Redox Reactions – Class 11 Chemistry | Free NEET Notes
1. Introduction with Learn Sufficient Notes😍
Chemistry is the study of different types of matter and how one kind of matter can change into another. These changes happen through different types of reactions. One major type of these reactions is called Redox Reactions. Many physical and biological processes involve redox reactions. These reactions are widely used in pharmaceutical, biological, industrial, metallurgical, and agricultural fields. Their importance can be seen in everyday examples. For instance, burning of various fuels to produce energy for homes, transport, and other commercial uses involves redox processes. Electrochemical processes used to extract highly reactive metals and non-metals, making of chemical compounds like caustic soda, the working of dry and wet batteries, and the corrosion of metals are all related to redox reactions. Recently, some environmental topics like the Hydrogen Economy (using liquid hydrogen as fuel) and the creation of the Ozone Hole are also being linked with redox reactions.
2. Traditional Concept of Redox Reactions: Understanding Oxidation and Reduction
- In the classical definition, a redox reaction is a type of chemical reaction that involves both oxidation and reduction happening together. At first, the term oxidation meant only the addition of oxygen to a substance. Since oxygen makes up about 20% of the air, many elements combine with it easily. That is why most elements on Earth are found as oxides. For example, when magnesium reacts with oxygen, it forms magnesium oxide. Similarly, sulphur reacts with oxygen to form sulphur dioxide. These are simple examples of oxidation. In another case, methane reacts with oxygen and produces carbon dioxide and water. This reaction also shows oxidation because oxygen replaces hydrogen. So, scientists expanded the meaning of oxidation to also include removal of hydrogen from a substance. For instance, when hydrogen sulphide reacts with oxygen, it forms sulphur and water. This shows removal of hydrogen and is also called oxidation. Later, chemists started including reactions where other electronegative elements like fluorine, chlorine, and sulphur combine with substances, even when oxygen is not involved. For example, magnesium reacts with fluorine, chlorine, and sulphur to form magnesium fluoride, magnesium chloride, and magnesium sulphide, respectively. These reactions also came under oxidation because they involve addition of electronegative elements. Another type of oxidation is when a positive metal like potassium is removed from a compound, such as in the reaction of potassium ferrocyanide with hydrogen peroxide, which forms potassium ferricyanide. So overall, oxidation can now mean addition of oxygen or any electronegative element, or removal of hydrogen or any electropositive element.
- Originally, the term reduction meant removal of oxygen from a compound. Now, its meaning has also become wider. Reduction now includes removal of oxygen or electronegative element, or addition of hydrogen or electropositive element. For example, when mercuric oxide is heated, it breaks down to form mercury and oxygen. This is reduction by removal of oxygen. Another example is when ferric chloride reacts with hydrogen gas to form ferrous chloride and hydrochloric acid, showing the removal of chlorine, an electronegative element. Similarly, when ethylene reacts with hydrogen, it forms ethane, showing addition of hydrogen, which is also reduction. One more example is when mercuric chloride reacts with stannous chloride, producing mercurous chloride and stannic chloride. Here, mercury is added to mercuric chloride, which is reduction. In the same reaction, the stannous chloride is oxidised because it gains chlorine, an electronegative element. From all these examples, it became clear that oxidation and reduction always happen together in a chemical reaction. That’s why the combined term “redox” is now used to describe such reactions.
3. Electron Transfer Perspective of Redox Reactions
- We already know that some reactions, like the ones where sodium reacts with chlorine, oxygen, or sulphur, are redox reactions. In these reactions, sodium gets oxidised because it combines with oxygen or with an electronegative element. At the same time, elements like chlorine, oxygen, and sulphur get reduced because they gain sodium, which is an electropositive element. Based on what we know about chemical bonding, the compounds formed—like sodium chloride, sodium oxide, and sodium sulphide—are ionic compounds. These are better written as Na⁺Cl⁻, (Na⁺)₂O²⁻, and (Na⁺)₂S²⁻. These charges tell us that we can think of these reactions in two steps: one part where electrons are lost and another where electrons are gained. Let’s take an example: the reaction where sodium reacts with chlorine. First, sodium atoms each lose one electron and become Na⁺ ions. This is the oxidation half-reaction. Then, chlorine molecules each gain electrons and become Cl⁻ ions. This is the reduction half-reaction. Each part is called a half-reaction, and together they make up the full redox reaction.
