You probably know by now (through your studies of chemistry) that electrons are the “magic” behind chemical reactions. They do all the hard work – moving here, moving there, joining this atom, leaving that atom, forming bonds, etc.

Understanding chemistry, at its very roots, means understanding how electrons behave. One of the most important concepts in chemistry is understanding what happens to an atom when it gains or loses an electron. Beyond that, there’s a whole class of chemical reactions where these things happen simultaneously – one atom loses electrons and another atom gains them. Because this is a large class of reactions, it’s worthwhile to study these in detail and to really understand how they operate. Don’t worry! You’ve probably heard the expression, “Follow the money?” Well, chemistry can sound confusing, but for this subject at least, it’s all about following the electrons. That’s easy – we can do that!

Our study will look at three different definitions of oxidations and reductions. We’ll start with two of the older, lesser-used definitions, which are still applicable. We’ll then finish with the more modern, most advanced definition of oxidation and reduction. In the end, we’ll use the principles we learn to identify the various components of a redox reaction, including writing “half-reaction”. Half-reactions are a great method to approach redox chemistry because they simplify things, and that’s always good!

Redox Definition One: Oxygen Transfer

One of the oldest definitions of oxidation and reduction involves oxygen transfer. Oxidation is simply the gain of oxygen, while reduction is the loss of oxygen. For example, consider the following chemical reaction:

\(2 \mathrm{Mg}+\mathrm{O}_{2} \rightarrow 2 \mathrm{MgO}\)

The magnesium metal ends up as Magnesium (II) ion.  You may recall that all Group II metals have a stable +2 valence state.  Magnesium metal has been oxidized; it’s gained oxygen.  The \({O}_{2}\) molecule has been reduced; it’s lost an oxygen atom.

Something which causes another molecule to be gain oxygen (be oxidized) is called an oxidizing agent.  Something which causes another molecule to lose oxygen (be reduced) is called a reducing agent.  Because in this reaction we have both reduction and oxidation happening at the same time, this type of reaction is known as a redox reaction.

So, that’s one type of a redox reaction.  Let’s do another practice one, so you can try for yourself!  Consider the following (unbalanced) chemical reaction:

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}+\mathrm{H}_{2} \mathrm{CrO} 4 \rightarrow \mathrm{CH}_{3} \mathrm{C}(\mathrm{O}) \mathrm{OH}+\mathrm{H}_{2} \mathrm{CrO}_{3}\)

Can you tell which starting material is being oxidized, and which is being reduced?  Remember that under this first definition, something which is oxidized gains oxygen.  Something which is reduced loses oxygen.  Notice that the first molecule in the reaction (ethanol) ends up as acetic acid as a product.  Acetic acid has two oxygens.  So, you can say for sure that ethanol has been oxidized!

Because this is a redox reaction, in order for oxidation to take place, reduction also has to be taking place.  Under this definition of redox chemistry, the reduction is the loss of oxygen.  Notice that the second molecule in the reaction above (chromic acid) loses oxygen when it ends up on the product side as chromous acids.  So, ethanol is oxidized by chromic acid, which is then itself reduced.  See how easy that was?  All you had to do is follow the oxygens!

What else can we follow in order to identify a redox reaction?  Well, it turns out we can also follow the number of hydrogens!

Redox Definition Two: Hydrogen Transfer

While redox reactions involving the transfer of oxygens are easy to spot, sometimes redox reactions involve hydrogen transfer, instead.  In this class of redox reactions, oxidation is defined as the loss of hydrogen.  The reduction is defined as the gain of hydrogen.  You probably won’t run across these definitions very often.  They’re a little old-fashioned, but they can still be found in chemistry and so it’s important that you be able to recognize them!

Take the example of propanol being transformed to propanal.

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}\)

Because the starting material has lost a hydrogen, it’s been oxidized.  It can’t accomplish this by itself – in a redox reaction, for something to be oxidized, something has to act as an oxidizing agent (which is in turn, itself, reduced).  Don’t panic if this is starting to get you confused.  It’s really easy.  Some molecules cause other molecules to be oxidized.  These are called oxidizing agents, and in the process of oxidizing another molecule, they become reduced.  Some molecules cause other molecules to be reduced.  These are called reducing agents, and in the process of reducing another molecule, they become oxidized.

What if we wanted to go backwards, and transform our propanal back into propanol?

\(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO} \rightarrow \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\)

This would be a reduction reaction.  The starting material is gaining hydrogen, which by our definition, means it’s being reduced.  We’d need a reducing agent to bring this about.  Lithium aluminium hydride is a good choice.  It likes to donate hydrogens, making it a good reducing agent.  In the process of reducing the propanal, the lithium aluminium hydride becomes oxidized.

