CBSE Class 11 Chemistry -Chapter 8 Redox Reactions- Study Materials

Subtopics of Class 11 Chemistry Chapter 8 Redox Reactions

  1. Classical Idea Of Redox Reactions – Oxidation And Reduction Reactions
  2. Redox Reactions In Terms Of Electron Transfer Reactions
    • Competitive Electron Transfer Reactions
  3. Oxidation Number
    • Types Of Redox Reactions
    • Balancing Of Redox Reactions
    • Redox Reactions As The Basis For Titrations
    • Limitations Of Concept Of Oxidation Number
  4. Redox Reactions And Electrode Processes.

Redox Reactions Class 11 Notes Chemistry Chapter 8

• Oxidation
Oxidation is defined as the addition of oxygen/electronegative element to a substance or rememoval of hydrogen/ electropositive element from a susbtance.
For example,
• Reduction
Reduction is defined as the memoval of oxygen/electronegative element from a substance or addition of hydrogen or electropositive element to a substance.
For example,
• Redox Reaction in Terms of Electron Transfer Reaction
A few examples of redox reaction on the basis of electronic concept are given below:
According to electronic concept every redox reaction consists of two steps known as half reactions.
(i) Oxidation reaction: Half reactions that involve loss of electrons are called oxidation reactions.
(ii) Reduction reaction: Half reactions that involve gain of electrons are called reduction reactions.
Oxidising agent: Acceptor of electrons.
Reducing agent: Donar of electrons.
• Competitive Electron Transfer Reactions
To understand this concept let us do an experiment.
Place a strip of metallic zinc in an aqueous solution of copper nitrate as shown in Fig. After one hour following changes will be noticed.
(i) Strips becomes coated with reddish metallic copper.
(ii) Blue colour of the solution disappears.
(iii) If hydrogen sulphide gas is passed through the solution appearance of white ZnS can be – seen on making the solution alkaline with ammonia.
• Oxidation Number
It is the oxidation state of an element in a compound which is the charge assigned to an atom of a compound is equal to the number of electrons in the valence shell of an atom that are gained or lost completely or to a large extent by that atom while forming a bond in a compound.
• Rules for Assigning Oxidation Numbers
(i) The oxidation number of an element in its elementary form is zero.
For example, H2, 02, Netc. have oxidation number equal to zero.
(ii) In a single monoatomic ion, the oxidation number is equal to the charge on the ion. For example, Na+ ion has oxidation number of +1 and Mg2+ ion has +2.
(iii) Oxygen has oxidation number -2 in its compounds. However, there are some exceptions.
Compounds such as peroxides. Na202, H202
oxidation number of oxygen = – 1 In OF2
O.N. of oxygen = +2 02F2
O.N. of oxygen = +1
(iv) In non-metallic compounds of hydrogen like HCl, H2S, H2O oxidation number of hydrogen = + 1 but in metal hydrides oxidation number of hydrogen = -1
[LiH, NaH, CaH2 etc.]
(v) In compounds of metals and non-metals metals have positive oxidation number while non-metals have negative oxidation number. For example, In NaCl. Na has +1 oxidation number while chlorine has -1.
(vi) If in a compound there are two non-metallic atoms the atoms with high electronegativity is assigned negative oxidation number while other atoms have positive oxidation number.
(vii) The algebraic sum of the oxidation number of all atoms in a compound is equal to zero.
(viii) In poly atomic ion the sum of the oxidation no. of all the atoms in the ion is equal to the net charge on the ion.
For example, in (C03)2—Sum of carbon atoms and three oxygen atoms is equal to -2.
Fluorine (F2) is so highly reactive non-metal that it displaces oxygen from water.
Disproportionation Reaction. In a disproportionation reaction an element in one oxidation state is simultaneously oxidises and reduced.
For example,
Hence, the oxygen of peroxide, which is present in -1 oxidation state is connected to zero oxidation state and in 02 and in H2O decreases to -2 oxidation state.
• Fractional Oxidation Numbers
Elements as such do not have any fractional oxidation numbers. When the same element are involved in different bonding in a species, their actual oxidation states are whole numbers but an average of these is fractional.
For example, In C302
Fractional O.N. of a particular element can be claculated only if we know about the structure of the compound or in which it is present.
• Balancing of Redox Reactions
(i) Oxidation Number Method. Following steps are involved:
(ii) Write the correct formula for each reactant and product.
(b) By assigning the oxidation change in oxidation number can be identified.
(c) Calculate the increase and decrease in oxidation number per atom with respect to the reactants. If more than one atom is present then multiply by suitable coefficient.
(d) Balance the equation with respect to all atoms. Balance hydrogen and oxygen atoms also.
(e) If the reaction is carried out in acidic medium, use H+ ions in the equation. If it is in basic medium use OH ions.
(f) Hydrogen atoms in the expression can be balanced by adding (H20) molecules to the reactants or products.
If there are the same number of oxygen atoms on the both side of equation then it represents the balanced redox reaction.
(ii) Half Reaction Method. In this method two half equation are balanced separately and than added together to give balanced equation.
• Redox Reactions as the Basis for Titration
Potassium Permanganate Titration: In these titrations potassium permanganate (pink in colour) acts as an oxidising agent in the acidic medium while oxalic acid or some ferrous salts acts as a reducing agents.
The ionic equation can be written as:
These are the examples of redox titration.
On both these titrations, potassium permanganate itself acts as indicator. It is commonly known as self indicator. The appearance of pink colour in the solution represents the end points.
Potassium Dichromate Titration: In place of potassium permanganate, potassium dichromate can also be used in the presence of dil. H2S04. The ionic equation for the redox reaction with FeS04 (Fe2+ ions) is given.
• Limitation of Concept of Oxidation Number
According to the concept of oxidation number, oxidation means increase in oxidation number – by loss of electrons and reduction means decrease in oxidation number by the gain of electrons. However, during oxidation there is decrease in electron density while increase in electron density around the atom undergoing reduction.
• Redox Reactions and Electrode Processes—Electrochemical Cells
A device in which the redox reaction is carried indirectly and the decrease in energy appears as the electrical energy are called electrochemical cell.
Electrolytic Cell. The cell in which electrical energy is converted into chemical energy. Example, when lead storage battery is recharged, it acts as electrolytic cell.
Redox Reactions and Electrode Processes. When zinc rod is dipped in copper sulphate solution redox reaction begins hence, zinc is oxidised to Zn2+ ions and Cu2+ ions are reduced to metal.
• Redox reaction. Reactions in which oxidation and reduction occur simultaneously are called redox reactions.
• Oxidation. Involves loss of one or more electrons.
• Reduction. Involves gain of one or more electrons.
• Oxidising agent. Accepting electrons.
• Reducing agent. Losing electrons.
• Electrochemical cell. It is a device in which redox reaction is carried indirectly and decrease in energy gives electrical energy.
• Electrode potential. It is the potential difference between the electrode and its ions in solution.
• Standard electrode potential. It is the potential of an electrode with respect to standard hydrogen electrode.
• Electrochemical series. It is activity series. It has been formed by arranging the metals in order of increasing standard reduction potential value.

