Redox Reactions Important Questions

Very Short Answer Questions (1 Mark Questions)

Ques. What are redox reactions? Explain with an example.

Ans. A redox reaction is the kind of reaction where reduction and oxidation take place simultaneously.

Ques. In terms of electrons, what is oxidation and what is reduction?

Ans. In terms of electrons, oxidation involves the loss of electrons and reduction involves the gain of electrons.

Ques. What is an oxidizing agent? Name the best oxidizing agent.

Ans. An oxidizing agent can be called a substance that gains electrons easily, therefore reducing itself in the process. F2 is the best oxidizing agent.

Ques. What is a reducing agent? Name the best reducing agent.

Ans. A reducing agent can be called a substance that loses electrons easily, therefore oxidizing itself in the process. Li is the best reducing agent.

Ques. What is a redox couple?

Ans. The redox couple consists of the oxidized and reduced form of the same substance taking part in an oxidation or reduction half-reaction.

Ques. What is the oxidation number of O in

• OF₂
• O₂F₂

Ans. The oxidation number of oxygen in OF_{2 i}s +2 and the oxidation number of oxygen in O₂F₂ is +1

Ques. Define oxidation and reduction in terms of oxidation numbers.

Ans. In terms of numbers, reduction involves a decrease in oxidation number while oxidation involves an increase in oxidation number. Sn²⁺ + 2Hg²⁺ \( \rightarrow\) Sn₄+ + Hg²⁺ is an example where Sn²⁺ gets oxidized while Hg²⁺ gets reduced.

Ques. Name one compound in which the oxidation number of Cl is +4.

Ans. ClO₂ is a compound in which the oxidation number of Cl is +4.

Read More: Types of Redox Reactions

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Short Answer Questions (2 Marks Questions)

Ques. In the reaction: Cl₂₍g) + 2OH– (aq) \( \rightarrow\) CIO– (aq) + Cl– (aq) + H₂O(l), identify the bleaching substances and explain the oxidation process.

Ans. In the reaction,Cl₂(g) + 2OH– (aq) \( \rightarrow\) CIO– (aq) + Cl– (aq) + H₂O(l), Cl₂ is both oxidized and reduced in ClO– and Cl–, respectively. Since Cl– cannot act as an oxidising agent (O.A.). Therefore, Cl₂ bleaches substances due to oxidising action of hypochlorite ClO– ion.

Ques. Nitric acid is an oxidizing agent and it reacts with PbO, but not with PbO₂. Explain why.

Ans. PbO, being a basic oxide, goes through a reaction with HNO3 also known as the acid-base reaction. While in the case of PbO₂, Pb is in the +4 oxidation state and thus cannot be oxidized anymore given the conditions. Therefore, no reaction occurs. This is why HNO3 reacts only with PbO and not with PbO₂.

Ques. Identify the reaction: 2H₂0₂ (aq) \( \rightarrow\)  2H₂0 (e) + O2 (g)

Ans. The decomposition of hydrogen peroxide is an example of a reaction called a disproportionation reaction. In this kind of reaction, oxygen is the element that experiences a disproportionation reaction. In this example, this happens

Ques. Which gas is liberated when less reactive metals like Mg and Fe react with steam?

Ans. Less reactive metals such as Mg and Fe react with steam (H₂O) to produce dihydrogen or H2 gas. The reaction is given below.

Mg + 2H₂O \(\to\)Mg (OH)₂ + H₂ Fe + 3H₂O\(\to\) Fe₂ O3 + 3H₂.

Ques. Is it possible for us to store copper sulphate in an iron vessel?

Ans. We cannot store copper sulphate in an iron vessel because iron is more reactive than copper and it can create holes in the iron vessel. The reaction is as follows.

Cu²⁺ (aq) + Fe(S)\(\to\)Fe²⁺ (aq) + Cu(S)

Ques. Why does ClO4- not show a disproportionation reaction whereas ClO^-, ClO₂^- and ClO₃^- show?

Ans. ClO4- does not show a disproportionation reaction because, in this compound, chlorine is present in its highest oxidation state which is +7. However, in the case of ClO^-, ClO₂^- and ClO₃^-, chlorine exists in + 1, +3 and +5 oxidation states respectively.

Ques. How would you come to know that a redox reaction is taking place in an acidic/alkaline or neutral medium?

Ans. If H+ or any acid appears on either side of the chemical equation, then the reaction has taken place in the acidic solution. If OH- or in fact any base appears on either side of the equation, the solution is basic. If neither of them (H⁺, OH^-, acids or alkali) appear on any side of the equation, the solution is neutral. 

Read More: Balanced Chemical Equations

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Long Answer Questions (3 Marks Questions)

Ques. Calculate the oxidation numbers of each sulphur atom in the following compounds:

1. Na₂S₂O₃
2. Na₂S₄O₆
3. Na₂SO₃
4. Na₂SO₄

Ans. (1) Sodium thiosulfate has the formula Na₂S₂O₃xH₂O. In the chemical compound, one sulphur molecule possesses +6 oxidation state., whereas the other has -2.

