Galvanic Cells

Why Redox Reactions Are So Important

Let’s think about the redox reaction here:

Remember that we balanced this by separating it into two half-reactions and balancing those half-reactions separately. Here are the balanced half-reactions before we matched the number of electrons:

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

\(Fe^{+2} \rightarrow Fe^{+3} + e^{-1}\)

The fact that this reaction can be broken into two separate halves tells us two very important, and related things.

1. Redox reactions can violate the collision theory
2. The manganese reaction could happen with ANY reaction that fed it electrons, not just the iron reaction here, and vice versa.

Let’s examine those two issues, one at a time.

Redox reactions can violate the collision theory

Remember that the collision theory (explained fully here) states that “In order to react, particles must collide with enough energy and with correct orientation.”

We’re going to focus our discussion here on the first part of that statement; “particles must collide”. As explained here, this is because in order to form the bonds necessary in the products, the atoms must be together.

What makes redox reactions different is that NONE of the elements from the first half-reaction are required to form the products in the second half-reaction and vice versa. That means that the need for collision has been eliminated.

In fact the only thing that is transferred from one half reaction to the other is electrons.

The Manganese reaction could happen with ANY reaction

What this tells us is that the two half-reactions that make up an overall redox reaction can be swapped and exchanged in a virtually infinite number of reactions. As long as one half-reaction takes electrons and the other loses them, the pairing will work.

Putting These Ideas Together

The product of these ideas has two different names. It is sometimes called a Galvanic Cell, after Luigi Galvani, who first demonstrated the idea by using redox reactions to cause contraction in a frog’s leg. Alternatively, it is called a Voltaic Cell, named after Allesandro Volta, who demonstrated the same idea in an inorganic setting with a force meter. Neither man fully understood what was happening however.

Let’s use the reaction above to understand how such a cell works.

First, we’ll need to picture the construction of the cell. Although there are several ways that a cell can be constructed, we’ll begin with a class cell construction. This consists of two beakers, each filled with a solution that contains all of the ingredients for one of the half reactions.

Into each beaker, we’ll submerge a wire,and then connect those two wires with a third. This will allow the transfer of electrons from one half-reaction to the other.

Now, if the reactions occur, as shown below, electrons will move from the iron beaker to the manganese beaker.

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

\(Fe^{+2} \rightarrow Fe^{+3} + e^{-1}\)

As you know, electrons moving through a wire is electricity. That means that we can use that flow of electrons to do electric work, like lighting a light bulb or running a cell phone.

Of course, there is a hitch. As shown, this cell won’t work. We are missing a crucial piece.

The Problem

Before we discuss what is missing, it is important to understand the problem with our system as shown.

The simplistic way to understand the issue is to remember that in order for electricity to flow, you must have a complete circuit. The cell, as drawn, does not have that. Keep in mind that we can’t just put another wire running between the beakers. Given that the reaction on the left is taking electrons and the reaction on the right is giving them, another wire would just allow more electrons to flow in the same direction. It would NOT complete the circuit.

To really understand what is going on, we need to think about those moving electrons and the charge they carry. As electrons move into the beaker on the left, the charge in that beaker will become more and more negative. As a result, electrons will become less and less likely to continue moving into the beaker. Very quickly in fact, electrons will cease to enter at all. At the same time, as electrons leave the beaker on the right, it will become more and more positive, making it less and less likely that electrons will continue to leave.

The Solution

What is needed then, is a way to balance charge. Specifically, we need a way to keep the beaker on the left from becoming too negative and the beaker on the right from becoming too positive.

We do this by adding a salt bridge. This is a piece of glass tubing, bent into a U-shape. It is then filled with a solution of a salt (an ionic compound) and capped at each end by a porous plug (we can use a piece of cotton for this purpose).

If we use potassium chloride (KCl) for the salt, we know that, in solution, we will find separate ions of potassium (\(K^{+1}\)) and chloride (\(Cl^{-01}\)).

The cotton plugs serve two purposes: 1) they allow us to turn the salt bridge upside down without spilling the solution and 2) they will allow ions to slowly travel in and out of the tube.

