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

8.2 What are the oxidation number of the underlined elements in each of the following and

how do you rationalise your results ?

8.3 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) + 3AlCl3 (s)

(d) 2K(s) + F2(g) → 2K+F- (s)

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

8.4 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.

8.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O72-and

NO3-.

Suggest structure of these compounds. Count for the fallacy.

8.6 Write formulas 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

8.7 Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4

and nitrogen from -3 to +5.

8.8 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 ?

8.9 Consider the reactions:

(a) 6CO2(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.

8.10 The compound AgF2 is unstable compound. However, if formed, the compound acts as

a very strong oxidising agent. Why ?

8.11 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.

8.12 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 ?

8.13 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) + 5OH-(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)

8.14 Consider the reactions :

2S2O32-(aq) + I2(s) → S4O6

2-(aq) + 2I-(aq) S2O32-(aq) + 2Br2(l) + 5H2O(l) → 2SO4

2-(aq) + 4Br-(aq)

+ 10H+(aq)

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

8.15 Justify giving reactions that among halogens, fluorine is the best oxidant and among

hydrohalic compounds, hydroiodic acid is the best reductant.

8.16 Why does the following reaction occur ?

XeO64- (aq) + 2F- (aq) + 6H+(aq) → XeO3(g)+ F2(g) + 3H2O(l)

What conclusion about the compound Na4XeO6 (of which XeO64- is a part) can be drawn from

the reaction.

8.17 Consider the reactions:

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

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

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

2H2O(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 ?

8.18 Balance the following redox reactions by ion – electron method :

(a) MnO4-(aq) + I- (aq) → MnO2(s) + I2(s) (in basic medium)

(b) MnO4-(aq) + SO2 (g) → Mn2+ (aq) + HSO4

- (aq) (in acidic solution)

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

(d) Cr2O72- + SO2(g) → Cr3+ (aq) + SO4

2- (aq) (in acidic solution)

8.19 Balance the following equations in basic medium by ion-electron method and oxidation

number methods and identify the oxidising agent and the reducing agent.

(a) P4(s) + OH-(aq) → PH3(g) + HPO2- (aq)

(b) N2H4(l) + ClO3-(aq) → NO(g) + Cl-(g)

(c) Cl2O7 (g) + H2O2(aq) → ClO2-(aq) + O2(g) + H+

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

(CN)2(g) + 2OH-(aq) → CN-(aq) + CNO-(aq) + H2O(l)

8.21 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.

8.22 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.

8.23 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.

8.24 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.

8.25 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 ?

8.26 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).

8.27 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.

8.28 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.

8.29 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.

8.30 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

Answer to Some Selected Problems

8.25 15 g

I. Multiple Choice Questions (Type-I)

1. Which of the following is not an example of redox reaction?

(i) CuO + H2 -> Cu + H2O

(ii) Fe2O3 + 3CO -> 2Fe + 3CO2

(iii) 2K + F2 -> 2KF

(iv) BaCl2 + H2SO4 -> BaSO4 + 2HCl

2. The more positive the value of EΘ, the greater is the tendency of the species to get

reduced. Using the standard electrode potential of redox couples given below find out which

of the following is the strongest oxidising agent.

EΘ values: Fe3+/Fe2+ = + 0.77; I2(s)/I– = + 0.54;

Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80V

(i) Fe3+

(ii) I2(s)

(iii) Cu2+

(iv) Ag+

3. EΘ values of some redox couples are given below. On the basis of these values choose

the correct option.

EΘ values : Br2/Br– = + 1.90; Ag+ /Ag(s) = + 0.80

Cu2+/Cu(s) = + 0.34; I2(s)/I– = + 0.54

(i) Cu will reduce Br–

(ii) Cu will reduce Ag

(iii) Cu will reduce I–

(iv) Cu will reduce Br2

4. Using the standard electrode potential, find out the pair between which redox reaction is

not feasible.

EΘ values : Fe3+/Fe2+ = + 0.77; I2/I– = + 0.54;

Cu2+/Cu = + 0.34; Ag+/Ag = + 0.80 V

(i) Fe3+ and I–

(ii) Ag+ and Cu

(iii) Fe3+ and Cu

(iv) Ag and Fe3+

5. Thiosulphate reacts differently with iodine and bromine in the reactions given below:

2S2O32– + I2 → S4O6

2– + 2I–

S2O32– + 2Br2 + 5H2O → 2SO4

2– + 2Br– + 10H+

Which of the following statements justifies the above dual behaviour of thiosulphate?

