Chapter 7: Redox Reactions Case Study Questions | chemistry Class 11 CBSE

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Case Set 1

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A student places a strip of metallic zinc into a beaker containing aqueous copper sulphate solution and observes it for about one hour. Gradually, the blue colour of the solution fades and a reddish deposit forms on the zinc strip. The student then passes hydrogen sulphide gas through the colourless solution remaining after the blue colour has completely disappeared and observes a white precipitate. In a parallel experiment, a copper strip is placed in zinc sulphate solution and observed for the same duration. No visible change is noticed, and attempts to detect Cu2+ ions by passing H2S gas through this solution to produce black CuS also fail. The student concludes that the equilibrium for the first reaction greatly favours the products.

Question 1: Write the ionic equation for the reaction occurring when zinc is placed in copper sulphate solution and identify which species is oxidised.

The ionic equation for the reaction is: Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s).
Zinc is oxidised in this reaction because it loses two electrons, and its oxidation number increases from 0 to +2.
The reddish deposit on the zinc strip is metallic copper, and the disappearance of the blue colour confirms the consumption of Cu2+ ions from the solution.

Question 2: Explain why passing H2S through the colourless solution gives a white precipitate of ZnS, and what this observation confirms about the products of the reaction.

After the reaction between Zn and CuSO4 is complete, the colourless solution contains Zn2+ ions (formed by oxidation of zinc). When H2S is passed through it, Zn2+ reacts with S2- to form white zinc sulphide (ZnS) precipitate.
This confirms that Zn2+ ions are indeed present as a product of the reaction, supporting the conclusion that zinc was oxidised to Zn2+ during the reaction with Cu2+.
The fact that H2S does not produce black CuS in the colourless solution also confirms that all Cu2+ has been consumed, meaning the reaction went essentially to completion, greatly favouring the products.

Question 3: Based on both experiments described in the passage, what conclusion can be drawn about the relative electron-releasing tendency of zinc and copper? How does this observation connect to the concept of the electrochemical series?

The first experiment (Zn displaces Cu2+) shows that zinc can spontaneously reduce copper ions, meaning zinc has a greater tendency to release electrons than copper. The second experiment (Cu does not displace Zn2+) confirms the reverse reaction is non-spontaneous, establishing that the electron-releasing order is Zn > Cu.
In the electrochemical series, zinc has a standard electrode potential of E° = -0.76 V and copper has E° = +0.34 V. A more negative E° corresponds to a greater tendency to lose electrons (stronger reducing agent), which is consistent with Zn > Cu.
This comparison between metals forms the basis of the electrochemical (activity) series, which ranks metals by their tendency to lose electrons. Such competitive electron transfer reactions, when designed in separate compartments connected by a wire and salt bridge, can be used to harness electrical energy in galvanic cells like the Daniell cell.

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Case Set 2

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In a Class XII practical examination, students are asked to set up a Daniell cell. One beaker contains copper sulphate solution with a copper rod immersed in it, and another beaker contains zinc sulphate solution with a zinc rod. The two solutions are connected by a salt bridge made of potassium chloride solution solidified with agar-agar. The copper and zinc rods are connected by a metallic wire fitted with an ammeter and a switch. When the switch is in the off position, the ammeter reads zero. As soon as the switch is turned on, the ammeter deflects, indicating current flow. Students observe that the zinc rod gradually loses mass while the copper rod gains mass over time.

Question 1: Identify the anode and cathode in the Daniell cell and write the half-reaction occurring at each electrode.

The zinc rod is the anode (negative electrode) where oxidation takes place: Zn(s) → Zn2+(aq) + 2e-. This explains why the zinc rod gradually loses mass.
The copper rod is the cathode (positive electrode) where reduction takes place: Cu2+(aq) + 2e- → Cu(s). This explains why the copper rod gains mass as copper ions are deposited.
Electrons flow from the anode (zinc) through the external metallic wire to the cathode (copper); conventional current flows in the opposite direction to electron flow.

Question 2: Explain the role of the salt bridge in the Daniell cell. What would happen if the salt bridge were replaced by a simple wire connecting the two solutions?

The salt bridge (KCl in agar-agar) provides electrical contact between the two half-cell solutions by allowing ions to migrate: anions move toward the anode compartment and cations toward the cathode compartment, maintaining electrical neutrality in both beakers.
Without the salt bridge, the anode compartment would accumulate excess positive charge (Zn2+ ions) and the cathode compartment would develop excess negative charge (depletion of Cu2+), quickly stopping the cell reaction.
A simple wire connecting the two solutions would allow them to mix directly, which would allow direct electron transfer between Zn and Cu2+ without going through the external circuit, and no current would flow through the wire; the Daniell cell would fail to convert chemical energy into electrical energy.

