Q: Arrange the following sulphur-containing anions in order of increasing oxidation state of sulphur: SO3²⁻, S2O4²⁻ and S2O6²⁻.

S2O4²⁻ < SO3²⁻ < S2O6²⁻. S oxidation states: S2O4^2- = +3, SO3^2- = +4, S2O6^2- = +5, so increasing order is S2O4^2- < SO3^2- < S2O6^2- (option 2). The stored order is wrong.

Q: Which of the following species is unable to act as an oxidising agent?

I⁻. I- has iodine in its lowest state (-1), so it cannot be reduced and cannot act as an oxidising agent (only as a reducing agent). Elemental S (0) can be reduced to S2-, so it can act as an oxidiser.

Q: During the iodometric standardization of sodium thiosulfate (Na2S2O3) with potassium dichromate (K2Cr2O7), what is the equivalent weight of K2Cr2O7?

Molecular mass divided by 6. The equivalent weight of potassium dichromate (K2Cr2O7) is calculated based on its ability to provide 6 moles of electrons in redox reactions, which is why its molecular mass is divided by 6 to find the equivalent weight.

Q: For the reaction H2SO3(aq) + Sn4+(aq) + H2O(l) → Sn2+(aq) + HSO4−(aq) + 3H+(aq), which statement is true?

H2SO3 acts as the reducing agent since it is oxidized. Sn4+ is reduced to Sn2+ (it is the oxidising agent), while sulfur in H2SO3 is oxidised from +4 to +6, so H2SO3 is the reducing agent because it is oxidised (option 2).

Q: In the unbalanced redox equation below, how many electrons are transferred overall? Cr2O7²− + Fe2+ + C2O4²− → Cr3+ + Fe3+ + CO2

6. In this redox reaction, chromium is reduced from Cr2O7²− to Cr3+, which involves a transfer of 6 electrons, while iron is oxidized from Fe2+ to Fe3+, which involves a transfer of 1 electron. The overall electron transfer is determined by balancing the half-reactions, leading to a total of 6 electrons being transferred.

Question 1

What volume of 20-volume hydrogen peroxide is needed to produce 5 L of oxygen gas at STP?

Accepted Answer

250 mL. To make 5000 mL O2 you need 5000/20 = 250 mL of 20-volume H2O2. Stored 125 mL is wrong.

Question 2

Which one of the following reactions shows sulfuric acid acting as an oxidizing agent?

Accepted Answer

2HI + H2SO4 → I2 + SO2 + 2H2O. In 2HI + H2SO4 -> I2 + SO2 + 2H2O, sulfur goes +6 to +4 (reduced) while I- is oxidised to I2, so H2SO4 acts as the oxidising agent. The stored reaction (with PCl5) involves no change in sulfur's oxidation state.

Question 3

Which of the following statements about redox behavior are true?
(i) A metal M whose standard reduction potential for the half-reaction Mⁿ⁺ + ne⁻ ⇌ M is highly negative acts as a strong reducing agent.
(ii) The oxidizing strength of the halogens falls in the order chlorine to iodine.
(iii)

Accepted Answer

(i), (ii) and (iii). All three statements are correct: a highly negative standard reduction potential indicates a strong tendency to lose electrons, making the metal a strong reducing agent; the oxidizing strength of halogens decreases from chlorine to iodine due to their decreasing electronegativity; and hydrogen halides become stronger reducing agents as the bond strength decreases from HCl to HI.

Question 4

For the balanced redox equation
X MnO4⁻ + Y C2O4²⁻ + Z H⁺ → X Mn²⁺ + 2Y CO2 + Z/2 H2O,
what are the values of X, Y, and Z, respectively?

Accepted Answer

2, 5, 16. The balanced equation is 2 MnO4- + 5 C2O4^2- + 16 H+ -> 2 Mn2+ + 10 CO2 + 8 H2O, so X=2, Y=5, Z=16. The stored 3,1,6 is wrong.

