Chapter 7: Redox Reactions Long Questions | chemistry Class 11 CBSE

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Chapter 7: Redox Reactions Long Questions | chemistry Class 11 CBSE | ByteNotes.in

Long Answer

Hard difficulty • Structured explanation

Hard

Question 1

Long Form

Trace the evolution of the concept of oxidation from its classical definition to the electronic definition, explaining how each stage was necessitated by new observations.

Originally, oxidation was defined only as the addition of oxygen to an element or compound, such as 2Mg + O2 → 2MgO or S + O2 → SO2. This definition was limited to reactions involving atmospheric oxygen.
Chemists then broadened the definition to include removal of hydrogen from a substance, based on observations like 2H2S + O2 → 2S + 2H2O, where hydrogen is removed from H2S.
The definition was extended further when magnesium was found to react with electronegative non-metals like F2, Cl2, and S without involving oxygen (e.g., Mg + Cl2 → MgCl2); this added 'addition of electronegative element' to the definition.
The removal of electropositive elements was also included; for example, oxidation of K4[Fe(CN)6] to K3[Fe(CN)6] involves removal of potassium (an electropositive element).
The electronic definition emerged from ionic compounds like NaCl (Na+Cl-): sodium loses electrons (2Na → 2Na+ + 2e-) and chlorine gains them (Cl2 + 2e- → 2Cl-). This led to defining oxidation as the loss of electrons by any species.
The oxidation number approach extends this to covalent compounds where complete electron transfer does not occur, treating electron pair sharing as a complete transfer to the more electronegative atom for bookkeeping purposes.

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Long Answer

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Medium

Question 2

Long Form

State all the rules for assigning oxidation numbers to elements in compounds and ions, with one example for each rule.

Rule 1: In the free or uncombined elemental state, each atom has an oxidation number of zero. Examples: each atom in H2, O2, Cl2, O3, P4, S8, Na, Mg has oxidation number 0.
Rule 2: For a monoatomic ion, the oxidation number equals the charge on the ion. Examples: Na+ is +1, Mg2+ is +2, Fe3+ is +3, Cl- is -1, O2- is -2.
Rule 3: Oxygen is -2 in most compounds; exceptions: -1 in peroxides (H2O2), -1/2 in superoxides (KO2), and +2 in OF2 where it is bonded to the more electronegative fluorine.
Rule 4: Hydrogen is +1 in most compounds but -1 in binary metal hydrides such as LiH, NaH, and CaH2.
Rule 5: Fluorine is always -1 in all its compounds. Other halogens are -1 as halide ions but can have positive oxidation numbers when combined with oxygen (in oxoacids and oxoanions).
Rule 6: The algebraic sum of oxidation numbers of all atoms in a neutral compound is zero; in a polyatomic ion, it equals the charge on the ion. Example: in SO42-, 1(x) + 4(-2) = -2, giving S as +6.

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Long Answer

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Medium

Question 3

Long Form

Classify and explain the four types of redox reactions with one balanced equation and the change in oxidation number for each type.

Combination reactions: A + B → C, where at least one reactant must be in elemental form. Example: C(s) + O2(g) → CO2(g); carbon is oxidised from 0 to +4 and oxygen is reduced from 0 to -2.
Decomposition reactions: A compound breaks down into simpler products, at least one in elemental form. Example: 2KClO3(s) → 2KCl(s) + 3O2(g); chlorine is reduced from +5 to -1 and oxygen is oxidised from -2 to 0.
Displacement reactions: An ion or atom in a compound is replaced by another. Metal displacement example: CuSO4(aq) + Zn(s) → Cu(s) + ZnSO4(aq); zinc is oxidised from 0 to +2 and copper is reduced from +2 to 0.
Non-metal displacement example: Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(l); chlorine is reduced from 0 to -1 and bromine is oxidised from -1 to 0.
Disproportionation reactions: A single element in an intermediate oxidation state is simultaneously oxidised and reduced. Example: 2H2O2(aq) → 2H2O(l) + O2(g); oxygen in the -1 state (H2O2) is reduced to -2 (H2O) and oxidised to 0 (O2).
All four types share the fundamental feature that oxidation and reduction occur simultaneously, with changes in oxidation numbers identifying the species acting as oxidant and reductant.

