🎓 Class 11ChemistryCBSETheoryCh 7 – Redox Reactions⏱ ~14 min
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Classical Concept and Electron Transfer in Redox Reactions
Introduction: Redox — the Chemistry of Electron Exchange
Strike a matchstick, breathe in a lungful of air, bite into an apple and let it brown on the kitchen counter — you have just witnessed three redox reactions. A leaf quietly turning sunlight and water into sugar (photosynthesis), your mitochondria burning that sugar back into carbon dioxide and water (respiration), the iron railings of your balcony flaking into rust, a lithium-ion cell powering your phone, and the blast furnace that converted iron ore into steel for your school bus — all belong to the same family of reactions in which electrons hop from one atom to another.
Redox is also the beating heart of the Hydrogen Economy — the clean-energy vision where water is split into H2 and O2 by electricity, and the H2 is later recombined with O2 in a fuel cell to give back electricity plus pure water. This chapter builds the language you need to read and write every such reaction with confidence.
7.1 Classical Ideas of Redox Reactions — Oxidation and Reduction
Long before electrons were known, chemists had already described oxidation and reduction by watching what went in and what came out of a reaction. These "classical" definitions still work perfectly in everyday descriptions.
Classical Definitions of Oxidation L1 Remember
Oxidation (classical view): a process in which a substance gains oxygen / any electronegative element, or loses hydrogen / any electropositive element.
Type
Example reaction
What happens
(i) Addition of oxygen
2 Mg(s) + O2(g) → 2 MgO(s)
Mg is oxidised — it picks up O.
(i) Addition of oxygen
S(s) + O2(g) → SO2(g)
S is oxidised — gains O.
(ii) Addition of electronegative element
2 Fe(s) + 3 Cl2(g) → 2 FeCl3(s)
Fe is oxidised — gains Cl (electronegative).
(iii) Removal of hydrogen
H2S(g) + Cl2(g) → 2 HCl(g) + S(s)
H2S is oxidised — loses H.
(iv) Removal of electropositive element
2 K4[Fe(CN)6] + H2O2 → 2 K3[Fe(CN)6] + 2 KOH
Ferrocyanide loses electropositive K — oxidised.
Classical Definitions of Reduction L1 Remember
Reduction (classical view): the mirror image — a substance loses oxygen / any electronegative element, or gains hydrogen / any electropositive element.
Type
Example reaction
What happens
Removal of oxygen
CuO(s) + H2(g) → Cu(s) + H2O(l)
CuO is reduced — loses O.
Removal of electronegative element
2 FeCl3(aq) + H2(g) → 2 FeCl2(aq) + 2 HCl(aq)
FeCl3 is reduced — loses Cl.
Addition of hydrogen
CH2=CH2(g) + H2(g) → CH3–CH3(g)
Ethene is reduced — gains H.
Addition of electropositive element
HgCl2(aq) + Hg(l) → Hg2Cl2(s)
HgCl2 is reduced — gains Hg.
Key idea: Oxidation and reduction never travel alone. Wherever one occurs, the other must occur at the same time — the combined process is a redox reaction. In 2 Mg + O2 → 2 MgO, Mg is oxidised and O2 is reduced simultaneously.
Fig 7.1 — Oxidation and reduction are two faces of every redox coin.
7.2 Redox Reactions in Terms of Electron Transfer
The classical picture cannot describe a reaction like Zn + Cu2+ → Zn2+ + Cu, where neither oxygen nor hydrogen is involved. The modern electron-transfer definition repairs this gap and explains why all the classical cases actually fit too.
Modern definitions:
Oxidation = loss of electrons — remember OIL (Oxidation Is Loss).
Reduction = gain of electrons — remember RIG (Reduction Is Gain).
A Familiar Example — the Zinc–Copper Displacement
Drop a zinc strip into blue copper sulphate solution. Within minutes a pink-brown deposit of copper appears on the zinc, and the solution fades from blue to colourless:
Zn(s) + CuSO4(aq) → ZnSO4(aq) + Cu(s)
Writing only the species that actually change (ignoring SO42−, the spectator ion):
Fig 7.2 — Two electrons leave Zn and are captured by Cu2+.Fig 7.3 — Splitting a redox equation into its two half-reactions.
Oxidising and Reducing Agents L2 Understand
Oxidising agent (oxidant): the species that accepts electrons. It is reduced itself while forcing another substance to be oxidised. Reducing agent (reductant): the species that donates electrons. It is oxidised itself while forcing another substance to be reduced.
In Zn + Cu2+ → Zn2+ + Cu, Zn is the reducing agent (it is oxidised) and Cu2+ is the oxidising agent (it is reduced). Never the other way around!