(i) Bromine is a stronger oxidant than iodine.

(ii) Bromine is a weaker oxidant than iodine.

(iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these

reactions.

(iv) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.

6. The oxidation number of an element in a compound is evaluated on the basis of certain 

rules. Which of the following rules is not correct in this respect?

(i) The oxidation number of hydrogen is always +1.

(ii) The algebraic sum of all the oxidation numbers in a compound is zero.

(iii) An element in the free or the uncombined state bears oxidation number zero.

(iv) In all its compounds, the oxidation number of fluorine is – 1.

7. In which of the following compounds, an element exhibits two different oxidation states.

(i) NH2OH

(ii) NH4NO3

(iii) N2H4

(iv) N3H

8. Which of the following arrangements represent increasing oxidation number of the central

atom?

(i) CrO2– , ClO3

– , CrO42–, MnO4

–

(ii) ClO3–, CrO4

2– , MnO4– , CrO2

–

(iii) CrO2– , ClO3

– , MnO4– , CrO4

2–

(iv) CrO42–, MnO4

– , CrO2– , ClO3

–

9. The largest oxidation number exhibited by an element depends on its outer electronic

configuration. With which of the following outer electronic configurations the element will

exhibit largest oxidation number?

(i) 3d14s2

(ii) 3d34s2

(iii) 3d54s1

(iv) 3d54s2

10. Identify disproportionation reaction

(i) CH4 + 2O2 → CO2 + 2H2O

(ii) CH4 + 4Cl2 → CCl4 + 4HCl

(iii) 2F2 + 2OH– → 2F– + OF2 + H2O

(iv) 2NO2 + 2OH– → NO2– + NO3

– + H2O

11. Which of the following elements does not show disproportionation tendency?

(i) Cl

(ii) Br

(iii) F

(iv) I

II. Multiple Choice Questions (Type-II)

In the following questions two or more options may be correct.

12. Which of the following statement(s) is/are not true about the following decomposition

reaction.

2KClO3 → 2KCl + 3O2

(i) Potassium is undergoing oxidation

(ii) Chlorine is undergoing oxidation

(iii) Oxygen is reduced

(iv) None of the species are undergoing oxidation or reduction

13. Identify the correct statement (s) in relation to the following reaction:

Zn + 2HCl → ZnCl2 + H2

(i) Zinc is acting as an oxidant

(ii) Chlorine is acting as a reductant

(iii) Hydrogen ion is acting as an oxidant

(iv) Zinc is acting as a reductant

14. The exhibition of various oxidation states by an element is also related to the outer

orbital electronic configuration of its atom. Atom(s) having which of the following outermost

electronic configurations will exhibit more than one oxidation state in its compounds.

(i) 3s1

(ii) 3d14s2

(iii) 3d24s2

(iv) 3s23p3

15. Identify the correct statements with reference to the given reaction

P4 + 3OH– + 3H2O → PH3 + 3H2PO2–

(i) Phosphorus is undergoing reduction only.

(ii) Phosphorus is undergoing oxidation only.

(iii) Phosphorus is undergoing oxidation as well as reduction.

(iv) Hydrogen is undergoing neither oxidation nor reduction.

16. Which of the following electrodes will act as anodes, when connected to Standard 

Hydrogen Electrode?

(i) Al/Al3+ EΘ = –1.66

(ii) Fe/Fe2+ EΘ = – 0.44

(iii) Cu/Cu2+ EΘ = + 0.34

(iv) F2(g)/2F–(aq) EΘ = + 2.87

III. Short Answer Type

17. The reaction

Cl2(g) + 2OH–(aq) → ClO–(aq) + Cl–(aq) + H2O (l)

represents the process of bleaching. Identify and name the species that bleaches the

substances due to its oxidising action.

18. MnO42– undergoes disproportionation reaction in acidic medium but MnO4

– does not. Give

reason.

19. PbO and PbO2 react with HCl according to following chemical equations :

2PbO + 4HCl → 2PbCl2 + 2H2O

PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O

Why do these compounds differ in their reactivity?

20. Nitric acid is an oxidising agent and reacts with PbO but it does not react with PbO2.

Explain why?

