Question Video: Calculating the Change in the Oxidation State of Nitrogen in the Oxidation of Graphite Chemistry

Nitric acid is a powerful oxidizing agent and will even react with elemental carbon as the allotrope graphite. The reaction is shown. 3 C graphite + 4 HNO₃ ⟶ 3 CO₂ + 4 NO + 2 H₂O. By what value does the oxidation state of nitrogen change during this reaction?

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Video Transcript

Nitric acid is a powerful oxidizing agent and will even react with elemental carbon as the allotrope graphite. The reaction is shown below. Three C graphite plus four HNO3 produces three CO2 plus four NO plus two H2O. By what value does the oxidation state of nitrogen change during this reaction?

This is a question about oxidation states. The oxidation state indicates the degree of oxidation of an atom, which generally corresponds to the number of electrons lost. As a simple example, we can look at the water molecule in the equation. The oxygen atom has an oxidation state of minus two because it has partially gained two electrons. We still count partially lost or gained electrons when determining the oxidation number. Meanwhile, the hydrogen atoms in this molecule each have an oxidation state of plus one because they’ve each partially lost one electron. Note that a positive oxidation state means electrons have been lost, while a negative oxidation state means electrons have been gained.

In order to answer this question, we need to determine the oxidation state, or in other words the oxidation number, of nitrogen in HNO3 and NO. Thankfully, there are some rules we can follow to help us along the way.

The first rule is that neutral compounds have a combined oxidation number of zero. For example, in our water molecule, combining the two plus one oxidation numbers of the hydrogen atoms with the minus two oxidation number of the oxygen atom gives us zero. Next, the oxidation state of an ion is equal to its charge. Lastly, oxygen’s oxidation state is minus two, except in special cases, such as in peroxides or when bonded to fluorine.

Let’s apply these rules to determine the oxidation state of nitrogen in these two compounds. As a reminder, the phrases “oxidation state” and “oxidation number” are interchangeable. Let’s start with nitric acid. The nitrate ion has a charge of minus one, so it must have an oxidation state of minus one as well. This is not a special case. So each oxygen atom has an oxidation state of minus two. In this ion, there are three oxygen atoms, each with an oxidation state of minus two. When we combine these with the unknown oxidation number of the nitrogen atom, we should end up with minus one, the charge and therefore the oxidation state of the nitrate ion.

Nitrogen’s oxidation state in this compound is plus five. The oxidation states of the nitrogen and the oxygen combine to give the ion an overall oxidation state of minus one. Combined with the plus one from the hydrogen atom, this neutral compound has a combined oxidation number of zero. But we’re most concerned with nitrogen’s oxidation number, plus five. So that’s the number we’ll hold on to as we move on to the second compound.

The second compound is a little bit simpler. This is not a special case for oxygen, so its oxidation state is minus two. This is a neutral compound, so the first rule applies. Its combined oxidation number should be zero. An oxidation number of plus two does the trick. The two oxidation numbers combine to equal zero. So we have obtained our second key number. The oxidation number of nitrogen in the product of this reaction is plus two. Because it is bonded to different atoms, the oxidation number of nitrogen changes from the beginning to the end of the reaction, from plus five to plus two. We can say that the oxidation state changes by a value of negative three; it decreases by three. This is the correct answer.

Since oxidation number is a general indication of the number of electrons lost, a change in oxidation number of negative three means that three electrons have been added to nitrogen from the beginning of the reaction to the end of the reaction. So by what value does the oxidation state of nitrogen change during this reaction? It changes by negative three.

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