Balance the following equation by the oxidation number method. $PbS + H_{2} O_{2} \to {PbSO}_{4} + H_{2} O$

Open in App

Solution

Step 1: Writing the oxidation number of the atoms:
The given equation is:
$PbS + H_{2} O_{2} \to {PbSO}_{4} + H_{2} O$
Thus the oxidation number of each atom are:
$\overset{+ 2 - 2}{PbS} + \overset{+ 1 - 2}{H_{2} O_{2}} \to \overset{+ 2}{Pb} \overset{+ 6}{S} \overset{- 2}{O_{4}} + \overset{+ 1 - 2}{H_{2} O}$
Step 2: Observing the oxidation number of Sulphur and Oxygen
The oxidation number of Sulphur increases from -2 to +6
$Pb \overset{- 2}{S} \to Pb \overset{+ 6}{S} O_{4} . . . . . . . . . . (i)$
On the other hand, the oxidation number of oxygen reduces from -1 to -2
$H_{2} \overset{- 1}{O_{2}} \to H_{2} \overset{- 2}{O} . . . . . . . . (ii)$
$∴$ The increase in the oxidation number of Sulphur is 8 units per $PbS$ molecule
And the decrease in the oxidation number of Oxygen is 1 unit per $\frac{1}{2} H_{2} O_{2}$ molecule $= 2$ unit per $2 H_{2} O_{2}$ molecule
Step 3: Balancing the oxidation numbers
Now to balance the increase and the decrease in the oxidation number we multiply equation (ii) by 4
Thus the resultant reaction becomes:
$PbS + 4 H_{2} O_{2} \to {PbSO}_{4} + 4 H_{2} O$
Therefore, the balanced equation using the oxidation method is $PbS + 4 H_{2} O_{2} \to {PbSO}_{4} + 4 H_{2} O$

Suggest Corrections

Similar questions

C r_{(S)} + P b (N O_{3})_{2 (a q)} \to C r (N O_{3})_{3 (a q)} + P b_{(S)}

Balance the following equation by oxidation number method-

Fe²⁺ + Cr₂O₇^2- + H⁺ ---- Fe³⁺ + Cr^3- + H₂O

C_{6} H_{12} O_{6} + H_{2} S O_{4} \to C O_{2} + S O_{2} + H_{2} O

P b S + H_{2} O_{2} \to P b S O_{4} + H_{2} O

In the given reactions,

$1) I_{2} + H_{2} O_{2} + 2 {OH}^{-} \to 2 I^{-} + 2 H_{2} O + O_{2}$

$2) H_{2} O_{2} + HOCl \to {Cl}^{-} + H_{3} O^{+} + O_{2}$

Join BYJU'S Learning Program