Inverting the salt bridge into our two beakers, will allow the cell to operate.

As the electrons move from the right to the left the ions in the salt bridge will migrate into the beakers to maintain neutral charge. For each electron that moves into the left beaker, a potassium ion (\(K^{+1}\)) moves into the beaker maintaining the charge. For each electron that leaves the right beaker, a chloride ion (\(Cl^{_1}\)) moves into the beaker maintaining that charge.

There are, of course, a few more details about the  construction of galvanic cells, the notation we use for Galvanic cells and the amount of power we can get from various cells.

Redox Reactions - Example 2

 

Let’s balance the simple redox reaction here:

\(Cr_2O_7^{-2} + Cl^{-1} \rightarrow Cr(OH)_3 + Cl_2\)

Once again, here are the steps for balancing redox reactions: 

1. Break the reaction into half-reactions 
2. Balance everything except H and O 
3. Balance O by adding \(H_2O\) 
4. Balance H by adding \(H^{+1}\) ions 
5. IF the reaction is in basic solution, add enough \(OH^{-1}\) ions to both sides to combine with the \(H^{+1}\) ions 
6. Balance the charge by adding electrons 
7. Match the number of electrons between the two half-reactions 
8. Add the half-reactions and cancel (if possible) 

 Now let’s apply those steps to our reaction.

Step 1 - Break the reaction into half-reactions

This is simple. We choose one thing on the left side of the reaction and match it with the obvious pair on the right side. Then repeat.

\(Cr_2O_7^{-2} \rightarrow Cr(OH)_3\)

\(Cl^{-1} \rightarrow Cl_2\)

Step 2 - Balance everything except H and O

For this reaction, that means making sure that the Cr and Cl are balanced. In this case, neither is. So we balance those elements, giving us:

\(Cr_2O_7^{-2} \rightarrow 2~Cr(OH)_3\)

\(2~Cl^{-1} \rightarrow Cl_2\)

Step 3 - Balance O by adding \(H_2O\)

Here, the chlorine reaction has no oxygen atoms, so there is nothing to do. The chromium reaction however has oxygen atoms. Since the \(Cr_2O_7^{-2}\) has 7 oxygens and the 2 \(Cr(OH)_3\) molecules have 6 oxygens between them, we will add 1 \(H_2O\) molecule to the right side of the reaction. That will give us:

\(Cr_2O_7^{-2} \rightarrow 2~Cr(OH)_3 + H_2O\)

\(2~Cl^{-1} \rightarrow Cl_2\)

Step 4 - Balance H by adding \(H^{+1}\) ions

Again, the chlorine reaction has no hydrogen atoms, so we can ignore that reaction. There are 8 H atoms on the right side of the chromium reaction (thanks to the 2 \(Cr(OH)_3\) molecules and the water molecule we just added). So, we’ll add 8 \(H^{+1}\) ions to the left side of that reaction. That will give us:

\(Cr_2O_7^{-2} + 8~H^{+1} \rightarrow 2~Cr(OH)_3 + H_2O\)

\(2~Cl^{-1} \rightarrow Cl_2\)

Step 5 - IF the reaction is in basic solution, add enough \(OH^{-1}\) ions to both sides to combine with the \(H^{+1}\) ions. 

In this case our reaction is basic. There are two ways that we could know this. This first is that you could be told “this reaction is in basic solution.” The second way to know that a reaction is basic is to see hydroxide ions (\(OH^{-1}\)) in the reaction. Since this reaction contains \(Cr(OH)_3\), we know that this is a basic reaction.