- From reactions like these, we understand that a half-reaction where electrons are lost is called oxidation, and one where electrons are gained is called reduction. This new definition of oxidation and reduction is based on the transfer of electrons, and it matches the older, classical idea of redox changes. In the above examples, sodium gives away electrons and therefore acts as a reducing agent—it helps reduce the other element. On the other hand, chlorine, oxygen, and sulphur each accept electrons, so they act as oxidising agents—they help oxidise sodium. To sum it all up: Oxidation means losing electrons, reduction means gaining electrons, an oxidising agent is one that accepts electrons, and a reducing agent is one that donates electrons.
4. Reactivity-Based Electron Transfer Reactions
- Take a strip of zinc metal and place it in a solution of copper nitrate. After about an hour, you will see that the zinc strip gets coated with a reddish layer of copper metal, and the blue colour of the solution disappears. This blue colour was because of Cu²⁺ ions in the solution. Their disappearance means that Zn²⁺ ions are now present in the solution. To confirm this, you can pass hydrogen sulphide (H₂S) gas through the solution. If you make the solution alkaline using ammonia, a white solid called zinc sulphide (ZnS) appears, confirming the presence of Zn²⁺. The chemical reaction between zinc and copper ions is written as: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s). In this reaction, zinc loses electrons and forms Zn²⁺, so it is oxidised. At the same time, Cu²⁺ gains electrons and becomes copper metal, so it is reduced. This is a redox reaction involving electron transfer.
- To check if the reverse reaction happens, place a strip of copper in zinc sulphate solution. No visible reaction takes place. Even if you try to test for Cu²⁺ ions using H₂S gas (which should form black CuS if Cu²⁺ is present), there is no result. This means that the reaction does not go backwards and the products are highly favoured in the zinc-copper reaction. Now, let us look at another example. If you place copper metal in a solution of silver nitrate, a reaction occurs where copper becomes Cu²⁺ and enters the solution, while Ag⁺ ions are reduced to silver metal. The solution turns blue due to the formation of Cu²⁺ ions, and this reaction also favours the products.
- Now, compare another case: put cobalt metal in a solution of nickel sulphate. The reaction that takes place forms Co²⁺ and Ni metal, but both reactants and products are found in moderate amounts. This means that neither side is strongly favoured. This situation, where different metals compete to release electrons, is similar to how acids compete to release protons (H⁺ ions). Just like we rank acids by their strength, we can also rank metals by how easily they give away electrons. From our earlier comparisons, we know that zinc gives electrons to copper, and copper gives electrons to silver. So the order of electron-releasing ability is: Zn > Cu > Ag. Using this idea, we can create a larger list called the metal activity series or electrochemical series. This comparison of metals is very useful because it helps us design special types of chemical setups called Galvanic cells, where chemical energy is converted into electrical energy.
5. Numerical Representation of Oxidation State
A less clear example of electron transfer happens when hydrogen reacts with oxygen to form water. The reaction is: 2H₂(g) + O₂(g) → 2H₂O(l). Here, the hydrogen atom goes from 0 (neutral) in H₂ to +1 in H₂O, and oxygen goes from 0 in O₂ to –2 in H₂O. This suggests that electrons are transferred from hydrogen to oxygen, so hydrogen is oxidised and oxygen is reduced. But this is only partial transfer of electrons, not complete. It is better to say that electrons shift instead of being fully gained or lost. This same idea is true for many other reactions involving covalent compounds. For example, when hydrogen reacts with chlorine to form HCl, or when methane reacts with chlorine to give carbon tetrachloride and HCl, we assume electron movement based on electronegativity. To clearly follow these changes, we use a concept called the oxidation number. This method assumes that electrons are completely transferred from the less electronegative atom to the more electronegative one. This helps us describe redox changes more easily.
The oxidation number tells us the oxidation state of an element in a compound. It is calculated by following a few rules. These are the rules:
- Elements in their natural form like H₂, O₂, Cl₂, P₄, S₈, Na, Mg, Al all have an oxidation number of 0.
- For single-atom ions, the oxidation number equals the charge on the ion. For example, Na⁺ is +1, Mg²⁺ is +2, Cl⁻ is –1, and O²⁻ is –2. Also, all alkali metals have +1, all alkaline earth metals have +2, and aluminium has +3 in all its compounds.
