Video Transcript
In this video, we will learn about oxidation numbers, what they are, how to calculate them, and how to use them to identify redox reactions.
Electrons in chemistry can be quite hard to keep track of. We have a good understanding of ionic bonding where anions and cations attract one another, and we have an understanding of covalent bonding where shared electrons sit between atoms bonding them together. However, it’s common for complex ions to also contain covalent bonds, for example, the nitrate anion. And it’s also common that the types of bonding can change because of a reaction. It’s easy enough keeping track of electrons when we’re dealing with purely ionic or purely covalent bonding. But what if we have a mixture? It would help to have a simple system that will keep track of electrons in all circumstances.
This is a carbon atom. A carbon atom can react with an oxygen atom to produce a molecule of carbon monoxide. This is clearly an oxidation reaction, since we’re adding oxygen to the carbon. And carbon monoxide can undergo a further oxidation to form carbon dioxide. But what’s not clear from this is which atoms have more electrons than normal and which have fewer. For this, chemists invented an accounting system that works for covalent bonding and ionic bonding. This system relies on the effective position of electrons in covalent bonds. If the electrons are closer to one atom than another, the atom they’re closest to is considered reduced and the atom they’re further from is considered oxidized.
We can remember this relationship using the mnemonic “OILRIG.” Oxidation is the loss of electrons, and reduction is the gain of electrons. If the electrons in a covalent bond are closer to one atom, we can consider that atom to have gained electrons and therefore be reduced.
We can use the electronegativities of the atoms of the elements in a covalent bond to predict which atom electrons will probably be closest to. The electronegativity of carbon is much lower than the electronegativity of oxygen. This means we’d expect the electrons in carbon oxygen bonds, for instance, in carbon monoxide and carbon dioxide to be closer to the oxygen atoms. In this accounting system that chemists came up with, we imagine that the atom the electrons are closest to actually gains those electrons. So, instead of atoms, we imagine anions and cations.
Using this method, we can identify the carbon in these reactions is going from a neutral atom to an effective two plus ion to an effective four plus ion. This system is not supposed to tell us what atoms or ions are like in reality. Instead, it helps us differentiate between different states of atoms or ions in different circumstances.
The system that chemists came up with is the oxidation state of an atom or ion, which uses oxidation numbers. The term “oxidation state” is more commonly used when we’re referring to the atom or ion itself, while the phrase “oxidation number” is more commonly used to refer to the actual value. However, they’re often used interchangeably; then that’s fine. An unbonded atom, like an atom of carbon in the gas state, is assigned the oxidation number zero. A simple anion is given the oxidation number equivalent to its charge, and we do the same for a simple cation.
It’s crucial to remember that an oxidation state is not identical to a charge. The charge of an atom or ion is its overall charge based on the relative number of electrons and protons, while oxidation state is part of an accounting system that assumes we’re dealing with 100 percent ionic bonding. We write oxidation numbers with the sign on the left-hand side of the number, and we write charges with a sign on the right-hand side. Of course, there are plenty of situations where the oxidation number and the charge of an atom or ion are the same. But we need to distinguish between the two because they have different meanings.
Now, let’s have a look at the oxidation states of atoms or ions in different substances. The elemental forms of the elements can be solid, liquid, or gas, and you can have different allotropes. An atom in an element, bonded or unbonded, is given the oxidation state of zero. If there’s a covalent bond between atoms of the same element, we consider the atoms to be identical. There’s no sensible reason we treat one atom as oxidized and the other as reduced, so we simply give both atoms the oxidation state of zero. The next easiest oxidation states to assign are those of simple ions. First, we examine the charge, identify the value, and then convert to the oxidation state by moving the sign to the opposite side. So, a sodium ion has an oxidation state of positive one, an aluminum ion has an oxidation state of positive three, and a nitride ion has the oxidation state of negative three.
Now what if we’re dealing with simple ionic compounds where the charges of the ions aren’t given? We can use the same understanding we use when we figure out the formulas for ionic compounds using the charges we expect for metals and nonmetals. A neutral ionic compound must have a net charge of zero. Oxidation numbers are based on the same principles of conservation of electrons, so the same applies. For a neutral compound, the sum of the oxidation numbers of the individual components must be zero. We can use our understanding of the common ions formed by sodium, magnesium, and oxygen to figure out the other answers.
