Question

Balance the following equation by oxidation number method.
$PbS+H_2{O}_2\to PbS{O}_4+H_{2}O$.

Answer 1

Writing oxidation numbers of all the atoms,
$\stackrel{+2}{Pb}\stackrel{-2}{S}+\stackrel{+1}{H_2}\stackrel{-1}{O_2}\to \stackrel{+2}{Pb}\stackrel{+6}{S}\stackrel{-2}{O_4}+\stackrel{+1}{H_2}\stackrel{-2}{O}$
The oxidation number of S has increased and O has decreased.
$Pb\stackrel{-2}{S}\to Pb\stackrel{+6}{S}{O}_4$ ........(i)
$H_2\stackrel{-1}{O_2}\to H_2\stackrel{-2}{O}$ ........(ii)
Increase in Ox. no. of S$=8$ units per PbS molecule
Decrease in Ox. no. of O$=1$ unit per $\frac{1}{2}{H}_2{O}_2$ molecule
$=2$ units per $H_2{O}_2$ molecule
Multiplying eq. (ii) by $4$ as to make increase and decrease equal
$PbS+4H_2{O}_2\to PbS{O}_4+4H_{2}O$
This is the balanced equation.