Redox reactions are short for reduction-oxidation reactions and they are described as a reaction where electron(s) are exchanged between atoms and compounds. In a redox reaction, one reactant undergoes oxidation (loses electrons), while the other undergoes reduction (gains electrons). Understanding redox chemistry is central to inorganic chemistry because many reactions—including precipitation, complexation, and even cation analysis steps—are driven or influenced by changes in oxidation state. A key concept in redox reactions is assigning oxidation states. Oxidation states are the charges assigned to each atom when the bonds are assumed to be ionic. When talking about oxidation states we talk about the atoms separately. For example, we say that oxygen has an oxidation state of 1- instead of saying that hydrogen-peroxide has an oxidation state of 1-. When assigning oxidation states there are some common rules to follow and are listed below:
Reducing agents and oxidizing agents are compounds that oxidize or reduce other compounds well. Reducing agents reduce a compound and thus usually get oxidized. Oxidizing agents, on the other hand, are good at oxidizing other compounds and thus are usually reduced. In order to be able to predict what redox reactions take place we must be familiar with oxidizing and reducing agents.
One of the strongest oxidizing agents is Ozone. It gets reduced to form oxygen gas and water. Oxygen gas can
undergo further reduction too but it is a little less common.
O₃ + 2e⁻ + 2H⁺ → O₂ + H₂O
O₂ + 2e⁻ → 2O²⁻
Hydrogen peroxide is also another oxidizing agent that can change in its function based on the pH of the
solution.
H₂O₂ + 2e⁻ → 2 OH⁻ (basic)
H₂O₂ + 2H⁺ + 2e⁻ → 2H₂O (acidic)
Halides are also very common oxidizing agents. In the following reactions halides with oxygens are also shown
as very good oxidizing agents.
XO₄⁻ + 8e⁻ → X⁻ (halides; tends to be kinetically slow)
XO₃⁻ + 6e⁻ → X⁻ (halides)
XO₂⁻ + 4e⁻ → X⁻ (halides)
XO⁻ + 2e⁻ → X⁻ (halides; fast-acting)
X₂ + 2e⁻ → 2X⁻
Arguably one of the most famous oxidizing agents is permanganate. Remember that permanganate is a purple ion
and as it gets oxidized. There are 2 different cases in which the permanganate ion gets reduced in an acidic
or a basic solution. Because of the change in color permanganate is commonly used for redox titrations and
the change of color marks the endpoint of the titration.
MnO₄⁻ + 5e⁻ → Mn²⁺ (acidic solution)
MnO₄⁻ + 3e⁻ → MnO₂(s) (basic or neutral solution)
Another commonly used anion in redox titrations is the dichromate ion. As it gets reduced it changes from an
orange solution to a green solution marked by the Cr3+ ion. Chromate compounds in general function as good
oxidizing agents.
Cr₂O₇²⁻ + 6e⁻ → 2Cr³⁺ (acidic solution)
CrO₃ + 3e⁻ → 2Cr³⁺
CrO₄²⁻ + 3e⁻ → 2Cr³⁺
Nitrate and other compounds containing nitrogen also undergo reduction commonly as good oxidizing agents.
NO₃⁻ + 4H⁺ + 3e⁻ → NO + 2H₂O (nitrate, acidic)
NO₂⁻ + 2H⁺ + e⁻ → NO + H₂O (nitrite)
Compounds with sulfur can work as oxidizers with varying strength. The two most important and common ones are
Peroxydisulfate(strong oxidizing agent) and Sulfite(weak reducing agent).
S₂O₈²⁻ + 2e⁻ → 2SO₄²⁻
SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻
Transition metals are also good oxidizing agents because they have lots of d-electron orbitals that can
accept electrons easily causing there to be a large variation of oxidation states.
Ce⁴⁺ + e⁻ → Ce³⁺ (used in redox titrations because of the change of color from pale yellow to colorless)
Metal ions can also be great oxidizing agents.
Fe³⁺ + e⁻ → Fe²⁺
Cu²⁺ + e⁻ → Cu⁺
Ag⁺ + e⁻ → Ag(s)
Reducing agents are compounds that cause other compounds to undergo reduction. Metals are the most common
reducing agents as they have a low ionization energy. Alkaline and alkaline-earth metals are really strong
reducing agents so they are typically not found in water. What happens when you place sodium in water? It
explodes. In general metals follow this formula for their oxidation pattern.
M → Mⁿ⁺ + ne⁻
However, here are the most common reducing agents from metals that you must remember. A general theme is that
when elements are at low oxidation states they tend to be good reducing agents and when they are at high
oxidation states they tend to be good oxidizing agents.