Competitive Electron Transfer — the Activity Series
Place a copper strip in zinc sulphate solution and wait — nothing happens. Place zinc in copper sulphate — rapid displacement. Why the asymmetry? Because zinc has a greater tendency to lose electrons (it is more active). The reverse push is not thermodynamically favourable.
Metals can be arranged in order of decreasing reactivity — the activity (or electrochemical) series. A metal higher in this series will displace a metal lower down it from an aqueous solution of the lower metal's salt.
Fig 7.4 — Simplified activity series. Any metal displaces those below it from aqueous salt solutions.
Interactive: Half-Reaction Splitter
Type a simple electron-transfer equation (for example Zn+Cu2+->Zn2++Cu or Mg+2HCl->MgCl2+H2) and see the two halves.
Waiting for input…
Worked Examples L3 Apply
Example 7.1 — Identify the oxidising and reducing agent
Reaction: 2 Na(s) + Cl2(g) → 2 NaCl(s)
Step 1 — Electron book-keeping. Each Na loses 1 e⁻ to form Na⁺. Each Cl atom gains 1 e⁻ to form Cl⁻.
Fe is the reducing agent; H⁺ (from HCl) is the oxidising agent.
Example 7.3 — Classical view meets electron view
Reaction: CuO(s) + H2(g) → Cu(s) + H2O(l)
Classical: CuO loses O (reduced); H2 gains O (oxidised).
Electron view: Cu2+ (in CuO) + 2 e⁻ → Cu (reduced); H2 → 2 H⁺ + 2 e⁻ (oxidised). Same conclusion, deeper explanation.
Example 7.4 — Will the reaction happen?
Predict whether Cu(s) reacts with ZnSO4(aq).
Cu lies below Zn in the activity series — Cu is less active. Therefore Cu cannot displace Zn from its salt solution. No reaction.
Reverse case: Zn in CuSO4 — Zn is above Cu, reaction proceeds spontaneously (our original example).
Example 7.5 — Spotting both half-reactions in a single line
Reaction: 2 Na(s) + 2 H2O(l) → 2 NaOH(aq) + H2(g)
Oxidation: Na → Na⁺ + e⁻ (×2)
Reduction: 2 H2O + 2 e⁻ → 2 OH⁻ + H2 (H goes from +1 to 0)
Na = reducing agent; H2O (or rather H in it) = oxidising agent.
Activity 7.1 — The Zn strip in CuSO4L3 Apply
Materials: clean Zn strip, 0.5 M CuSO4 solution, small beaker, tweezers.
Pour ~30 mL of CuSO4 into the beaker and note the colour.
Dip the polished Zn strip halfway in. Observe continuously for 10 minutes.
Remove and rinse. Examine the strip.
Predict: Will the blue fade? What will coat the zinc?
The blue colour fades as Cu2+ ions are consumed. A pink-brown film of metallic copper coats the zinc. The solution now contains colourless Zn2+. Two electrons have travelled from each Zn atom to a Cu2+ ion.
Competency-Based Questions
A student sets up four test tubes. Tube A: Zn strip in CuSO4; Tube B: Cu strip in ZnSO4; Tube C: Fe nail in CuSO4; Tube D: Ag wire in FeSO4. The activity order is Zn > Fe > Cu > Ag.
Q1. In which tube(s) will a displacement reaction occur?
A. A and C only
B. A, B and C
C. A, B, C and D
D. Only A
A. A (Zn displaces Cu) and C (Fe displaces Cu) work. In B and D the incoming metal is below, not above, the metal in solution.
Q2. Name the oxidising and reducing agent in Tube C.
Cu2+ is the oxidising agent (it gains 2 e⁻ → Cu); Fe is the reducing agent (it loses 2 e⁻ → Fe2+).
Q3. Write both half-reactions for Tube A.
Ox: Zn → Zn2+ + 2 e⁻. Red: Cu2+ + 2 e⁻ → Cu.
Q4. True/False: In Tube B, the Cu strip will develop a layer of Zn over time.
False. Cu lies below Zn in the activity series, so it cannot reduce Zn2+ to Zn. No reaction.
Q5. (HOT) Iron nails in direct contact with copper plumbing pipes in moist air corrode faster than an isolated iron nail. Explain using the activity concept.
Fe and Cu form a tiny galvanic cell when bridged by moisture. Fe, being more active, is forced to become the anode and loses electrons (corrodes); Cu acts as the inert cathode. This is called galvanic or bimetallic corrosion.
