Class 11 Chemistry · NCERT chapter 8 · 6 topics · free to read
Oxidation & Reduction
Oxidation is loss of electrons (increase in oxidation number); reduction is gain of electrons (decrease in oxidation number). Three historical definitions evolved: classical (O₂/H₂ gain/loss), electronic (electron transfer), and oxidation number based. In any redox reaction, oxidation and reduction always occur simultaneously.
Oxidation: loss of e^- | Reduction: gain of e^-
Classical Definition
The earliest definition: oxidation is addition of oxygen or removal of hydrogen; reduction is addition of hydrogen or removal of oxygen.
The word 'oxidation' literally comes from oxygen. In the 18th century, Lavoisier noticed that metals combine with oxygen to form oxides — and called the process *oxidation*. The reverse — removing oxygen — was called *reduction* (the metal was 'reduced' back to its original form).
Oxidation = addition of oxygen OR removal of hydrogen
Reduction = addition of hydrogen OR removal of oxygen
In CuO + H_2 arrow Cu + H_2O : CuO is reduced (loses O), H_2 is oxidized (gains O)
This definition is limited — it cannot explain reactions that don't involve O or H at all (e.g., Na + Cl_2 )
Electron Transfer Definition
Oxidation is loss of electrons; reduction is gain of electrons. This definition works for ionic reactions but struggles with covalent ones.
The electronic definition is more general: oxidation = loss of electrons, reduction = gain of electrons. When sodium reacts with chlorine, Na *loses* an electron (oxidized) and Cl *gains* an electron (reduced). No oxygen or hydrogen needed!
Oxidation = loss of electrons (species becomes more positive)
Reduction = gain of electrons (species becomes more negative)
Oxidation and reduction ALWAYS occur together — you can't have one without the other
The species that gets oxidized is called the reducing agent (it reduces the other species by donating electrons)
The species that gets reduced is called the oxidizing agent (it oxidizes the other species by accepting electrons)
Oxidation Number Definition
Oxidation is increase in oxidation number; reduction is decrease in oxidation number. This definition works for ALL reactions — ionic and covalent alike.
The oxidation number (or oxidation state) is a hypothetical charge that an atom WOULD have if all bonds were purely ionic (electrons assigned to the more electronegative atom). It's a bookkeeping tool that lets us track electron flow even in covalent compounds.
Oxidation: increase in oxidation number Reduction: decrease in oxidation number
In H_2 + Cl_2 arrow 2HCl : - H goes from 0 to +1 (oxidation number increases → oxidized) - Cl goes from 0 to -1 (oxidation number decreases → reduced)
Classical (O₂/H₂ gain/loss)
Fails for reactions without O or H (e.g., Na + Cl_2 )
Electronic (electron transfer)
Fails for covalent reactions (no complete electron transfer)
Oxidation number (ON change)
None — works universally for all reactions
All three definitions are consistent — they agree on which species is oxidized/reduced when applicable
Oxidation number definition is the most general and the one NCERT recommends for complex cases
A redox reaction involves simultaneous oxidation and reduction — one species is oxidized while another is reduced
The term redox = reduction + oxidation combined
Quick recall
In Zn + CuSO_4 arrow ZnSO_4 + Cu , identify the oxidized and reduced species using all three definitions.
Zn is oxidized (classical: gains O in ZnSO₄ partner; electronic: loses 2e⁻ → Zn²⁺; ON: 0 → +2). Cu²⁺ is reduced (classical: metal freed from oxide partner; electronic: gains 2e⁻ → Cu; ON: +2 → 0). All three definitions agree.
NEET 2019, 2021, 2023 all had questions requiring identification of oxidation/reduction using oxidation numbers. This is non-negotiable — you MUST master ON assignment.
Oxidation Number Rules
Oxidation number (ON) is assigned using a hierarchy of rules: free element = 0, monoatomic ion = charge, F = −1 always, O = −2 (except peroxides −1, OF₂ +2), H = +1 (except metal hydrides −1), and the sum rule. For polyatomic ions, the sum of all ONs equals the charge on the ion. Mastering these rules is essential for identifying redox reactions and balancing equations.