(2) Let the oxidation number of sulphur in Na₂S₄O₆ be x

 2 + 4x – 12 = 0

⇒ 4x = +10

⇒ x = 10/4

⇒ x = +2.5

Hence, the oxidation number of sulphur in Na₂S₄O₆ is given as +2.5.

(3) Let the oxidation number in Na₂SO₃ be x.

Na₂SO₃

⇒ 2×(+1)+x+(−2)×3 = 0

⇒ 2+x−6 = 0

⇒ x = +4

Thus, the oxidation number of sulphur in Na₂SO₃is +4.

(4) Let the oxidation number of sulphur in Na₂SO₄ be x.

⇒ 2×(+1)+x+(−2)×4 = 0

⇒ 2+x−8 = 0

⇒ x = +6

Thus, the oxidation number of sulphur in Na₂SO_{4 i}s +6.

Ques. Balance the equation by the oxidation number method.

Ans. In the above reaction, we can see that the oxidation number of copper has decreased from +2 to 0 and the oxidation number of nitrogen effectively increases from -3 to O. To balance the equation properly, there should be 3 atoms of copper and 2 atoms of nitrogen. A proper reaction is necessary to balance out the equation. 

Therefore, 3CuO+2NH₃= 3Cu+N₂+H₂O.

After balancing the hydrogen and oxygen atoms we can get, 3CuO+2NH₃ \( \rightarrow\) 3Cu+N₂+3H₂O.

Ques. Balance the equation by the oxidation number method.

Ans. In the above equation, the oxidation number of Mn decreases from +6 to +4 (MnO₂) and in another molecule, the oxidation number of Mn increases from +6 to +7 (KMnO₄). 1 molecule loses two electrons, and the other molecule loses 1 electron as we can see visually. In order to balance the oxidation number of manganese, K₂MnO₄ and KMnO₄ are multiplied by 2. 

Therefore, the reaction becomes K₂MnO₄ + 2K₂MnO₄ + H₂O \( \rightarrow\) MnO₂ + 2K₂MnO₄ +KOH. In order to balance the number of K and H atoms, KOH is multiplied by 4 and H₂O by 2. 

The reaction becomes: 3K₂MnO₄ + 2H₂O  \( \rightarrow\) MnO₂ + 2KMnO₄ + 4KOH. 

Thus, the reaction is balanced.

Read More: Chemical Reactions Formula

Ques. Write the balanced half-reaction and overall equations for the following equations.

• NO₃^- + Bi(S) \( \rightarrow\) Bi₃⁺ + NO₂ (in acid solution)
• Fe (OH)₂ (S) + H₂O₂ \( \rightarrow\)  Fe (OH)₃(S) + H₂O (in basic medium)

Ans. In the first reaction, H+ ions can be seen.

Therefore, oxidation half-reaction is: Bi (S) Bi₃⁺+ 3e^-

Reduction half-reaction is: [NO₃^- + 2H⁺ + e- NO₂ + H₂O ] x³

Balanced equation is: Bi (S) + 3NO₃^- + 6H+ Bi3⁺+ 3NO₂ + 3H₂O

In the second reaction, Fe (OH)² (S) + H₂O₂ \( \rightarrow\)  Fe (OH)₃(S) + H₂O.

The solution is basic, which means that OH- ions are involved in the reaction. Therefore, 

Oxidation half reduction reaction is: [Fe (OH)₂ + OH^- Fe (OH)₃ + e^-] x²

Reduction half-reaction is: H₂O₂+ 2e\(\to\)^- 2OH-

Balanced equation is: 2Fe (OH)₂ + H₂O₂ \(\to\)2Fe(OH)₃.

Ques. Identify the redox reactions from the two reactions and identify the oxidizing and reducing agents in them.

• 3HCl (aq)+HNO₃ (aq) \( \rightarrow\) Cl₂ (g)+ NOCl (g)+ 2H₂O (l)
• HgCl₂ (aq)+ 2KI (aq) \( \rightarrow\)  HgI₂ (s)+2KCl (aq)

Ans. In the first reaction, writing the oxidation number of each atom, we can get 

3HCl (aq)+HNO₃ (aq) \( \rightarrow\)  Cl₂ (g)+ NOCl (g)+ 2H₂O (l)

In HCl, the oxidation number of Cl goes from -1 to 0 in Cl₂. As a result of oxidation of Cl-, HCl is used as a reducing agent. HNO₃, on the other hand, works as an oxidizing agent because the oxidizing number of N reduces from +5 in HNO₃ to +3 in NOCl. The reaction is a redox reaction.

In the second reaction, there is no change in the oxidation numbers of any of the atoms. Therefore, it is not a redox reaction. 

Read More: Hydrogen Bonding

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Very Long Answer Questions (5 Marks Questions)

Ques. Which method can be used to find the strength of an oxidant/reductant in a reaction? Explain with an example.

Ans. The relative electrode potential can be measured when a reducing agent or an oxidizing agent is linked to a solution using an electrode cell. An electrode with a conventional cell with a known electrode potential can be chosen as the reference for performing the reaction. If the reductant is positive and the oxidant is negative, the given system works without any fault.

