Redox Reactions
Oxidation–reduction, oxidation numbers, balancing and redox types for NEET
Concepts and Oxidation Numbers
Oxidation, Reduction and Redox AgentsTopic 1
Redox reactions are reactions in which electrons are transferred from one species to another. The word 'redox' combines reduction and oxidation, because the two always happen together — one substance cannot lose electrons unless another gains them. This electron-transfer view is the modern definition NEET expects, and it replaces the older oxygen/hydrogen-based descriptions.
The classical definitions are still useful starting points: oxidation was originally the addition of oxygen or removal of hydrogen, and reduction the addition of hydrogen or removal of oxygen. The electronic definition is broader and more powerful: oxidation is the loss of electrons and reduction is the gain of electrons. The classic mnemonic is OIL RIG — Oxidation Is Loss, Reduction Is Gain — which NEET students lean on to avoid mixing the two up.
Because oxidation and reduction are coupled, every redox reaction has an oxidising agent and a reducing agent. The oxidising agent is the species that accepts electrons; in doing so it is itself reduced. The reducing agent is the species that donates electrons and is itself oxidised. A frequent NEET trap is to confuse 'agent' with 'what happens to it' — the oxidising agent gets reduced, not oxidised. Strong oxidising agents include $\text{KMnO}_4$ and $\text{K}_2\text{Cr}_2\text{O}_7$; common reducing agents include $\text{H}_2$, carbon and active metals.
Redox processes are everywhere in biology and daily life. Cellular respiration oxidises glucose, transferring electrons step by step to oxygen and capturing the energy as ATP; photosynthesis runs the reverse. Rusting of iron, bleaching, combustion, the working of batteries, and the action of antioxidants in the body are all redox phenomena — context NEET often uses to frame questions.
| Process | Electrons / agent |
|---|---|
| Oxidation | loss of electrons (done by reducing agent) |
| Reduction | gain of electrons (done by oxidising agent) |
| Oxidising agent | accepts e⁻, is itself reduced |
| Reducing agent | donates e⁻, is itself oxidised |
In $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, identify the oxidising and reducing agents.
Show solution
Zn loses electrons (oxidised) so Zn is the reducing agent. $\text{Cu}^{2+}$ gains electrons (reduced) so $\text{Cu}^{2+}$ is the oxidising agent.
Using OIL RIG, state what happens to a species whose oxidation number increases.
Show solution
An increase in oxidation number means loss of electrons, i.e. oxidation (Oxidation Is Loss). That species acts as the reducing agent.
Oxidation is the:
An oxidising agent is itself:
The mnemonic OIL RIG stands for:
Which is a strong oxidising agent?
A reducing agent during a reaction:
NEET trap: the oxidising agent is reduced and the reducing agent is oxidised — never the other way round. Increase in oxidation number = oxidation = reducing agent.
Oxidation Number and Balancing Redox EquationsTopic 2
To track electron transfer in complex molecules we use the oxidation number (or oxidation state) — a hypothetical charge an atom would have if all bonds were fully ionic. A set of rules assigns it. The oxidation number of a free element is $0$; of a monatomic ion equals its charge; oxygen is usually $-2$ (but $-1$ in peroxides such as $\text{H}_2\text{O}_2$); hydrogen is usually $+1$ (but $-1$ in metal hydrides like $\text{NaH}$); fluorine is always $-1$. For any neutral molecule the oxidation numbers sum to $0$, and for an ion they sum to the ionic charge. Applying these rules quickly is a guaranteed NEET skill.
Oxidation numbers let us spot redox at a glance: if an element's oxidation number increases during a reaction it has been oxidised, and if it decreases it has been reduced. Many transition metals show several oxidation states (e.g. Mn from $+2$ to $+7$), which is why they form so many oxidising and reducing agents.
Balancing redox equations is best done by the ion-electron (half-reaction) method. Split the reaction into an oxidation half and a reduction half. In each half: balance the main atoms, then balance oxygen by adding $\text{H}_2\text{O}$, balance hydrogen by adding $\text{H}^+$ (in acidic medium), and balance charge by adding electrons. Multiply the halves so the electrons cancel, then add them. In basic medium, add $\text{OH}^-$ to neutralise the $\text{H}^+$ at the end.
