An oxidation state is the sum of - and + charges in an atom, which represents the number of electrons it has donated or accepted. Atoms have an oxidation number of zero, which means they are neutral. The protons (+) in the nucleus balance out the electron cloud (-) that surrounds it. When an atom loses an electron it will have more protons than electrons and will become positive. This ion has an oxidation number of +1. If an atom accepts an electron it will become negative, and will gain an oxidation number of -1.

Here are the rules for finding oxidation numbers:

Here's an easy method to determine oxidation states. Set up the charges as an algebraic expression. Let's find the oxidation states of the different elements in KMnO₄. Using the rules listed above, we know that potassium has an oxidation state of +1. Let's assign x to manganese, since manganese has several oxidation states. We have 4 oxygens with -2 charges each. Set the overall charge equal to zero:

The algebraic expression will be:

Solve for x:

x - 7 = 0

x = 7

Suppose we're working with a polyatomic ion. What would be the oxidation state for chromium in Cr₂O₇^2-, (dichromate ion)? Assign -2 as the oxidation state for oxygen. We don't know chromium's oxidation state, and we have two chromium atoms, so assign them the value of 2x.

Set up your algebraic equation and solve for x. The overall charge for the ion is -2, set your equation equal to -2 instead of 0.

The oxidation state for each chromium in the ion is +6.

Let's do one more example with a polyatomic ion. Find the oxidation states for all of the atoms found in Fe₂(CO₃)₃. In this example, iron and carbon have more than one oxidation state each. What if you do not know the oxidation state for carbon in the carbonate ion? It doesn't matter because we know that the charge of the carbonate ion is -2. Set up your algebraic expression.

Each iron ion has a +3 oxidation state.

Now let's look at the carbonate ion separate from the iron(III) ion:

Carbon has an oxidation state of +4 and oxygen is has an oxidation state of -2.

The Oxidizing and Reducing Agents

Now that you are experienced with determining oxidation states, you can now figure out the oxidizing and reducing agents in a oxidation-reduction reaction. Let's define oxidizing and reducing agents. An "oxidizing agent" oxidizes another substance in electrochemistry. In doing so, the oxidizing agent, sometimes called an oxidizer, becomes reduced. Because the oxidizing agent receives electrons it is also known as an electron acceptor. A "reducing agent" is a substance that will reduce another substance. In doing so, it becomes oxidized. Another definition can be formulated as - a reduction agent is the electron donor in a oxidation-reduction reaction. Reducing agents need to be protected from air because they react with oxygen. Another good way to rmember this is the word "OIL-RIG", meaning Oxidation Is Loss, Reduction Is Gain (of electrons).

Using a number line to help identify substances reduced or oxidized and whether they are oxidizing or reducing agents.

Let's examine a redox reaction and identify which substance is oxidized, which substance is reduced, and the oxidizing and reducing agents.

Determine the oxidation states for the atoms in both the reactants and products.

Cu(s), elemental copper, has a zero oxidation state.

For HNO₃, nitric acid:

For Cu(NO₃)₂, copper(II) nitrate:

Copper is +2, because two nitrates are each -1.

For nitrate ion:

For NO₂, nitrogen dioxide:

For H₂O, water:

Now find the compounds that have changed oxidation states. In our example, copper changed from 0 to +2. The oxidation number increased, which means copper has been oxidized. Since it has been oxidized, copper is our reducing agent. Copper reduced the nitric acid by giving away its electrons to nitrogen in nitric acid. Since copper lost these electrons, it was oxidized (oxidation is loss).

The nitrogen found in HNO₃ went from +5 to +4 in NO₂. We know that the oxidation number of N decreased, so HNO₃ has been reduced. Since nitric acid has been reduced (reduction is gain), it is the oxidizing agent. HNO₃ took the electrons from copper to oxidize the Cu, gaining those electrons.