Balance the following equation by oxidation number method.
$ PbS + {H_2}{O_2} \to PbS{O_4} + {H_2}O $
Answer
568.5k+ views
Hint :Chemical equations are the reaction equation in which the reactants convert into the products. These chemical equations need to be balanced in order to follow the “Law of conservation of mass” which states that the mass can neither be created nor destroyed. In this solution, we will balance the given chemical equation using the oxidation number method.
Complete Step By Step Answer:
We know that in chemical equations the reactants get converted into the products.
In the oxidation number method, first we identify the compounds undergoing oxidation and reduction individually. Then we write separate equations for them such as one equation of oxidation reaction and second equation for reduction reaction. The oxidation states of each element is monitored in the reactant as well as product side and we see the change in the oxidation number of each element.
Now, we will see the given chemical equation:
$ P{b^{ + 2}}{S^{ - 2}} + H_2^{ + 1}O_2^{ - 1} \to P{b^{ + 2}}{S^{ + 6}}O_4^{2 - } + H_2^{ + 1}{O^{ - 2}} $
In this equation, we can see that Sulphur is being oxidised and oxygen is being reduced.
Oxidation half reaction is:
$ P{b^{ + 2}}{S^{ - 2}} \to P{b^{ + 2}}{S^{ + 6}}O_4^{2 - } $
Increase in oxidation number is $ + 8. $
Reduction half reaction is:
$ H_2^{ + 1}O_2^{ - 1} \to H_2^{ + 1}{O^{ - 2}} $
Decrease in oxidation number for oxygen is one unit per half peroxide molecule. Hence, the decrease in oxidation number is $ 2. $
So, we will multiply the reduction reaction equation by $ 4 $ and add it with the oxidation equation.
Hence, the balanced chemical equation is:
$ PbS + 4{H_2}{O_2} \to PbS{O_4} + 4{H_2}O $ .
Note :
We should remember that while balancing an equation by the oxidation number method, we should take care of the number of moles or units of compounds present on the reactant and product sides. So, we need to balance these chemical equations to follow the “Law of conservation of mass” which states that the mass can neither be created nor destroyed.
Complete Step By Step Answer:
We know that in chemical equations the reactants get converted into the products.
In the oxidation number method, first we identify the compounds undergoing oxidation and reduction individually. Then we write separate equations for them such as one equation of oxidation reaction and second equation for reduction reaction. The oxidation states of each element is monitored in the reactant as well as product side and we see the change in the oxidation number of each element.
Now, we will see the given chemical equation:
$ P{b^{ + 2}}{S^{ - 2}} + H_2^{ + 1}O_2^{ - 1} \to P{b^{ + 2}}{S^{ + 6}}O_4^{2 - } + H_2^{ + 1}{O^{ - 2}} $
In this equation, we can see that Sulphur is being oxidised and oxygen is being reduced.
Oxidation half reaction is:
$ P{b^{ + 2}}{S^{ - 2}} \to P{b^{ + 2}}{S^{ + 6}}O_4^{2 - } $
Increase in oxidation number is $ + 8. $
Reduction half reaction is:
$ H_2^{ + 1}O_2^{ - 1} \to H_2^{ + 1}{O^{ - 2}} $
Decrease in oxidation number for oxygen is one unit per half peroxide molecule. Hence, the decrease in oxidation number is $ 2. $
So, we will multiply the reduction reaction equation by $ 4 $ and add it with the oxidation equation.
Hence, the balanced chemical equation is:
$ PbS + 4{H_2}{O_2} \to PbS{O_4} + 4{H_2}O $ .
Note :
We should remember that while balancing an equation by the oxidation number method, we should take care of the number of moles or units of compounds present on the reactant and product sides. So, we need to balance these chemical equations to follow the “Law of conservation of mass” which states that the mass can neither be created nor destroyed.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

What do you mean by retardation What is its SI uni class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Difference between physical and chemical change class 11 chemistry CBSE

