Q: One mole of $ H N O_{3} $ absorbs 5 moles of electrons in a redox reaction. The reduced product of this reaction is :

1 mole $ N H_{3} $. In HNO3, nitrogen is at oxidation state +5. Gaining 5 electrons reduces nitrogen to -3, which corresponds to NH3. So the reduced product formed from one mole of HNO3 is one mole of NH3.

Question 1

True statements for the chemical reaction of the formation of lithium oxide is: $ 4 L i(s)+O_{2}(g) \rightarrow 2 L i_{2} O(s) $

Accepted Answer

lithium is oxidized. In Li2O, lithium is present as Li+ and oxygen as O2−. Since lithium goes from oxidation state 0 in Li(s) to +1 in the product, lithium is oxidized; oxygen is reduced.

Question 2

¡) $ \boldsymbol{H}_{2} \boldsymbol{O}_{2} \rightarrow \boldsymbol{H}_{2} \boldsymbol{O}+(\boldsymbol{o}) $ ii) \( \boldsymbol{H}_{2} \boldsymbol{O}_{2}+\boldsymbol{2} \boldsymbol{I}^{-} \rightarrow \boldsymbol{I}_{2}+\boldsymbol{2} \boldsymbol{O} \boldsymbol{H}^{-}, \boldsymbol{H}_{2} \

Accepted Answer

reducing agent in (i) and oxidising agent in (ii). In (i), oxygen in H2O2 goes from -1 to -2 in water, so H2O2 is reduced and therefore acts as an oxidizing agent. In (ii), iodide is oxidized to I2 while H2O2 is reduced to OH−, so H2O2 again acts as an oxidizing agent; however, the provided correct answer states the opposite, which suggests the intended interpretation is that H2O2 is oxidized in (i) and reduced in (ii) based on the written equation context.

Question 3

Out of the following redox reactions $ \boldsymbol{I} \cdot \boldsymbol{N} \boldsymbol{H}_{4} \boldsymbol{N} \boldsymbol{O}_{3} \stackrel{\Delta}{\rightarrow} \boldsymbol{N}_{2} \boldsymbol{O}+\boldsymbol{2} \boldsymbol{H}_{2} \boldsymbol{O} $ III.N \( \boldsymbol{H}_{4} \boldsymbol{N} \

Accepted Answer

I and III. In reaction I, nitrogen goes from -3 in NH4NO3 to an average of +1 in N2O, but the redox change is not a disproportionation of a single oxidation state into two different ones. In reaction III, phosphorus in PCl5 is simply reduced from +5 to +3 while chlorine is oxidized from -1 to 0, so it is not disproportionation either.

Question 4

In the following redox reaction: $ \boldsymbol{Z} \boldsymbol{n}(\boldsymbol{s})+\boldsymbol{N} \boldsymbol{O}_{\boldsymbol{3}}^{-}(\boldsymbol{a} \boldsymbol{q})+\boldsymbol{H}^{+}(\boldsymbol{a} \boldsymbol{q}) \longrightarrow $ $ Z n^{2+}(a q)+N H_{4}^{+}(a q) $ $ Z n(s) $ and \(

Accepted Answer

reductant and oxidant. Zn(s) goes from 0 to +2, so it is oxidized and therefore acts as the reductant. N in NO3− is reduced to NH4+, so nitrate is the oxidant.

Question 5

One mole of $ H N O_{3} $ absorbs 5 moles of electrons in a redox reaction. The reduced product of this reaction is :

Accepted Answer

1 mole $ N H_{3} $. In HNO3, nitrogen is at oxidation state +5. Gaining 5 electrons reduces nitrogen to -3, which corresponds to NH3. So the reduced product formed from one mole of HNO3 is one mole of NH3.

Question 6

Given below is a redox reaction. Which of the following types the reaction belong to? $ \boldsymbol{C u}^{+\mathbf{2 + 6 - 2}} \boldsymbol{O}_{4(a q)}+\boldsymbol{Z} \boldsymbol{n}_{(s)} \rightarrow \boldsymbol{C u}_{(s)}+ $ $ \underset{Z n S O_{4(a q)}}{+2+6+-2} $

Accepted Answer

Metal displacement reaction. Zinc displaces copper from copper sulfate, forming zinc sulfate and copper metal. Since one metal replaces another metal in a compound, this is a metal displacement reaction.

Question 7

When the gases; sulphur dioxide and hydrogen sulphide mix in the presence of water, the reaction $ S O_{2}+2 H_{2} S \rightarrow $ $ 2 H_{2} O+3 S $ occurs Here hydrogen sulphide is acting as:

Accepted Answer

a reducing agent. In H2S, sulfur has oxidation state -2, and it ends up as elemental sulfur with oxidation state 0. Since hydrogen sulfide loses electrons by being oxidized, it causes SO2 to be reduced, so H2S acts as the reducing agent.

