Question

Rewrite the given equation in its balanced form and find the n factor of ${\mathrm{H}}_2{\mathrm{SO}}_4$ :

${\mathrm{PbO}}_2+\mathrm{Pb}+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{PbSO}}_4+{\mathrm{H}}_2\mathrm{O}$?

Answer 1

Balancing the equation:

Step 1: Given the reaction:

The given reaction with mentioning the physical state of the compounds is shown as:
$\mathrm{Pb}\left(s\right)+\mathrm{Pb}{\mathrm{O}}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\left(\mathrm{aq}\right)\to \mathrm{PbS}{\mathrm{O}}_4\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

Step 2: Assign oxidation number:
$\stackrel{0}{\mathrm{Pb}}\left(s\right)+\stackrel{+4}{\mathrm{Pb}}{\mathrm{O}}_2\left(\mathrm{aq}\right)+\stackrel{+6}{{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\left(\mathrm{aq}\right)}\to \stackrel{+2}{\mathrm{Pb}}S{\mathrm{O}}_4\left(\mathrm{s}\right)+\stackrel{+1}{{\mathrm{H}}_2}\mathrm{O}\left(\mathrm{l}\right)$

Step 3: Writing down two half-cell reactions:

This reaction is an example of a redox reaction where both oxidation reaction and reduction reaction take place simultaneously., for which the half cell reactions can be given as:

Reduction:$\stackrel{+4}{\mathrm{Pb}}+2e^{-}\to \stackrel{+2}{\mathrm{Pb}}$

Oxidation:$Pb\to P{b}^{+2}+2e^{-}$

Step 4: Balancing the number of electrons gained and lost:
$\stackrel{+4}{\mathrm{Pb}}+2e^{-}\to \stackrel{+2}{\mathrm{Pb}}$
$Pb\to P{b}^{+2}+2e^{-}$

Step 5: Balancing both the sides by balancing the coefficients:

Here The number of moles $\mathrm{Pb}$ on the reactant side is 2 , whereas it is given as 1 on the product side, hence we need to add 2 beside ${\mathrm{PbSO}}_4$ to balance the equation as:
$\underset{\left(\mathrm{Lead}\right)}{\mathrm{Pb}}+\underset{\left(\mathrm{Lead}\mathrm{oxide}\right)}{\mathrm{Pb}{\mathrm{O}}_2}+\underset{\left(\mathrm{Sulphuric}\mathrm{acid}\right)}{2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}\to \underset{\left(\mathrm{Lead}\mathrm{sulfate}\right)}{2\mathrm{PbS}{\mathrm{O}}_4}+\underset{\left(\mathrm{Water}\right)}{2{\mathrm{H}}_2\mathrm{O}}$

Step 6: Writing the balanced equation:

The balanced equation can be given as:
$\underset{\left(\mathrm{Lead}\right)}{\mathrm{Pb}}+\underset{\left(\mathrm{Lead}\mathrm{oxide}\right)}{\mathrm{Pb}{\mathrm{O}}_2}+\underset{\left(\mathrm{Sulphuric}\mathrm{acid}\right)}{2{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}\to \underset{\left(\mathrm{Lead}\mathrm{sulfate}\right)}{2\mathrm{PbS}{\mathrm{O}}_4}+\underset{\left(\mathrm{Water}\right)}{2{\mathrm{H}}_2\mathrm{O}}$

In this reaction, one mole of Lead reacts with one mole of Lead oxide and two moles of Sulphuric acid to give the product as two moles of Lead sulfate and two moles of water.

Calculation of n factor of ${\mathrm{H}}_2{\mathrm{SO}}_4$:

• In this case, as salt has been formed, hence the n-factor can be given as the total number of cationic or anion charges replaced by 1 mole of salt.
• In the given reaction, lead loses its two electrons to form lead sulfate, and an oxidation reaction takes place whereas the Lead oxide gains two electrons to form Lead sulfate and a reduction reaction takes place.
• Hence, for 2 moles of${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$ the value of the n-factor is 2.
  and for 1 mole of ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$, the n-factor is given as 1.

Thus, the n factor of ${\mathrm{H}}_2{\mathrm{SO}}_4$is calculated to be $1.$