Let us take an example. We can consider Fe3+/Fe with a standard hydrogen electrode. For Fe and H, the half-life reaction is given below.

H⁺ + e^– \( \rightarrow\) H₂E^o = 0.0V

Fe₃⁺ + e– \( \rightarrow\) Fe₂+E^o= 0.77

Any element that needs to be calculated properly using SHE can be used as an electrode. Electric potential is the quantity of an emf that an element produces in the cell. The reaction is given below.

E^ocell = Eocathode – Eoanode

E^ocell= 0 – Eoanode

E^ocell = 0 – 0.77

E^ocell= -0.77

Fe3+ has a higher chance to undergo reduction when compared to hydrogen. Therefore, the previously assumed Fe anode configuration can be reversed and as a result, the strength of Fe as a reductant can be established. Therefore, the oxidant’s strength can be defined. This is how the strength of an oxidant/reductant can be determined easily in a chemical reaction.

Read more:

Ques. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.

NaClO4, NaClO₃, NaCIO, KCIO₂, Cl₂O₇, ClO₃, Cl₂O, NaCl, Cl₂, CIO₂. 

Also, mention the oxidation state which is not present in any of the following compounds.

Ans. Let the oxidation number of Chlorine be ‘x’. Therefore, 1 + x + 4 * (-2) = 0

∴ x – 7 = 0

∴ x = +7

Na⁽⁺¹⁾Cl⁽⁺⁷⁾O₄^(-2)

Here, the oxidation number of chlorine is +7.

From the above method it can also be inferred that

• In Na⁽⁺¹⁾Cl⁽⁺¹⁾O₃^(-2), the oxidation number of chlorine is +5
• In Na⁽⁺¹⁾Cl⁽⁺⁷⁾O^(-2), the oxidation number of chlorine is +1
• In K⁽⁺¹⁾Cl⁽⁺³⁾O₂^(-2), the oxidation number of chlorine is +3
• In Cl₂⁽⁺⁷⁾O₇^(-2) the oxidation number of chlorine is +7
• In Cl⁽⁺⁶⁾O₃^(-2) the oxidation number of chlorine is +6
• In Cl₂⁽⁺¹⁾O^(-2) the oxidation number of chlorine is +1
• In Na⁽⁺¹⁾Cl^(-1) the oxidation number of chlorine is -1
• In Cl₂ the oxidation number of chlorine is 0
• In Cl⁽⁺⁴⁾O₂^(-2) the oxidation number of chlorine is +4

The following compounds are jotted down in increasing order of chlorine oxidation number:

NaCl, Cl₂, Cl₂O, NaCIO, KCIO₂, CIO₂, NaClO₃, ClO₃, Cl₂O₇, NaClO₄.

Ques. Explain redox reactions based on electron transfer methods. Give an example to justify it.

Ans. An oxidation-reduction reaction or a redox reaction can be defined as a reaction in which electron transfer occurs. A chemical reaction where one substance loses electrons while the other one gains electrons can be classified as a redox reaction. We can take the following examples.

2Na(s) + Cl₂(g) \( \rightarrow\)  2NaCl(s)

4Na(s) + O₂(g) \( \rightarrow\) 2Na₂(s)

In these reactions, loss and gain of electrons happens simultaneously. Therefore, this can be defined as a redox reaction. In this reaction, oxygen and chlorine are depleted whereas sodium, an electropositive element, has been added. NaCl and Na₂O₂ can also be described as Na+Cl–(s) and Na2+O2, respectively to make the reaction easier to understand. The reaction can be broken down as given below: 

One part of the reaction shows electron loss, while the other part shows electron gain. One of these reactions can be further broken down as.

2Na(s) \( \rightarrow\) 2Na+(g) + 2e⁻

Cl2 + 2e− \( \rightarrow\) 2Cl⁻(g)

Each of the steps before can also be called half-reaction since it clearly shows the involvement of electrons. The total reaction is the addition of the half-reactions taken together. Half reactions involving electron loss are oxidation reactions while half-reactions involving electron gain are reduction reactions. This is how a redox reaction overall works.

Electron Transfer in Redox Reaction

Read More: Redox Titration

Ques. On the basis of standard electrode potential values, suggest which of the following reactions should take place: 

Ans. The cell’s net cell EMFis helpful in determining whether or not a reaction will occur. The formula is as follows:

E^ocell= E^ocathode- E^oanode

The reaction occurs in option (ii) because manganese has a lower Eo cell value. As a result, Mg is oxidized by losing an electron. And iron gains electrons, it gets reduced successively.

Mg+F²⁺= Mg²⁺ + Fe

Fe goes through reduction and Mg goes through oxidation. 

It can be now stated that Fe goes through reduction and Mg goes through oxidation.

Taking the Eo values,

Eocathode = -0.44V

Eoanode = -2.36V

Eocell = -0.44 – (-2.36)V

E^ocell = +1.92V 

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