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?
The element that is oxidized is the potassium (K), which has an oxidation state of +1 on the reactant side and +2 on the product side. The element that is reduced is the chlorine (Cl), which has an oxidation state of -1 on the reactant side and 0 on the product side. The oxygen (O) has an oxidation state of -2 on both sides of the equation.
To assign oxidation states to each element on each side of the equation, we need to understand some rules:
1. The oxidation state of an element in its elemental form is always 0.
2. The sum of the oxidation states of all the atoms in a compound must be equal to the charge of the compound.
3. In most compounds, oxygen has an oxidation state of -2.
4. The sum of the oxidation states of all the atoms in a polyatomic ion must be equal to the charge of the ion.
5. The oxidation state of hydrogen is usually +1, except when it is bonded to a metal in a binary compound, where it is -1.
Now, let's assign the oxidation states:
Reactant:
- K has an oxidation state of +1 since it is a Group 1 element.
- Cl has an oxidation state of +7 since it is combined with oxygen, which has an oxidation state of -2 (Rule 3).
Product:
- K has an oxidation state of +1 since it is a Group 1 element.
- Cl has an oxidation state of -1 since it is combined with K (Rule 2).
- Each O atom has an oxidation state of -2 since it is combined with K (Rule 3).
Now, to determine which element is oxidized and which is reduced:
- Oxidation is the process of losing electrons, which corresponds to an increase in oxidation state.
- Reduction is the process of gaining electrons, which corresponds to a decrease in oxidation state.
In the given reaction:
- The oxidation state of Cl is reduced from +7 to -1, so Cl is reduced.
- The oxidation state of O is increased from -2 to 0, so O is oxidized.
To assign oxidation states to each element on each side of the equation, we need to understand a few rules:
1. The oxidation state of an element in its pure form (e.g., K, Cl, O) is always zero.
2. The oxidation state of a monatomic ion is equal to its charge (e.g., K+, Cl-).
3. Oxygen is typically assigned an oxidation state of -2, except in peroxides (e.g., H2O2) where it is assigned -1.
4. Hydrogen is typically assigned an oxidation state of +1, except when it is bonded to metals where it is assigned -1.
5. The sum of oxidation states in a neutral compound is zero, while in ions it is equal to the ion's charge.
Let's apply these rules to assign the oxidation states for each element on each side of the equation:
Reactant:
K: Since it is a pure form, the oxidation state of K is 0.
Cl: Since it is a pure form, the oxidation state of Cl is 0.
O: Typically, oxygen has an oxidation state of -2. Therefore, in KClO4, Cl has an oxidation state of +4 because the sum of oxidation states in KClO4 is equal to zero (K: 1 * 0 + Cl: 1 * X + O: 4 * -2 = 0).
Product:
K: Since it is a pure form, the oxidation state of K is 0.
Cl: Since it is a pure form, the oxidation state of Cl is 0.
O: In KCl, Cl still has an oxidation state of 0. Therefore, for each O2 molecule, the oxidation state of O is -2.
Now, let's answer the questions:
Which element is oxidized?
Oxygen (O) is oxidized. Its oxidation state changes from -2 in KClO4 to -2 (two molecules) in the product side, where O2 is formed. The oxidation state of oxygen increases from -2 to 0.
Which element is reduced?
Chlorine (Cl) is reduced. Its oxidation state changes from +4 in KClO4 to 0 in KCl. The oxidation state of chlorine decreases from +4 to 0.