NCERT Solutions for Class 11 Chemistry Chemistry Part-2 Chapter 8

Redox Reactions Class 11

Chapter 8 Redox Reactions Exercise Solutions

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Exercise : Solutions of Questions on Page Number : 272

Q1 :  

Assign oxidation numbers to the underlined elements in each of the following species:

(a) NaH2PO4 (b) NaHSO4 (c) H4P2O7 (d) K2MnO4

(e) CaO2 (f) NaBH4 (g) H2S2O7 (h) KAl(SO4)2.12 H2O


Answer :

(a)

Let the oxidation number of P be x.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = –2

Then, we have

Hence, the oxidation number of P is +5.

(b)

Then, we have

Hence, the oxidation number of S is + 6.

(c)

Then, we have

Hence, the oxidation number of P is + 5.

(d)

Then, we have

Hence, the oxidation number of Mn is + 6.

(e)

Then, we have

Hence, the oxidation number of O is – 1.

(f)

Then, we have

Hence, the oxidation number of B is + 3.

(g)

Then, we have

Hence, the oxidation number of S is + 6.

(h)

Then, we have

Or,

We can ignore the water molecule as it is a neutral molecule. Then, the sum of the oxidation numbers of all atoms of the water molecule may be taken as zero. Therefore, after ignoring the water molecule, we have

Hence, the oxidation number of S is + 6.

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Q2 :  

What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?

(a) KI3 (b) H2S4O6 (c) Fe3O4 (d) CH3CH2OH (e) CH3COOH


Answer :

(a) KI3

In KI3, the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is . However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3to find the oxidation states. 

In a KI3molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

Hence, in a KI3molecule, the O.N. of the two I atoms forming the I2molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

(b) H2S4O6

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

The O.N. of two of the four S atoms is +5 and the O.N. of the other two S atoms is 0.

(c)

On taking the O.N. of O as –2, the O.N. of Fe is found to be. However, O.N. cannot be fractional. 

Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3.

(d)

2 (x) + 6 (+1) + 1 (-2) = 0

or, 2x + 4 = 0

or, x = -2

Hence, the O.N. of C is –2.

(e)

2 (x) + 4 (+1) + 2 (-2) = 0

or, 2x = 0

or, x = 0

However, 0 is average O.N. of C. The two carbon atoms present in this molecule are present in different environments. Hence, they cannot have the same oxidation number. Thus, C exhibits the oxidation states of +2 and –2 in CH3COOH.

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Q3 :  

Justify that the following reactions are redox reactions:

(a) CuO(s) + H2(g) → Cu(s) + H2O(g)

(b) Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)

(c) 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s)

(d) 2K(s) + F2(g) → 2K+F- (s)

(e) 4 NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(g)


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Q4 :  

Fluorine reacts with ice and results in the change:

H2O(s) + F2(g) → HF(g) + HOF(g)

Justify that this reaction is a redox reaction.


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Q5 :  

Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5,–. Suggest structure of these compounds. Count for the fallacy.


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Q6 :  

Write the formulae for the following compounds:

(a) Mercury(II) chloride                        (b) Nickel(II) sulphate

(c) Tin(IV) oxide                                 (d) Thallium(I) sulphate

(e) Iron(III) sulphate                           (f) Chromium(III) oxide


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Q7 :  

Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.


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Q8 :  

While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?


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Q9 :  

Consider the reactions:

(a) 6 CO2(g) + 6H2O(l) C6H12O6(aq) + 6O2(g)

(b) O3(g) + H2O2(l) H2O(l) + 2O2(g)

Why it is more appropriate to write these reactions as:

(a) 6CO2(g) + 12H2O(l) C6H12O6(aq) + 6H2O(l) + 6O2(g)

(b) O3(g) + H2O2(l) H2O(l) + O2(g) + O2(g)

Also suggest a technique to investigate the path of the above (a) and (b) redox

reactions.


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Q10 :  

The compound AgF2 is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?


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Q11 :  

Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.


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Q12 :  

How do you count for the following observations?

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?