The chlorine reaction has no \(H^{+1}\) ions, so we don’t need to do anything to it. The chromium reaction has 8 \(H^{+1}\) ions, so we will add 8 \(OH^{-1}\) ions to BOTH sides. We add to both sides to maintain the balance of O and H atoms that we had already achieved. This will give us:

\(Cr_2O_7^{-2} + 8~H^{+1} + 8~OH^{-1} \rightarrow 2~Cr(OH)_3 + H_2O + 8~OH^{-1}\)

\(2~Cl^{-1} \rightarrow Cl_2\)

We know that \(H^{+1} + OH^{-1} \rightarrow H_2O\), so we can rewrite the chromium reaction to be:

\(Cr_2O_7^{-2} + 8~H_2O \rightarrow 2~Cr(OH)_3 + H_2O + 8~OH^{-1}\)

We have \(H_2O\) on both sides of the reaction, so we can simplify by subtracting 1 \(H_2O\) from each side, giving:

\(Cr_2O_7^{-2} + 7~H_2O \rightarrow 2~Cr(OH)_3 + 8~OH^{-1}\)

Our two half-reactions now look like this:

\(Cr_2O_7^{-2} + 7~H_2O \rightarrow 2~Cr(OH)_3 + 8~OH^{-1}\)

\(2~Cl^{-1} \rightarrow Cl_2\)

Step 6 - Balance the charge by adding electrons

The left side of the chromium reaction has a charge of -2, while the right side has a charge of -8. To balance those, we’ll need to add 6 electrons to the left side.

\(Cr_2O_7^{-2} + 7~H_2O + 6~e^{-1} \rightarrow 2~Cr(OH)_3 + 8~OH^{-1}\)

The charges for the chlorine reaction are -2 and 0, so we’ll add two electrons to the right.

\(2~Cl^{-1} \rightarrow Cl_2 + 2~e^{-1}\)

Step 7 - Match the number of electrons between the two half-reactions

The chromium reaction requires 6 electrons, but the chlorine reaction only gives 2 electrons. To make that work, we’ll multiply the chlorine reaction by 3. That gives us:

\(Cr_2O_7^{-2} + 7~H_2O + 6~e^{-1} \rightarrow 2~Cr(OH)_3 + 8~OH^{-1}\)

\(6~Cl^{-1} \rightarrow 3~Cl_2 + 6~e^{-1}\)

Step 8 - Add the half-reactions and cancel (if possible)

\(Cr_2O_7^{-2} + 7~H_2O + 6~e^{-1} + 6~Cl^{-1} \rightarrow 2~Cr(OH)_3 + 8~OH^{-1} + 3~Cl_2 + 6~e^{-1}\)

Then cancelling the electrons that appear on both sides, we get the final, balanced reaction:

\(Cr_2O_7^{-2} + 7~H_2O + 6~Cl^{-1} \rightarrow 2~Cr(OH)_3 + 8~OH^{-1} + 3~Cl_2\)

Redox Reaction - Example 1

 

Let’s balance the simple redox reaction here:

\(MnO_4^{-1} + Fe^{+2} \rightarrow Mn^{+2} + Fe^{+3}\)

Once again, here are the steps for balancing redox reactions:

1. Break the reaction into half-reactions
2. Balance everything except H and O
3. Balance O by adding \(H_2O\)
4. Balance H by adding \(H^{+1}\) ions
5. IF the reaction is in basic solution, add enough \(OH^{-1}\) ions to both sides to combine with the \(H^{+1}\) ions
6. Balance the charge by adding electrons
7. Match the number of electrons between the two half-reactions
8. Add the half-reactions and cancel (if possible)

Now let’s apply those steps to our reaction.

Step 1 - Break the reaction into half-reactions

This is simple. We choose one thing on the left side of the reaction and match it with the obvious pair on the right side. Then repeat.

\(MnO_4^{-1} \rightarrow Mn^{+2} \)

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

Step 2 - Balance everything except H and O

For this reaction, that means making sure that the Mn and the Fe are balanced. However, I chose this reaction to be easy, so both are already balanced. So, we still have:

\(MnO_4^{-1} \rightarrow Mn^{+2} \)

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

Step 3 - Balance O by adding \(H_2O\)

Here, the iron reaction has no oxygen atoms, so there is nothing to do. The manganese reaction however has oxygen atoms. Since the \(MnO_4^{-1}\) has 4 oxygens, we will add 4 \(H_2O\) molecules to the right side of the reaction. That will give us:

\(MnO_4^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

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

Step 4 - Balance H by adding \(H^{+1}\) ions

Again, the iron reaction has no hydrogen atoms, so we can ignore that reaction. There are 8 H atoms on the right side of the manganese reaction (thanks to the water molecules we just added). So, we’ll add 8 \(H^{+1}\) ions to the left side of that reaction. That will give us:

\(MnO_4^{-1} + 8~H^{+1} \rightarrow Mn^{+2} + 4~H_2O \)

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

Step 5 - IF the reaction is in basic solution, add enough \(OH^{-1}\) ions to both sides to combine with the \(H^{+1}\) ions. 

