In my chemistry lab we decomposed lead (iv) oxide by heating it and then placing a wooden splint(test the evloved gas) into the mouth of the test tube. The color changed into red lead ( Pb304).Then we reheated the sample again and it turned to PbO. I do not understand why the compound accumulated more atoms and then lost atoms. PbO2---->Pb304----->PbO
The two-step decomposition you described is:
(a) PbO2 --> Pb3O4 + O2
Balancing it,
3PbO2 ---> Pb3O4 + O2
(b) Pb3O4 ---> PbO + O2
or, balanced,
2Pb3O4 ---> 6PbO + O2
In each step, all atoms are accounted for. However the first step must be doubled to reconcile it with the 2nd step.
The chemical reactions you observed in your lab involving lead (IV) oxide (PbO2) can be explained by the process of oxidation and reduction.
1. Decomposition of lead (IV) oxide (PbO2):
When lead (IV) oxide is heated, it decomposes into its constituent elements, lead (Pb) and oxygen (O2). This is an example of a decomposition reaction. The balanced equation is:
PbO2(s) -> Pb(s) + O2(g)
In this reaction, lead (IV) oxide decomposes to form lead metal and oxygen gas. The wooden splint you placed into the test tube is commonly used to test for oxygen. When the splint is inserted into the mouth of the test tube, it ignites due to the presence of oxygen, indicating that oxygen gas was produced in the reaction.
2. Formation of red lead (Pb3O4):
After decomposing lead (IV) oxide, the resulting lead metal reacts with a portion of the oxygen gas present to form red lead oxide (Pb3O4). This is an example of a redox reaction, where oxidation and reduction occur simultaneously. The balanced equation is:
4Pb(s) + O2(g) -> 2Pb2O3(s)
The red lead oxide formed in this reaction has the chemical formula Pb3O4. It is called "red lead" due to its characteristic color.
3. Reduction of red lead to lead (II) oxide (PbO):
When red lead oxide is reheated, it undergoes a reduction reaction, where it loses oxygen atoms and is converted back to lead (II) oxide (PbO). The balanced equation is:
2Pb3O4(s) -> 6PbO(s) + O2(g)
In this reaction, the red lead oxide decomposes to form lead (II) oxide and oxygen gas is released.
In summary, during the decomposition of lead (IV) oxide, it goes through two intermediate steps: the formation of red lead oxide (Pb3O4) and its subsequent reduction to lead (II) oxide (PbO). These reactions are a result of the interplay between the oxidation and reduction processes occurring between the different forms of lead oxides and oxygen.