- The oxidation number of oxygen is generally –2. But there are exceptions. In peroxides like H₂O₂ or Na₂O₂, oxygen has –1. In superoxides like KO₂, oxygen has –½. Another exception is when oxygen bonds with fluorine. In OF₂, oxygen is +2, and in O₂F₂, oxygen is +1. So, oxygen can also have a positive oxidation number depending on the situation.
- Hydrogen usually has an oxidation number of +1, but when bonded with metals in compounds like LiH, NaH, CaH₂, it becomes –1.
- Fluorine always has an oxidation number of –1. Other halogens like chlorine, bromine, and iodine also have –1 when in halide form. But when they are in oxides or oxyacids, they can have positive oxidation numbers.
- The sum of oxidation numbers of all atoms in a compound must be zero. In a polyatomic ion, the total oxidation numbers must equal the charge of the ion. For example, in carbonate ion (CO₃²⁻), the sum of oxidation numbers of one carbon and three oxygen atoms must be –2.
Using these rules, we can find the oxidation number of any element in a compound or ion. Metals usually have positive oxidation numbers, while nonmetals can have positive or negative values. Transition metals often show more than one positive oxidation state. The maximum oxidation number that an element can show is the same as its group number for groups 1 and 2, and group number minus 10 for the rest (based on the long form of periodic table). So, as you move left to right in a period, the highest oxidation number increases. For example, in the third period, the highest oxidation numbers for elements increase from +1 to +7 in their compounds.
The concept of oxidation number is used to explain the terms oxidation, reduction, oxidising agent, reducing agent, and redox reaction.
Oxidation means there is an increase in the oxidation number of an element in the substance.
Reduction means there is a decrease in the oxidation number of an element in the substance.
An oxidising agent is a chemical that increases the oxidation number of an element in another substance. It is also called an oxidant.
A reducing agent is a chemical that decreases the oxidation number of an element in another substance. It is also called a reductant.
Redox reactions are chemical reactions where there is a change in the oxidation number of the elements involved.
6. Different Forms of Redox Processes
1. Combination Reactions
A combination reaction happens when two substances combine to form one product. It can be written as:
A + B → C
For this reaction to be a redox reaction, at least one of the reactants (A or B or both) must be in its elemental form.
All combustion reactions (burning reactions) are redox reactions. These use elemental oxygen (O₂).
Reactions involving other elements (not just oxygen) in their elemental form are also redox reactions.
Some important examples of combination redox reactions are:
- C(s) + O₂(g) → CO₂(g)
Here, carbon and oxygen are in elemental form (oxidation number 0). After the reaction, carbon becomes +4 and oxygen becomes –2. So, it is a redox reaction. - 3Mg(s) + N₂(g) → Mg₃N₂(s)
Magnesium and nitrogen are in elemental form (oxidation number 0). After reaction, magnesium becomes +2, and nitrogen becomes –3. This is also a redox reaction. - CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
In this combustion reaction, carbon in methane changes from –4 to +4, and oxygen changes from 0 to –2. So, both oxidation and reduction happen.
2. Decomposition Reactions
Decomposition reactions are the opposite of combination reactions.
In a decomposition reaction, a compound breaks down into two or more parts.
At least one of the products must be in its elemental form for it to be called a redox reaction.
Some examples of decomposition redox reactions are:
- 2H₂O(l) → 2H₂(g) + O₂(g)
In this reaction, water breaks into hydrogen and oxygen, both in elemental form. So, it’s a redox reaction. - 2NaH(s) → 2Na(s) + H₂(g)
Sodium hydride breaks down into sodium metal and hydrogen gas. Both sodium and hydrogen appear in elemental form, so it is also a redox reaction. - 2KClO₃(s) → 2KCl(s) + 3O₂(g)
Here, potassium chlorate breaks into potassium chloride and oxygen gas. Oxygen is in elemental form, so it is a redox reaction.
It is important to notice that not all decomposition reactions are redox reactions.
For example, in the reaction:
CaCO₃(s) → CaO(s) + CO₂(g)
There is no change in oxidation numbers. So, this is not a redox reaction.
Also, in earlier examples, the oxidation number of hydrogen in methane and potassium in potassium chlorate did not change. This tells us that not every element must undergo a change for the reaction to be redox.
3. Displacement Reactions
A displacement reaction happens when an atom or ion in a compound is replaced by an atom or ion of another element. This kind of reaction is written as X + YZ → XZ + Y.
