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 Cl₂ + 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. F₂ is 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.

Ans.ClO₂

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 + 2H₂O Mg (OH)₂ + H₂ Fe + 3H₂O Fe₂ 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)Cr₂O₇^2- + C₂O₄^2- Cr³⁺ + CO₂ (in presence of acid)

Sn²⁺ + Cr₂O₇^2- Sn⁴⁺ + Cr³⁺ (in presence of acid)

Ans . (i) Cr₂O₇^2- + 14H⁺ + 6e^- 2Cr³⁺ + 7H₂O

[C₂O₄^2- 2CO₂ + 2e^-] x 3

Cr₂O₇^2- 14H⁺ + 3C₂O₄^2- 2Cr³⁺ + 6 CO₂ + 7H₂O

(ii) Cr₂O₇^2- + 14H⁺ + 6e^- 2Cr³⁺ + 7H₂O

[Sn²⁺ Sn⁴⁺ +2e^-] x3

Cr₂O₇^2- + 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.
Therefore,
Oxidation half reactionBi (S) Bi³⁺ + 3e^-
Reduction half reaction[NO₃^- + 2H⁺ + e^- NO₂ + 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)₂ + OH^- Fe (OH)₃ + e^-] x2
Reduction half reaction H₂O₂ + 2e^- 2OH^-
Balanced equation 2Fe (OH)₂ + 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 ClO₄^-does 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 + NH₃ Cu + N₂ + H₂O
K₂ MnO₄ + H₂O MnO₂ + 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 + H₂O MnO₂ + 2KMnO₄ +KOH
In order to balance the number of K and H atoms KOH is multiplied by 4 and H2O by 2. 3K₂MnO₄ + 2H₂O MnO₂ + 2KMnO₄ + 4KOH