The simpler oxidation-number method works too: find the total increase and total decrease in oxidation number, make them equal by suitable coefficients, then balance the rest by inspection. NEET problems usually only need you to balance one or two half-reactions or to find a coefficient, so being fluent with the half-reaction steps and the oxidation-number rules covers almost every question.
| Species | Oxidation number rule |
|---|---|
| Free element | $0$ |
| Oxygen | $-2$ (peroxide $-1$) |
| Hydrogen | $+1$ (metal hydride $-1$) |
| Sum (molecule / ion) | $0$ / ionic charge |
Find the oxidation number of Mn in $\text{KMnO}_4$.
Show solution
K is $+1$, each O is $-2$ (four O = $-8$). Sum $= 0$: $(+1) + x + (-8) = 0 \Rightarrow x = +7$. So Mn is $+7$.
In acidic medium, balance the oxygen in the half-reaction $\text{MnO}_4^- \rightarrow \text{Mn}^{2+}$.
Show solution
Add $4\,\text{H}_2\text{O}$ to the right for the 4 oxygens, then $8\,\text{H}^+$ to the left for the hydrogens: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$.
The oxidation number of a free element is:
Oxygen in $\text{H}_2\text{O}_2$ has oxidation number:
The oxidation number of S in $\text{H}_2\text{SO}_4$ is:
In the ion-electron method (acidic), oxygen is balanced by adding:
If an element's oxidation number increases, it has been:
NEET tip: learn the oxidation-number rules cold (O = $-2$ except peroxides; H = $+1$ except metal hydrides; sum = charge). In acidic half-reactions: balance O with $\text{H}_2\text{O}$, H with $\text{H}^+$, charge with $e^-$.
Types and Applications of Redox
Types of Redox ReactionsTopic 3
Redox reactions fall into four standard categories, and NEET frequently asks you to classify a given reaction. The first is the combination reaction, where two elements or simple species combine, with at least one element changing oxidation state, e.g. $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ (carbon $0 \rightarrow +4$). Many combustion and synthesis reactions belong here.
The second is the decomposition reaction, the reverse idea: a single compound breaks into two or more products with changes in oxidation state, e.g. $2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$. Not every decomposition is redox — only those where oxidation numbers actually change (the decomposition of $\text{CaCO}_3$ into $\text{CaO}$ and $\text{CO}_2$ is not redox, a common trap).
The third is the displacement reaction, where one element displaces another from its compound. In a metal displacement, a more reactive metal pushes out a less reactive one (e.g. $\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$); in non-metal displacement, a more reactive halogen displaces a less reactive halide (e.g. chlorine displacing bromine). These follow the activity/electrochemical series.
The fourth and most distinctive is disproportionation, where the same element in a single species is simultaneously oxidised and reduced, ending up in both a higher and a lower oxidation state. A classic example is chlorine in cold dilute alkali: $\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}$, where chlorine ($0$) goes to $-1$ and $+1$. Hydrogen peroxide and many oxoacids of intermediate oxidation state also disproportionate. Recognising disproportionation — one element, two fates — is a favourite NEET question.
| Type | Example |
|---|---|
| Combination | $\text{C} + \text{O}_2 \rightarrow \text{CO}_2$ |
| Decomposition | $2\text{H}_2\text{O} \rightarrow 2\text{H}_2 + \text{O}_2$ |
| Displacement | $\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$ |
| Disproportionation | $\text{Cl}_2 + 2\text{OH}^- \rightarrow \text{Cl}^- + \text{ClO}^- + \text{H}_2\text{O}$ |
Classify: $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$.
Show solution
Oxygen in $\text{H}_2\text{O}_2$ is $-1$; it becomes $-2$ in $\text{H}_2\text{O}$ (reduced) and $0$ in $\text{O}_2$ (oxidised). The same element goes both ways, so this is disproportionation.
Is $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ a redox reaction?
Show solution
Check oxidation numbers: Ca stays $+2$, C stays $+4$, O stays $-2$ throughout. No change, so it is not a redox reaction (just thermal decomposition).
$\text{Zn} + \text{CuSO}_4 \rightarrow \text{ZnSO}_4 + \text{Cu}$ is a:
In disproportionation, the same element is:
Which is NOT a redox reaction?
