Oxidation-Reduction Reactions

Oxidation-Reduction Reactions Video

Hi, and welcome to this video on oxidation–reduction reactions (or redox reactions, for short).

These reactions involve the transfer of electrons between chemical species and are ubiquitous in nature and synthetic chemistry. When you burn propane in your barbecue, that combustion reaction is a redox reaction. The pesky rust on your car is also formed through a redox reaction. Even your own cells generate energy through redox reactions. So, quite literally, you are surrounded by them.

Let’s start by understanding the terminology. Redox reactions are titled as such because during the reaction one chemical species is oxidized while another is simultaneously reduced. But what does that mean?

When something is oxidized, it has lost electrons and when something is reduced, it has gained electrons. A simple mnemonic to remember this is OIL-RIG: Oxidized Is Lost, Reduced Is Gained.

Oxidation and reduction reactions are always paired. There’s no such thing as just an oxidation reaction. When a chemical species is oxidized, it loses electrons and those electrons have to go somewhere, so they go to reducing another chemical species. Thus, the species that has been oxidized is also called the reducing agent because it provides the electrons for reduction. Similarly, the species that is reduced is called the oxidizing agent because it takes the electrons for oxidation to occur. This terminology can often be confusing, so you definitely want to get comfortable with it.

Let’s walk through a redox reaction with a familiar product, sodium chloride, which can be formed from sodium metal and chlorine gas.

We know that sodium chloride is an ionic compound, which means sodium has transferred an electron to chlorine to form bulk Na plus and Cl minus. Given what we’ve just learned about redox reactions, we can now say that because sodium loses an electron, it is oxidized, and because the chlorine gains an electron, it is reduced.

In a simple reaction like this, it’s easy to track the transfer of 1 electron. However, in unfamiliar and more complex reactions, it becomes challenging to follow the flow of electrons. So, chemists assign an oxidation number (known as “O.N.” or “oxidation state”) to each atom, which generally reflects how electron-rich or -poor that atom is. We simplify the process by assuming that all bonds, even covalent ones, are ionic, meaning the electrons are 100% transferred. Because of this assumption, oxidation numbers do not represent the actual charge on an atom. But by tracking the change in oxidation numbers, we can follow the flow of electrons in a chemical reaction.

To assign oxidation numbers, we’ll need to learn a set of rules. Let’s start by learning the ones necessary to assign the oxidation numbers for the formation of sodium chloride.

Rule 1: The oxidation of any free element is 0. So, in our example, our sodium metal and chlorine gas both have an oxidation number of 0.

Rule 2: The oxidation number of a Group 1A element in a compound is +1. This rule applies to sodium in sodium chloride, so we can assign a +1 oxidation number to sodium.

Rule 3: The oxidation number of a Group 7A(17) element in a compound is -1 unless it is paired with an atom with a larger electronegativity, in which case, it takes on a value to balance the charge of the compound. In our case, chlorine is paired with sodium, which has a much lower electronegativity, so chlorine has an oxidation number of -1.

Now that we have assigned an oxidation number for each element of our reaction, let’s do a final check to make sure the assignments are right using the following rule:

Rule 4: The sum of oxidation numbers for the atoms within a compound must equal the charge of the compound. In our example, sodium chloride is the only compound, and it has an overall charge of 0, which checks out because the oxidation numbers for Na, +1, and Cl, -1, do indeed sum to 0.

By assigning the oxidation numbers to each species, we can see that the sodium atoms were oxidized because their oxidation number increased whereas the chlorine atoms were reduced because their oxidation number decreased. We can also say that sodium acts as the reducing agent and chlorine acts as the oxidizing agent in this reaction.

Now, let’s consider a slightly more complex reaction. What happens when a solution of sodium chloride is added to a solution of silver nitrate? Sodium chloride and silver nitrate are both water-soluble, which means they split into solvated ions.

When the two solutions are mixed, the silver and chloride ions find each other and form silver chloride, a water-insoluble solid. The overall reaction can be written as sodium chloride, which is water-soluble, plus silver nitrate, which is also water-soluble, reacts to form sodium nitrate, which is water-soluble, plus silver chloride, which is a solid.

Is this a redox reaction? To answer that question, we need to know if electrons have been transferred between species. To do that, let’s assign oxidation numbers.

We can begin with sodium chloride as we’ve already assigned these oxidation numbers. Sodium is +1 and chlorine is -1. To assign silver nitrate, we need to know this rule:

Rule 5: The oxidation number of oxygen is -2 unless it is bonded to fluorine or another oxygen.

In the nitrate ion, the three oxygens are all bonded to nitrogen, so each oxygen has an oxidation number of -2. What this says chemically, is that even though the electrons in the bond between nitrogen and oxygen are shared, they spend more of their time with oxygen, so we count them as “belonging” to the oxygen.