Assertion–Reason Questions
Options: A both true and R correctly explains A · B both true but R does not explain A · C A true R false · D A false R true.
A: In the reaction 2 Na + Cl2 → 2 NaCl, sodium is the reducing agent.
R: Sodium loses electrons and is itself oxidised.
A. Both statements are true and R explains A correctly.
A: A copper spoon placed in silver nitrate solution develops a silver deposit.
R: Silver is placed higher than copper in the activity series.
C. Assertion is true (Cu is above Ag, so it displaces Ag from AgNO3), but R is false — silver is below copper, not above it.
A: Every oxidation must be accompanied by a reduction.
R: Electrons lost by one species must be gained by another — they cannot float free.
A. Both true and R is the reason for A.
Frequently Asked Questions — Classical Concept and Electron Transfer in Redox Reactions
What is the classical concept of oxidation and reduction?
The classical concept of redox in NCERT Class 11 Chemistry Chapter 7 defines oxidation as the addition of oxygen or an electronegative element, or the removal of hydrogen or an electropositive element. Reduction is the opposite: addition of hydrogen or an electropositive element, or removal of oxygen or an electronegative element. Examples: 2Mg + O₂ → 2MgO (Mg oxidised); CuO + H₂ → Cu + H₂O (CuO reduced). This original definition is limited because many redox reactions do not involve oxygen or hydrogen — leading to the modern electron-transfer concept.
What is the electron transfer concept of redox?
The modern electron transfer concept defines oxidation as loss of electrons and reduction as gain of electrons. Useful mnemonic: OIL RIG (Oxidation Is Loss, Reduction Is Gain) of electrons. NCERT Class 11 Chemistry Chapter 7 uses this concept because it includes all redox reactions, not just those involving oxygen. Example: Zn + Cu²⁺ → Zn²⁺ + Cu — Zn loses 2 electrons (oxidised) and Cu²⁺ gains 2 electrons (reduced). Every redox reaction can be split into two half-reactions, one oxidation and one reduction, which always occur simultaneously.
What are oxidising and reducing agents?
An oxidising agent (oxidant) is a substance that accepts electrons from another substance, thereby getting reduced itself while oxidising the other. Examples in NCERT Class 11 Chemistry Chapter 7: F₂, O₃, KMnO₄, K₂Cr₂O₇, HNO₃ (concentrated). A reducing agent (reductant) donates electrons, getting oxidised itself while reducing the other substance. Examples: H₂, C, CO, metals (Na, Mg, Al, Zn). The strongest oxidising agent in chemistry is fluorine and the strongest reducing agent is lithium (based on standard reduction potentials).
What are combination and decomposition redox reactions?
Combination redox reactions occur when two substances combine to form one product, accompanied by change in oxidation states. Examples in NCERT Class 11 Chemistry Chapter 7: 2H₂ + O₂ → 2H₂O; 4Na + O₂ → 2Na₂O; C + O₂ → CO₂. Decomposition redox reactions occur when a single compound breaks down into two or more simpler substances, with changes in oxidation states. Examples: 2KClO₃ → 2KCl + 3O₂ (Cl reduced, O oxidised); 2HgO → 2Hg + O₂ (Hg reduced from +2 to 0). Both types involve simultaneous oxidation and reduction.
What is a displacement reaction?
A displacement reaction is one in which a more reactive element displaces a less reactive element from its compound. NCERT Class 11 Chemistry Chapter 7 distinguishes two types: (1) metal displacement — a more reactive metal displaces a less reactive metal from its salt solution, e.g., Zn + CuSO₄ → ZnSO₄ + Cu (Zn oxidised, Cu reduced); (2) non-metal displacement — Cl₂ + 2KBr → 2KCl + Br₂ (Cl reduced, Br oxidised). The reactivity series of metals (K > Na > Ca > Mg > Al > Zn > Fe > Cu > Ag) determines whether displacement occurs.
What is a disproportionation reaction?
A disproportionation reaction is a special redox reaction in which the same element is simultaneously oxidised and reduced. The element must be in an intermediate oxidation state to be both reduced (to a lower state) and oxidised (to a higher state) at the same time. Examples in NCERT Class 11 Chemistry Chapter 7: 2H₂O₂ → 2H₂O + O₂ (O goes from −1 to −2 and 0); 3Cl₂ + 6NaOH → 5NaCl + NaClO₃ + 3H₂O (Cl goes from 0 to −1 and +5); P₄ + 3NaOH + 3H₂O → PH₃ + 3NaH₂PO₂ (P goes from 0 to −3 and +1). Disproportionation occurs in alkaline conditions for many halogens.
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