There are two types of displacement reactions: metal displacement and non-metal displacement.
(a) Metal Displacement
In metal displacement reactions, a metal in a compound is replaced by another metal that is in its pure elemental state. These reactions are commonly used in metallurgy to extract pure metals from their ores.
One example is: CuSO₄(aq) + Zn(s) → Cu(s) + ZnSO₄(aq). Here, zinc replaces copper in the solution, and copper comes out as a free element. This happens because zinc is more reactive and a better reducing agent than copper.
Another example is: V₂O₅(s) + 5Ca(s) → 2V(s) + 5CaO(s). In this reaction, calcium replaces vanadium from vanadium oxide, producing vanadium metal and calcium oxide.
In the reaction TiCl₄(l) + 2Mg(s) → Ti(s) + 2MgCl₂(s), magnesium displaces titanium, forming titanium metal and magnesium chloride. This again shows magnesium’s higher ability to lose electrons.
Another similar reaction is: Cr₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Cr(s). Here, aluminium replaces chromium in chromium oxide to give aluminium oxide and free chromium.
In all these examples, the metal that replaces the other has a greater tendency to lose electrons, making it a stronger reducing agent.
(b) Non-Metal Displacement
In non-metal displacement reactions, typically hydrogen or oxygen is replaced. These reactions mainly involve hydrogen displacement.
Alkali metals and some alkaline earth metals like calcium, strontium, and barium can displace hydrogen from cold water, because they are strong reducing agents. For example, 2Na(s) + 2H₂O(l) → 2NaOH(aq) + H₂(g) shows that sodium reacts with water to release hydrogen gas.
In another reaction, Ca(s) + 2H₂O(l) → Ca(OH)₂(aq) + H₂(g), calcium replaces hydrogen in water and forms calcium hydroxide and hydrogen gas.
Less reactive metals such as magnesium and iron require steam instead of cold water for the reaction. For example, Mg(s) + 2H₂O(g) → Mg(OH)₂(s) + H₂(g) shows that magnesium reacts with steam to produce magnesium hydroxide and hydrogen.
In another similar case, 2Fe(s) + 3H₂O(g) → Fe₂O₃(s) + 3H₂(g), iron reacts with steam and forms iron oxide and hydrogen gas.
Many metals that do not react with water can still displace hydrogen from acids. For example, Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g) is a reaction where zinc replaces hydrogen in hydrochloric acid.
Another reaction is Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Here, magnesium reacts rapidly, showing high reactivity.
In the reaction Fe(s) + 2HCl(aq) → FeCl₂(aq) + H₂(g), iron reacts slowly and produces hydrogen gas.
These acid-based hydrogen displacement reactions are commonly used in laboratories to produce hydrogen gas. The speed of gas production shows how reactive the metal is—magnesium reacts fastest, and iron reacts the slowest.
Very unreactive metals like silver (Ag) and gold (Au) do not react with hydrochloric acid at all. These metals are often found in native state in nature.
In the reactivity order of some common metals, we have: Zn > Cu > Ag, based on their ability to lose electrons and act as reducing agents.
Halogen Displacement Reactions
Just like metals, halogens also follow a reactivity series. In Group 17, the ability of halogens to act as oxidising agents decreases as we go down the group, i.e., from fluorine to iodine.
This means fluorine (F₂) is the most reactive halogen and can displace chloride (Cl⁻), bromide (Br⁻), and iodide (I⁻) ions from their solutions.
In fact, fluorine is so reactive that it even attacks water and releases oxygen gas, as seen in the reaction:
2H₂O(l) + 2F₂(g) → 4HF(aq) + O₂(g)
Because fluorine reacts with water, we usually don’t use fluorine in aqueous displacement reactions involving other halogens.
On the other hand, chlorine can displace bromide and iodide ions in water. For example:
Cl₂(g) + 2KBr(aq) → 2KCl(aq) + Br₂(l) shows that chlorine displaces bromine.
Similarly, in the reaction Cl₂(g) + 2KI(aq) → 2KCl(aq) + I₂(s), chlorine displaces iodine.
Bromine (Br₂) and iodine (I₂) are coloured substances. When they dissolve in carbon tetrachloride (CCl₄), their colour makes it easy to identify them.