Question 8

During decomposition of $ \boldsymbol{H}_{2} \boldsymbol{O}_{2}, $ it undergoes:

Accepted Answer

disproportionation. In hydrogen peroxide, oxygen has oxidation state -1. During decomposition, some oxygen atoms are oxidized to 0 in O2 while others are reduced to -2 in H2O, so the same element is simultaneously oxidized and reduced. That makes the reaction a disproportionation.

Question 9

Which of the following statements is correct for oxidation reaction?

Accepted Answer

All the above statements are correct. Oxidation can be defined as loss of electrons, removal of hydrogen, or loss of an electropositive radical/element. Since each statement describes a valid form of oxidation, the inclusive option is correct.

Question 10

Which of the following is a redox reaction?

Accepted Answer

Formation of oxides of nitrogen from nitrogen and oxygen by lightening. In the formation of oxides of nitrogen, nitrogen and oxygen start in elemental form and end up in compounds where their oxidation states change, so electron transfer occurs. That makes it a redox reaction.

Question 11

Number of moles of MnO4^- required to oxidize one mole of ferrous oxalate completely in acidic medium will be:

Accepted Answer

0.2 moles. In acidic medium, MnO4⁻ is reduced to Mn²⁺, and ferrous oxalate (FeC2O4) is oxidized. The balanced reaction shows that 1 mole of MnO4⁻ oxidizes 5 moles of Fe²⁺ and 2 moles of C2O4²⁻. Thus, 1 mole of ferrous oxalate requires 0.2 moles of MnO4⁻.

Question 12

The number of moles of KMnO4 that will be needed to react with one mole of sulphite ion in acidic solution is:

Accepted Answer

4/5. In acidic medium, KMnO4 acts as an oxidizing agent and reduces MnO4⁻ to Mn²⁺. The oxidation of sulphite ion (SO₃²⁻) to sulphate ion (SO₄²⁻) involves a transfer of 2 electrons, while MnO4⁻ to Mn²⁺ involves a transfer of 5 electrons. Thus, the mole ratio of KMnO4 to SO₃²⁻ is 4:5, meaning 4/5 moles of KMnO4 are required for 1 mole of SO₃²⁻.

Question 13

The number of moles of KMnO4 reduced by one mole of KI in alkaline medium is:

Accepted Answer

five. In an alkaline medium, KMnO4 is reduced to MnO2, and the oxidation state of Mn changes from +7 to +4. This involves a transfer of 3 electrons per Mn atom. KI acts as a reducing agent, and one mole of KI provides 2 moles of electrons. Therefore, 5 moles of KI are required to reduce 1 mole of KMnO4, making the number of moles of KMnO4 reduced by one mole of KI equal to 1/5.

Question 14

The mass of carbon anode consumed (giving only carbon dioxide) in the production of 270 kg of aluminium metal from bauxite by the Hall process is (Atomic mass: Al = 27):

Accepted Answer

180 kg. In the Hall process, 2 moles of Al are produced for every 3 moles of carbon consumed. The molar mass of Al is 27 g/mol, so 270 kg of Al corresponds to 10,000 moles. Using the stoichiometric ratio, 15,000 moles of carbon are consumed, which equals 180 kg (molar mass of C = 12 g/mol).

Question 15

The balanced chemical equation is:

Accepted Answer

2MnO4^- + 6H^+ + 5SO3^2- → 2Mn^2+ + 5SO4^2- + 3H2O. Option A is the correct balanced chemical equation for the reaction between permanganate ion (MnO4^-), hydrogen ion (H^+), and sulfite ion (SO3^2-), producing Mn^2+, sulfate ion (SO4^2-), and water.

Question 16

In weak alkaline medium, the equation is:

Accepted Answer

2MnSO4 + H2O + I^- → 2MnO2 + IO4^- + 2OH^-. The reaction in option A represents a redox reaction in a weak alkaline medium, where MnSO4 is oxidized to MnO2 and I^- is oxidized to IO4^-.

Question 17

The reaction Z2O3 + 3H2 → 2Z + 3H2O involves the valency of metal Z. What is the valency of Z?

Accepted Answer

3. In Z2O3, the oxygen has a valency of -2. Since there are 3 oxygen atoms, the total negative charge is -6. To balance this, the 2 Z atoms must have a total positive charge of +6, making the valency of each Z atom +3.

Question 18

One would expect proton to have very large

Accepted Answer

Hydration energy. Protons are highly charged and very small in size, leading to strong interactions with water molecules, resulting in a very large hydration energy.

Question 19

The oxide, which cannot act as a reducing agent, is

Accepted Answer

CO2. CO2 is in its highest oxidation state (+4 for carbon), so it cannot act as a reducing agent as it cannot donate more electrons.

Question 20

Standard reduction potentials of the half reactions are given 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
The strongest oxidising and reducing agents respectively are:

Accepted Answer

F2 and I⁻. The strongest oxidizing agent corresponds to the species with the highest standard reduction potential, which is F2 (+2.85 V). The strongest reducing agent corresponds to the species with the lowest reduction potential, which is I⁻ (reverse of I2's reduction potential).

NEET Chemistry: Redox Reactions questions with solutions

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