cells which are also known as storage batteries or rechargeable batteries. Molarity is defined as the number of moles of solute per litres of the solution.