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Long Answer

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Hard

Question 4

Long Form

Balance the equation Cr2O72-(aq) + SO32-(aq) → Cr3+(aq) + SO42-(aq) in acidic medium using the oxidation number method, showing all five steps.

Step 1 — Skeletal ionic equation: Cr2O72-(aq) + SO32-(aq) → Cr3+(aq) + SO42-(aq). Assign oxidation numbers: Cr is +6 in Cr2O72- and +3 in Cr3+; S is +4 in SO32- and +6 in SO42-.
Step 2 — Identify changes: Cr decreases from +6 to +3 (decrease of 3 per Cr atom, decrease of 6 for 2 Cr atoms). S increases from +4 to +6 (increase of 2 per S atom).
Step 3 — Equalise changes: to make total increase equal to total decrease, multiply S by 3 (total increase = 6) and keep Cr2O72- with coefficient 1 (total decrease = 6). Place 2 before Cr3+ and 3 before SO42-: Cr2O72-(aq) + 3SO32-(aq) → 2Cr3+(aq) + 3SO42-(aq).
Step 4 — Balance charges (acidic medium): left side has charge 2-(-2) + 3(-2) = -8; right side has charge 2(+3) + 3(-2) = 0. Add 8H+ on the left: Cr2O72- + 3SO32- + 8H+ → 2Cr3+ + 3SO42-.
Step 5 — Balance hydrogen with water: left side has 8 H; add 4H2O on the right. Final balanced equation: Cr2O72-(aq) + 3SO32-(aq) + 8H+(aq) → 2Cr3+(aq) + 3SO42-(aq) + 4H2O(l).
Verification: Cr — 2 on each side; S — 3 on each side; O — 7+9+0=16 left and 12+4=16 right; H — 8 on each side; charge — (-2-6+8) = 0 left and (+6-6) = 0 right. Equation is fully balanced.

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Long Answer

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Medium

Question 5

Long Form

Describe the construction, working, and electrode reactions of the Daniell cell, including the role of the salt bridge.

Construction: The Daniell cell consists of two beakers — one with copper sulphate solution containing a copper rod (cathode, positive electrode), and one with zinc sulphate solution containing a zinc rod (anode, negative electrode). The beakers are connected by a salt bridge and the rods by a metallic wire with an ammeter.
Salt bridge: A U-tube containing KCl or NH4NO3 solution solidified with agar-agar; it provides electrical contact between the two solutions by allowing ion migration without mixing the solutions, completing the internal circuit.
Anode reaction (oxidation): Zn(s) → Zn2+(aq) + 2e-; zinc is oxidised, releasing electrons into the external circuit, which causes the zinc rod to gradually dissolve.
Cathode reaction (reduction): Cu2+(aq) + 2e- → Cu(s); copper ions gain electrons from the external circuit and are deposited as metallic copper on the copper rod.
Electron flow: Electrons flow from the anode (zinc) through the external wire to the cathode (copper); conventional current flows in the opposite direction.
The two redox couples in this cell are Zn2+/Zn (E° = -0.76 V) and Cu2+/Cu (E° = +0.34 V). The cell converts chemical energy from the spontaneous redox reaction into electrical energy.

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Long Answer

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Hard

Question 6

Long Form

Explain the half-reaction method for balancing redox equations in basic medium, using the reaction MnO4-(aq) + I-(aq) → MnO2(s) + I2(s) as an example.

