For the following reaction

KClO2->KCl+O2
assign oxidation states to each element on each side of the equation.

Reactants
K=?
Cl=?
O=?

Which element is reduced?

I got K=5, Cl=-1, and O=-4 . The element that reduced is K, but its says its wrong. Help, Please!

If you intended KClO2 (and not KClO3), here is he scoop.

K is +1. It's ALWAYS +1 because it's in group I of the periodic table. Cl is +3 on the left (remember all compounds are 0 so K + Cl + 2O = 0; therefore, 1 + ? + 2*-2 = 0. That makes Cl +3) and O is -2 for EACH oxygen and as you have it as -4 for the total. USUALLY, you want oxidation states for EACH atom. On the right K is +1, Cl is -1 and O2 is 0.
What is reduced? Reduction is the gain of electrons.
So Cl goes from +3 on the left to -1 on the right or gain of 4 e for each Cl.
O goes from -4 total on the left to 0 on the right or loss of 4e.
So Cl is reduced and O is oxidized.
It is important that you know the rules. Here they are.
http://www.chemteam.info/Redox/Redox-Rules.html

K in KClO2: +1

Cl in KClO2: +3
O in KClO2: -2

K in KCl: +1
Cl in KCl: -1
O in O2: 0

O is oxidized, Cl is reduced

Oh, dear! It seems like I may have made a mistake. Let's take another look at it, shall we?

In the reactants:

K = +1 (group 1 elements are always +1)
Cl = +4 (+1 from KClO2 and -1 from ClO2)
O = -2 (usually, oxygen has an oxidation state of -2)

Now, let's check if the net change in oxidation state is consistent on each side of the equation:

On the reactants' side, the total oxidation state is +1 + (+4) + (-2 * 2) = 0.

Now, let's examine the products:

K = +1 (still group 1)
Cl = -1 (group 7 elements are usually -1)
O = 0 (Oxygen generally has an oxidation state of -2, but in O2, it has an oxidation state of 0)

So, it appears that the oxidation state of potassium (K) remains the same on both sides of the equation. The element reduced in this reaction is actually chlorine (Cl), as its oxidation state decreases from +4 to -1.

I apologize for the mix-up earlier. Keep up the good work!

To assign oxidation states to each element on each side of the equation, we need to follow a set of rules:

1. The oxidation state of an element in its elemental form is always zero. Therefore, for the reactants:

- K: The reaction is not in the elemental form, so the oxidation state is not zero.
- Cl: The reaction is not in the elemental form, so the oxidation state is not zero.
- O: The reaction is not in the elemental form, so the oxidation state is not zero.

2. The sum of the oxidation states in a neutral compound is zero. Therefore, for the products:

- KCl: The oxidation state of Cl is -1 in KCl. Since the sum of the oxidation states in a neutral compound is zero (K + Cl = 0), the oxidation state of K is +1 in KCl.
- O2: O2 is a diatomic molecule, and each oxygen atom in O2 has the same oxidation state. Let's assign it as x. Since O2 is a stable molecule, the sum of the oxidation states should be zero. We have 2x = 0, so the oxidation state of O in O2 is 0.

Therefore, the oxidation states for each element on each side of the equation are:

Reactants:
- K: Unknown
- Cl: Unknown
- O: Unknown

Products:
- KCl: K = +1, Cl = -1
- O2: O = 0

To determine which element is reduced, we look for the element that decreases its oxidation number. In this reaction, oxygen (O) is reduced from an oxidation state of 0 on the reactant side to an oxidation state of -2 in KCl on the product side. Therefore, oxygen is the element that is reduced.

Please note that the oxidation states can vary depending on the specific reaction conditions.

To assign oxidation states to each element in a chemical equation, you need to follow a set of rules:

1. The oxidation state of an element in its elemental form is always 0. In this case, since KClO2 is a compound, we don't assign an oxidation state of zero to any element.

2. The sum of the oxidation states of all the atoms in a neutral compound is always 0. In this case, KCl is neutral, so the sum of the oxidation states of K and Cl should be 0.

3. The oxidation state of an alkali metal, such as potassium (K), is typically +1 in compounds.

4. The oxidation state of oxygen (O) in most compounds (except peroxides) is -2.

Using these rules, let's assign the oxidation states for each element:

Reactants:
1. K: We assign it an oxidation state of +1 since it's an alkali metal.

2. Cl: We assign it an oxidation state of -1 since KCl is neutral, and the sum of the oxidation states should be 0.

3. O: We assign it an oxidation state of -2.

So for the reactants in the equation KClO2 → KCl + O2, we have:
K = +1
Cl = -1
O = -2

Now, to determine which element is reduced, we compare the oxidation states of the elements on the reactant side and the product side of the equation. In this case, the oxidation state of K goes from +1 in KClO2 to 0 in KCl, which means it has been reduced.

Therefore, the correct answer is that potassium (K) is the element that is reduced.