Question 3: The standard electrode potential of Zn2+/Zn is -0.76 V and of Cu2+/Cu is +0.34 V. Using these values, explain why the Daniell cell reaction is spontaneous and calculate the standard cell potential.

Standard cell potential E°cell = E°cathode - E°anode = E°(Cu2+/Cu) - E°(Zn2+/Zn) = (+0.34) - (-0.76) = +1.10 V.
A positive E°cell value confirms that the cell reaction is spontaneous in the direction written (Zn is oxidised and Cu2+ is reduced). The cell converts the chemical energy of this spontaneous redox reaction into electrical energy.
The positive E°cell also means that the overall reaction Zn(s) + Cu2+(aq) → Zn2+(aq) + Cu(s) has a negative Gibbs free energy change (ΔG = -nFE°cell), which is the thermodynamic criterion for a spontaneous process. This is why the Daniell cell can do useful electrical work.

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Case Set 3

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A chemistry teacher demonstrates three different reactions on the board. In Reaction I, magnesium burns in nitrogen gas to form magnesium nitride: 3Mg(s) + N2(g) → Mg3N2(s). In Reaction II, potassium chlorate decomposes on heating: 2KClO3(s) → 2KCl(s) + 3O2(g). In Reaction III, calcium carbonate decomposes: CaCO3(s) → CaO(s) + CO2(g). The teacher asks students to identify which reactions are redox reactions and which type each one is, by tracking oxidation numbers of all elements. One student notices that Reactions I and II involve changes in oxidation number while Reaction III does not.

Question 1: Identify which of the three reactions are redox reactions and state the type of each redox reaction.

Reaction I (3Mg + N2 → Mg3N2) is a redox reaction of the combination type. Both Mg and N2 are in the elemental state (oxidation number 0) and combine to form a compound.
Reaction II (2KClO3 → 2KCl + 3O2) is a redox reaction of the decomposition type. A compound breaks down and at least one product (O2) is in the elemental state, with changes in oxidation number.
Reaction III (CaCO3 → CaO + CO2) is NOT a redox reaction because there is no change in the oxidation number of any element: Ca remains +2, C remains +4, and O remains -2 throughout.

Question 2: For Reaction II, assign oxidation numbers to all elements and identify the oxidising agent and reducing agent.

In 2KClO3 → 2KCl + 3O2: K is +1 throughout (no change); O in KClO3 is -2 and in O2 is 0 (oxidation number increases from -2 to 0, so oxygen in chlorate is oxidised); Cl in KClO3 is +5 and in KCl is -1 (oxidation number decreases from +5 to -1, so chlorine is reduced).
KClO3 acts as both the oxidising agent (Cl is reduced from +5 to -1) and the source of the reductant (O is oxidised from -2 to 0). This makes Reaction II a self-redox (disproportionation-like within one compound) decomposition.
Cl (in KClO3) accepts electrons and is reduced — it is the oxidising agent. O (in KClO3) loses electrons and is oxidised — it is the reducing agent. Note that potassium does not change oxidation state and plays no role in the redox process.

Question 3: Explain the principle used by the student to determine that Reaction III is not a redox reaction, and generalise the condition that must be met for a decomposition reaction to qualify as a redox reaction.

The student applied the oxidation number rule: a reaction is a redox reaction only if the oxidation number of at least one element changes from reactant to product. In CaCO3 → CaO + CO2, Ca is +2 in both CaCO3 and CaO; C is +4 in both CaCO3 and CO2; O is -2 in all three compounds. Since no oxidation number changes, no redox occurs.
For a decomposition reaction to qualify as a redox reaction, at least one of the products must be in the elemental state (with oxidation number 0) while the corresponding element was in a non-zero oxidation state in the reactant, or there must be a redistribution of oxidation states between elements in the products.
The general condition is: a decomposition reaction is redox only if it leads to a change in the oxidation number of at least one element involved. Reactions that only break bonds between different elements without changing their individual oxidation numbers (as in CaCO3 decomposition) are not redox reactions — they are better classified as simple bond-breaking or acid-base decomposition reactions.

