Oxidation states (also called oxidation numbers) tell us which chemicals have lost or gained electrons in a chemical reaction. If an atom loses electrons, its oxidation number will increase. However, if an atom gains electrons, its oxidation number will decrease.
Oxidation states are numbers assigned to ions that show how many electrons the ion has lost or gained, compared to the element in its uncombined state. A positive oxidation state shows that the element lost electrons, whilst a negative oxidation state shows that it gained electrons. They can also be referred to as oxidation numbers
For example, when magnesium oxide (MgO) forms, two electrons from the magnesium atom transfer to the oxygen atom. So magnesium oxide contains the ions Mg2+ and O2-. We say that since the magnesium atom lost two electrons, it has an oxidation state of +2. In the same way, the oxygen atom gains two electrons, so it has an oxidation state of -2.
Unlike ionic charge, we write oxidation states with the sign before the number. For example, +2, or -6.
Variable oxidation state definition
The number of electrons available for bonding in an atom determines its oxidation state. In transition metals, the 4s and 3d electrons are available for bonding. This gives transition metals a unique characteristic: variable oxidation states.
A variable oxidation state is a number that determines the charge on an atom depending on certain conditions.
To understand variable oxidation states, you must imagine that only ionic bonding is possible. If an atom could only lose up to three electrons to make bonds with another atom, you can say that it has 3 variable oxidation states, depending on the atom it bonds to.
- We will explore why transition metals have variable oxidation states.
- We will learn why the redox potential of a transition metal when going from a high oxidation state to a lower one depends on the pH and ligand.
- We will discover the five oxidation states of vanadium through a reaction between vanadate(V) and zinc.
- Finally, we will carry out the ‘silver mirror’ test, or the Tollens' test.
Variable oxidation states of transition metals
As we've already mentioned, the 4s and 3d electrons in transition metals are available for bonding (visit Transition Metals for lots more information). The reason is that there is not a considerable energy gap between the 3d orbital and the 4s orbital. The 4s electrons are lost first. In transition metals, the ionisation energy required to remove the third electron (on the 3d sub-shell) is not much bigger than the ionisation energies required to remove the first two (on the 4s sub-shell).
The graph below shows the first ionisation energies of vanadium (a transition metal), calcium, and scandium (non-transition metals). Notice how the first ionisation energies of vanadium are closer together than in scandium and calcium? As you can see, the small difference between the ionisation energies makes it easy for transition metals like vanadium to have variable oxidation states. Not much energy is required to remove a small number of electrons from transition metals.
Fig. 1 - Ionisation energies of vanadium, calcium, and scandium
The table below shows examples of transition metals and their common oxidation states.
| Ti | V | Cr | Mn | Fe | Co | Ni | Cu |
| | | | | | | +1 |
| +2 | +2 | +2 | +2 | +2 | +2 | +2 | +2 |
| +3 | +3 | +3 | +3 | +3 | +3 | | |
| +4 | +4 | | +4 | | | | |
| +5 | | | | | | |
| | +6 | +6 | | | | |
| | | +7 | | | | |
- The atoms with lower oxidation states can exist as simple ions like Mn2+ or Mn3+. The higher oxidation states can only exist as complex ions covalently bonded to significantly electronegative elements like oxygen. For example, MnO4-.
- The higher oxidation states are oxidising while the lower ones are reducing. Examples include the MnO4- and the Cr2O72- ions which make excellent oxidising agents. On the other hand, Cr2+ and Fe2+ are reducing.
Transition metals are not the only elements on the periodic table with variable oxidation states. Most elements have variable oxidation states!
Oxidation states can be crucial in helping a chemist balance an equation. But it can be tricky to figure out the oxidation state of an element with variable oxidation states, so chemists follow several rules to make it easier. The most straightforward rule is that the sum of the oxidation states of all the elements in a compound must always equal zero. Some other oxidation state rules that chemists follow are listed below. You may already be familiar with some of them.
- Metals always have a positive oxidation state.
- Metals cannot have an oxidation state below zero.
- The oxidation state of a single ion is the same as its ionic charge.
- The lowest oxidation state for non-metals is 8 minus the group number of the element.
