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Question: In the reaction: \(Pb + Pb{O_2} + {H_2}S{O_4} \to PbS{O_4} + {H_2}O\), the equivalent weight of \(...

In the reaction:
Pb+PbO2+H2SO4→PbSO4+H2OPb + Pb{O_2} + {H_2}S{O_4} \to PbS{O_4} + {H_2}O, the equivalent weight of H2SO4{H_2}S{O_4} is?
A. M
B. M/2
C. 2M
D. M/4

Explanation

Solution

Equivalent weight of the compound is calculated by dividing the molecular weight of the compound with the n-factor. To calculate the equivalent weight first determine the n-factor. This reaction is a type of redox reaction.

Complete step by step answer:
The given reaction is shown below.
Pb+PbO2+H2SO4→PbSO4+H2OPb + Pb{O_2} + {H_2}S{O_4} \to PbS{O_4} + {H_2}O
Balance the given reaction.
Pb+PbO2+2H2SO4→2PbSO4+2H2OPb + Pb{O_2} + 2{H_2}S{O_4} \to 2PbS{O_4} + 2{H_2}O
In this reaction, one mole of lead reacts with one mole of lead oxide and two mole of sulphuric acid to form two mole of lead sulphate and two mole of water.
This reaction is an example of redox reaction where both oxidation reaction and reduction reaction takes place.
The redox reaction is defined as the reaction where transfer of electrons takes place.
In oxidation reaction, a reactant loses its electrons to form a new product.
In reduction reaction, the reactant gains electrons from the other reactant to form a new product.
In the given reaction, lead loses its two electrons to form lead sulphate. Here oxidation reaction takes place and the lead oxide gains two electrons to form lead sulphate. Here reduction reaction takes place.
The equivalent weight is calculated by dividing the molecular weight of the compound by its n-factor.
For 2 moles of H2SO4{H_2}S{O_4} the value of n-factor is 2.
For 1 moles of H2SO4{H_2}S{O_4}, the n-factor is 1.
The formula for the calculation of equivalent weight is shown below.
E.W=MnE.W = \dfrac{M}{n}
Where,
E.W is the equivalent weight.
M is the molecular weight.
n is the number of charges.
To calculate the equivalent weight substitute, the value of n-factor in the above equation.
⇒E.W=M1\Rightarrow E.W = \dfrac{M}{1}
⇒E.W=M\Rightarrow E.W = M
So, the correct answer is “Option A”.

Note:
In this reaction salt is formed therefore for salt the n-factor is defined as the total number of cationic or anion charge which is replaced by 1 mole of salt.