Polyatomic ions, ions that contain more than one nucleus, can be more tricky, for instance, the nitrate anion, which has a net charge of one minus. Even though we’ve got a net charge, we can apply the same principle as we did with neutral compounds. The oxidation numbers of the components must sum to the equivalent value of the overall charge. The overall charge is one minus, so the oxidation numbers must sum to negative one. On the left-hand side, we have the unknown oxidation number of nitrogen and three lots of the unknown oxidation number for oxygen. Oxygen is more electronegative than nitrogen. So, in a 100 percent ionic system, we predict that the oxygen’s in this system would have the bonding electrons. Oxygen typically forms two minus ions. So, we can assign the oxygens an oxidation number of negative two.
If we evaluate our expression and rearrange, we end up with an oxidation number for nitrogen of positive five. In equations, you may see oxidation numbers written as numbers or as Roman numerals above the element’s symbol. Just be careful not to go backwards; just because we’ve assigned an oxidation number of positive five to this nitrogen does not mean in reality we’re dealing with a nitrogen five plus ion. Oxidation numbers allow us to compare atoms or ions in different circumstances. They don’t necessarily reflect the charge of atoms or ions in reality.
Let’s do one more example, the permanganate ion. When a metal and a nonmetal react, we expect the nonmetal to form anions and the metal to form cations. So, here, we’d expect oxygen to be in the form of O2−. Again, we’re dealing with an ion with an overall charge of one minus. So the sum of the oxidation numbers must be negative one. With four oxygens, we have four lots of the negative two oxidation number. When we evaluate and rearrange, we end up with an oxidation number for the manganese of positive seven.
Next up, polyatomic ions can form part of ionic compounds just like simple ions. For instance, calcium carbonate, the starting point is to know that the common ion for calcium is calcium two plus. So, to balance, the charge of the carbon ion must be two minus. Doing this simple ion is easy, and we can do the carbonate just like how we’ve done every other polyatomic ion. As before, we’ll assume that oxygen has an oxidation number of negative two. When we evaluate and rearrange, we end up with an oxidation number for the carbon of positive four.
Now that’s all the easy categories for assigning oxidation numbers, let’s move on to the hard one. That’s covalent substances like carbon, carbon monoxide, and carbon dioxide. For pure carbon, the oxidation state is set at zero, but we have to figure out the oxidation numbers for carbon in carbon monoxide and carbon dioxide. To figure those out, we need to remember that the typical oxidation state of oxygen in a compound is negative two. For carbon and carbon dioxide, the oxidation number is positive two. Here, we relied on the fact that the sum of the oxidation numbers is zero. When we do the same for carbon dioxide, we need to be careful to multiply the oxidation number of oxygen by two. This gives us an oxidation number for the carbon of positive four.
But there’s an unusual example, hydrogen peroxide. If the oxygen and hydrogen peroxide had the oxidation number of negative two, then the hydrogens would have to have an oxidation number of positive two. Hydrogen atoms only have one electron. So, it doesn’t make sense to talk about hydrogen in a plus two oxidation state, since that would be the equivalent of losing two electrons. So, what’s actually going on here?
This is the structure of a hydrogen peroxide molecule. Oxygen is much more electronegative than hydrogen, so the electrons in the oxygen hydrogen bonds end up closer to the oxygens. But the electrons in the oxygen-oxygen bond are exactly in the middle. If we try to turn this system 100 percent ionic, this oxygen-oxygen bond poses a problem. It’s not reasonable to say that one atom loses electrons and one atom gains electrons. So in this case, we keep the bond. What we end up with are H+ ions and O− ions. This is why we see oxygen with a negative one oxidation state when it’s in a peroxide.
This is a lot to remember. Thankfully, there’s a system of rules that can help cut to the answer. They can be easily remembered using 123FHOC. The number one refers to group one. In a compound, we assume group one metals will have an oxidation number of positive one. Next, the group two metals, which we assign the oxidation number of positive two. And we assign positive three to members of group three. Remember that if a rule comes first in the sequence, that rule gets priority.
The next rule is that fluorine in a compound will always have the oxidation number of negative one. And the hydrogen rule says the oxidation number for hydrogen will be positive one with the exception of hydrides where we see the H− ion. In this example, it’s easy to figure out we’re dealing with a hydride since we use the group one rule first. If sodium is positive one, then hydrogen must be negative one. In all hydrides, the oxidation number for the hydrogen is negative one. Next is the oxygen rule, which says generally oxygen in compounds will have an oxidation number of negative two, except in peroxides where we see an oxidation state of negative one. And the C in 123FHOC isn’t for carbon; it’s for chlorine. In compounds, chlorine will have an oxidation state of negative one unless it’s bonded to fluorine or oxygen.