In this case, the reaction is NOT basic. Our assumption is that reactions are acidic unless we have specific reason to know that it is basic. Those reasons will be discussed in the second example. Since this reaction is NOT basic, we skip step 5.

Step 6 - Balance the charge by adding electrons

Not only is matter conserved, but so is charge. The left side of the manganese reaction has a charge of +7 (\(MnO_4^{-1}\) and 8 \(H^{+1}\)). The charge on the right side is +2 (\(Mn^{+2}\)). To make those equal, we will add 5 electrons (\(e^{-1}\)) to the left side giving:

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

The charges for the iron reaction are +2 and +3, so we’ll add one electron to the right.

\(Fe^{+2} \rightarrow Fe^{+3} + e^{-1}\)

At this point you should have added electrons to OPPOSITE sides of each reaction. If you haven’t, then you have done something very wrong.

Step 7 - Match the number of electrons between the two half-reactions

The manganese reaction requires 5 electrons, but the iron reaction only gives 1 electron. The only way these can work together is if the iron reaction happens 5 times (to provide the 5 electrons needed for the manganese). That gives us:

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} \rightarrow Mn^{+2} + 4~H_2O \)

\(5~Fe^{+2} \rightarrow 5~Fe^{+3} + 5~e^{-1}\)

Step 8 - Add the half-reactions and cancel (if possible)

Now, we’ll add the two half reactions together and cancel anything that appears on both sides of the reaction. Note: you should ALWAYS be able to cancel the electrons. If you can't, you've done something wrong. So, adding them together we get:

\(MnO_4^{-1} + 8~H^{+1} + 5~e^{-1} + 5~Fe^{+2} \rightarrow Mn^{+2} + 4~H_2O + 5~Fe^{+3} + 5~e^{-1}\)

Then cancelling the electrons that appear on both sides, we get the final, balanced reaction:

\(MnO_4^{-1} + 8~H^{+1} + 5~Fe^{+2} \rightarrow Mn^{+2} + 4~H_2O + 5~Fe^{+3}\)

Now, let's look at a more complex example.

Balancing Redox Reactions with the Half-Reaction Method

 

Many redox reactions also fall into another category of reaction. Combustion reactions, single displacement reactions and most decomposition and synthesis reactions all involve electron theft. Because they can be understood as one of the basic types of reactions discussed elsewhere, we can balance them simply.

However, now that we are looking closely at the transfer of electrons within reactions, we will come across electron theft reactions that do not fit one of those old patterns. You will recognize these in several ways. For instance, here is a redox reaction that doesn’t fit anything you’ve seen before:

\(MnO_4^{-1} + Fe^{+2} \rightarrow Fe^{+3} + Mn^{+2}\)

Right away you should see some problems. Here are the big ones that jump out at me:

• Oxygen seems to have disappeared from the reaction, which violates the conservation of matter
• There are ions all over the place and we have only seen ions in reactions as part of acid/base chemistry or as the net-ionic equation of a displacement reaction.
• There is a “connection” issue. Remember that one of the lessons of balancing reactions is that you should balance any element that is alone last. But here, there are two elements that are alone (Fe on BOTH sides of the reaction and Mn on the right). Although it may not seem obvious to you now, this presents a huge problem. - it means that you could have ANY amount of iron and it would seem not to affect the amount of Mn needed.