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Case Set 4

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A student performing a qualitative analysis experiment adds chlorine water to a test tube containing potassium bromide solution, followed by a few drops of carbon tetrachloride (CCl4). On shaking, the CCl4 layer turns orange-brown. In a second test tube, chlorine water is added to potassium iodide solution with CCl4; the CCl4 layer turns violet. In a third test tube, bromine water is added to potassium iodide solution with CCl4; the CCl4 layer again turns violet. However, when bromine water is added to potassium chloride solution with CCl4, no colour change is observed in the CCl4 layer. This test is commonly used in inorganic qualitative analysis and is known as the Layer Test.

Question 1: Write the ionic equations for the reactions in the first two test tubes and identify the oxidising agent in each.

Test tube 1 (Cl2 + KBr): Cl2(g) + 2Br-(aq) → 2Cl-(aq) + Br2(l). The orange-brown colour of the CCl4 layer is due to Br2 dissolved in it. Cl2 is the oxidising agent (reduced from 0 to -1).
Test tube 2 (Cl2 + KI): Cl2(g) + 2I-(aq) → 2Cl-(aq) + I2(s). The violet colour of the CCl4 layer is due to I2 dissolved in it. Cl2 is again the oxidising agent (reduced from 0 to -1).
In both reactions, the halide ion (Br- or I-) is the reducing agent, being oxidised from -1 to 0 (released as free halogen molecules). Cl2 acts as the oxidant because it is a stronger oxidising agent than both Br2 and I2.

Question 2: Explain why bromine water does not displace chloride from KCl solution in the third test tube, based on the relative oxidising power of halogens.

The oxidising power of halogens decreases down Group 17: F2 > Cl2 > Br2 > I2. This means Br2 is a weaker oxidising agent than Cl2 and cannot oxidise Cl- ions to Cl2.
For bromine to displace chlorine from KCl, the reaction would be Br2 + 2Cl- → 2Br- + Cl2. For this to be spontaneous, Br2 would need to be a stronger oxidising agent than Cl2, which contradicts the electrochemical series (E°Br2/Br- = +1.09 V < E°Cl2/Cl- = +1.36 V).
Since Cl2 has a higher reduction potential than Br2, it is a stronger oxidant and can displace Br- but not vice versa. Therefore, no reaction occurs and no colour develops in the CCl4 layer when Br2 is added to KCl.

Question 3: Using the Layer Test observations, establish the complete order of oxidising power of the halogens Cl2, Br2, and I2. Explain why fluorine's displacement reactions cannot be performed in aqueous solution in this test.

From the observations: Cl2 displaces both Br- (test tube 1) and I- (test tube 2), showing Cl2 is a stronger oxidant than both Br2 and I2. Br2 displaces I- (test tube 3), showing Br2 is a stronger oxidant than I2. The complete order of oxidising power is: Cl2 > Br2 > I2.
Fluorine (F2) is the strongest oxidising agent among halogens and can displace Cl-, Br-, and I- from their salts. However, F2 is so reactive that it attacks water itself: 2H2O(l) + 2F2(g) → 4HF(aq) + O2(g). This means F2 cannot be handled in aqueous solution at all.
Because F2 reacts with the water solvent rather than selectively with the halide ions, the Layer Test cannot be performed with F2 in aqueous medium. Furthermore, F- ions cannot be oxidised to F2 by any chemical oxidising agent since F2 is the strongest oxidant known, making F- the weakest reducing agent among the halide ions.

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Case Set 5

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A chemist is studying the disproportionation reactions of chlorine oxoanions in alkaline medium. She observes that when Cl2 gas is passed into NaOH solution, hypochlorite ions (ClO-) and chloride ions (Cl-) are formed. She also notes that when three different chlorine oxoanions — ClO-, ClO2-, and ClO3- — are treated under appropriate conditions, each undergoes further disproportionation. However, when the experiment is repeated with ClO4-, no disproportionation is observed. The chemist concludes that the inability of ClO4- to disproportionate is related to the oxidation state of chlorine in this ion.

Question 1: Write the equation for the reaction of Cl2 with NaOH and identify it as a disproportionation reaction by tracking the oxidation state of chlorine.

The reaction is: Cl2(g) + 2OH-(aq) → ClO-(aq) + Cl-(aq) + H2O(l).
In Cl2, chlorine has oxidation number 0 (intermediate state). In ClO-, chlorine has oxidation number +1 (oxidised). In Cl-, chlorine has oxidation number -1 (reduced).
Since the same element (Cl) in one oxidation state (0) is simultaneously oxidised to +1 and reduced to -1, this is a disproportionation reaction. The hypochlorite ion formed is the active ingredient in household bleaches and acts as an oxidising agent to decolourise stains.