Question 5

Balance the redox equation below and identify the coefficients X, Y and Z in order:
XNa2HAsO3 + YNaBrO3 + ZHCl → NaBr + H3AsO4 + NaCl

Accepted Answer

3, 1, 6. Balanced: 3 Na2HAsO3 + NaBrO3 + 6 HCl -> NaBr + 3 H3AsO4 + 6 NaCl, so X=3, Y=1, Z=6. The stored 2,1,2 doesn't balance.

Question 6

Arrange the following sulphur-containing anions in order of increasing oxidation state of sulphur: SO3²⁻, S2O4²⁻ and S2O6²⁻.

Accepted Answer

S2O4²⁻ < SO3²⁻ < S2O6²⁻. S oxidation states: S2O4^2- = +3, SO3^2- = +4, S2O6^2- = +5, so increasing order is S2O4^2- < SO3^2- < S2O6^2- (option 2). The stored order is wrong.

Question 7

Which of the following species is unable to act as an oxidising agent?

Accepted Answer

I⁻. I- has iodine in its lowest state (-1), so it cannot be reduced and cannot act as an oxidising agent (only as a reducing agent). Elemental S (0) can be reduced to S2-, so it can act as an oxidiser.

Question 8

During the iodometric standardization of sodium thiosulfate (Na2S2O3) with potassium dichromate (K2Cr2O7), what is the equivalent weight of K2Cr2O7?

Accepted Answer

Molecular mass divided by 6. The equivalent weight of potassium dichromate (K2Cr2O7) is calculated based on its ability to provide 6 moles of electrons in redox reactions, which is why its molecular mass is divided by 6 to find the equivalent weight.

Question 9

For the reaction
H2SO3(aq) + Sn4+(aq) + H2O(l) → Sn2+(aq) + HSO4−(aq) + 3H+(aq),
which statement is true?

Accepted Answer

H2SO3 acts as the reducing agent since it is oxidized. Sn4+ is reduced to Sn2+ (it is the oxidising agent), while sulfur in H2SO3 is oxidised from +4 to +6, so H2SO3 is the reducing agent because it is oxidised (option 2).

Question 10

In the unbalanced redox equation below, how many electrons are transferred overall?
Cr2O7²− + Fe2+ + C2O4²− → Cr3+ + Fe3+ + CO2

Accepted Answer

6. In this redox reaction, chromium is reduced from Cr2O7²− to Cr3+, which involves a transfer of 6 electrons, while iron is oxidized from Fe2+ to Fe3+, which involves a transfer of 1 electron. The overall electron transfer is determined by balancing the half-reactions, leading to a total of 6 electrons being transferred.

Question 11

Hydrogen sulfide behaves only as a reducing agent, whereas sulfur dioxide can function as both a reducing agent and an oxidizing agent, because

Accepted Answer

sulfur in H2S has an oxidation state of −2. H2S can only reduce because sulfur is in its lowest oxidation state (-2) and can only be oxidised; SO2 (S at +4, intermediate) can do both. The correct reason is option 3, not the hydrogen-character statement.

Question 12

In the reaction below, which species acts as the reducing agent?
14H+ + Cr2O7²− + 3Ni → 2Cr3+ + 7H2O + 3Ni2+

Accepted Answer

Ni. Nickel (Ni) is the reducing agent because it donates electrons during the reaction, causing the chromium in Cr2O7²− to be reduced to Cr3+. This electron transfer is characteristic of a reducing agent.

Question 13

Which of the following equations shows the oxidizing action of H2SO4?

Accepted Answer

2HI + H2SO4 → I2 + SO2 + 2H2O. In 2HI + H2SO4 -> I2 + SO2 + 2H2O sulfur is reduced (+6 to +4) while I- is oxidised, so H2SO4 acts as the oxidiser (option 3). The stored NaCl reaction involves no change in oxidation state.

Question 14

Which of the following reactions is an example of disproportionation?

Accepted Answer

2KOH + Cl2 → KCl + KClO + H2O. 2KOH + Cl2 -> KCl + KClO + H2O has chlorine simultaneously reduced to -1 and oxidised to +1, the definition of disproportionation. The stored option (Ca3P2 + H2O) involves no oxidation-state change.

Question 15

Which statement is correct for sodium tetrathionate, Na2S4O6?

