Question

Balance the following equation :
$PbS+H_2{O}_2\to PbS{O}_4+H_{2}O$

• A. $2PbS+4H_2{O}_2\to 2PbS{O}_4+4H_{2}O$
• B. $PbS+4H_2{O}_2\to PbS{O}_4+4H_{2}O$
• C. $PbS+2H_2{O}_2\to PbS{O}_4+2H_{2}O$
• D. $2PbS+2H_2{O}_2\to 2PbS{O}_4+4H_{2}O$

Answer 1

The correct option is B $PbS+4H_2{O}_2\to PbS{O}_4+4H_{2}O$

Writing oxidation numbers of all the atoms,
$\stackrel{+2-2}{PbS}+\stackrel{+1-1}{H_2{O}_2}\to \stackrel{+2+6-2}{PbS{O}_4}+\stackrel{+1-2}{H_{2}O}$
The oxidation number of $S$ has increased and $O$ has decreased.
$\stackrel{-2}{PbS}\to \stackrel{+6}{PbS{O}_4}$ ...(i)
$\stackrel{-1}{H_2{O}_2}\to \stackrel{-2}{H_{2}O}$ ...(ii)
Increase in oxidation number of $S$ = $8unitsperPbSmolecule$
Decrease in oxidation number of $O=1unitper\frac{1}{2}{H}_2{O}_{2}molecule$
$=2unitsper{H}_2{O}_{2}molecule$
Multiplying eq. (ii) by $4$ as to make increase and decrease equal.
$PbS+4H_2{O}_2\to PbS{O}_4+4H_{2}O$