21. Write balanced chemical equation for the following reactions:

(i) Permanganate ion (MnO4–) reacts with sulphur dioxide gas in acidic medium to produce

Mn2+ and hydrogensulphate ion.(Balance by ion electron method)

(ii) Reaction of liquid hydrazine (N2H4) with chlorate ion (ClO3–) in basic medium produces

nitric oxide gas and chloride ion in gaseous state.(Balance by oxidation number method)

(iii) Dichlorine heptaoxide (Cl2O7) in gaseous state combines with an aqueous solution of 

hydrogen peroxide in acidic medium to give chlorite ion (ClO2–) and oxygen gas.(Balance by

ion electron method)

22. Calculate the oxidation number of phosphorus in the following species.

(a) HPO32– and (b) PO4

3–

23. Calculate the oxidation number of each sulphur atom in the following compounds:

(a) Na2S2O3

(b) Na2S4O6

(c) Na2SO3

(d) Na2SO4

24. Balance the following equations by the oxidation number method.

(i) Fe2+ + H+ + Cr2O72– → Cr3+ + Fe3+ + H2O

(ii) I2 + NO3– → NO2 + IO3

–

(iii) I2 + S2O32– → I– + S4O6

2–

(iv) MnO2 + C2O42– → Mn2+ + CO2

25. Identify the redox reactions out of the following reactions and identify the oxidising and

reducing agents in them.

(i) 3HCl(aq) + HNO3(aq) → Cl2(g) + NOCl (g) + 2H2O (l)

(ii) HgCl2(aq) + 2KI (aq) → HgI2(s) + 2KCl (aq)

(iv) PCl3 (l) + 3H2O (l) → 3HCl (aq) + H3PO3 (aq)

(v) 4NH3 + 3O2 (g) → 2N2 (g) + 6H2O (g)

26. Balance the following ionic equations

(i) Cr2O72– + H+ + I– → Cr3+ + I2 + H2O

(ii) Cr2O72– + Fe2+ + H+ → Cr3+ + Fe3+ + H2O

(iii) MnO4– + SO3

2– + H+ → Mn2+ + SO42– + H2O (iv) MnO4

– + H+ + Br– → Mn2+ + Br2 + H2O

IV. Matching Type

27. Match Column I with Column II for the oxidation states of the central atoms.

Column I Column II

(i) Cr2O72- (a) + 3

(ii) MnO4– (b) + 4

(iii) VO3– (c) + 5

(iv) FeF63– (d) +6

(e) + 7

28. Match the items in Column I with relevant items in Column II.

Column I Column II

(i) Ions having positive charge (a) +7

(ii)

The sum of oxidation number of all atoms in a

neutral molecule (b) –1

(iii) Oxidation number of hydrogen ion (H+) (c) +1

(iv) Oxidation number of fluorine in NaF (d) 0

(v) Ions having negative charge (e) Cation

(f) Anion

V. Assertion and Reason Type

In the following questions a statement of assertion (A) followed by a statement of reason (R)

is given. Choose the correct option out of the choices given below each question.

29. Assertion (A) : Among halogens fluorine is the best oxidant.

Reason (R) : Fluorine is the most electronegative atom.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

30. Assertion (A): In the reaction between potassium permanganate and potassium iodide,

permanganate ions act as oxidising agent.

Reason (R) : Oxidation state of manganese changes from +2 to +7 during the reaction.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

31. Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen is an

example of disproportionation reaction.

Reason (R) : The oxygen of peroxide is in –1 oxidation state and it is converted to zero

oxidation state in O2 and –2 oxidation state in H2O.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

32. Assertion (A) : Redox couple is the combination of oxidised and reduced form of a

substance involved in an oxidation or reduction half cell.

Reason (R) : In the representation EΘFe

3+/ Fe2+ and EΘCu

2+ / Cu , Fe3+/ Fe2+ and Cu2+ / Cu are redox

couples.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) Both A and R are false.

VI. Long Answer Type

33. Explain redox reactions on the basis of electron transfer. Give suitable examples.

34. On the basis of standard electrode potential values, suggest which of the following

reactions would take place? (Consult the book for EΘ value).

(i) Cu + Zn2+ → Cu2+ + Zn

(ii) Mg + Fe2+ → Mg2+ + Fe

(iii) Br2 + 2Cl– → Cl2 + 2Br–

(iv) Fe + Cd2+ → Cd + Fe2+

35. Why does fluorine not show disporportionation reaction?

36. Write redox couples involved in the reactions (i) to (iv) given in question 34.

37. Find out the oxidation number of chlorine in the following compounds and arrange them

in increasing order of oxidation number of chlorine.

