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

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

Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 1

Applied Concept

A student places a zinc rod in a beaker containing copper sulphate solution. With reference to competitive electron transfer reactions, explain the observations and identify the oxidising and reducing agents.

The blue colour of the copper sulphate solution gradually fades as Cu2+ ions are consumed, and a reddish deposit of metallic copper forms on the zinc rod as zinc dissolves to form Zn2+ ions.
The reaction is Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s); zinc loses electrons (oxidised, oxidation number 0 to +2) and copper ions gain electrons (reduced, oxidation number +2 to 0).
Zinc is the reducing agent (reductant) because it donates electrons, and Cu2+ ion is the oxidising agent (oxidant) because it accepts electrons.
The equilibrium greatly favours the products, indicating that zinc has a far greater tendency to release electrons than copper; this is consistent with Zn > Cu in the electrochemical series.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 2

Applied Concept

In a laboratory, a chemist uses permanganate (MnO4-) as the titrant in a redox titration of Fe2+ ions in acidic solution. Explain how the end point is detected and why no separate indicator is needed.

MnO4- is an intensely pink/purple coloured ion that acts as a self-indicator. During the titration, MnO4- oxidises Fe2+ to Fe3+ and is itself reduced to the colourless Mn2+ ion; as long as Fe2+ remains, each drop of MnO4- added is immediately decolourised.
The end point is reached when the last mole of Fe2+ has been oxidised; the very next drop of MnO4- added remains unreduced and imparts a permanent faint pink colour to the solution.
Since MnO4- is detectable at concentrations as low as 10-6 mol L-1, the colour change is perceptible at the true equivalence point with minimal overshoot.
No external indicator is needed because the reactant itself (MnO4-) provides a dramatic colour signal — this is the defining characteristic of a self-indicator in redox titrations.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 3

Applied Concept

With reference to the Daniell cell described in the chapter, explain what would happen if the salt bridge is removed, and justify your answer in terms of the role of the salt bridge.

If the salt bridge is removed, the flow of current through the external circuit would stop almost immediately, even though the zinc rod and copper sulphate solution are still present in their respective beakers.
The salt bridge provides the internal circuit path by allowing ions to migrate between the two half-cell solutions — anions migrate toward the anode compartment and cations toward the cathode compartment — thus maintaining electrical neutrality.
Without the salt bridge, the anode compartment would develop a positive charge (excess Zn2+ from oxidation of Zn) and the cathode compartment a negative charge (deficit of Cu2+ from reduction), quickly creating a potential that opposes the cell reaction and stopping it.
The salt bridge thus completes the circuit inside the cell and sustains the flow of current; without it, the Daniell cell cannot function as a source of electrical energy.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 4

Applied Concept

A piece of metallic sodium is dropped into water. Identify the type of redox reaction, write the balanced equation, and explain the role of water as an oxidising agent.

This is a metal displacement reaction (non-metal displacement — hydrogen displacement from water by an active metal). The reaction is: 2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g).
Oxidation numbers: Na changes from 0 to +1 (oxidised); H in H2O changes from +1 to 0 in H2 (reduced); O remains -2 in both H2O and NaOH.
Water acts as the oxidising agent in this reaction because the hydrogen of water gains electrons from sodium and is reduced to molecular dihydrogen.
This reaction is vigorous and exothermic; it occurs because sodium is a very active metal with a highly negative standard electrode potential (E° = -2.71 V), giving it a strong tendency to lose electrons and displace hydrogen from water.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 5

Applied Concept

Household bleaching powder acts by liberating hypochlorite ions. With reference to the disproportionation of Cl2 in alkaline medium, explain how bleach is produced and how it removes stains.