The Assumptions We’ll Need to Make

To deal with these issues, we’ll need to make a few assumptions

• These reactions occur in aqueous solution. (This will solve the problem of ions, since we know that ions can float freely in water as described here.) In addition, the water will solve our “oxygen problem”
• These reactions generally occur in acidic solution. Although this is not universal, the presence of \(H^{+1}\) ions will also help with our missing matter.
• It's OK to use \(H^{+1}\) instead of \(H_3O^{+1}\)

The Steps of the Half-Reaction Method

We balance redox reactions (that are not also one of the 5 basic types) through a process called the half-reaction method. This involves breaking the redox reaction into the reduction and oxidation, balancing each of those separately and then recombining them by matching the electrons transferred.

Here are the steps we follow:

1. Break the reaction into half-reactions
2. Balance everything except H and O
3. Balance O by adding \(H_2O\)
4. Balance H by adding \(H^{+1}\) ions
5. IF the reaction is in basic solution, add enough \(OH^{-1}\) ions to both sides to combine with the \(H^{+1}\) ions
6. Balance the charge by adding electrons
7. Match the number of electrons between the two half-reactions
8. Add the half-reactions and cancel (if possible)

Let's look at two examples. The first example is relatively straightforward and simple, using only some of the rules. The second is more complex, but uses all of the rules. If you understand these two examples, you should be able to balance anything.

The Vocabulary of Electron Theft

 

Oxidation and Reduction

Chemists use a new vocabulary to discuss reactions that involve electron theft. To understand that vocabulary, we’ll look at a relatively simple reaction:

\(2 ~ AgNO_3 + Cu \rightleftharpoons Cu(NO_3)_2 + 2 ~ Ag\)

As discussed in the opening page of this unit, this reaction involves the silver ions stealing electrons from the copper atom. We can think about these as separate occurrences:

The copper atom loses two electrons and becomes a +2 ion

\(Cu \rightarrow Cu^{+2} + 2~e^{-1}\)

Each silver ion takes in one electron and becomes a silver atom

\(Ag^{+1} + e^{-1} \rightarrow Ag\)

Let’s look at the silver first. The charge on the silver goes from +1 down to 0. This is called reduction, which is easy to remember since the charge was reduced. This reduction occurred because the silver took in an electron. Thus the official definition of reduction is the gain of one or more electrons.

The copper charge starts at 0 and rises to +2. This process is called oxidation. The source of this name is less obvious because it comes from a time when we didn’t even know that electrons existed. If a piece of copper (charge = 0) is left exposed to the air for a long time (think of an old statue) it will turn green. This is the color of copper oxide. Since the process involves the copper reacting with oxygen, chemists called the process oxidation. Later, we realized that the reaction involves the copper losing electrons and the definition of oxidation became the broader statement that it is today. Oxidation is the loss of one or more electrons.

Two mnemonics to remember the definitions:

LEO the lion says GER

• LEO: Loss of Electrons is Oxidation
• GER: Gain of Electrons is Reduction

OIL RIG

• OIL: Oxidation Is Loss (of electrons)
• RIG: Reduction Is Gain (of electrons)

Since no atom loses electrons unless they are stolen, all reactions involve BOTH reduction and oxidation. As a short-cut, chemists call these redox reactions.

Oxidizing and Reducing Agents

The other piece of vocabulary looks at cause and effect. Let’s continue using the reaction

\(2 ~ AgNO_3 + Cu \rightleftharpoons Cu(NO_3)_2 + 2 ~ Ag\)

The copper was oxidized, that is, the copper lost electrons. But why? No atom ever gives away electrons. The cause is simple. The silver ions took them. Since the silver ions caused the copper to be oxidized, we call the silver ions the oxidizing agent.

We can also view the reaction from the other direction. The silver ions were reduced (they gained electrons). The source of those electrons was the copper, so the copper is the reducing agent.

Simply put,

• The oxidizing agent causes something else to be oxidized (by taking electrons)
• The reducing agent causes something else to be reduced (by giving electrons)

Be sure to notice that the oxidizing agent is the thing that is reduced and the reducing agent is the thing that is oxidized.

Of course, this example was designed to be simple, but redox reactions can be much more complex.

