Question

What is the density of solution of sulphuric acid used as an electrolyte in lead accumulator\[?\]
A) $1.5\,g\,{L^{ - 1}}$
B) $1.2\,g\,{L^{ - 1}}$
C) $1.8\,g\,{L^{ - 1}}$
D) $2.0\,g\,{L^{ - 1}}$

Answer 1

Hint: Lead accumulators are generally present in lead acid batteries and in these accumulators, electrodes are made of lead and the electrolyte consists of dilute sulphuric acid. The mass percentage of sulphuric acid present in the accumulators is $38\% $ by weight, hence calculate the density.

Complete step-by-step answer:
Lead accumulators are generally present in lead acid batteries .These batteries use sponge lead and lead peroxide for the conversion of the chemical energy into electrical power. These lead acid batteries are most commonly used in power stations and substations because they have higher cell voltage and lower cost.

The total chemical reaction can be written as:
$Pb\left( s \right) + Pb{O_2}\left( s \right) + 2{H_2}S{O_4}\left( {aq} \right) \to 2PbS{O_4}\left( s \right) + 2{H_2}O\left( l \right)$

As we can see from the equation, the electrolyte present in these kinds of batteries is dilute sulphuric acid. The mass percentage of sulphuric acid present in the accumulators is $38\% $ by weight. Since only sulphuric acid is present in the aqueous medium, the density of the solution will only depend upon the mass percentage of sulphuric acid. Now, for $100\% $ by weight sulphuric acid, the density is $1.83\,g\,{L^{ - 1}}$. Therefore, for $38\% $ by weight sulphuric acid, the density is $1.28\,g\,{L^{ - 1}}$.

Hence the correct answer is (B) $1.2\,g\,{L^{ - 1}}$.

Additional Information: In the charged state, the chemical energy of the battery is stored as the potential difference between pure lead at the negative side and $Pb{O_2}$on the positive side, plus the aqueous sulphuric acid. The electrical energy produced by a discharging lead acid battery can be attributed to the energy released when the strong chemical bonds of water molecules are formed from ${H^ + }$ ions of sulphuric acid and ${O^{2 - }}$ ions of$Pb{O_2}$. Conversely, during charging, the battery acts as a water-splitting device.

Note: The most important step of the question is to know the percentage by weight value of the amount of sulphuric acid present as the electrolyte. If the percentage by weight value is known you can easily calculate the density of the solution present in the lead accumulators of lead acid batteries.