Accepted Answer

Two sulfur atoms have oxidation number 0 each, while the other two have oxidation number +5 each.. In sodium tetrathionate, Na2S4O6, the structure includes two sulfur atoms in the zero oxidation state, which are part of a thiosulfate group, and two sulfur atoms that are in the +5 oxidation state, typically found in the sulfate group. This distribution of oxidation states accurately reflects the overall charge balance and molecular structure.

Question 16

Thiosulfate shows different reactions with iodine and bromine, as represented below:
2S2O3²− + I2 → S4O6²− + 2I−
S2O3²− + 2Br2 + 5H2O → 2SO4²− + 2Br− + 10H+
Which statement best explains this contrasting behaviour of thiosulfate?

Accepted Answer

Bromine has greater oxidising power than iodine. Bromine's greater oxidizing power compared to iodine allows it to react more vigorously with thiosulfate, leading to different reaction products. This difference in oxidizing strength explains why thiosulfate behaves differently in reactions with these two halogens.

Question 17

The standard reduction potentials for the following half-reactions are listed below:
F2(g) + 2e− → 2F−(aq); E° = +2.85 V
Cl2(g) + 2e− → 2Cl−(aq); E° = +1.36 V
Br2(l) + 2e− → 2Br−(aq); E° = +1.06 V
I2(s) + 2e− → 2I−(aq); E° = +0.53 V
From these, identify the most powerful oxidizing agent an

Accepted Answer

F2 and I−. F2 has the highest reduction potential (+2.85 V) so it is the strongest oxidising agent, and I- (from the lowest-E couple) is the strongest reducing agent. The pair is F2 and I-, not Cl2 and Br-.

Question 18

Zinc liberates hydrogen gas with dilute H2SO4 and HCl, but not with HNO3, because:

Accepted Answer

NO3− gets reduced in preference to hydronium ion. With dilute HNO3 the nitrate ion is reduced (to NO/NO2) in preference to H+, so H2 is not liberated. The stored 'weaker acid' reason is incorrect; the correct reason is option 3.

Question 19

Magnesium ribbon continues to burn in carbon dioxide gas because

Accepted Answer

magnesium acts as a reducing agent and takes oxygen away from CO2. Magnesium keeps burning in CO2 because it is a strong reducing agent and strips oxygen from CO2 (2Mg + CO2 -> 2MgO + C). So the correct reason is option 2, not 'none of the above'.

Question 20

At 25°C, a gas X at 1 atm is passed through a solution that contains both 1 M Y and M Z. If the standard reduction potentials satisfy Z > Y > X, which statement is correct?

Accepted Answer

Y will oxidize X but not Z. A species oxidizes another only if its own reduction potential is higher. With Z > Y > X, Y has a higher reduction potential than X (so Y oxidizes X) but lower than Z (so Y cannot oxidize Z). Hence 'Y will oxidize X but not Z.'

Question 21

Standard reduction potentials help judge whether an oxidizing agent is appropriate in a redox titration. The following half-reactions and their standard potentials are given:

MnO4-(aq) + 8H+(aq) + 5e- → Mn2+(aq) + 4H2O(l) E° = 1.51 V
Cr2O7²-(aq) + 14H+(aq) + 6e- → 2Cr3+(aq) + 7H2O(l) E° =

Accepted Answer

MnO4- may be employed in aqueous HCl. MnO4- (E=1.51 V) is strong enough to oxidize Cl- (1.40 V), so in HCl it consumes acid by liberating Cl2, giving erroneous results. Cr2O7^2- (1.38 V) cannot oxidize Cl-, so it is safe in HCl. Hence 'MnO4- may be employed in aqueous HCl' is the incorrect statement.

Question 22

When thiosulfate is oxidized by iodine, the product formed is:

Accepted Answer

tetrathionate ion. When thiosulfate is oxidized by iodine, it undergoes a reaction that leads to the formation of tetrathionate ion, as the oxidation process involves the combination of two thiosulfate ions.

Question 23

When sodium thiosulphate reacts with silver nitrate, the black precipitate produced is

Accepted Answer

silver sulphide (Ag2S). The reaction between sodium thiosulphate and silver nitrate produces silver sulphide (Ag2S) as a black precipitate, which is a characteristic result of the interaction between silver ions and sulfide ions.