Ammonium Nitrate Disasters

  Most chemical reactions involve the transfer of electrons from one element or compound to another. In most of these reactions (combustion for example) we can ignore the transfer of electrons as a primary cause and still understand what happened.

However, there are some reactions that can ONLY be understood in terms of electron transfer, and one of these in particular has made headlines a disturbing number of times.

Let’s take a look at the compound ammonium nitrate (\(NH_4NO_3\)). If we carelessly assign oxidation numbers to this compound, we might do the following:

H = +1 (Rule 3), and O = -2 (Rule 4), N = x

\(0 = 4(1) + 2(x) + 3(-2)\), which when rearranged would give \(N=+1\). However, that is really WRONG.

The reason it is wrong is that this compound is ionic and each ion has a separate “identity”. If we re-assign oxidation numbers to the two separate ions, we get something very different.

\(NH_4^{+1}\) : H = +1, N = x

\(N = -3\)

\(NO_3^{-1}\): O = -2, N = x

\(N=+5\)

That means that in this compound, there are some nitrogen ions that have 3 extra electrons and some which are missing 5 electrons. To appreciate this, imagine two toddlers. You take 5 toys away from the first and give three extra toys to the other. Then lock the toddlers in a room together. It is, as you can imagine, a recipe for disaster.

With a little kick of activation energy, the nitrogen atoms will exchange electrons with a HUGE release of energy.

Why is this (dangerous) compound around?

Nitrogen is a key element in plant growth (in both the +5 and -3 forms), however nitrogen in the atmosphere is “locked” up as \(N_2\). This molecule is held together by a triple bond which most plants cannot break. As a result, most plants cannot get the nitrogen they need from the atmosphere. Thus, there is a problem that farmers need to solve.

A few plants can obtain the nitrogen they need from the atmosphere - a process called “fixing” nitrogen. Beans, and other legumes, can do this, as can some soil fungi. As a result, one solution is to mix crops (say beans and corn). These can be mixed in the same field or mixed over time (beans one year and corn the next). However, beans are not nearly as lucrative as corn and other “cash” crops, so farmers, who make barely enough money as it is to survive cannot afford to give fields over to beans on a regular basis. Planting beans in among another crop makes harvesting difficult and expensive.

The other choice is to fertilize fields with nitrogen compounds in the forms that the plants need. The simplest (and therefore least expensive) way to add both forms of nitrogen to a field is the compound ammonium nitrate. As a result, ammonium nitrate is produced, stored, transported and applied in vast quantities.

Disasters involving ammonium nitrate

Texas City Disaster (Wikipedia link)

Galveston County Daily News

On April 16, 1947, a fire started on a ship filled with about 2300 tons of ammonium nitrate in Galveston Harbor. The resulting explosion started a chain reaction of other fires and explosions including oil storage facilities and another ammonium nitrate-containing ship.

The explosion leveled nearly 1,000 buildings on land, blew two small planes out of the sky and shattered windows in Galveston, nearly 16 miles away.

The disaster killed 567 people and is considered the worst industrial accident in US history.

Oklahoma City Bombing - Murrah Federal Office Building (Wikipedia link)

ABC News

On the morning of April 19, 1995, Timothy McVeigh parked a Ryder Rental Truck filled with ammonium nitrate and other explosives in front of the Murrah Federal Office building in Oklahoma City. The bomb was detonated shortly after 9 in the morning, destroying more than a third of the building and killing 168 people and injuring 680 others.

McVeigh and his partner Terry Nichols were anti-government extremists, and chose the date to coincide with the anniversary of the fire that ended the federal siege of the Branch Davidian Complex in Waco. Texas, an event that many in their movement considered the ultimate example of the danger posed by the federal government.

West Fertilizer Company Explosion (Wikipedia link)

NBC News

On April 17, 2013, a fire at the West Fertilizer Company Factory led to an explosion that left a 93 foot wide crater where the factory had been. The explosion destroyed or damaged 150 buildings including a 50 unit apartment building and a nearby nursing home. 15 people were killed and more than 160 were injured. 