Step 1 — Write skeletal equation and identify half-reactions: Oxidation half: I-(aq) → I2(s); Reduction half: MnO4-(aq) → MnO2(s). Oxidation numbers: I goes from -1 to 0; Mn goes from +7 to +4.
Step 2 — Balance atoms other than O and H: Oxidation: 2I-(aq) → I2(s). Reduction: MnO4-(aq) → MnO2(s) (Mn atoms already balanced).
Step 3 — Balance O and H for basic medium (first balance as for acidic): MnO4-(aq) + 4H+(aq) → MnO2(s) + 2H2O(l). Then add 4OH- to both sides: MnO4-(aq) + 4H2O(l) → MnO2(s) + 4OH-(aq).
Step 4 — Balance charges by adding electrons: Oxidation: 2I-(aq) → I2(s) + 2e-. Reduction: MnO4-(aq) + 2H2O(l) + 3e- → MnO2(s) + 4OH-(aq).
Step 5 — Equalise electrons: multiply oxidation half by 3 and reduction half by 2: 6I-(aq) → 3I2(s) + 6e-; 2MnO4-(aq) + 4H2O(l) + 6e- → 2MnO2(s) + 8OH-(aq).
Step 6 — Add half-reactions (cancel 6e- on each side): 6I-(aq) + 2MnO4-(aq) + 4H2O(l) → 3I2(s) + 2MnO2(s) + 8OH-(aq). Verification confirms atoms and charges are balanced on both sides.

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Long Answer

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Medium

Question 7

Long Form

Explain competitive electron transfer reactions among metals, leading to the concept of the electrochemical series. How is this series useful in predicting redox reactions?

When a zinc strip is placed in copper nitrate solution, zinc displaces copper: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s). The reverse reaction (copper placed in zinc sulphate solution) shows no visible reaction, confirming that zinc has a greater tendency to lose electrons than copper.
When copper is placed in silver nitrate solution, copper displaces silver: Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s), confirming that copper has a greater tendency to lose electrons than silver.
These comparisons establish the order of electron-releasing tendency: Zn > Cu > Ag. This competition for electrons among metals is analogous to the competition for protons among acids.
This concept is extended to all metals to design the electrochemical series (metal activity series), in which metals are listed in increasing order of their electrode potentials (standard reduction potentials).
Metals higher in the series (more negative E°) are stronger reducing agents and can displace metals with more positive E° from their salt solutions.
The electrochemical series allows prediction of the feasibility of redox reactions; if the reducing agent has a more negative E° than the oxidising agent, the reaction is spontaneous. It also forms the basis for designing galvanic cells (such as the Daniell cell) in which chemical energy is converted to electrical energy.

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Long Answer

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Hard

Question 8

Long Form

Analyse how disproportionation reactions are different from other redox reactions, explaining with examples of Cl2 in alkaline medium and H2O2 decomposition.

In ordinary redox reactions, one species is oxidised and a different species is reduced. In disproportionation reactions, a single species containing one element in an intermediate oxidation state simultaneously undergoes both oxidation and reduction in the same reaction.
A key condition for disproportionation is that the element must be capable of existing in at least three oxidation states — a lower state, the intermediate (reacting) state, and a higher state.
In the reaction 2H2O2(aq) → 2H2O(l) + O2(g), oxygen in H2O2 is in the -1 state; it is simultaneously reduced to -2 (in H2O) and oxidised to 0 (in O2). The same element in the same compound undergoes both changes.
In the reaction Cl2(g) + 2OH-(aq) → ClO-(aq) + Cl-(aq) + H2O(l), chlorine in Cl2 is in the 0 oxidation state; it is simultaneously oxidised to +1 (in ClO-) and reduced to -1 (in Cl-). This reaction produces household bleaching agents.
Fluorine cannot show disproportionation because, being the most electronegative element, it cannot achieve a positive oxidation state — a necessary requirement for the oxidised product.
ClO4- does not disproportionate because chlorine is already in its highest possible oxidation state (+7) and cannot be further oxidised. This rule — that an element in its highest oxidation state cannot be oxidised further — helps identify which species cannot disproportionate.

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Long Answer

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Hard

Question 9

Long Form

Compare the oxidation number method and the half-reaction method for balancing redox equations. When is each method preferred, and what are the key steps in each?