Question 24

When thiosulphate is oxidized quantitatively by KMnO4 in a neutral or slightly alkaline medium, the product formed is

Accepted Answer

SO4²−. In neutral/faintly alkaline medium KMnO4 oxidises thiosulphate quantitatively to sulphate: 8MnO4- + 3S2O3^2- + H2O -> 8MnO2 + 6SO4^2- + 2OH-. The product ion is SO4^2-.

Question 25

When ammonium perchlorate reacts with dilute nitric acid, it forms X and perchloric acid. On heating X, a gaseous product Y is obtained. Identify gas Y.

Accepted Answer

N2O. When ammonium perchlorate reacts with dilute nitric acid, it produces nitrogen gas (N2) and water, but upon heating, the nitrogen gas can further react to form nitrous oxide (N2O), which is the gaseous product Y.

Question 26

A solution containing SeO3²− was treated with 70 mL of M/60 KBrO3. The bromine produced was expelled by boiling, and the unreacted KBrO3 was then back-titrated with 12.5 mL of M/25 NaAsO2. Using the reactions below, determine the amount of SeO3²− present in millimoles.

(1) SeO3²− + BrO3−

Accepted Answer

2.5. The correct option is 2.5 millimoles because the amount of KBrO3 used in the reaction can be calculated, and the stoichiometry of the reactions indicates that each mole of SeO3²− reacts with one mole of BrO3−. By determining the moles of unreacted KBrO3 through back-titration with NaAsO2, we can find that 2.5 millimoles of SeO3²− were initially present.

Question 27

The oxidising ability of chlorine in water can be assessed from the energy change for the following sequence:

1/2 Cl2(g) —ΔdissH°→ Cl(g) —ΔegH°→ Cl−(g)

1/2 Cl2(g) —ΔhydH°→ Cl−(aq)

Using the values ΔdissH°(Cl2) = 240 kJ mol−1, ΔegH°(Cl−) = −349 kJ mol−1, and ΔhydH°(Cl−) = −381 kJ mol−1,

Accepted Answer

−610 kJ mol−1. The enthalpy change for converting 1/2 Cl2(g) into Cl−(aq) is calculated by combining the dissociation energy of Cl2, the electron gain enthalpy of Cl, and the hydration enthalpy of Cl−. The values provided yield a net change of −610 kJ mol−1, indicating that the process is exothermic and demonstrates chlorine's strong oxidizing ability.

Question 28

Identify the reaction in which oxidation and reduction both occur.

Accepted Answer

Zn + 2AgCN → 2Ag + Zn(CN)2. This reaction involves the transfer of electrons where zinc is oxidized (losing electrons) and silver ions are reduced (gaining electrons), demonstrating both oxidation and reduction processes.

Question 29

Why are pieces of magnesium attached to the underside of a ship?

Accepted Answer

To protect it from the effects of water and dissolved salts. Magnesium is used as a sacrificial anode on ships to prevent corrosion caused by electrolysis in seawater, effectively protecting the metal parts of the ship from damage.

Question 30

Which of the following equations shows sulfuric acid acting as an oxidizing agent?

Accepted Answer

2HI + H2SO4 → I2 + SO2 + 2H2O. In the reaction 2HI + H2SO4 → I2 + SO2 + 2H2O, sulfuric acid acts as an oxidizing agent because it facilitates the oxidation of iodide ions (I-) to iodine (I2), while itself being reduced to sulfur dioxide (SO2). This demonstrates its ability to accept electrons, characteristic of oxidizing agents.

Question 31

For the balanced redox equation X MnO4− + Y C2O4²− + Z H+ → X Mn²+ + 2Y CO2 + Z/2 H2O, what are the values of X, Y and Z in that order?

Accepted Answer

2, 5, 16. The correct values of X, Y, and Z are determined by balancing the number of atoms and charges in the redox reaction. In this case, 2 MnO4− ions reduce to 2 Mn²+, while 5 C2O4²− ions oxidize to produce 10 CO2, and 16 H+ ions balance the reaction, ensuring both mass and charge are conserved.