When Cl2 reacts with alkali (OH-), it undergoes disproportionation: Cl2(g) + 2OH-(aq) → ClO-(aq) + Cl-(aq) + H2O(l). Chlorine in the zero oxidation state is simultaneously oxidised to +1 (in ClO-) and reduced to -1 (in Cl-).
The hypochlorite ion (ClO-) produced in this reaction is the active bleaching component. It is a strong oxidising agent that oxidises the colour-bearing organic compounds (chromophores) in stains to colourless products.
Bromine and iodine show the same trend as chlorine in this reaction, while fluorine cannot undergo this disproportionation because it cannot attain the +1 oxidation state due to being the most electronegative element.
The bleaching action is therefore fundamentally a redox process — the stain molecules are oxidised (lose electrons) to colourless species by the hypochlorite ion, which is itself reduced in the process.

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Application Question

Medium difficulty • Concept in a practical situation

Medium

Question 6

Applied Concept

A forensic chemist uses the 'Layer Test' to identify halide ions in a water sample. With reference to halogen displacement reactions, explain the principle of this test for Br- and I-.

The Layer Test is based on halogen displacement reactions. When Cl2 water is added to the sample and then CCl4 (a non-polar organic solvent) is added and shaken, a two-layer system forms.
If Br- is present: Cl2(g) + 2Br-(aq) → 2Cl-(aq) + Br2(l). The liberated Br2, being coloured and more soluble in CCl4 than in water, colours the CCl4 layer orange-brown.
If I- is present: Cl2(g) + 2I-(aq) → 2Cl-(aq) + I2(s). The liberated I2 dissolves in the CCl4 layer giving a violet/purple colour.
The test works because Cl2 is a stronger oxidising agent than both Br2 and I2 (higher E° value), and therefore spontaneously oxidises Br- and I- to their respective molecular forms; the distinct colours of Br2 (orange) and I2 (violet) in the organic layer allow easy identification.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 7

Applied Concept

In Ostwald's process for the industrial manufacture of nitric acid, ammonia is oxidised by oxygen. Write the balanced equation for the first step and identify the changes in oxidation number.

The first step in Ostwald's process is the catalytic oxidation of ammonia: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g).
In NH3, nitrogen has oxidation number -3 (combined with three H atoms each at +1, and the molecule is neutral). In NO, nitrogen has oxidation number +2.
Nitrogen is therefore oxidised from -3 to +2 (increase of 5 per N atom, total increase = 4 × 5 = 20), making NH3 the reducing agent.
Oxygen changes from 0 in O2 to -2 in both NO and H2O (decrease of 2 per O atom). Oxygen is reduced and O2 is the oxidising agent. The total decrease = 5 × 2 × 2 = 20, confirming the balance.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 8

Applied Concept

With reference to the concept of oxidation number, explain why SO2 can act as both an oxidising and a reducing agent, while ozone (O3) can act only as an oxidant.

In SO2, sulphur has an oxidation number of +4, which is an intermediate state between the lowest (-2 in H2S) and highest (+6 in H2SO4) values for sulphur. Since sulphur can be both oxidised to +6 and reduced to lower states, SO2 can act as both an oxidising and a reducing agent.
As a reducing agent (sulphur oxidised from +4 to +6): SO2 reduces KMnO4 and reduces Cl2 to HCl (SO2 + Cl2 + 2H2O → H2SO4 + 2HCl).
As an oxidising agent (sulphur reduced from +4 to lower states): SO2 oxidises H2S in the reaction 2H2S + SO2 → 3S + 2H2O.
Ozone (O3) contains oxygen in the 0 oxidation state — the same as in elemental O2. Since oxygen in ozone cannot be further oxidised to a positive state (due to its high electronegativity), ozone can only act as an oxidant, not a reductant.

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 9

Applied Concept

A student is asked to balance the reaction MnO4-(aq) + SO2(g) → Mn2+(aq) + HSO4-(aq) in acidic medium using the oxidation number method. Outline the key steps.

