For the following reaction kclo4 → kcl + 2o2 assign oxidation states to each element on each side of the equation. reactant product k cl o which element is oxidized? which element is reduced?

Oxidation states number (OSN) show how many electrons gained, lost or shared during chemical reactions. 

Here are some rules of oxidation state numbers.

1. OSN of 1A elements in compounds is +1
2. OSN of Oxygen in compounds is -2
3. Sum of OSN of elements equals zero in neutral compounds.
4. OSN of free elements is zero. ex. F2, N2, O2.. all have 0 OSN.

If an element loses electrons during reaction, it is oxidized. That element is called as reducing agent.

If an element gains electrons during reaction, it is reduced. That element is called as oxidizing agent.

In the given reaction, potassium perchlorate decompose into potassium chlorate and oxygen.

a) KClO4.
K (potassium) is 1A element, OSN +1,
Oxygen in the compound has OSN -2.
Let OSN of Cl be x 
1+ x+(4*-2)=0 (since compound is neutral)
Solve the equation. X is equal to +7. 

b)Potassium perchlorate is the reactant, Potassium chloride and Oxygen are the products.
 Reactants > Products

c) Check out change of OSN of elements.
in KCl, OSN number of K is +1, Cl is -1. So Cl+7>Cl-1. It takes 8 electrons. It was reduced.
in KClO4, OSN number of O is -2, in O2 it becomes O. So O-2>O. It gives electrons. It was oxidized.
Shortly,
LOSS OF ELECTRONS=OXIDIZED 
GAIN OF ELECTRONS=REDUCED

Sorry i dont understand the question can you rephase it?