ON of all atoms = charge on species
The Rule Hierarchy
A systematic set of rules applied in order of priority to assign oxidation numbers to every atom in any compound or ion.
Oxidation number assignment follows a strict priority order. Apply the highest-priority rule first, then use the remaining rules to find unknowns.
Rule 1: Free (uncombined) element → ON = 0 . Examples: Na , O_2 , P_4 , S_8 all have ON = 0
Rule 2: Monoatomic ion → ON = charge. Na^+ = +1 , Cl^- = -1 , Fe^ 3+ = +3
Rule 3: Fluorine → ON = -1 always (most electronegative element, no exceptions)
Rule 4: Oxygen → ON = -2 in most compounds
Rule 5: Hydrogen → ON = +1 with non-metals, -1 with metals (metal hydrides like NaH, CaH₂)
Rule 6: Sum of ONs in a neutral molecule = 0
Rule 7: Sum of ONs in a polyatomic ion = charge on the ion
Important Exceptions
Oxygen is not always −2 and hydrogen is not always +1. Knowing the exceptions is critical because NEET loves testing them.
The standard rules have well-defined exceptions that NCERT explicitly mentions. These are NEET favourites — expect at least one question testing an exception.
O in peroxides ( H_2O_2 , Na_2O_2 , BaO_2 )
-1
O in superoxides ( KO_2 , NaO_2 )
- 1 2
O in OF_2 (oxygen difluoride)
+2
O in O_2F_2 (dioxygen difluoride)
+1
H in metal hydrides ( NaH , CaH_2 , LiAlH_4 )
-1
Peroxides have an O–O single bond: each O is -1
Superoxides have an O_2^- unit: each O is -1/2 (fractional ON is allowed!)
In OF_2 : F must be -1 (rule 3 priority) → O must be +2 (sum rule: +2 + 2(-1) = 0 )
In metal hydrides, the metal is more electropositive → H gets the electrons → H is -1
Fractional oxidation numbers ARE valid — they represent the average ON per atom
Worked Examples
Applying the rules to common NEET compounds: KMnO₄, K₂Cr₂O₇, H₂SO₄, Na₂S₂O₃, and more.
The method is always the same: assign known ONs first (F, O, H, alkali/alkaline earth metals), then use the sum rule to find the unknown.
K_2Cr_2O_7: 2(+1) + 2x + 7(-2) = 0 x = +6 (Cr)
H_2SO_4: 2(+1) + x + 4(-2) = 0 x = +6 (S)
Na_2S_2O_3: 2(+1) + 2x + 3(-2) = 0 x = +2 (S)
For Cr_2O_7^ 2- : 2x + 7(-2) = -2 → Cr = +6
For MnO_4^- : x + 4(-2) = -1 → Mn = +7
For SO_4^ 2- : x + 4(-2) = -2 → S = +6
For NO_3^- : x + 3(-2) = -1 → N = +5
For C_2O_4^ 2- (oxalate): 2x + 4(-2) = -2 → C = +3
Quick recall
What is the ON of Cr in CrO_5 (blue peroxide of chromium)?
CrO_5 has one regular O ( -2 ) and two peroxide linkages ( O_2^ 2- , each O = -1 ). So: x + (-2) + 4(-1) = 0 → x = +6 . Even in CrO_5 , Cr is +6 ! This is a classic NEET trick question.
NCERT Table 8.1 lists common oxidation states of elements. Memorize at minimum: Mn ( +2 to +7 ), Cr ( +2 to +6 ), S ( -2 to +6 ), N ( -3 to +5 ), C ( -4 to +4 ).
Types of Redox Reactions
Redox reactions are classified into combination (two or more substances → one product), decomposition (one substance → two or more products), displacement (more reactive element displaces a less reactive one), disproportionation (same element is simultaneously oxidized AND reduced), and comproportionation (same element in two different ONs converges to one intermediate ON). Not all combination/decomposition reactions are redox — only those with a change in oxidation number.
Disproportionation: same element higher ON + lower ON
Combination Reactions
Two or more reactants combine to form a single product. A combination reaction is redox only if oxidation numbers change.
In a combination reaction, two or more substances combine to form a single product. When at least one element changes its oxidation number, it's a redox combination reaction.
