Hint: we use a catalyst that will help us to liberate oxygen at a much lesser temperature and a faster rate as well. It is an oxide of transition element and often used in dry cells, the so-called Leclanche cell. This oxide is surrounded within the positive electrode carbon.
Complete answer:
-A catalyst is basically a substance which causes an increase in the rate of the reaction by decreasing the activation energy without itself being used up in the reaction. They do not alter the end result of the reaction; they only increase the speed with which the reaction occurs. The end product of the reaction remains the same with or without the catalyst.
-Let us now see the decomposition reaction of the potassium chlorate ($KCl{O_3}$). It decomposes into potassium chloride ($KCl$) and oxygen (${O_2}$) in the presence of catalyst manganese dioxide ($Mn{O_2}$).
The reaction is: $2KCl{O_3}\xrightarrow[\Delta ]{{Mn{O_2}}}3{O_2} + KCl$.
-Here we have used manganese dioxide ($Mn{O_2}$) as a catalyst because it helps us to give off oxygen at a much lower temperature and much more readily.
Instead if we had not used $Mn{O_2}$ then also ${O_2}$ would have been released but only at a very high temperature of ${400^ \circ }C$ and that too very slowly.
-$Mn{O_2}$ acts as a very good catalyst because it reduces the activation energy to a much lower level than usual and helps to complete the reaction at a much lower temperature and at a faster rate also.
-So, the catalyst used in the decomposition of potassium chlorate ($KCl{O_3}$) to get potassium chloride (KCl) and oxygen (${O_2}$) is manganese dioxide ($Mn{O_2}$).
Note: Always remember that the presence or absence of a catalyst does not affect the final product of the reaction. A catalyst only increases the rate of the reaction without being involved in it.