Redox potential
The redox potential of a substance tells us the likelihood that a species will be reduced or oxidised. In other words, it tells us how easily a species accepts electrons or how easily it gives up electrons.
When we talk about redox reactions, we are usually only interested in the transfer of electrons, so we only write what we call the half-reaction. In the example of the half-reaction below, the zinc 2+ ion accepts two electrons and gets reduced to zinc with an oxidation number of zero.
Zn2+(aq) + 2e- ⇌ Zn(s) Eº = -76 V
You will also notice we have given this half-reaction an Eº value of -76 V. Voltage (V) is the unit of measure for the electrical potential of a half-reaction. Essentially, it is the difference between the demand for electrons in one half of a half-reaction and the tendency to lose electrons in the other half. E is what we call the electrode potential (or reduction potential). It shows how easily a substance gets reduced. (You can learn more about standard electrode potentials in Electrochemical Cells.)
Half-reactions with positive Eº values go towards the right while half-reactions with negative Eº values move to the left.
The redox potential of transition metal ions when going from a high oxidation state to a lower one depends on two things:
pH
What do we mean by pH? When transition metal ions in an aqueous solution go through a redox reaction, it usually involves hydrogen ions. In other words, it requires acidic conditions. You can see this clearly in the example of manganese(VII).
The half-reaction below shows that the manganate(VII) ion easily readily reduces in an acidic solution. As you can see from the positive Eº value, the process moves towards the right. You can also see that manganese gets reduced from the +8 to +2 oxidation state.
MnO4-(aq) + 8H+(aq) + 5e- ⇌ Mn2+(aq) + 4H2O(l) Eº = +1.51 V
We use an excess of acid to ensure this reaction goes to completion.
On the other hand, if you use a neutral solution like water, manganese ions only get reduced to the +4 oxidation state. Notice how the Eº value of the following reaction is much lower? The reason is that the manganese ions are less willing to accept electrons in a neutral solution.
MnO4- (aq) + 2H2O (l) + 3e- ⇌ MnO2 (s) + 4OH- (aq) Eº = +0.59 V
Ligand
How do ligands affect the redox potential of transition ions? Notice the Eº values of the half-reactions below. They separately show the reduction of the nickel(II) ion and the hexaaminenickel(II) ion. What can you conclude about the ligands by looking at the Eº values?
[Ni(H2O)6]2+ (aq) + 2e- ⇌ Ni (s) + 6H2O (l) Eº = -0.26 V
[Ni(NH3)6]2+ (aq) + 2e- ⇌ Ni (s) + 6NH3 (aq) Eº = -0.49 V
By comparing the above two equations, we can state the following:
Eº becomes increasingly negative when ammonia replaces the water ligands.
The ammonia ligands are more firmly attached to the nickel ions than the water ligands.
The process with nickel(II) is more positive than with ammonia. So the equilibrium lies slightly more to the right.
Variable oxidation state of vanadium
You can observe the variable oxidation states of vanadium in a reaction between vanadate(V) ions and zinc.
Vanadium has four oxidation states: vanadium (II), (III), (IV), and (V). Vanadium(IV) is the most stable oxidation state. We can form the vanadium species in various oxidation states through a redox reaction between vanadate(V) ions and zinc. In that reaction, the vanadate ions get reduced by zinc in an acidic solution.
For the reaction, we use ammonium vanadate, a white powder, that dissolves in hydrochloric acid to produce a yellow-coloured solution. When we add zinc to the solution, vanadium gets reduced from +5 to +2. You can tell this is happening because the solution changes colour from yellow to blue to green to violet.
The image below shows the colours of the solution as the vanadium gets reduced from the +5 to +2 oxidation state.
You can see the species of vanadium present in the solution as it changes colour from yellow to blue to green to violet.
You will also notice that the higher oxidation states of vanadium do not exist as simple ions such as V5+(aq) or V4+(aq).
Fig. 2 - Variable oxidation states of vanadium
- The vanadium(II) ions quickly oxidise in the air when you remove the zinc, because they are unstable.
- If you experiment, you may briefly observe a pale green colour as the solution goes from yellow (+5) to blue (+4). What you see is not a new oxidation state, but a mixture of the two colours as the vanadium(V) ions get reduced to vanadium(IV).