These reactions can also be written in ionic form, like:
Cl₂(g) + 2Br⁻(aq) → 2Cl⁻(aq) + Br₂(l) and
Cl₂(g) + 2I⁻(aq) → 2Cl⁻(aq) + I₂(s)
These reactions are used in laboratories to identify Br⁻ and I⁻ ions, and this method is called the ‘Layer Test’.
Also, bromine can displace iodide ions in the reaction:
Br₂(l) + 2I⁻(aq) → 2Br⁻(aq) + I₂(s)
Industrial Use of Halogen Displacement
These halogen displacement reactions are used in industry to recover halogens from their halide compounds.
The general form of this recovery reaction is:
2X⁻ → X₂ + 2e⁻, where X is any halogen.
Chlorine, bromine, and iodine can be recovered using chemical methods.
However, fluoride ions (F⁻) cannot be oxidised to fluorine gas (F₂) using chemical methods. The only way to do this is through electrolysis, which you will learn in later chapters.
Disproportionation Reactions
Disproportionation reactions are a special kind of redox reactions in which the same element is both oxidised and reduced in a single chemical reaction. In such reactions, an element starts from an intermediate oxidation state and gets converted to both higher and lower oxidation states simultaneously. For this to happen, the reacting element must be capable of existing in at least three different oxidation states. One common and simple example of this type of reaction is the decomposition of hydrogen peroxide (H₂O₂).
In the decomposition of hydrogen peroxide, the oxygen atom in H₂O₂ is initially in the –1 oxidation state. During the reaction, some of it gets oxidised to O₂, where the oxidation state of oxygen becomes 0, and some gets reduced to H₂O, where oxygen has an oxidation state of –2. The reaction is:
2H₂O₂(aq) → 2H₂O(l) + O₂(g)
Here, you can clearly see that the same element, oxygen, undergoes both oxidation and reduction, which is the key characteristic of disproportionation.
Elements such as phosphorus, sulphur, and chlorine also show disproportionation reactions, especially in alkaline medium. For example, when white phosphorus (P₄) reacts with hydroxide ions (OH⁻) in the presence of water, it forms phosphine (PH₃) and hypophosphite ion (H₂PO₂⁻). In this reaction, phosphorus is initially in the zero oxidation state, but it changes to –3 in PH₃ (reduction) and to +1 in H₂PO₂⁻ (oxidation), showing both types of redox processes at the same time. The reaction is:
P₄(s) + 3OH⁻(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂⁻(aq)
A similar reaction occurs with sulphur. When S₈ (elemental sulphur) reacts with alkaline OH⁻ ions, it produces sulfide ions (S²⁻) and thiosulfate ions (S₂O₃²⁻). In this case, sulphur in the 0 oxidation state is simultaneously oxidised to +2 in thiosulfate and reduced to –2 in sulphide. The balanced reaction is:
S₈(s) + 12OH⁻(aq) → 4S²⁻(aq) + 2S₂O₃²⁻(aq) + 6H₂O(l)
Chlorine also undergoes a classic disproportionation reaction in an alkaline medium. When chlorine gas (Cl₂) reacts with OH⁻ ions, it forms chloride ions (Cl⁻) and hypochlorite ions (ClO⁻). In this reaction, chlorine in the 0 oxidation state gets oxidised to +1 in ClO⁻ and reduced to –1 in Cl⁻, again satisfying the condition for disproportionation. The reaction is:
Cl₂(g) + 2OH⁻(aq) → Cl⁻(aq) + ClO⁻(aq) + H₂O(l)
The hypochlorite ion (ClO⁻) formed in this reaction is a key ingredient in household bleaching agents, as it has the ability to oxidise coloured stains and convert them into colourless compounds, making it useful in cleaning and disinfection. Interestingly, bromine and iodine also behave like chlorine in such reactions and show similar disproportionation patterns in alkaline media.
However, fluorine behaves differently. Being the most electronegative element, fluorine cannot exhibit positive oxidation states, and hence does not undergo disproportionation reactions like other halogens. Instead, when fluorine reacts with alkali, it forms fluoride ions (F⁻) and oxygen difluoride (OF₂). The reaction is:
2F₂(g) + 2OH⁻(aq) → 2F⁻(aq) + OF₂(g) + H₂O(l)
Even here, fluorine attacks water, and it may also produce some molecular oxygen (O₂) due to its high reactivity.