C(s) + O_2(g) arrow CO_2(g)
Here C goes from 0 to +4 (oxidized) and O goes from 0 to -2 (reduced). This IS a redox reaction.
CaO(s) + CO_2(g) arrow CaCO_3(s) (NOT redox — no ON change)
Combination of an element with O_2 or a halogen is almost always redox
Combination of two compounds (like CaO + CO₂) may NOT be redox — check ONs!
NCERT highlights: combination of elements is always redox; combination of compounds may or may not be
Decomposition Reactions
A single compound breaks down into two or more simpler substances. Redox decomposition occurs when oxidation numbers change during the breakdown.
In a decomposition reaction, a single compound breaks down into two or more products, usually triggered by heat, light, or electricity.
2KClO_3 2KCl + 3O_2
Cl goes from +5 (in KClO_3 ) to -1 (in KCl) → reduced. O goes from -2 to 0 → oxidized. This IS redox.
CaCO_3 CaO + CO_2 (NOT redox — no ON change)
Decomposition producing a free element ( O_2 , metal, etc.) is almost always redox
Decomposition giving only compounds may NOT be redox (e.g., CaCO_3 arrow CaO + CO_2 )
A more reactive element displaces a less reactive element from its compound. Includes metal displacement, non-metal displacement, and the activity series.
In a displacement reaction, a more reactive element displaces a less reactive element from its compound. ALL displacement reactions are redox — the displacing element is oxidized, and the displaced ion is reduced.
Zn(s) + CuSO_4(aq) arrow ZnSO_4(aq) + Cu(s)
Zn ( 0 arrow +2 , oxidized) displaces Cu ( +2 arrow 0 , reduced). Zn is more reactive than Cu — it's higher in the activity series (reactivity series).
Metal displacement: A more reactive metal displaces a less reactive metal from solution. Fe + CuSO_4 arrow FeSO_4 + Cu
Hydrogen displacement: Active metals displace H_2 from acids. Zn + H_2SO_4 arrow ZnSO_4 + H_2
Non-metal displacement: More reactive halogen displaces less reactive halogen. Cl_2 + 2KBr arrow 2KCl + Br_2
Reactivity order (metals): K > Na > Ca > Mg > Al > Zn > Fe > Ni > Sn > Pb > H > Cu > Ag > Au
Reactivity order (halogens): F_2 > Cl_2 > Br_2 > I_2
Disproportionation
In disproportionation, the SAME element in ONE oxidation state is simultaneously oxidized and reduced to two different oxidation states. The reverse process is comproportionation.
Disproportionation is a special redox reaction where the same element in a single oxidation state is simultaneously oxidized AND reduced. One portion goes to a higher ON, the other to a lower ON.
2H_2O_2 arrow 2H_2O + O_2
Oxygen in H_2O_2 has ON = -1 . In the products: O in H_2O is -2 (reduced) and O in O_2 is 0 (oxidized). Same element ( O at -1 ) splits into two different ONs ( -2 and 0 ).
2H_2O_2 arrow 2H_2O + O_2 : O goes from -1 to -2 (reduced) AND -1 to 0 (oxidized)
Cl_2 + 2NaOH arrow NaCl + NaOCl + H_2O : Cl goes from 0 to -1 (reduced) AND 0 to +1 (oxidized)
3Cl_2 + 6NaOH hot 5NaCl + NaClO_3 + 3H_2O : Cl goes from 0 to -1 and +5
2Cu^+ arrow Cu^ 2+ + Cu : Cu goes from +1 to +2 (oxidized) and +1 to 0 (reduced)
Phosphorus: P_4 + 3NaOH + 3H_2O arrow PH_3 + 3NaH_2PO_2 : P goes from 0 to -3 and +1
Oxidizing & Reducing Agents
An oxidizing agent (oxidant) is the substance that gets reduced — it accepts electrons and its oxidation number decreases. A reducing agent (reductant) is the substance that gets oxidized — it donates electrons and its oxidation number increases. Common oxidizing agents include KMnO₄, K₂Cr₂O₇, HNO₃, conc. H₂SO₄, and halogens. Common reducing agents include H₂S, C₂O₄²⁻, FeSO₄, metals (Zn, Na), and carbon.