Both methods give the same final balanced equation. The Oxidation Number Method is based on tracking changes in oxidation state, while the Half-Reaction (Ion-Electron) Method explicitly shows electron gain and loss in separate partial equations.
Key steps in the Oxidation Number Method: (1) Write correct formulas, (2) Assign oxidation numbers and identify changes, (3) Multiply atoms to equalise total increase and decrease, (4) Add H+ or OH- to balance ionic charge, (5) Add H2O to balance hydrogen atoms and verify oxygen balance.
Key steps in the Half-Reaction Method: (1) Write unbalanced ionic equation, (2) Separate into oxidation and reduction half-reactions, (3) Balance atoms other than O and H, (4) Balance O using H2O and H using H+, (5) Balance charges by adding electrons, (6) Equalise electrons in both half-reactions and add them, (7) Verify the final equation.
For basic medium, the Half-Reaction Method requires an additional step: for each H+ present after balancing, add an equal number of OH- to both sides, then combine H+ and OH- on the same side to give H2O.
The Oxidation Number Method is often quicker for straightforward equations, while the Half-Reaction Method is more systematic and transparent in showing electron flow, making it useful for complex reactions and for understanding electrode processes.
Both methods require identifying the species that changes oxidation number and ensuring that the total increase in oxidation number equals the total decrease, reflecting the conservation of electrons in a redox reaction.

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Long Answer

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Hard

Question 10

Long Form

Discuss how redox reactions are used in titrations, describing three different indicator situations with examples from the chapter.

Redox titrations are used to determine the concentration (strength) of an oxidant or reductant using a redox-sensitive indicator. The end point is the point at which stoichiometrically equal amounts of oxidant and reductant have reacted.
Situation 1 — Self-indicator: Permanganate ion (MnO4-) is intensely coloured (pink/purple) and is reduced to colourless Mn2+ during the titration. No separate indicator is needed; the end point is marked by the first permanent faint pink colour when all the reductant (e.g., Fe2+ or C2O42-) has been consumed. The detection threshold is as low as 10-6 mol L-1.
Situation 2 — External oxidation indicator: Dichromate ion (Cr2O72-) is not a self-indicator. Diphenylamine is used as an external indicator, which is oxidised by excess Cr2O72- just after the equivalence point, producing an intense blue colour to signal the end point.
Situation 3 — Iodometric titration: Some oxidants (e.g., Cu2+) can oxidise I- to I2: 2Cu2+(aq) + 4I-(aq) → Cu2I2(s) + I2(aq). The liberated I2 is then titrated with thiosulphate (S2O32-) in the presence of starch as indicator; the intense blue colour of the starch-I2 complex disappears at the end point when all I2 has been reduced.
The starch-I2 colour disappears when I2 is consumed: I2(aq) + 2S2O32-(aq) → 2I-(aq) + S4O62-(aq). I2 remains in KI solution as KI3.
These three methods illustrate how the visible colour change or colour disappearance tied to a redox reaction is used to accurately locate the equivalence point in redox titrations.

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Chapter 7: Redox Reactions | Long Questions

Trace the evolution of the concept of oxidation from its classical definition to the electronic definition, explaining how each stage was necessitated by new observations.

State all the rules for assigning oxidation numbers to elements in compounds and ions, with one example for each rule.

Classify and explain the four types of redox reactions with one balanced equation and the change in oxidation number for each type.

Balance the equation Cr2O72-(aq) + SO32-(aq) → Cr3+(aq) + SO42-(aq) in acidic medium using the oxidation number method, showing all five steps.

Describe the construction, working, and electrode reactions of the Daniell cell, including the role of the salt bridge.

Explain the half-reaction method for balancing redox equations in basic medium, using the reaction MnO4-(aq) + I-(aq) → MnO2(s) + I2(s) as an example.

Explain competitive electron transfer reactions among metals, leading to the concept of the electrochemical series. How is this series useful in predicting redox reactions?

Analyse how disproportionation reactions are different from other redox reactions, explaining with examples of Cl2 in alkaline medium and H2O2 decomposition.

Compare the oxidation number method and the half-reaction method for balancing redox equations. When is each method preferred, and what are the key steps in each?

Discuss how redox reactions are used in titrations, describing three different indicator situations with examples from the chapter.

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