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Q13 :  

Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions:

(a) 2AgBr (s) + C6H6O2(aq) 2Ag(s) + 2HBr (aq) + C6H4O2(aq)

(b) HCHO(l) + 2[Ag (NH3)2]+(aq) + 3OH-(aq) 2Ag(s) + HCOO-(aq) + 4NH3(aq) + 2H2O(l)

(c) HCHO (l) + 2Cu2+(aq) + 5 OH-(aq) Cu2O(s) + HCOO-(aq) + 3H2O(l)

(d) N2H4(l) + 2H2O2(l) N2(g) + 4H2O(l)

(e) Pb(s) + PbO2(s) + 2H2SO4(aq) 2PbSO4(s) + 2H2O(l)


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Q14 :  

Consider the reactions:

(aq) + I2(s) → (aq) + 2I–(aq)

aq) + 2Br2(l) + 5 H2O(l) → (aq) + 4Br–(aq) + 10H+(aq)

Why does the same reductant, thiosulphate react differently with iodine and bromine?


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Q15 :  

Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.


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Q16 :  

Why does the following reaction occur?

(aq) + 2F–(aq) + 6H+(aq) → XeO3(g) + F2(g) + 3H2O(l)

What conclusion about the compound Na4XeO6(of which is a part) can be drawn from the reaction.


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Q17 :  

Consider the reactions:

(a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) H3PO4(aq) + 4Ag(s) + 4HNO3(aq)

(b) H3PO2(aq) + 2CuSO4(aq) + 2 H2O(l) H3PO4(aq) + 2Cu(s) + H2SO4(aq)

(c) C6H5CHO(l) + 2[Ag (NH3)2]+(aq) + 3OH-(aq) C6H5COO-(aq) + 2Ag(s) + 4NH3(aq) + 2 H2O(l)

(d) C6H5CHO(l) + 2Cu2+(aq) + 5OH-(aq) No change observed.

What inference do you draw about the behaviour of Ag+and Cu2+from these reactions?


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Q18 :  

Balance the following redox reactions by ion-electron method:

(a) (aq) + I–(aq) → MnO2(s) + I2(s) (in basic medium)

(b) (aq) + SO2(g) → Mn2+(aq) + (aq) (in acidic solution)

(c) H2O2(aq) + Fe2+(aq) → Fe3+(aq) + H2O (l) (in acidic solution)

(d)+ SO2(g) → Cr3+(aq) + (aq) (in acidic solution)


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Q19 :  

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.


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Q20 :  

What sorts of informations can you draw from the following reaction ?


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Q21 :  

The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2, and H+ ion. Write a balanced ionic equation for the reaction.


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Q22 :  

Consider the elements:

Cs, Ne, I and F

(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only postive oxidation state.

(c) Identify the element that exhibits both positive and negative oxidation states.

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.


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Q23 :  

Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.


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Q24 :  

Refer to the periodic table given in your book and now answer the following questions:

(a) Select the possible non metals that can show disproportionation reaction.

(b) Select three metals that can show disproportionation reaction.


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Q25 :  

In Ostwald's process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?


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Q26 :  

Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:

(a) Fe3+(aq) and I-(aq)

(b) Ag+(aq) and Cu(s)

(c) Fe3+ (aq) and Cu(s)

(d) Ag(s) and Fe3+(aq)

(e) Br2(aq) and Fe2+(aq)


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Q27 :  

Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes

(ii) An aqueous solution AgNO3 with platinum electrodes

(iii) A dilute solution of H2SO4 with platinum electrodes

(iv) An aqueous solution of CuCl2 with platinum electrodes.


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Q28 :  

Arrange the following metals in the order in which they displace each other from the solution of their salts.

Al, Cu, Fe, Mg and Zn.


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Q29 :  

Given the standard electrode potentials,

K+/K = -2.93V, Ag+/Ag = 0.80V,

Hg2+/Hg = 0.79V

Mg2+/Mg = -2.37V. Cr3+/Cr = -0.74V

Arrange these metals in their increasing order of reducing power.


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Q30 :  

Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s) takes place, further show:

(i) which of the electrode is negatively charged,

(ii) the carriers of the current in the cell, and

(iii) individual reaction at each electrode.


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<< Previous Chapter 7 : Equilibrium Next Chapter 9 : Hydrogen >>

Chemistry Part-2 - Chemistry : CBSE NCERT Exercise Solutions for Class 11th for Redox Reactions will be available online in PDF book form soon. The solutions are absolutely Free. Soon you will be able to download the solutions.

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