$\text{Cl}_2$ reacting with cold dilute NaOH is an example of:
Burning of carbon in oxygen is a:
NEET trap: not every decomposition/combination is redox — check whether oxidation numbers actually change ($\text{CaCO}_3 \to \text{CaO}+\text{CO}_2$ is non-redox). Disproportionation = one element, two oxidation states.
Redox Couples, the Activity Series and TitrationsTopic 4
Whether a redox reaction will actually happen depends on the relative tendencies of the species to lose or gain electrons. These tendencies are ranked in the activity series (for metals) and, more quantitatively, the electrochemical series, which orders elements by their standard electrode potentials. A metal higher in the series (more reactive, stronger reducing agent) can displace the ions of a metal lower down — which is exactly why zinc displaces copper but not the reverse.
Each redox reaction involves two redox couples, written as oxidised/reduced forms (e.g. $\text{Zn}^{2+}/\text{Zn}$ and $\text{Cu}^{2+}/\text{Cu}$). The couple with the greater tendency to be reduced acts as the oxidising half; the other supplies the electrons. This couple language connects redox chemistry to electrochemistry, where the same reactions are separated into half-cells to generate electricity — the basis of batteries, a topic NEET builds on later.
A major practical use of redox is redox titration, used to find the concentration of an oxidising or reducing agent. Permanganometry uses $\text{KMnO}_4$, which is its own indicator: the end point is the first permanent pink tinge once all the reducing agent is consumed. Dichrometry uses $\text{K}_2\text{Cr}_2\text{O}_7$ with a redox indicator. Iodometric/iodimetric titrations involve iodine and thiosulphate with starch as indicator.
Redox also explains everyday and biological processes: corrosion (the oxidation of metals, e.g. rusting of iron), the action of antioxidants that protect cells from oxidative damage, the bleaching action of chlorine and hydrogen peroxide, and the energy-yielding redox chain of respiration. Knowing the activity-series displacement rule, the role of $\text{KMnO}_4$/$\text{K}_2\text{Cr}_2\text{O}_7$ in titrations, and the meaning of a redox couple covers the application-based NEET questions in this chapter.
| Idea | Key point |
|---|---|
| Activity series | more reactive metal displaces less reactive |
| Redox couple | oxidised/reduced form, e.g. $\text{Cu}^{2+}/\text{Cu}$ |
| $\text{KMnO}_4$ titration | self-indicating (pink end point) |
| Corrosion | oxidation of metal (rusting of iron) |
Will silver displace copper from copper sulphate solution? (Activity: Cu more reactive than Ag.)
Show solution
No. Silver is less reactive than copper, so it cannot displace copper ions. Displacement only works when the added metal is higher in the activity series.
Why does $\text{KMnO}_4$ not need a separate indicator in titration?
Show solution
$\text{KMnO}_4$ is intensely purple and is decolourised as it reacts. Once the reducing agent is used up, the next drop stays pink — that first permanent pink colour marks the end point, so it is self-indicating.
A metal can displace another metal's ions if it is:
A redox couple is written as:
$\text{KMnO}_4$ in titration acts as its own:
Rusting of iron is an example of:
The electrochemical series ranks elements by:
NEET tip: use the activity/electrochemical series to predict displacement (reactive displaces less reactive). $\text{KMnO}_4$ is self-indicating; $\text{K}_2\text{Cr}_2\text{O}_7$ needs an indicator. Rusting/corrosion = oxidation of metal.
Quick Revision — Redox Reactions
- Oxidation = loss of electrons / increase in oxidation number; reduction = gain of electrons / decrease in oxidation number (remember OIL RIG).
- The oxidising agent is itself reduced; the reducing agent is itself oxidised.
- Oxidation number assigned by rules: free element $0$, O usually $-2$ (peroxide $-1$), H usually $+1$ (metal hydride $-1$); sum = charge.
- Balance redox in steps by the ion-electron (half-reaction) method — balance atoms, then O with $\text{H}_2\text{O}$, H with $\text{H}^+$, charge with electrons.
- Four types: combination, decomposition, displacement, disproportionation.
- Disproportionation: the same element is simultaneously oxidised and reduced (e.g. $\text{Cl}_2$ in cold alkali).
- The activity / electrochemical series ranks elements by reducing power; a metal displaces ions of a less reactive metal.
- Redox underlies titrations (KMnO₄, K₂Cr₂O₇), respiration, corrosion and batteries.
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