Question

During the discharge of a lead storage battery, density of ${H_2}S{O_4}$ fell from 1.294 to 1.139g /mL. Sulphuric acid of density 1.294 is 39% ${H_2}S{O_4}$ by weight and that of density 1.139 g/ mL is 20% ${H_2}S{O_4}$ by weight. The battery holds 3.5 litre of the acid and volume remains practically constant during discharge. Calculate ampere-hour of which the battery must have been used. The charging and discharging reactions are:
$Pb + SO_4^{2 - } \to PbS{O_4} + 2{e^ - }$ (charging)
$Pb{O_2} + 4{H^ + } + SO_4^{2 - } + 2{e^ - } \to PbS{O_4} + 2{H_2}O$ (discharging)

Answer 1

Hint: In this question first find out the final and initial number of moles of ${H_2}S{O_4}$by adding the charging and discharging equations. By finding the remaining number of moles we can then use it directly to find out the Ampere-hour of which the battery has been used. This approach can be used to navigate the question further.

Complete step-by-step answer:
So adding the charging and discharging equations, we get:
$Pb + Pb{O_2} + 4{H^ + } + 2SO_4^{2 - } \to 2PbS{O_4} + 2{H_2}O$
As we know, Molarity is defined as the number of moles of solute per one liter of solution, whereas Normality is defined as the number of gram equivalents per one liter of a solution.
Here, we can see in the equation that the moles solute is equal to the number of molar equivalents.
So, we can say that:
Molarity of ${H_2}S{O_4}$=Normality of ${H_2}S{O_4}$(As $2SO_4^{2 - }$ requires only 2 electrons)
Now to calculate the number of moles, we will find out the molarity of ${H_2}S{O_4}$before and after electrolysis.

Using Formula: ${M_{{H_2}S{O_4}}} = \dfrac{{Density \times Percentage \times 1000}}{{Molar\,Mass}}$
Now before electrolysis we are given:
Density=1.294, Percentage=39, Molar Mass=98
Putting these values in molarity formula :
M= $\dfrac{{39 \times 1.294 \times 1000}}{{100 \times 98}}$ = 5.15
Since, Molarity = number of moles of solute per one liter of solution
Hence, number of moles(n) =$Molarity \times volume$ (volume of solution is 3.5 liter)
n= $5.15 \times 3.5$= 18.025
Similarly after electrolysis, we have :
Density=1.139, Percentage=20, Molar Mass=98
M= $\dfrac{{20 \times 1.139 \times 1000}}{{100 \times 98}}$ = 2.325
n= $2.325 \times 3.5$= 8.1375
So the remaining number of moles or equivalents of ${H_2}S{O_4}$ is:
18.025-8.1375= 9.8875
Now we have the number of moles remaining as well and we are asked to find out the ampere-hour for which the battery has been used.
So using the formula:
$i \times t = n \times F$
Where, n is the remaining number of moles or equivalents, and F is faraday’s constant.
So putting the values we get:
$i \times t = \dfrac{{9.8875 \times 96500}}{{3600}}$ (Dividing by 3600 to get the units of Ampere-hour)
=265.04 Amp-hr.

Note: Different types of batteries are rated in measurements depending on the operations they are expected to perform. For instance, batteries rated as ampere-hours (AH, also called amp hours) are developed to deliver low current for a long period. We can also use a multimeter to find out the Amp hours of a battery.