If you remember these rules, you should be able to figure out unknown oxidation numbers in unfamiliar compounds. Just beware that these rules don’t work in all cases. Sometimes, we need to do high-level calculations. However, these rules will apply in most circumstances. You’re most likely to need to work out oxidation numbers when you’re looking at reactions. We can analyze the oxidation numbers to work out whether a particular reaction is a redox process or a non–redox process.
Here’s the reaction of magnesium oxide with hydrochloric acid. We can use the mnemonic 123FHOC to remind us which rules we need to observe first. We don’t have any group one metals here, but we can use the two rule to assign the oxidation number for magnesium. Since we’re dealing with magnesium compounds, the oxidation number of magnesium will be positive two. There aren’t any group three elements, and there’s no fluorine. So, we can skip those rules and just look at hydrogen. In HCl and H2O, hydrogen is the first element to have its oxidation number assigned, so we can be confident in assigning it the oxidation number of positive one.
Next, we have the oxygen rule. But in both cases, oxygen is the last element to be assigned in the compound. So, we can ignore the oxygen rule and instead work out the oxidation number of oxygen based on the total oxidation number for the compound. In magnesium oxide, oxygen will have an oxidation number of negative two to counteract the positive two oxidation number of magnesium. And in water, oxygen will again have the oxidation number of negative two balancing the two positive ones of the hydrogens. We need to account for the oxidation numbers of both hydrogens, but we usually only write one above the element’s symbol.
Now we can move on to the chlorine rule. Again, in both compounds where we have chlorine, chlorine is the last element to have its oxidation number assigned. So, in HCl, chlorine must have an oxidation number of negative one, and it’s the same in magnesium chloride. We need each chlorine to have an oxidation number of negative one to balance the positive two of magnesium. But by convention, we only write negative one once. Now, let’s see whether any reduction or oxidation has taken place. The oxidation number for magnesium hasn’t changed. It stayed at positive two, and there’s no change for oxygen, hydrogen, or chlorine. This is, therefore, a non–redox reaction.
Next, we can look at the reaction between elemental calcium and sulfuric acid. We can skip rule one because there are no group one elements and jump to rule two with calcium. But the rules only apply to substances in compounds. Calcium in its elemental form will have an oxidation number of zero. But calcium and calcium sulfate will have an oxidation number of positive two. Next, we jump to the hydrogen rule. We have H2 which is hydrogen in an elemental form. So, the oxidation number of hydrogen here is zero, and in sulfuric acid, it’s positive one.
Next, we have the oxygen rule. In sulfuric acid and calcium sulfate, oxygen is not the last element to have its oxidation number assigned. So, we can use the rule. So, in both cases, the oxidation number for oxygen is negative two. We have no more rules left. So, we have to figure out the oxidation number for sulfur in the old-fashioned way. The oxidation numbers for sulfuric acid must sum to zero. So, we can expand all the oxidation numbers for hydrogen and oxygen. We have an excess of negative six. So, the oxidation state for sulfur must be positive six.
Now, the sulfate group in calcium sulfate is identical to that in sulfuric acid. So, sulfur in calcium sulfate must also have the oxidation number of positive six. In this equation, we can see calcium going from an oxidation number of zero to an oxidation number of positive two. This is an oxidation. And we can also see that hydrogen is reduced, going from an oxidation number of positive one to zero. So, we’ve demonstrated we’re dealing with a redox reaction.
There’s also a special type of redox reaction called a disproportionation. In a disproportionation reaction, we see a single substance undergo reduction and oxidation simultaneously. When hydrogen peroxide degrades, one of the oxygens goes from an oxidation state of negative one to an oxidation state of zero. So, it’s oxidized, while the other one is reduced, going from an oxidation state of negative one to an oxidation state of negative two. Reduction and oxidation are occurring simultaneously to the same chemical.
Now, let’s finish up with the key points. We can think of oxidation number simply as an accounting number equivalent to the charge of an atom or ion in a 100 percent ionic system. There are simple rules we can follow for compounds to figure out oxidation numbers. Group one, group two, and group three elements in compounds will have oxidation numbers of positive one, positive two, and positive three, respectively.
Fluorine will have an oxidation state of negative one. Hydrogen will have an oxidation number of positive one, except in hydrides, where it’ll be negative one. Oxygen will have an oxidation number of negative two except in peroxides, where it will be negative one. And chlorine will have an oxidation state of negative one, unless it’s bonded to fluorine or oxygen. And elements in their pure forms will have oxidation states of zero.