Beirut, Lebanon (Wikipedia link)

CNN

On August 4, 2020, a warehouse in the harbor of Beirut, where more than 2750 metric tons was stored without proper safety measures exploded. The United States Geological Survey measured the blast as a 3.3 on the Richter scale while the Jordan Seismological Survey measured it at 4.5. The explosion was heard more than 150 miles away.

The explosion killed at least 207 people, injured more than 7500 and did at least $15 billion in property damage.

The blast left a crater 407 feet wide and 141 feet deep. Homes, as far as 6 miles away, were destroyed or made uninhabitable and an estimated 300,000 people were left homeless. More than half of Beirut city was damaged by the explosion and the city’s largest medical center (Saint George Hospital) was so damaged that the injured had to be treated outside.

Using the McAfoos Rules for Oxidation Numers

Here are the rules in use. Remember that Rule 1 is not only the most important, it is also the LAST thing you do each time.

Rule #1 – The sum of the oxidation numbers = the total charge. 

This rule assigns oxidation numbers to pure elements, and single element ions. Here are some examples:

Rule 1 is also used to assign the "last" oxidation number. That does NOT mean the oxidation of the element on the right (or the left for that matter). It means that once we have assigned oxidation numbers for all but one of the element in the compound or ion, we use math to assign the last.

Let's look at the compound \(CoF_2\). If we know that the F has an oxidation number of -1 (Rule 2 would allow us to know that), we can determine the the Co must have a charge of +2 according to the math below:

\(charge=sum~of~oxidation~numbers\)

The charge on this (or any) compound =  0, and each F is -1. Let's set the oxidation number of Co = x. That gives us:

\(0=x + 2(-1)\)

Rearranging gives us \( x=+2\), so the oxidation number of Co is +2.

For another example, let's look at the compound \(K_2O_2\). We can know (again from rule 2) that the potassium has an oxidation number of +1. We know that the charge is 0, so let's set the oxidation number of oxygen to "y". That gives us:

\(0 = 2(+1) + 2(y)\)

which, when rearranged, gives us \(y=-1\). This is an odd charge for oxygen, but it's common enough that we gave it a name ⎼ this is a peroxide.

Rule #2 – Single charge elements

This rule assigns oxidation numbers to those elements that only have one non-zero charge. The list is short: alkali metals are always +1 (this doesn't include H), alkaline earth metals are always +2, F is always -1. There are some other elements that have only one (non-zero) charge, but you can probably get away without thinking about them. 

Immediately after applying this rule, you should apply Rule #1 (if possible).

Here are some examples:

Rule #3 - Hydrogen is always +1

Remember that this rule is a LOWER priority than the ones above, so we would not apply it for the following:

\(H_2\):    Hydrogen's oxidation # = 0 (Rule 1)

KH:    Hydrogen's oxidation # = -1 (Rule 2)

Now let's look at some examples where Rule 3 does matter:

Rule #4 - Oxygen is always -2

Just like Rule 3, this seems to contradict some of the things we've already done. However, as long as you remember that these rules are in RANKED order, you'll be fine. 

For instance we would NOT use Rule 4 in the following cases:

\(O_2\): Oxygen's oxidation # = 0

\(CaO_2\):    Oxygen's oxidation # = -1

Now, let's look at some examples for which we would use Rule #4:

Rule #5 - The most electronegative element gets its most logical negative charge

This rule is only used occasionally. In fact, you may not even realize that you are using it.

Here's a simple example:

\(NiCl_3\)

None of the previous rules can be applied. We have more than one element (so no Rule 1), Both of these elements have more than one possible charge (Rule 2), there is no Hydrogen (Rule 3) and no Oxygen (Rule 4).

We do know that Chlorine is more electronegative than Ni. The logical charge on Chlorine is -1 (since that makes it isoelectronic with Argon.

That makes our math easy: \(0 = x + 3(-1)\) which gives \(x=+3\). So this is nickel III chloride.

A word of caution: Complex ionic compounds (those with polyatomic ions and especially with polyatomic ions that contain the same element) have to be treated carefully - both in terms of assigning oxidation numbers and physically. This should explain.