Question

For the reaction KClO2⟶KCl+O2 KClO2⟶KCl+O2 assign oxidation numbers to each element on each side of the equation. K in KClO2:K in KClO2: K in KCl:K in KCl: Cl in KClO2:Cl in KClO2: Cl in KCl:Cl in KCl: O in KClO2:O in KClO2: O in O2:O in O2: Which element is oxidized? KK OO ClCl Which element is reduced? OO KK Cl

Answer 1

Answer:

K in KClO2 = +1

Cl in KClO2 = +3

O in KClO2 = -2

K in KCl = +1

Cl in KCl = -1

O in O2 = 0

Chlorine is going from +3 to -1 so it is being reduced

Oxygen is going from -2 to 0 so it is being oxidized

Explanation:

Potassium is a constant +1

Chlorine could be -1, +1, +3, +5, or +7

Oxygen can be either -2 or +2

Every reactant has to equal zero if there is not a given charge.

Each element can only have certain charges, for example in the previous answer K = +5. Potassium can only be the charge of +1

Answer 2

Answer:

Explanation:

The formula of the reaction:

            KClO₂ → KCl + O₂

To assign oxidation numbers, we have to obey some rules:

1. Elements in an uncombined state or one whose atoms combine with one another to form molecules have an oxidation number of zero.
2. The charge on simple ions signifies their oxidation number.
3. The algebraic sum of all the oxidation number of all atoms in a neutral compound is zero. For radicals with charges, their oxidation number is the charge.

The oxidation number of K in KClO₂:

                                   K + (-1) + 2(-2) = 0

                                    K-5 = 0

                                    K = +5

The oxidation number of K in KCl:

                                K + (-1) = 0

                                K = +1

The oxidation number Cl in KClO₂ is -1

For Cl in KCl, the oxidation number is -1

For O in KClO₂, the oxidation number is (2 x -2) = -4

For O in O₂, the oxidation number is 0

K moves from an oxidation state of +5 to +1. This is a gain of electrons and K has undergone reduction. We then say K is reduced.

O moves from an oxidation state of -4 to 0. This is a loss of electrons and O has undergone oxidation. We say O is oxidized.