Now that you know what is happening between the molecules, consider the half-reactions that show the stages of the reaction between vanadate(V) and zinc:
Stage 1
2VO2+ (aq) + 4H+ (aq) + Zn (s) ⟶ 2VO2+ (aq) + 2H2O (l) + Zn2+ (aq)
The half-equations for the above reaction are below. You get the ionic equation above by putting the two half-reactions for vanadate(V) and zinc together and balancing them out. Notice how the half-equation for zinc moves towards the left because of its negative Eº value.
VO2+ (aq) + 2H+ (aq) + e- ⇌ VO2+ (aq) + H2O (l) Eº = +1.00 V
Zn (s) ⇌ Zn2+ (aq) + 2e- Eº = -0.76 V
2VO2+ (aq) + 4H+ (aq) + Zn (s) ⟶ 2VO2+ (aq) + 2H2O (l) + Zn2+ (aq)
Find out how to write an ionic equation in Balancing Equations!
Stage 2
VO2+ (aq) + 2H+ (aq) + e- ⇌ V3+ (aq) + H2O (l) Eº = +0.34 V
Stage 3
V3+ (aq) + e- ⇌ V2+ (aq) Eº = -0.26 V
We can also use tin as the reducing agent instead of zinc to achieve the same results. As long as vanadium has the more positive Eº value, the reaction will proceed in the direction where the vanadium(V) ion gets reduced to the vanadium(II) ion.
The 'silver mirror' test is another reaction that uses the variable oxidation states of a complex ion. Read on to discover why we call it that!
Variable oxidation state of dichromate ions
As mentioned previously, transition metal ions can have a variety of oxidation states. We have gone through this for vanadium, so now we will be exploring dichromate ions.
Firstly, let's explore how the dichromate(VI) ion, Cr2O72− can be reduced to Cr3+ and Cr2+ ions. This can be done using zinc and a dilute acid such as sulphuric acid or hydrochloric acid. Cr2O72− is orange and when it is reduced using zinc and a dilute acid, it can form Cr3+ (green), which can be further reduced to Cr2+ (blue). This is represented with the following equations.
This equation shows the reduction from +6 to +3.
Cr2O72- + 14H+ + 3Zn → 2Cr3+ + 7H2O + 3 Zn2+
This shows the reduction from +3 to +2.
2Cr3+ + Zn → 2Cr2+ + Zn2+
Now we shall explore how dichromate ions can be produced from the oxidation of Cr3+. This is done using hydrogen peroxide in alkaline conditions which is then followed by acidification. When a transition metal in a low oxidation state is in an alkaline solution, it is more easily oxidised than when it is in an acidic solution. This can be seen in the following equation:
[Cr(H2O)6]3+ (aq) → [Cr(OH)6]3- (aq) in excess sodium hydroxide (NaOH)
[Cr(OH)6]3- can then be oxidised by warming it with a hydrogen peroxide solution to produce a yellow solution of of chromate ions.
The reduction can be seen here:H2O2 + 2e- → 2OH-The oxidation can be seen here: [Cr(OH)6]3- + 2OH- → CrO42-+ 3e- + 4H2O
This then leads to the following equation:
2[Cr(OH)6]3- + 3H2O2 → 2CrO42- + 2OH- + 8H2O
Finally, let us explore how dichromate(VI) ion Cr2O72−, can be converted into chromate(VI). Chromate can be converted to dichromate using this equilibrium equation:
2CrO42- + 2H+ ⇌ Cr2O72- + H2O
It is important to note that this reaction is not a redox reaction. This is because the oxidation number is alway +6. is instead an acid base reaction.
CrO42- is a yellow solution and can be turned into Cr2O72-, an orange solution by adding dilute sulphuric acid. To change from the orange solution to the yellow solution, we need the addition of sodium hydroxide.
Reduction of Tollens' reagent
We carry out a test to distinguish between an aldehyde or ketone by using the complex ion diamminesilver(I). This test, also called Tollens' test, is one of the ways we use to identify the functional group in an unknown organic compound. Diamminesilver(I) [Ag(NH3)2]+ is also known as Tollens' reagent after the German chemist Bernard Tollens.