Assign oxidation numbers: Mn in MnO4- is +7; Mn in Mn2+ is +2 (decrease of 5 — Mn is reduced). S in SO2 is +4; S in HSO4- is +6 (increase of 2 — S is oxidised). MnO4- is the oxidant; SO2 is the reductant.
To equalise total increase and decrease: LCM of 5 and 2 is 10; place coefficient 2 before MnO4- and Mn2+ (total decrease = 10), and coefficient 5 before SO2 and HSO4- (total increase = 10). Equation: 2MnO4- + 5SO2 → 2Mn2+ + 5HSO4-.
Balance charges in acidic medium: left side charge = 2(-1) = -2; right side charge = 2(+2) + 5(-1) = -1. Add H+ ions to the left: 2MnO4- + 5SO2 + ?H+ → 2Mn2+ + 5HSO4-. Right charge = -1; left must be -1; so add 1H+ on the right is not needed; try: 2MnO4- + 5SO2 + 2H+ → 2Mn2+ + 5HSO4-; left = -2+2=0; right = +4-5=-1. Not equal. Final: 2MnO4- + 5SO2 + 2H2O → 2Mn2+ + 5HSO4-; check charges: left = -2; right = +4-5 = -1. Add H+: 2MnO4- + 5SO2 + 2H+ + ... The balanced equation is 2MnO4-(aq) + 5SO2(g) + 2H2O(l) → 2Mn2+(aq) + 5HSO4-(aq).
Verify: Mn: 2 = 2; S: 5 = 5; O: 8+10+2 = 20 left; 20 right. H: 4 left; 5 right — add H+: final balanced: 2MnO4-(aq) + 5SO2(g) + 2H2O(l) + H+(aq) → 2Mn2+(aq) + 5HSO4-(aq).

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Application Question

Hard difficulty • Concept in a practical situation

Hard

Question 10

Applied Concept

An iron nail is placed in a solution of FeCl3. With reference to the concept of oxidising and reducing agents in excess, predict and explain the product formed.

This is a displacement redox reaction: Fe(s) + 2Fe3+(aq) → 3Fe2+(aq). The iron nail (Fe, oxidation state 0) reduces Fe3+ (oxidation state +3) to Fe2+ (oxidation state +2), and the nail itself is oxidised from 0 to +2.
The principle from the chapter states: when a reducing agent is in excess, a compound of lower oxidation state is formed; when the oxidising agent is in excess, a compound of higher oxidation state is formed.
If the iron nail (reducing agent) is in excess: Fe3+ is completely reduced to Fe2+, and the product is FeCl2 (lower oxidation state).
If FeCl3 (oxidising agent) is in excess: the iron nail dissolves completely and is oxidised; under excess oxidant, Fe is oxidised to its higher oxidation state +3, so the product would be FeCl3 (higher oxidation state, i.e., Fe remains as Fe3+).

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Chapter 7: Redox Reactions | Application Questions

A student places a zinc rod in a beaker containing copper sulphate solution. With reference to competitive electron transfer reactions, explain the observations and identify the oxidising and reducing agents.

In a laboratory, a chemist uses permanganate (MnO4-) as the titrant in a redox titration of Fe2+ ions in acidic solution. Explain how the end point is detected and why no separate indicator is needed.

With reference to the Daniell cell described in the chapter, explain what would happen if the salt bridge is removed, and justify your answer in terms of the role of the salt bridge.

A piece of metallic sodium is dropped into water. Identify the type of redox reaction, write the balanced equation, and explain the role of water as an oxidising agent.

Household bleaching powder acts by liberating hypochlorite ions. With reference to the disproportionation of Cl2 in alkaline medium, explain how bleach is produced and how it removes stains.

A forensic chemist uses the 'Layer Test' to identify halide ions in a water sample. With reference to halogen displacement reactions, explain the principle of this test for Br- and I-.

In Ostwald's process for the industrial manufacture of nitric acid, ammonia is oxidised by oxygen. Write the balanced equation for the first step and identify the changes in oxidation number.

With reference to the concept of oxidation number, explain why SO2 can act as both an oxidising and a reducing agent, while ozone (O3) can act only as an oxidant.

A student is asked to balance the reaction MnO4-(aq) + SO2(g) → Mn2+(aq) + HSO4-(aq) in acidic medium using the oxidation number method. Outline the key steps.

An iron nail is placed in a solution of FeCl3. With reference to the concept of oxidising and reducing agents in excess, predict and explain the product formed.

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