CBSE Class 11 Chemistry Chapter-8 Important Questions

1 Marks Questions

1.Define oxidation reaction?

Ans.Addition of oxygen /electronegative element to a substance or removal of hydrogen / electropositive element from a substance.

2.Define reduction reaction?

Ans.Removal of oxygen / electronegative element form a substance or addition of hydrogen / electropositive element to a substance.

3.In the reactions given below, identify the species undergoing oxidation and reduction.

H2S (g) + Cl2 (g) 2HCl (g) + S (S)

Ans.H2S is oxidized because a more electronegative element, Chlorine is added to hydrogen (or more electropositive element hydrogen has been removed from S). Chlorine is reduced due to addition of hydrogen to it.

4.What is the most essential conditions that must be satisfied in a redox reaction?

Ans.In a redox reaction, the total number of electrons lost by the reducing agent must be equal to the number of electrons gained by the oxidizing agent.

5.In the reaction MnO2 + 4HCl Mn Cl+ Cl2 +2H2O Which species is oxidized?

Ans.HCl is oxidized to Cl2.

6.Why the following reaction is an example of oxidation reaction?

CH4 (g) +2O2 (g)CO2 (g) + 2H2O

Ans.Methane is oxidized owing to the addition of oxygen to it.

7.Define oxidation in terms of electron transfer.

Ans.Oxidation is a process in which loss of electrons takes place.

8.What is meant by reduction?

Ans.Reduction is a process in which gain of electrons take place.

