Question

Equivalent mass of sulphuric acid in lead storage battery is:
A. 49
B. 98
C. 24.5
D. None of these

Answer 1

First calculate the number of electrons per molecule of sulphuric acid in the chemical reaction. This will give you a factor. Then divide molecular mass with n factor to obtain the equivalent mass.

Complete answer:
Lead storage battery is a secondary voltaic cell that can be recharged. The reactions taking place during the recharging process are reverse of the reactions taking place during the discharge. During discharge, chemical energy is converted to electrical energy. During recharging, electrical energy is converted to chemical energy.
Anode is made from a series of lead plates with spongy lead. Anode is a negative electrode. Cathode is a series of lead plates coated with lead dioxide. Cathode is a positive electrode. The aqueous sulphuric acid solution (density around ) is the electrolyte.
During discharge, the following reactions take place.
Sulphuric acid ionizes to give protons and sulphate ions. At anode, lead reacts with sulphate ions to form lead sulphate and two electrons. At cathode, lead dioxide reacts with protons and sulphate ions to form lead sulphate and water. The net cell reaction is the reaction between lead, lead dioxide and sulphuric acid to form lead sulphate and water.
${\rm{2 }}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_{\rm{4}}}\left( {aq} \right){\rm{ }} \to {\rm{ 4 }}{{\rm{H}}^ + }\left( {aq} \right){\rm{ + 2 SO}}_4^{2 - }\left( {aq} \right)\\\ {\rm{Pb}}\left( s \right){\rm{ + SO}}_4^{2 - }\left( {aq} \right){\rm{ }} \to {\rm{ PbS}}{{\rm{O}}_4}\left( s \right){\rm{ + 2 }}{{\rm{e}}^ - }{\rm{ }}\left( {{\rm{anode}}} \right)\\\ {\rm{Pb}}{{\rm{O}}_2}\left( s \right){\rm{ + 4 }}{{\rm{H}}^ + }\left( {aq} \right){\rm{ + SO}}_4^{2 - }\left( {aq} \right){\rm{ + 2 }}{{\rm{e}}^ - }{\rm{ }} \to {\rm{ PbS}}{{\rm{O}}_4}\left( s \right){\rm{ + }}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right){\rm{ }}\left( {{\rm{cathode}}} \right)\\\ {\rm{Pb}}\left( s \right){\rm{ + Pb}}{{\rm{O}}_2}\left( s \right){\rm{ + 2 }}{{\rm{H}}_2}{\rm{S}}{{\rm{O}}_{\rm{4}}}\left( {aq} \right){\rm{ }} \to {\rm{ 2 PbS}}{{\rm{O}}_4}\left( s \right){\rm{ + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O}}\left( l \right){\rm{ }}\left( {{\rm{overall reaction}}} \right){\rm{ }}$
When a lead storage battery is discharged, then sulphuric acid is consumed.
In the above reactions you can say that two electrons are involved for 2 molecules of sulphuric acid. In other words, for each molecule of sulphuric acid, one electron is involved. Hence, the n factor for sulphuric acid is 1.
${\text{Equivalent mass = }}\dfrac{{{\text{molecular mass}}}}{{{\text{n factor}}}} = \dfrac{{98}}{1} = 98$

**Hence, the option B ) is the correct answer.

Note:**
During recharging, sulphuric acid is formed. The overall reaction during charging is the reaction between lead sulphate and water to form lead, lead dioxide and sulphuric acid.