Oxidizing agent: gets reduced (ON decreases) | Reducing agent: gets oxidized (ON increases)
Identification Method
To identify oxidizing and reducing agents, assign oxidation numbers to every element on both sides, find which element's ON increased (reducing agent) and which decreased (oxidizing agent).
The method is mechanical and foolproof: 1. Assign ON to every atom in reactants and products 2. Find which element's ON increased → that species was oxidized → it's the reducing agent 3. Find which element's ON decreased → that species was reduced → it's the oxidizing agent
Worked example: MnO_2 + 4HCl arrow MnCl_2 + Cl_2 + 2H_2O - Mn: +4 arrow +2 (decreased → reduced → MnO_2 is the oxidizing agent) - Cl: -1 arrow 0 (increased → oxidized → HCl is the reducing agent) - H: +1 arrow +1 (no change) and O: -2 arrow -2 (no change)
The substance containing the element whose ON decreased = oxidizing agent
The substance containing the element whose ON increased = reducing agent
Some species can act as BOTH oxidizing and reducing agents depending on the reaction partner
H_2O_2 is the classic example: it oxidizes Fe^ 2+ to Fe^ 3+ (acting as oxidizing agent) but reduces KMnO_4 (acting as reducing agent)
Common Oxidizing Agents
A catalogue of the most important oxidizing agents for NEET, their active elements, and how they behave in reactions.
Oxidizing agents accept electrons (get reduced). They contain elements in high oxidation states that want to come down. Here are the most important ones for NEET.
N: +5 arrow +4 ( NO_2 ) or +2 ( NO ). Attacks even Cu, Ag
Conc. H_2SO_4 (sulphuric acid)
S: +6 arrow +4 ( SO_2 ). Acts as oxidizer only when hot and concentrated
H_2O_2 (as oxidizing agent)
O: -1 arrow -2 . Mild oxidizer in neutral/acidic medium
Halogens ( F_2, Cl_2, Br_2, I_2 )
F_2 is the strongest. ON: 0 arrow -1
KMnO_4 in acidic medium: Mn goes +7 arrow +2 (gains 5 electrons per Mn)
KMnO_4 in neutral/basic medium: Mn goes +7 arrow +4 ( MnO_2 , brown ppt)
K_2Cr_2O_7 always works in acidic medium: Cr goes +6 arrow +3 (gains 3 electrons per Cr)
Conc. HNO_3 oxidizes even noble metals like Cu and Ag (dilute HCl cannot!)
F_2 is the strongest oxidizing agent known — it can oxidize water to O_2
Common Reducing Agents
Important reducing agents for NEET include metals, H₂S, oxalic acid, FeSO₄, and carbon. They contain elements in low oxidation states that want to go up.
Reducing agents donate electrons (get oxidized). They contain elements in low oxidation states that can easily go up. The stronger the tendency to lose electrons, the stronger the reducing agent.
Active metals (Na, K, Mg, Al, Zn)
ON: 0 arrow +n . Na is stronger than Zn (higher in activity series)
H_2S (hydrogen sulphide)
S: -2 arrow 0 (or higher). Rotten-egg gas, S is at minimum ON
C_2O_4^ 2- / H_2C_2O_4 (oxalic acid)
C: +3 arrow +4 ( CO_2 ). Used in KMnO₄ titrations
FeSO_4 / Fe^ 2+ (ferrous)
Fe: +2 arrow +3 . Classic Mohr's salt titration
H_2O_2 (as reducing agent)
O: -1 arrow 0 ( O_2 released). Dual nature!
Carbon (C, CO)
C: 0 arrow +4 ( CO_2 ). Used in metallurgy to reduce metal oxides
Metals are reducing agents because they readily lose electrons to form positive ions
Alkali metals (Na, K) are the strongest metallic reducing agents — they have the lowest ionization energies
H_2S is a reducing agent because S is at its minimum ON ( -2 ) — it can only go up
Oxalic acid ( H_2C_2O_4 ) is the standard reducing agent in KMnO_4 titrations ( C: +3 arrow +4 )
Carbon reduces metal oxides in blast furnaces: Fe_2O_3 + 3C arrow 2Fe + 3CO
Quick recall
In the reaction 2FeCl_3 + H_2S arrow 2FeCl_2 + S + 2HCl , identify the oxidizing and reducing agents.