9.Define an oxidizing agent. Name the best reducing agent.

Ans. Oxidising agent is a substance which can gain electrons easily. Fis the best oxidizing agent.

10.What is meant by reducing? Name the best reducing agent.

Ans. Reducing agent is a substance which can lose electrons easily. Li is the best reducing agent.

11.What is the oxidation number of Mn in KMnO4?

Ans. Let oxidation number of Mn be x

1 + x + 4 (-2) = 0

X = 

12.What happens to the oxidation number of an element in oxidation?

Ans.It increases.

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


14.Indicate the oxidizing and reducing agents in the following reaction :

2Cu2+ + 4I 2CuI + I2.

Ans.Cu2+ : Oxidising agent

I-: Reducing agent.

15.A metal ion M3+ loses 3 electrons. What will be its oxidation number?

Ans. Oxidaton number changes from +3 to + 6.

16.Name the different types of redox reaction

Ans. The different types of redox reactions are

(i)Combination reactions

(ii)Decomposition reactions

(iii)Displacement reactions

(iv)Disproportionation reactions.

17.Identify the type of redox reaction this reaction follows.

3Mg (S) + N2 (g) Mg3 N2 (S)

Ans. The above equation represents a combination reaction.

18.The displacement reactions of Cl, Br, I using fluorine are not generally carried out in aqueous solution. Give reason.

Ans. Fluorine is so reactive that it can replace chloride bromide and iodide ions in solution and it attacks water and displaces the oxygen of water.

19.Which is the strongest oxidizing agent?

Ans. Fluorine is the strongest oxidizing agent.

20.Why F ions Cannot be converted to F2 by chemical means?

Ans. F- ions cannot be converted to F2 by chemical means because fluorine is the strongest oxidizing agent.

21.Define disproportionation reaction.

Ans. In a disproportionation reaction an element in one oxidation state is simultaneously oxidized and reduced.

22.Identify the reaction

2H2O2 (aq) 2H2O(e) + O2 (g)

Ans . The decomposition of hydrogen peroxide is an example of disproportionation reaction where oxygen experiences disproportionation reaction.

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

Ans. Less reactive metals such as Mg and fFe react with steam to produce dihydrogen gas

Mg + 2H2Mg (OH)2 + H2 Fe + 3H2Fe2 O3 + 3H2.

24.All decomposition reactions are not redox reactions. Give reason.

Ans. Decomposition of calcium carbonate is not a redox reaction


25.Complete the following redox reactions and balance the following equations-

(i)Cr2O72- + C2O42- Cr3+ + CO2 (in presence of acid)

Sn2+ + Cr2O72- Sn4+ + Cr3+ (in presence of acid)

Ans . (i) Cr2O72- + 14H+ + 6e– 2Cr3+ + 7H2O

[C2O42- 2CO2 + 2e] x 3

Cr2O72- 14H+ + 3C2O42- 2Cr3+ + 6 CO2 + 7H2O

(ii) Cr2O72- + 14H+ + 6e 2Cr3+ + 7H2O

[Sn2+ Sn4+ +2e] x3

Cr2O72- + 3Sn2+ + 14H+ 2Cr3+ + 3Sn4+ + 7H2O

26.Write correctly the balanced half – reaction and the overall equations for the following skeletal equations.

(i) NO3 + Bi(S) Bi3+ + NO2 (in acid solution)

(ii) Fe (OH)2 (S) + H2O2Fe (OH)3(S) + H2O (in basic medium)

Ans.(i) In this reaction, H+ ions are available.
Oxidation half reactionBi (S) Bi3+ + 3e
Reduction half reaction[NO3 + 2H+ + e NO+ H2O ] x3
Balanced equation Bi (S) + 3NO3 + 6H+ Bi3+ + 3NO2 + 3H2O

(ii) Fe(OH)2 (S) + H2O2 Fe (OH)3 (S) + H2O
The solution is basic. Therefore, OH- are involved in the reaction, Then
Oxidation half – reduction [Fe (OH)+ OH Fe (OH)3 + e] x2
Reduction half reaction H2O2 + 2e 2OH
Balanced equation 2Fe (OH)+ H2O2 2Fe(OH)3.

27.Define half – cell.

Ans. Combination of an electrode and the solution in which it is dipped is called a half – cell.