Definition Three: Electron Transfer

Oxidation refers to the loss of electrons.  If we take an atom and remove one of its electrons, that atom has been oxidized.  The classic example of this is iron rusting.  We start with iron metal, and then through reaction with oxygen in the air, the iron loses some of its electrons.  We say that the iron has been oxidized.

Reduction refers to the gain of electrons.  If we take an atom and give it an electron, that atom has been reduced.  We can go back to our example of iron rusting.  The iron loses electrons, and the oxygen gains electrons.  We say that oxygen has been reduced.

You can keep these two processes straight by using an old trick: remember the phrase “LEO the lion goes GER”.  LEO = lose electrons oxidized, and GER = gain electrons reduced.  So if you lose electrons, you become oxidized; if you gain electrons, you become reduced.

Often times these two processes take place at the same time.  You have two starting materials and two products.  One of the starting materials becomes oxidized, and the other becomes reduced.  Because reduction and oxidation are taking place in the same reaction, we call these “redox” reactions.  Sometimes one material is more active than the other.  A block of iron, for example, is not a very powerful chemical reagent.  However, oxygen is great at oxidizing other things.  Materials which are good at oxidizing other materials are called oxidizing agents.  Conversely, materials which are good at reducing other materials (causing them to gain electrons, GER) are called reducing agents.

In a redox reaction, an oxidizing agent becomes reduced in the process of oxidizing the other material.  A reducing agent becomes oxidized in the process of reducing the other material.  This is because the location of the electron in question is what determines the labelling.  A good oxidizing agent is something that can strip an electron from another material.  That other material becomes oxidized because it’s lost an electron (LEO).  However!  Because the electron has now joined the oxidizing agent, the oxidizing agent becomes reduced (GER).

Let’s take a look at a simple example, which is the reaction of copper wire with a silver nitrate solution.

\(\mathrm{Cu}+2 \mathrm{AgNO}_{3} \rightarrow \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}+2 \mathrm{Ag}\)

We can take the \(\mathrm{NO}_{3}{ }^{-}\) anions out, as they’re just spectators.  This leaves us with:

\(\mathrm{Cu}+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}\)

The copper started out neutral and ends up with a +2 charge.  Since an electron carries one negative charge, copper must have lost two electrons during the course of this reaction.  Since copper lost electrons, it has become oxidized (LEO).  The silver started out with a positive charge and ended up neutral.  It must have gained an electron during the course of the reaction.  Since silver gained electrons, it has become reduced (GER).

Before we move on to writing half-reactions, let’s review the rules and guidelines we’ve learned so far.

Oxidation means loss of electrons (LEO).  An oxidising agent removes electrons from another substance, causing that substance to become oxidized.  As a result, the oxidising agent gains those electrons, meaning that it becomes reduced (GER).  Or, you could think about it in reverse.  Reduction means the gain of electrons (GER).  A reducing agent gives electrons to another substance, causing that substance to become reduced.  As a result, the reducing agent becomes oxidized (LEO).

See how easy that is?  Just keep track of the electrons, and you’ll always know what’s being oxidized, and what’s being reduced.  The electrons will show you the way!

Now, let’s talk about writing ionic half-reactions.  This is will be the culmination of your skills in using and understanding redox equations.  Don’t worry – just keep your definitions straight, and pay attention to the guidelines coming up, and you’ll be a master of writing redox equations in no time!

Writing Ionic Equations: Half Reactions

Let’s go back to our example of copper wire and silver nitrate.

\(\mathrm{Cu}+\mathrm{Ag}^{+} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Ag}\)

We could split this reaction into two parts, and study both the copper and the silver separately.  This is why we call this technique writing “half-reactions”.  For example, we could split up the above reaction to:

\(\mathrm{Cu} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{e}^{-}\)


\(\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}\)

We have to keep each side neutral, so we add in electrons as needed to balance the charges.  So every redox reaction can be split into two half-reactions: one in which electrons are products (an oxidation process), and where were electrons are reactants (a reduction process).

Let’s now combine the two half-reactions into one reaction, and see what we get.

\(\mathrm{Cu}+\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Ag}+2 \mathrm{e}^{-}\)

Wait for a second…this is an indication that something is wrong.  Can you see it?  The number of electrons doesn’t match!  How can we get them to match?  If we can do that, then we can cross them off the paper, because they’ll be present in equal amounts on both the left and the right.  That would greatly simplify things.  If you look back at the half-reaction you wrote, you see that electrons enter the left side of the combined equation from the silver half-reaction.  Right now, that equation only uses one \(e^{-}\).  We need it to use two so that the total combined reaction will balance.  How can we do that?