Zn → Zn²⁺ + 2e⁻
Fe²⁺ → Fe³⁺ + e⁻
Sn²⁺ → Sn⁴⁺ + 2e⁻
Al → Al³⁺ + 3e⁻
Hydrogen also shows up as a common reducing agent.
H₂ → 2H⁺ + 2e⁻
However, One of the most important reducing agents are compounds with carbon as they show up all the time.
This happens typical when carbon is in a low oxidation state and goes up to a 2+ in CO or goes up to a 4+ in
CO2.
C → CO₂ + 4e⁻
C → CO + 2e⁻
C₂O₄²⁻ → 2CO₂ + 2e⁻
HCOO⁻ → CO₂ + 2e⁻ (when in basic conditions)
Another important element in reducing agents happens to be sulfur. When sulfur is in a low oxidation state it
will reduce a substance to achieve an oxidation state typically close to 6+.
H₂S → S(s) + 2H⁺ + 2e⁻
S²⁻ → S(s) + 2e⁻
SO₃²⁻ + H₂O → SO₄²⁻ + 2H⁺ + 2e⁻
S₂O₃²⁻ → S₄O₆²⁻ + 2e⁻
Nitrogen based reducing agents are also common.
N₂H₄ → N₂ + 4H⁺ + 4e⁻
NH₃ → N₂ (oxidized)
Balancing redox reactions is a very fundamental skill to have in inorganic chemistry and to do this we must keep one principle in mind, the amount of electrons lost is equal to the amount of electrons gained and the law of conservation of charge which states that charge cannot be created. Before we explore the reactions between reducing agents and oxidizing agents, we need to understand how redox half-reactions get balanced in an acidic or basic solution.
Half-reactions are reactions that only involve either reduction or oxidation. To balance these reactions
there are some rules that you should follow. Let's take the example of balancing the orange dichromate ion
being reduced into its yellow chromate ion in an acidic solution.
Cr₂O₇²⁻ (aq) → Cr³⁺ (aq)
The first step is to balance all the atoms on each side of the equation other than H and O. In this case
there are 2 Cr atoms on the left side and only one on the right side so we would balance this by multiplying
the left side by 2. This causes the reaction to actually become,
Cr₂O₇²⁻ (aq) → 2Cr³⁺ (aq)
The next step is to add H₂O to the equation to balance out the Oxygen atoms.
Cr₂O₇²⁻ (aq) → 2Cr³⁺ (aq) + 7H₂O
After that, we need to add H+ ions to balance out the H atoms on each side.
Cr₂O₇²⁻ (aq) + 14H⁺(aq) → 2Cr³⁺ (aq) + 7H₂O
Now notice that the charge on both sides is unbalanced. The sum of the charges on the left side is 14+ and
2- giving a total charge of 12+. While, on the left side the charge is only 2(3+) making 6+. This does not
go with the law of conservation of charge which states that charge cannot just be created but rather, it
must stay consistent. To balance these charges we can add electrons to a side, the side with the higher
charge. In this case we would add 6e- to the left side to make the charge on the left side 6+ and the charge
on the right side 6+. Now the Law of conservation of charge makes sense.
Cr₂O₇²⁻ + 14 H⁺ + 6 e⁻ → 2 Cr³⁺ + 7 H₂O
And that is all you need to do to balance half-redox reactions in an acidic medium but from these steps we
can also leverage this reaction to balance half-reactions in a basic medium. To balance in a basic medium we
just add an OH- for each H+ ion present. In this case we would add 14 OH- to each side. On the left side the
H+ and OH- together would form water.
Cr₂O₇²⁻ + 14 H₂O + 6 e⁻ → 2 Cr³⁺ + 7 H₂O + 14OH⁻
Then cancelling 7H₂O’s from both sides and we end up with
Cr₂O₇²⁻ + 7 H₂O + 6 e⁻ → 2 Cr³⁺ + 14OH⁻
And notice how the charges on both sides are equal, 8-. This is true for all redox reactions since the Law
of conservation of charge applies. This example was about a reduction half-reaction but the process is
almost identical for an oxidation half-reaction only that the electrons will be produced instead of
consumed. Now that we know how to balance Half-reactions we can move on to balancing Full redox reactions
which is just a combination of both half-reactions. For showcasing how to balance the whole reaction we will
use a very popular redox reaction of permanganate and Fe2+. As you may remember, permanganate is a very
strong oxidizing agent and Fe2+ is a reducing agent. When permanganate is reduced a Mn2+ ion is formed.
MnO₄⁻ → Mn²⁺
Balancing this in an acidic medium we get:
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Similarly, Fe(ii) undergoes oxidation to form the Fe3+ ion so balancing this is a very simple addition of an
electron.
