Question

For the following reaction KClO3---->KCl+1/2 O2 assign oxidation states to each element on each side of the equation...reactants and products. Which element is oxidized? Which element is reduced?

Reactants for K, Cl, and O please.

Answer 1

the given reaction is

KCl0 ---> KCl + 0.5 02

reactants :

consider KCl0

we know that

potassium always have an oxidation number of +1

also

in most oxides oxidation number of oxygen is -2

now

we can see that

there is no overall charge on KCl0

so

total oxidation number = 0

so

1 + oxidation number of Cl + ( -2) = 0

oxidation number of Cl = 2-1

oxidation number of Cl = +1

so

in reactant KCl0

oxidation number of K = +1

oxidation number of Cl = +1

oxidation number of each oxygen = -2

Products :

Consider 02

when an element binds to itself

the oxidation number is zero

so

oxidation number of oxygen is 0

now

consider KCl

K is always +1

so

1 + oxidation number of Cl = 0

oxidation number of Cl = -1

so

in KCl

K = +1

Cl = -1

now

we know that

oxidation is the increase in oxidation number

reduction is the decrease in oxidation number

so

oxygen is oxidized from -2 to 0

chlroine is reduced from +1 to -1