To assign the oxidation number for nitrogen, we apply the fourth rule that we discussed earlier. Because we know the overall charge of a nitrate ion is -1 and that the sum of the oxidation numbers must equal the charge of the compound, we can solve for nitrogen’s oxidation state: The overall charge equals three times the oxidation number of oxygen plus the oxidation number of nitrogen. Negative one equals three times negative 2 plus the oxidation number of nitrogen. Positive five equals the oxidation number of nitrogen.

We now know that the oxidation number of nitrogen is +5.

Next let’s assign the oxidation number for silver in silver nitrate, which leads to our sixth rule.

Rule 6: The oxidation number of any monatomic ion is the charge of the ion. When silver nitrate dissolves in water, it splits into a silver ion and a nitrate ion. We know the silver ion has a charge of +1, therefore, the silver must be in an oxidation state of +1 as well.

We’ve assigned the oxidation number to every element on the reactant side, now let’s quickly tackle the right side. Sodium nitrate is water-soluble, so this will exist as sodium and nitrate ions. Sodium is still in a +1 oxidation state and the nitrogen and oxygens are still +5 and -2, respectively, in the nitrate ion.

And lastly, per rule 3, chlorine in silver chlorine is -1, and because the overall compound is neutral, silver must be +1.

Notice that the oxidation state did not change for any of the elements. Thus, no electrons were transferred between species, so this is not a redox reaction!

This was an example of a double-displacement reaction, or metathesis reaction, which are not redox reactions. Along with precipitation reactions, this category includes acid-base reactions and takes the general form AB plus CD reacts to form AD plus CB.

Consequently, if you’re confronted with a reaction of this form, you can quickly identify it as NOT a redox reaction without even assigning oxidation states.

Lastly, let’s wrap up with one more example: the decomposition of potassium chlorate. If you’d like, pause the video and take a second to assign oxidation states on your own.

Alright, let’s try this together now.

Using rule 1, the oxygens on the product side have an oxidation number of 0.

Using rules 2 and 3, we can easily assign oxidation numbers of +1 and -1 to the potassium and chlorine in potassium chloride on the product side.

As there are no exceptions to rule 2, we can also assign an oxidation state of +1 to potassium in potassium chlorate.

Chlorine is bonded to oxygen, which is the more electronegative atom, which means the oxidation number for Cl is not -1, and we must solve for it instead. Knowing oxygen has a -2 oxidation state, we can set up the following equation: The overall charge of potassium chlorate equals the oxidation number of potassium plus the oxidation number of chlorine plus three times the oxidation number of oxygen. Zero equals positive one plus the oxidation number of chlorine plus three times negative 2. Positive 5 equals the oxidation number of chlorine.

In potassium chlorate, chlorine has a +5 oxidation number.

Now let’s look at the overall movement of electrons.

Chlorine’s oxidation number decreased from +5 to -1, so chlorine gained electrons and thus was reduced. Oxygen’s oxidation number increased from -2 to 0, so lost electrons and was oxidized. Notice that in a decomposition redox reaction, the reactant is the oxidizing and reducing agent.

Potassium chloride is described as a strong oxidizing agent, which means it is easily reduced. We can make sense of this by assessing the oxidation numbers. Chlorine, a relatively electronegative element, has a +5 oxidation state because it is bonded to 3 oxygen atoms. Consequently, potassium chlorate readily accepts electrons in order to reduce the chlorine.

Generally, elements on the left side of the periodic table, like group 1 and 2 metals, are easily oxidized because they have low electronegativities. Conversely, elements on the right side, like the oxygen family and halogens, are easily reduced, because they have high electronegativities.

But as you probably noticed, there are often exceptions to the rules.

For example, generally, hydrogen has a +1 oxidation state in compounds. So in water, H2O, hydrochloric acid, HCl, and methane, CH4, and basically every other organic compound, hydrogen has a +1 oxidation number. That’s because, in these cases, hydrogen is bonded to a more electronegative atom. However, hydrogen can form compounds with less electronegative atoms, in which case, its oxidation state is -1! For example, in sodium hydride, NaH, the electrons spend more time around the hydrogen, so we assign an oxidation state of -1 to hydrogen and +1 to sodium. While such compounds are rare, they pop up in synthetic chemistry, so it’s good to be aware of them.

To sum it up, the best approach to solving redox reactions is to learn the rules for assigning oxidation states, but to always stop and consider whether the assignments make chemical sense. Remember, the oxidation state should generally reflect whether the atom is electron-rich or poor, so make sure to look at each chemical bond and consider where the electrons would spend more of their time. This should help you avoid mistakes when working with oxidation states and redox reactions.

Thanks for watching, and happy studying!


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by Mometrix Test Preparation | This Page Last Updated: June 27, 2023

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