Fe²⁺ → Fe³⁺ + e⁻
Now, to combine the half-reactions, we must ensure that the number of electrons lost equals the number of
electrons gained. In the reaction between permanganate and iron(II) in acidic solution, the oxidation
half-reaction is
Fe²⁺ → Fe³⁺ + e⁻
and the reduction half-reaction is
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
Since the permanganate half-reaction involves 5 electrons while each Fe²⁺ only loses 1 electron, we multiply
the Fe²⁺ half-reaction by 5 to match the electrons:
5Fe²⁺ → 5 Fe³⁺ + 5 e⁻
Now we can add the two half-reactions together, including all reactants and products:
5Fe²⁺ → 5 Fe³⁺ + 5 e⁻
+
MnO₄⁻ + 8 H⁺ + 5 e⁻ → Mn²⁺ + 4 H₂O
---------------------------------------------------
5Fe²⁺ + MnO₄⁻ + 8 H⁺ + 5 e⁻ → 5 Fe³⁺ + 5 e⁻ + Mn²⁺ + 4 H₂O
The electrons on both sides cancel out, leaving the fully balanced overall reaction:
5Fe²⁺ + MnO₄⁻ + 8 H⁺ → 5 Fe³⁺ + Mn²⁺ + 4 H₂O
This shows that the electrons lost by the reducing agent (Fe²⁺) exactly equal the electrons gained by the oxidizing agent (MnO₄⁻), and the stoichiometric ratio of Fe²⁺ to MnO₄⁻ is 5:1, which is essential for calculating reactant amounts in titrations and lab experiments. Recall that the reaction involves the loss of the purple color of the permanganate ion and so it is very useful in titrations to calculate the concentration of the Fe2+ ion.
A disproportionation reaction is a special type of redox reaction in which the same substance is both oxidized and reduced at the same time. This means one element begins in a single oxidation state and forms two different products, one with a higher oxidation state and one with a lower oxidation state. In this type of reaction, the species acts as both the reducing agent and the oxidizing agent. Disproportionation reactions commonly occur when an element is in an intermediate oxidation state, meaning it is unstable and can either gain or lose electrons.
A classic example of a disproportionation reaction is hydrogen peroxide. In H₂O₂, oxygen has an oxidation state of −1 and can undergo both oxidation and reduction simultaneously.
Reduction half-reaction:
H₂O₂ + 2 H⁺ + 2 e⁻ → 2 H₂O
Oxidation half-reaction:
H₂O₂ → O₂ + 2 H⁺ + 2 e⁻
When these half-reactions are added together, the electrons cancel:
2 H₂O₂ → 2 H₂O + O₂
One of the disproportion reactions that you cannot miss is when halogens other than fluorine get reduced and
oxidized in a basic solution. Fun fact: when this happens with chlorine it gives a ClO- ion and this is how
bleach is made.
X₂ + 2 OH⁻ → X⁻ + XO⁻ + H₂O
On the same topic of bleach, ClO- also undergoes another disproportion reaction. This is why older bleach is
weaker than newer bleach. BrO- and IO- also go into disproportion reactions but since the BrO- ion and IO-
ions are more unstable this reaction happens almost instantly.
Another common disproportion reaction is the formation of nitric acid.
3 NO₂ + H₂O → 2 HNO₃ + NO
The production of this NO is immediately followed by oxidation.
2 NO + O₂ → 2 NO₂
This is why nitric acid plants are so efficient with their production of nitric acid. The reaction is almost
just a loop with little to no nitrogen waste!
Another reaction that disproportion is found in is when Cu+ is made. It almost instantly reacts to form Cu²⁺
ions and Cu(s). This is why Cu+ is never found in water.
2 Cu⁺ → Cu²⁺ + Cu(s)
Sulfur is an element that also undergoes tons of disproportion reactions. Some of the common ones are:
3 SO₂ + 2 H₂O → 2 H₂SO₄ + S
S₂O₃²⁻ + 2 H⁺ → S(s) + SO₂ + H₂O
A synproportion reaction occurs when two species of the same element in different oxidation states react to
form a product in an intermediate oxidation state. One species is oxidized while the other is reduced, so
the element “meets in the middle.” One common synproportion reaction stabilizes Fe²⁺ solutions in labs,
because the two forms of iron balance each other out.
Fe(s) + Fe³⁺ → 2 Fe²⁺
NO + NO₃⁻ + 2 H⁺ → 2 NO₂
This reaction shows how nitrogen oxides can adjust to a stable intermediate form — it’s actually used in
some redox nitrate chemistry in labs.
2 MnO₄²⁻ + MnO₄⁻ + H₂O → 3 MnO₂ + 2 OH⁻
This synproportion reaction explains why mixtures of permanganate and manganate can produce manganese
dioxide in solution — it’s a classic lab demonstration of comproportionation.