Question 32

Which one of the following chemical changes illustrates a redox process?

Accepted Answer

XeF4 + O2F2 → XeF6 + O2. This reaction involves the transfer of electrons, where xenon is oxidized from XeF4 to XeF6 and oxygen is reduced from O2F2 to O2, demonstrating the characteristics of a redox process.

Question 33

What are the oxidation states of potassium in K2O, K2O2, and KO2, in that order?

Accepted Answer

+1, +1, and +1. In K2O, K has an oxidation state of +1 because oxygen is -2, leading to a total of +2 from potassium to balance it. In K2O2, each potassium still has an oxidation state of +1, as the peroxide ion (O2²-) requires two potassiums to balance the -2 charge. Similarly, in KO2, potassium remains at +1 while the superoxide ion (O2⁻) accounts for the overall charge balance.

Question 34

What is the best description of the change that occurs when Na2O(s) is dissolved in water ?

Accepted Answer

Oxide ion donates a pair of electrons. When Na2O is dissolved in water, it reacts to form hydroxide ions, where the oxide ion (O2-) donates a pair of electrons to water molecules, facilitating the formation of hydroxide ions (OH-). This electron donation is a key characteristic of the oxide ion's behavior in this reaction.

Question 35

The standard reduction potentials E° for the M3+/M2+ couples of Cr, Mn, Fe and Co are −0.41, +1.57, +0.77 and +1.97 V, respectively. For which metal is oxidation from the +2 state to the +3 state most readily achieved?

Accepted Answer

Cr. The standard reduction potential indicates how easily a species can be reduced; a more negative value suggests a greater tendency to oxidize. Since Cr has the lowest reduction potential among the options, it is more readily oxidized from the +2 state to the +3 state.

Question 36

When aluminium oxide is electrolysed at about 1000°C to obtain aluminium metal, and the cathode process is Al3+ + 3e− → Al, how much electric charge is needed to produce 5.12 kg of aluminium? (Atomic mass of Al = 27 amu; 1 Faraday = 96,500 C)

Accepted Answer

5.49 × 10⁷ C of electricity. To produce aluminium, the electrolysis process requires a specific amount of electric charge based on the number of moles of Al produced. Given that 5.12 kg of aluminium corresponds to approximately 190 moles, and since each mole of Al requires 3 moles of electrons (3 Faradays), the total charge needed is calculated to be 5.49 × 10⁷ C, making this the correct option.

Question 37

Consider the standard reduction potentials:
E°(Cr3+/Cr) = −0.74 V; E°(MnO4−/Mn2+) = 1.51 V
E°(Cr2O7²−/Cr3+) = 1.33 V; E°(Cl2/Cl−) = 1.36 V
From the values above, which species is the most powerful oxidizing agent?

Accepted Answer

MnO4−. MnO4− is the most powerful oxidizing agent because it has the highest standard reduction potential of 1.51 V, indicating it has a strong tendency to gain electrons and be reduced.

Question 38

Oxidation number of Cl in CaOCl2 (bleaching powder) is

Accepted Answer

+1 and −1 since it contains ClO− and Cl−. In CaOCl2, chlorine exists in two different oxidation states: +1 in the hypochlorite ion (ClO−) and −1 in the chloride ion (Cl−), which is why the correct answer reflects both oxidation states.

Question 39

Excess of KI reacts with CuSO4 solution and then Na2S2O3 is added to it. Which of the statements is incorrect for this reaction ?

Accepted Answer

CuI2 is formed. CuI2 is formed because the excess of KI reacts with CuSO4 to produce copper(I) iodide (CuI), and in the presence of excess iodide, it can further react to form CuI2, which is a stable compound in this context.

Question 40

The products obtained when chlorine gas reacts with cold and dilute aqueous NaOH are:

Accepted Answer

Cl− and ClO−. When chlorine gas reacts with cold and dilute aqueous sodium hydroxide, it undergoes a disproportionation reaction, resulting in the formation of chloride ions (Cl−) and hypochlorite ions (ClO−). This reaction occurs under mild conditions, favoring the production of these specific products.