Simple!  Just multiply the silver half-reaction by two!

\(2\left(\mathrm{Ag}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{Ag}\right)=2 \mathrm{Ag}^{+}+2 \mathrm{e} . \rightarrow 2 \mathrm{Ag}\)

This now gets us what we need.  Now, combine this new Ag half-reaction with the one for copper, and let’s see what we get.

\(\mathrm{Cu}+2 \mathrm{Ag}^{+}+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}^{2+}+\mathrm{Ag}+2 \mathrm{e}^{-}\)

We can strike the 2 electrons on each side, making the final, balanced equation equal to:

\(\mathrm{Cu}+2 \mathrm{Ag}^{+} \rightarrow \mathrm{Cu}^{2+}+2 \mathrm{Ag}\)

And now you’re done!  You’ve successfully written a redox reaction.  You identified what was being oxidized and what was being reduced, you broke them up into half-reactions, and then you balanced the half-reactions so that the combined reaction balanced.  Great work!

Maybe that one was too easy.  Let’s try another one :).

Suppose you’re told on an exam that bromine liquid can oxidise iron (II) ions to iron (III) ions.  In the process, the bromine is reduced to bromide ions.  How could you write the balanced redox reaction for this?  Well, start with writing the half-reactions!

You’re told that the bromine liquid turns into bromide ions, so start with that.

\(\mathrm{Br}_{2} \rightarrow 2 \mathrm{Br}^{-}\)

Make sure to check that the atoms are balanced before you balance the electrons, or you could be thrown off.

Now, the rules for balancing redox reactions are simple: you can only add electrons, water, hydroxide ions (if the pH is basic), or hydrogen atoms (if the pH is acidic).  In this case with the bromine, we have the atoms balanced, but we need to balance the electrons.  We can add 2 electrons to the left-hand side to fix that!

\(\mathrm{Br}_{2}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}\)

Ok, that one is done.  What about the other half-reaction?  We’re told that iron (II) ions are oxidized to iron (III) ions.  So, let’s write that down.

\(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3^{+}}\)

The only thing we need to do to balance this is to add an electron to the right-hand side.  Then both sides will have an equal charge (+2).

\(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}\)

Now, let’s try combining the two half-reactions and see what we get!

\(\mathrm{Br}_{2}+\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \rightarrow 2 \mathrm{Br}^{-}+\mathrm{Fe}^{3+}+\mathrm{e}^{-}\)

Hmm.  The electrons aren’t balanced.  Like before, we’re going to have to multiply one of the half-reactions to make this right.  We need an extra electron on the right-hand side of the combined reaction so that we end up with two electrons on either side and can strike them both off.  We can do that by multiplying the iron half-reaction by two.

\(2\left(\mathrm{Fe}^{2+} \rightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}\right)=2 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-}\)

Now we can combine this new iron half-reaction with the bromine half-reaction.

\(\mathrm{Br}_{2}+2 \mathrm{e}^{-}+2 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Br}^{-}+2 \mathrm{Fe}^{3+}+2 \mathrm{e}^{-}\)

We can strike out the two electrons on either side and get the final reaction!

\(\mathrm{Br}_{2}+2 \mathrm{Fe}^{2+} \rightarrow 2 \mathrm{Br}^{-}+2 \mathrm{Fe}^{3+}\)

The number of atoms balance on each side, the electrical charge balances on both sides – congratulations, you’ve done it!  You’ve successfully gone through all the steps and have balanced a rather difficult redox reaction.

It’s possible you’ll come across a very difficult redox reaction where there aren’t spectator ions you can remove to make things easier.  In this case, you can use the same basic principles, but there are just a few more guidelines.  This is very uncommon at your level of chemistry, but just in case, here are the guidelines you would follow.

First, break things into the two half-reactions.  Then, balance all the atoms except for oxygen and hydrogen.  Then, you can balance the oxygens by adding water molecules to the side of the half-reaction that needs oxygens.  Because this adds hydrogens to that side (water has hydrogens in it), you can balance the hydrogens by adding hydrogen ions to the other side of that half-reaction.  Finally, balance all the charges in the half-reaction by adding electrons.

Then, combine the two half-reactions, just like before, and multiply one and / or the other by a constant if you need to, in order to the point where you can cross out all the things that are present in equal amounts on both sides.  The result will be the final redox reaction!

Redox reactions are extremely common in chemistry.  Since most chemical manipulations and calculations require a chemical reaction that is balanced, knowing how to balance these common equations is an essential skill.  You now have that skill!  Your life in chemistry has just become a whole lot easier.

Have fun!