FeCl_3 is the oxidizing agent (Fe: +3 arrow +2 , reduced). H_2S is the reducing agent (S: -2 arrow 0 , oxidized).
For NEET, memorize which common substances are ALWAYS oxidizing agents (KMnO₄, K₂Cr₂O₇) vs ALWAYS reducing agents (H₂S, metals) vs BOTH (H₂O₂). This saves time in MCQs.
Balancing: Oxidation Number Method
The oxidation number method balances redox equations by ensuring that the total increase in oxidation number equals the total decrease. Steps: write skeleton equation, assign ONs, identify which atoms change ON, equalize the increase and decrease by multiplying, then balance remaining atoms (O and H) by inspection. This method works in any medium and is often faster for simpler reactions.
Total increase in ON = Total decrease in ON
Step-by-Step Method
A clear, repeatable procedure to balance any redox equation using oxidation number changes.
The oxidation number method has a fixed sequence of steps. Follow them mechanically and you'll balance any equation.
Step 1: Write the unbalanced skeleton equation with correct formulas
Step 2: Assign oxidation numbers to ALL atoms on both sides
Step 3: Identify which atoms undergo a change in ON — note the increase and decrease
Step 4: Calculate the total increase and total decrease per formula unit
Step 5: Equalize total increase = total decrease by multiplying the species with appropriate coefficients
Step 6: Balance remaining atoms (usually O and H) by inspection
Step 7: Verify: count every atom on both sides and confirm charge balance
Worked Example 1
Balancing the classic reaction of copper with dilute nitric acid using the oxidation number method.
Reaction: Cu + HNO_3 (dilute) arrow Cu(NO_3)_2 + NO + H_2O
Step 1: Skeleton: Cu + HNO_3 arrow Cu(NO_3)_2 + NO + H_2O
Step 2: Cu: 0 arrow +2 . N in HNO_3 : +5 . N in Cu(NO_3)_2 : +5 (no change). N in NO: +2 (changed!)
Step 3: Cu increases by 2 (oxidized). N decreases by 3 (reduced, only the N that becomes NO)
Step 4: Increase = +2 per Cu. Decrease = -3 per N (in NO)
Step 5: LCM of 2 and 3 = 6 . Multiply Cu by 3 (total increase = 6 ), N(→NO) by 2 (total decrease = 6 )
The half-reaction (ion-electron) method splits the overall reaction into two half-reactions — one for oxidation and one for reduction. Each half-reaction is balanced separately for atoms, then for charge using electrons. Finally, the two are combined so electrons cancel. The method differs for acidic medium (balance O with H₂O, H with H⁺) and basic medium (add OH⁻ to neutralize H⁺). This is the most systematic method and is preferred for ionic equations.
Electrons lost in oxidation half = Electrons gained in reduction half
Steps in Acidic Medium
In acidic medium, oxygen atoms are balanced using H₂O and hydrogen atoms using H⁺. Charges are balanced using electrons.
The half-reaction method is the gold standard for balancing redox equations in solution. In acidic medium, follow these steps for EACH half-reaction separately.
Step 1: Split the reaction into two half-reactions — one for the species being oxidized, one for the species being reduced
Step 2: Balance all atoms OTHER than O and H first
Step 3: Balance O atoms by adding H_2O to the side that needs oxygen
Step 4: Balance H atoms by adding H^+ to the side that needs hydrogen
Step 5: Balance charge by adding electrons ( e^- ) to the more positive side
Step 6: Multiply the half-reactions by appropriate integers so electrons cancel
Step 7: Add the two half-reactions and simplify (cancel common species on both sides)
Step 8: Verify atom and charge balance
Worked Example (Acidic)
Balancing permanganate oxidizing oxalate in acidic medium — the most important half-reaction problem for NEET.
NCERT Chapter 8 provides worked examples in both acidic and basic media. The 2024 and 2023 NEET papers both had questions on half-reaction balancing. This is high-yield content.