NaClO4, NaClO3, NaClO, KClO2, Cl2O7, ClO3, Cl2O, NaCl, Cl2, ClO2.

Which oxidation state is not present in any of the above compounds?

38. Which method can be used to find out strength of reductant/oxidant in a solution?

Explain with an example.

ANSWERS

I. Multiple Choice Questions (Type-I)

1. (iv)      2. (iv)      3. (iv)       4. (iv)      5. (i)      6. (i)      7. (ii)      8. (i)      9. (iv)      10. (iv)

11. (iii)

II. Multiple Choice Questions (Type-II)

12. (i), (iv)      13. (iii), (iv)      14. (iii), (iv)      15. (iii), (iv)      16. (i), (ii)

III. Short Answer Type

17. Hypochloriteion

18. In MnO4– , Mn is in the highest oxidation state i.e. +7. Therefore, it does not undergo

disproportionation. MnO42– undergoes disproportionation as follows : 3MnO4

2– + 4H+ →

2MnO4– + MnO2 + 2H2O

19. 2PbO + 4HCl → 2PbCl2 + 2H2O (Acid base reaction) PbO2 + 4HCl → PbCl2 + Cl2 + 2H2O

(Redox reaction) (Hint : Note the oxidation number of lead in the oxides)

20. PbO is a basic oxide and simple acid base reaction takes place between PbO and

HNO3. On the other hand in PbO2 lead is in + 4 oxidation state and cannot be oxidised

further. Therefore no reaction takes place. Thus, PbO2 is passive, only PbO reacts with

HNO3. 2PbO + 4HNO3 → 2Pb (NO3)2 + 2H2O (Acid base reaction)

22. (a) +3,      (b) +5 23. (a) +2      (b) +5, 0, 0, +5      (c) +4      

(d) +6

Justification : Write Lewis structure of each ion then assign electron pair shared between

atoms of different electronegativity to more electronegative atom and distribute the electron

pair shared between atoms of same element equally. Now count the number of electrons

possessed by each atom. Find out the difference in number of electrons possessed by

neutral atom and that possessed by atom in the compound. This difference is the oxidation

number. If atom present in the compound possesses more electrons than the neutral atom,

the oxidation number is negative. If it possesses less electrons then oxidation number is

positive.

(i) Lewis structure of S2O42– can be written as follows :

Electron pair shared between sulphur and oxygen is assigned to oxygen atoms because of

more electronegativity of oxygen. Thus each sulphur atom is deficient of two electrons with

respect to neutral sulphur atom hence, each sulphur atom is in +2 oxidation state. Each

oxygen atom gets two excess electrons hence, it is in –2 oxidation state. Lewis structure of

S4O62– can be written as follows :

To find out oxidation state of each atom we distribute electrons of electron pair shared

between two sulphur atoms equally (i.e. one electron is assigned to each sulphur atom).

Both the electrons of electron pair shared between sulphur and oxygen atom are assigned

to oxygen as oxygen is more electronegative. Thus we find that each of the central sulphur

atoms obtain six electrons. This number is same as that in the outer shell of neutral sulphur

atom hence oxidation state of central sulphur atoms is zero. Each of the sulphur atoms

attached to oxygen atoms obtain only one electron as its share. This number is less by five

electrons in comparison to the neutral sulphur atom. So, outer sulphur atoms are in +5

oxidation state. Therefore average oxidation state of sulphur atoms is :

(5 + 0 + 0 + 5)/4 = 10/4 = 2.5

By using the formula we obtain average oxidation state of the particular type of atoms. Real

oxidation state can be obtained only by writing the complete structural formula. Similarly we

can see that each oxygen atom is in – 2 oxidation state.

In the same way one can find out the oxidation state of each atom in SO32– and SO4

2–ions.

Oxidation state of metal atoms will be +1 as these will lose one electron in each case.

IV. Matching Type

27. (i) → (d) (ii) → (e) (iii) → (c) (iv) → (a)

28. (i)→ (e) (ii) → (d) (iii) → (c) (iv) → (b) (v) → (f)

V. Assertion and Reason Type

29. (ii) 30. (iii) 31. (i) 32. (ii)