This deviation in behaviour of fluorine is expected because fluorine cannot be oxidised to a positive oxidation state, and thus disproportionation is not possible for it. This property distinguishes fluorine from other halogens like chlorine, bromine, and iodine.
7. Equalizing Redox Equations
There are two main methods to balance/equilize redox reactions. The first method uses the oxidation number of elements involved. The second method splits the overall reaction into two half-reactions: one for oxidation and the other for reduction. Both methods are correct, and anyone can use whichever they find easier.
(a) Oxidation Number Method
In the oxidation number method, you need to know the correct formulas of all reactants and products before starting.
In Step 1, write the correct chemical formula for every reactant and product involved in the reaction.
In Step 2, assign oxidation numbers to all the elements in the reaction. Then, identify the atoms whose oxidation number changes.
In Step 3, calculate how much the oxidation number increases or decreases per atom and for the entire molecule or ion. If the increase and decrease are not equal, multiply them by suitable numbers to make them equal. If you find that only oxidation or only reduction is happening, then either the formulas are wrong or the oxidation numbers are incorrect.
In Step 4, if the reaction happens in aqueous solution, check if it’s acidic or basic. In acidic solutions, add H⁺ ions to balance the charges. In basic solutions, use OH⁻ ions.
In Step 5, balance the number of hydrogen atoms by adding H₂O molecules to either the reactants or products. Then, check and balance the oxygen atoms. When the number of atoms and charges are equal on both sides, the redox reaction is balanced.
(b) Half Reaction Method
In the half-reaction method, the reaction is separated into two parts — one shows oxidation and the other shows reduction. These two parts are then balanced separately and finally combined.
Let’s take an example of Fe²⁺ being oxidised to Fe³⁺ by dichromate ions (Cr₂O₇²⁻) in acidic medium, where dichromate is reduced to Cr³⁺.
In Step 1, write the unbalanced ionic equation:
Fe²⁺(aq) + Cr₂O₇²⁻(aq) → Fe³⁺(aq) + Cr³⁺(aq)
In Step 2, split this equation into two half reactions.
Oxidation half:
Fe²⁺(aq) → Fe³⁺(aq)
Reduction half:
Cr₂O₇²⁻(aq) → Cr³⁺(aq)
In Step 3, balance all atoms except oxygen and hydrogen. In the reduction half, there are two chromium atoms, so we write:
Cr₂O₇²⁻(aq) → 2Cr³⁺(aq)
In Step 4, balance oxygen atoms by adding H₂O molecules, and balance hydrogen atoms by adding H⁺ ions. The reduction half becomes:
Cr₂O₇²⁻(aq) + 14H⁺(aq) → 2Cr³⁺(aq) + 7H₂O(l)
In Step 5, balance the charges by adding electrons (e⁻). In the oxidation half:
Fe²⁺(aq) → Fe³⁺(aq) + e⁻
In the reduction half, the left side has a total charge of +12 and the right side has +6. So, we add 6 electrons to the left side:
Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l)
To equalise the number of electrons in both half-reactions, multiply the oxidation half by 6:
6Fe²⁺(aq) → 6Fe³⁺(aq) + 6e⁻
In Step 6, add both half-reactions and cancel the 6 electrons from each side:
6Fe²⁺(aq) + Cr₂O₇²⁻(aq) + 14H⁺(aq) → 6Fe³⁺(aq) + 2Cr³⁺(aq) + 7H₂O(l)
In Step 7, check whether both sides have equal number of atoms and total charges. If yes, the redox reaction is now fully balanced.
If the reaction happens in a basic medium, first balance it as if it were in an acidic medium. Then, for every H⁺ ion, add the same number of OH⁻ ions to both sides. If H⁺ and OH⁻ appear on the same side, combine them to make H₂O molecules.
8. Redox-Based Volumetric Analysis
Just like in acid-base titrations/Volumetric analysis where a pH-sensitive indicator is used to find the strength of an acid or base, in redox reactions, titration methods are used to determine the strength of an oxidising or reducing agent. In redox titrations, instead of acid-base indicators, we use redox-sensitive indicators. Sometimes, the reagent itself acts as an indicator, and in other cases, we add a separate indicator to detect the end-point of the titration.
