Question

When lead-acid accumulator (battery) is discharged, the following reaction takes place. If $1.25A$ average current is drawn for a period of $1930{\text{mins}}$, the mass of ${H_2}S{O_4}$ consumed is:
A.$196gm$
B.$73.5gm$
C.$98gm$
D.$147gm$

Answer 1

Primary batteries: Those batteries which cannot be rechargeable , are known as primary batteries. For example: dry cells.
Secondary batteries: Those batteries which can be rechargeable, are known as secondary batteries. For example: lead acid.

Complete answer:
Let us first talk about batteries and its types.
Cell: It is defined as an energy source which can deliver dc voltage and current but in very small quantities.
Battery: It is defined as a device which consists of one or more electrochemical cells that are capable of converting chemical energy to electrical energy. The energy stored in batteries are measured in watt-hour i.e. product of voltage across the battery and current in the battery.
There are three components of batteries. Positive electrode which is known as cathode and negative electrode which is known as anode.
Charging current: It is defined as the maximum current which can be applied to charge the battery.
Charging voltage: It is defined as the maximum voltage which can be applied across the battery to charge the battery.
Discharging current: It is defined as the current which is delivered to the load at the time of discharging of the battery.
There are two types of batteries: primary batteries i.e. non-rechargeable and second batteries i.e. rechargeable batteries.
Primary batteries: Those batteries which cannot be rechargeable , are known as primary batteries. For example: dry cells. The cells used in remote are examples of primary cells because they cannot be recharged once they discharge.
Secondary batteries: Those batteries which can be rechargeable, are known as secondary batteries. For example: lead acid. The batteries used in our house are examples of secondary batteries i.e. the batteries which can be recharged.
Here in the question we are given with the secondary battery i.e. lead-acid accumulator.
We know that the mass of deposited sulphuric acid will be equal to the mass of consumed sulphuric acid.
We can calculate the mass of deposited sulphuric acid as $Zit$ where $Z$ is the molar mass of the compound divided by its n-factor and $96500$ $i$ is currently passed and $t$ is the time interval to which current is passed.
Here in the question $i = 1.25A$ and $t = 1930\min s$. And we know that the molecular mass of sulphuric acid is $98$ and its n-factor is $2$ so the value of $Z$ is $\dfrac{{49}}{{96500}}$. So the value of mass deposited or consumed is as:
$w = \dfrac{{49}}{{96500}} \times 1.25 \times 1930 \times 60 = 73.5gm$.

Hence option B is correct.

Note:
The value of n-factor for base is defined as the number of hydroxide ions replaced by one mole of base. For example: the n-factor for sodium hydroxide is $1$. Similarly for acids the number of hydrogen ions replaced by one mole of acid, is known as n-factor of acid. For example: the n-factor of sulphuric acid is $2$.