Question 2: Write the disproportionation reactions for ClO- and ClO3- as given in the chapter, and calculate the oxidation number of Cl in each species involved.

For ClO- (Cl in +1 state): 3ClO- → 2Cl- + ClO3-. Cl in ClO- is +1, in Cl- is -1 (reduced), and in ClO3- is +5 (oxidised). So +1 state disproportionates to -1 and +5.
For ClO3- (Cl in +5 state): 4ClO3- → Cl- + 3ClO4-. Cl in ClO3- is +5, in Cl- is -1 (reduced), and in ClO4- is +7 (oxidised). So +5 state disproportionates to -1 and +7.
In both cases, the element in the intermediate oxidation state simultaneously goes to a lower and a higher oxidation state, which is the defining feature of disproportionation. The disproportionation of ClO3- produces ClO4- (chlorine in its highest, +7, state).

Question 3: Explain why ClO4- cannot undergo disproportionation, and extend this reasoning to explain why fluorine also cannot undergo disproportionation in alkaline medium unlike the other halogens.

ClO4- cannot disproportionate because chlorine is in its highest possible oxidation state of +7 in this ion. Disproportionation requires the element to exist in at least three oxidation states — a lower, intermediate, and higher state. Since +7 is the maximum for chlorine, it cannot be further oxidised, making disproportionation impossible.
Fluorine cannot undergo disproportionation in alkaline medium for a different but related reason: fluorine is the most electronegative element and can never attain a positive oxidation state. Disproportionation of X2 in alkali requires X to be oxidised to a positive state (as in Cl2 → ClO- where Cl goes to +1). Fluorine cannot do this.
When F2 reacts with alkali, the reaction is: 2F2(g) + 2OH-(aq) → 2F-(aq) + OF2(g) + H2O(l). Here, oxygen (not fluorine) is oxidised to +2 in OF2; fluorine is only reduced from 0 to -1. This is not disproportionation — it is an oxidation of OH- by F2. Thus, the inability to achieve a positive oxidation state prevents fluorine from disproportionating.

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Chapter 7: Redox Reactions | Case Study Questions

Passage

Question 1: Write the ionic equation for the reaction occurring when zinc is placed in copper sulphate solution and identify which species is oxidised.

Question 2: Explain why passing H2S through the colourless solution gives a white precipitate of ZnS, and what this observation confirms about the products of the reaction.

Question 3: Based on both experiments described in the passage, what conclusion can be drawn about the relative electron-releasing tendency of zinc and copper? How does this observation connect to the concept of the electrochemical series?

Passage

Question 1: Identify the anode and cathode in the Daniell cell and write the half-reaction occurring at each electrode.

Question 2: Explain the role of the salt bridge in the Daniell cell. What would happen if the salt bridge were replaced by a simple wire connecting the two solutions?

Question 3: The standard electrode potential of Zn2+/Zn is -0.76 V and of Cu2+/Cu is +0.34 V. Using these values, explain why the Daniell cell reaction is spontaneous and calculate the standard cell potential.

Passage

Question 1: Identify which of the three reactions are redox reactions and state the type of each redox reaction.

Question 2: For Reaction II, assign oxidation numbers to all elements and identify the oxidising agent and reducing agent.

Question 3: Explain the principle used by the student to determine that Reaction III is not a redox reaction, and generalise the condition that must be met for a decomposition reaction to qualify as a redox reaction.

Passage

Question 1: Write the ionic equations for the reactions in the first two test tubes and identify the oxidising agent in each.

Question 2: Explain why bromine water does not displace chloride from KCl solution in the third test tube, based on the relative oxidising power of halogens.

Question 3: Using the Layer Test observations, establish the complete order of oxidising power of the halogens Cl2, Br2, and I2. Explain why fluorine's displacement reactions cannot be performed in aqueous solution in this test.

Passage

Question 1: Write the equation for the reaction of Cl2 with NaOH and identify it as a disproportionation reaction by tracking the oxidation state of chlorine.

Question 2: Write the disproportionation reactions for ClO- and ClO3- as given in the chapter, and calculate the oxidation number of Cl in each species involved.

Question 3: Explain why ClO4- cannot undergo disproportionation, and extend this reasoning to explain why fluorine also cannot undergo disproportionation in alkaline medium unlike the other halogens.

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