The test involves reducing Tollens' reagent, which contains silver(I) nitrate, to metallic silver. In order to carry out the test, you must first prepare Tollens' reagent. We prepare it for each test since Tollens' reagent is unstable in solution.
To prepare Tollens' reagent:
Add some sodium hydroxide to silver nitrate to produce silver(I) oxide, a brown precipitate.
Add concentrated ammonia solution to redissolve the silver(I) oxide back into diamminesilver(I).
Now the Tollens' reagent is ready for Tollens' test. To carry out the test:
Add a few drops of the unknown organic compound to Tollens' reagent.
Then gently warm in a water bath.
If the unknown compound is a ketone, you will observe no change in the colourless solution.
If the unknown compound is an aldehyde, you will get a grey silver precipitate. The Ag+ ions have reduced to Ag while the aldehyde oxidised to carboxylic acid.
We can observe the silver precipitate when the substance contains an aldehyde because aldehydes are reducing agents and reduce the silver(I) nitrate to metallic silver. The Tollens' test is also called the 'silver mirror' test because of the silver coating formed inside the test tube.
- Tollens' test can also be used to detect sugars like glucose.
- Previously, we have used this reaction to coat mirrors with silver.
You can find the half-equations for the reduction of Tollens' reagent by an aldehyde below, as well as the net ionic equation.
Reduction of Tollens' reagent
Ag(NH3)2+ + e- ⟶ Ag + 2NH3
Oxidation of aldehyde
RCHO + 3OH- ⟶ RCOO- + 2H2O + 2e-
Net ionic equation
2Ag(NH3)2+ + RCHO + 3OH- ⟶ 2Ag + 4NH3 + RCOO- + 2H2O
Redox titrations with manganese
Potassium manganate (VII) is a violently strong oxidiser in acidic conditions - it can even break carbon-carbon bonds! Today, we use it as an oxidising agent to synthesise organic compounds such as aldehydes, ketones, and carboxylic acids.
Potassium manganate (VII) is also frequently used to calculate the concentration of reducing agents in redox titrations. Essentially, potassium manganate (VII) oxidises any redox couple with an Eº value less than +1.33V. We will briefly examine two such processes with MnO4- ions:
- With iron (II) and iron (III) ions.
- With ethanedioate ions.
- A redox couple is an oxidising and reducing agent that appears on opposite sides of a half-equation.
- Potassium manganate (VII) is also commonly called potassium permanganate. Permanganate is the name we give to manganese in the +7 oxidation state.
In these redox reactions, manganese gets reduced from +7 to +2.
MnO4- + 8H+ + 5e- ➔ Mn2+ + 4H2O
We carry out these redox reactions as titrations. Always set up the experiment with potassium permanganate in the burette. We place the other solution in a flask with dilute sulphuric acid. You must use dilute sulphuric acid in these titrations because other acids can act as oxidising agents in side reactions that will give inaccurate readings.
Manganese(II) appear colourless in solution. As soon as one drop of excess MnO4- is added, the solution turns a visible pale pink, indicating the endpoint of the titration.
Would you like to know the procedure for these titrations? Keep reading in Titrations!
Variable Oxidation State of Transition Elements - Key takeaways
- The oxidation state (or oxidation number) of an atom is the same as the charge on the atom if you assume all the bonds are ionic.
- Transition elements show variable oxidation states because their 3d and 4s electrons are available for bonding.
- The redox potential of a substance tells us the likelihood that a species will be reduced or oxidised. In other words, it tells us how easily a species accepts electrons or how easily they give up electrons.
- The redox potential of a transition metal depends on the pH and the ligands when going from a high oxidation state to a lower one.
- Transition metal ions readily accept electrons in an acidic solution.
- Ligands that are more firmly attached to the central ion lower the redox potential.
- We can form the vanadium species in variable oxidation states through a redox reaction between vanadate (V) ions and zinc.
- 'Tollens' test' is one of the ways we use to identify the functional group in an unknown organic compound. It involves reducing Tollens' reagent, which contains silver(I) nitrate, to metallic silver. An aldehyde will turn the Ag+ ions to Ag molecules to show a grey-silver precipitate.
- Potassium manganate(VII) acts as a strong oxidising agent in acidic conditions.
- In redox titrations, potassium manganate(VII) is reduced from +7 to +2.
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