Question 41

The correct order of the oxidation states of nitrogen in NO, N2O, NO2 and N2O3 is:

Accepted Answer

N2O < NO < N2O3 < NO2. Oxidation number of N: N2O = +1, NO = +2, N2O3 = +3, NO2 = +4. Increasing order is therefore N2O < NO < N2O3 < NO2.

Question 42

The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is:

Accepted Answer

+3. In the reaction between KI and acidified potassium dichromate, chromium is reduced from its +6 oxidation state in dichromate to +3 in the final product, which is a common outcome in redox reactions involving dichromate ions.

Question 43

Given E°Cr3+/Cr = -0.74 V; E°MnO4-/Mn2+ = 1.51 V; E°Cr2O7²-/Cr3+ = 1.33 V; E°Cl2/Cl- = 1.36 V. Based on the data given above, strongest oxidizing agent will be -
(1) Mn2+
(2) MnO4-
(3) Cl-
(4) Cr3+

Accepted Answer

MnO4-. MnO4- is the strongest oxidizing agent because it has the highest standard reduction potential (1.51 V), indicating a greater tendency to gain electrons compared to the other species listed.

Question 44

In which of the following reactions H2O2 acts as a reducing agent ?
(a) H2O2 + 2H+ + 2e− → 2H2O
(b) H2O2 − 2e− → O2 + 2H+
(c) H2O2 + 2e− → 2OH−
(d) H2O2 + 2OH− − 2e− → O2 + 2H2O

Accepted Answer

(b), (d). In reactions (b) and (d), H2O2 is oxidized to O2, which indicates that it is losing electrons and thus acting as a reducing agent. A reducing agent donates electrons in a chemical reaction, facilitating the reduction of another species.

Question 45

Iodine reacts with concentrated HNO3 to yield Y along with other products. The oxidation state of iodine in Y is -

Accepted Answer

5. In the reaction between iodine and concentrated nitric acid, iodine is oxidized to a higher oxidation state. The product Y, which contains iodine in the form of iodate (IO3-), has iodine in the +5 oxidation state.

Question 46

On heating, lead (II) nitrate gives a brown gas (A). The gas (A) on cooling changes to a colourless solid/liquid (B). (B) on heating with NO changes to a blue solid (C). The oxidation number of nitrogen in solid (C) is

Accepted Answer

+3. Pb(NO3)2 -> NO2 (brown gas A); NO2 cools to N2O4 (colourless B); N2O4 + NO -> N2O3 (blue solid C). In N2O3 the oxidation number of nitrogen is +3.

Question 47

Oxidation number of potassium in K2O, K2O2 and KO2, respectively, is:

Accepted Answer

+1, +1 and +1. Potassium has oxidation number +1 in K2O, K2O2 (peroxide) and KO2 (superoxide); only oxygen varies (-2, -1, -1/2). So +1, +1, +1.

Question 48

The redox reaction among the following is:

Accepted Answer

combination of dinitrogen with dioxygen at 200 K. Combination of N2 with O2 to give NO is a redox reaction (N: 0 -> +2, O: 0 -> -2). Ozone formation from O2 keeps oxygen at 0 (no redox), H2SO4 + NaOH is acid-base, and [Co(H2O)6]Cl3 + AgNO3 is precipitation. So the redox reaction is the dinitrogen-dioxygen combination.

Question 49

The compound that cannot act both as oxidising and reducing agent is:

Accepted Answer

H3PO4. H3PO4, or phosphoric acid, does not have the ability to either gain or lose electrons in redox reactions, which is a requirement for a substance to act as both an oxidizing and reducing agent. In contrast, the other compounds listed can participate in such reactions.

Question 50

In the given chemical reaction, colors of the Fe2+ and Fe3+ ions are respectively:
5Fe2+ + MnO4− + 8H+ → Mn2+ + 4H2O + 5Fe3+

Accepted Answer

Green, Yellow. Fe2+ ion in solution is pale green and Fe3+ is yellow (yellow-brown). Taken respectively, the colours are Green, Yellow.

JEE Main Chemistry: Redox Reactions questions with solutions

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