28.Set up an electrochemical cell for the redox reaction
Ni2+ (aq) + Fe(S) Ni(S) + Fe2+ (aq)

Ans. Fe (S) / Fe2+(aq) || Ni2+ (aq) / Ni(S)

29.Can we store copper sulphate in an iron vessel?

Ans. We cannot store CuSO4 in an iron vessel because iron is more reactive than Cu and thus holes will be developed in iron vessel.
Cu2+ (aq) + Fe(S) Fe2+ (aq) + Cu(S)

30.What is the role of a salt bridge in an electro chemical cell?

Ans. To complete the electric circuit without mixing the two solution of two half cells. It avoids the accumulation of electric charges in two half – cells.

31.Which reaction occurs at cathode in a galvanic cell?sss

Ans . Reduction.

2 Marks Questions

1.Why ClO4does not show disproportionation reaction where as ClO, ClO2 , ClO3– shows?

Ans. ClO4 does not disproportionate because in this oxoanion chlorine is present in its highest oxidation state that is +7 whereas in ClO, ClO2 and ClO3-, chlorine exists in + 1, +3 and +5 respectively.

2.How would you know whether 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, the reaction takes place in the acidic solution.

If OH or any base, appears on either side of the chemical equation, the solution is basic. If neither H+, OH nor any acid or base is present in the chemical equation, the solution is neutral.

3.Write the following redox reactions in the oxidation and reduction half reaction reactions in the oxidation and reduction half reactions.

(i) 2K(S) + Cl2(g) 2KCl (S)

2Al (S) +3Cu2+ (aq) 2Al3+ (aq) + 3Cu(S)

Ans.(i) K(S) K+ (aq) + e(oxidation)

Cl2(g) 2e 2Cl (reduction)

(ii) Al (S) Al3+ (aq) + 3e (oxidation)

Cu2+ + 2e Cu (S) (reduction)

4.An electrochemical cell is constituted by combining Al electrode (E0 = – 1.66v) and Cu electrode (E0 = + 0.34v). Which of these electrodes will work as cathode and why?

Ans.Since the electrode potential of Cu is higher than that of Al, therefore, Cu has a higher tendency to get reduced and hence Cu electrode acts as a cathode.

5.The E0 of Cu2+ / Cu is + 0.34V. What does it signify?

Ans.Cu lies below hydrogen in the activity series.

6.If reduction potential of an electrode is 1.28V. What will be its oxidation potential?

Ans. – 1. 28V.

7.What is the electrode potential of a standard hydrogen electrode?

Ans . Zero.

8.Define a redox couple.

Ans. A redox couple is defined as having together oxidized and reduced forms of a substance taking part in an oxidation and reduction half – reaction.

9.Explain why 3Fe3O4 (S) +8Al(S) -> 9Fe (S) +4Al2O3. Is an oxidation reaction. ?

Ans .Aluminum is oxidized because oxygen is added to it Ferrous ferric oxide (Fe3O4) is reduced because oxygen has been removed from it.

4 Marks Questions

1.Balance the following equations by oxidation number method:

(i)CuO + NHCu + N+ H2O
K2 MnO4 + H2MnO2 + KMnO4 + KOH

Ans.(i) Skeleton of equation 

Oxidation number of copper decreases from +2 to O and ox no of Nitrogen increases from – 3 to 0.
In order to balance the increase of O.N with decease of O. N there should be three atoms of copper and two atoms of nitrogen. Hence 3CuO + 2NH3 3Cu + N2 + H2O

Balancing hydrogen and oxygen atoms we have 3 CuO + 2 NH3 3Cu + N2 + 3H2O

(ii) Writing K2 MnO4 twice O.N of Mn, we have the skeleton of the equation
O.N of Mn in 1 mol k2MnO4 decreases from + 6 to + 4 (MnO2) and in the other mol increases from +6 to +7 (KMnO4) i.e. 1 mol acquires two electrons while the other loses 1 electrons .
In order to balance the O. N of Mn, 1 mol. K2MnO4 and kMnO4 are multiplied by 2. Hence K2MnO4 + 2K2MnO4 + H2MnO2 + 2KMnO4 +KOH
In order to balance the number of K and H atoms KOH is multiplied by 4 and H2O by 2. 3K2MnO4 + 2H2MnO2 + 2KMnO4 + 4KOH

Leave a Reply