In the first type of redox titration, the reagent itself is intensely coloured and works as a self-indicator. A common example is the permanganate ion (MnO₄⁻), which is purple in colour. During titration, as long as a reducing agent like Fe²⁺ or C₂O₄²⁻ is present, MnO₄⁻ gets reduced and the purple colour disappears. When all of the reductant has been oxidised, even a small excess of MnO₄⁻ — as little as 10⁻⁶ mol/L — gives a lasting pink colour, signalling the end-point. This kind of reaction helps achieve a sharp end-point with minimal colour overshoot.
In the second type, when the reagent does not show any visible colour change by itself, a separate redox-sensitive indicator is used. A good example is dichromate ion (Cr₂O₇²⁻), which is not a self-indicator. In this case, an external indicator such as diphenylamine is used. After all of the reducing agent is consumed, the Cr₂O₇²⁻ starts oxidising diphenylamine, producing a deep blue colour, which marks the end-point of the titration.
The third method is very interesting and commonly used when the oxidising agent can oxidise iodide ions (I⁻). A good example is copper(II) ions (Cu²⁺), which oxidise iodide to form iodine (I₂) according to the reaction:
2Cu²⁺(aq) + 4I⁻(aq) → Cu₂I₂(s) + I₂(aq)
The liberated iodine reacts with starch to give a deep blue complex. This colour disappears when iodine is reduced by thiosulphate ions (S₂O₃²⁻) in a redox reaction:
I₂(aq) + 2S₂O₃²⁻(aq) → 2I⁻(aq) + S₄O₆²⁻(aq)
Though iodine is insoluble in water, it forms a soluble complex (KI₃) in the presence of KI. When iodine is present, adding starch produces a blue colour. This colour fades as soon as all the iodine is consumed by thiosulphate, clearly indicating the end-point. After that, stoichiometric calculations are done to determine the concentration of the substance being analysed.
9. Drawbacks of the Oxidation Number Concept
The concept of oxidation number has been an essential tool to understand redox reactions, but it comes with certain limitations. As scientific understanding has progressed, the explanation of oxidation and reduction has also evolved. Traditionally, oxidation was defined as an increase in oxidation number, and reduction as a decrease in oxidation number. However, modern chemistry interprets oxidation as a decrease in electron density around an atom, and reduction as an increase in electron density. This updated view helps explain redox processes more accurately, especially in cases where changes in oxidation number do not clearly represent what is happening at the electron level. Therefore, while the oxidation number is a useful concept, it cannot fully explain the electron transfer or electron density changes involved in all redox reactions.
10. Redox Changes in Electrochemical Processes
In an experiment, when a zinc rod is placed in copper sulphate solution, a redox reaction happens. In this process, zinc loses electrons and forms zinc ions, while copper ions gain those electrons and become metallic copper. This is a direct transfer of electrons, and the reaction releases heat.
Now, if we modify the setup so that the electron transfer happens indirectly, we need to separate the zinc metal from the copper sulphate solution. We take two beakers — one has copper sulphate solution with a copper strip, and the other has zinc sulphate solution with a zinc strip. In both setups, the metal and its salt solution are in contact, forming what we call a redox couple. A redox couple includes both the oxidized and reduced forms of a substance, and it is written like Zn²⁺/Zn or Cu²⁺/Cu, where the oxidized form is written first.
To connect the two beakers, we use a salt bridge, which is a U-shaped tube containing a solution like potassium chloride or ammonium nitrate. This allows ions to move between the solutions without mixing them directly. The metal strips (zinc and copper) are connected by a wire with a switch and an ammeter. This whole setup is called a Daniell cell.
When the switch is off, no reaction occurs, and no current flows. But when the switch is on, the following happens:
- Electrons flow through the wire from zinc to copper ions. So, the electron transfer is now indirect.
- The ions move through the salt bridge, maintaining the electric flow between the two solutions.
The zinc and copper strips are called electrodes. The ability of each electrode to push or pull electrons is called its electrode potential. When all species in the reaction are at standard conditions (1 mol/L for solutions, 1 atm for gases, and 298 K temperature), the electrode potential is called the standard electrode potential (E°).
By agreement, the standard electrode potential of the hydrogen electrode is set to 0.00 volts. The value of E° for any electrode shows how easily it gives or accepts electrons:
- A negative E° means that the redox couple is a stronger reducing agent than the H⁺/H₂ couple.
- A positive E° means that the redox couple is a weaker reducing agent than the H⁺/H₂ couple.
These standard electrode potentials are very useful in understanding and predicting the behavior of redox reactions.
