Redox Reactions Class 11 NCERT Solutions: Download Topic-wise Questions & Answers PDF

Q: Q 8.2. What is the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

A 8.2 (a) In Kl3, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K+[I —I <— I]–. Here, a coordinate bond is formedbetween I2 molecule and I– ion. The oxidation number of two iodine atoms forming the I2 molecule is zero while that of iodine forming the coordinate bond is -1. Thus, the oxidation number of three I atoms, atoms in Kl3 is 0, 0 and -1 respectively. (b) By conventional method O.N. of S in H2S4O6is calculated as: 2 (+1) +4x + 6) (-2) = 0 Or x = +2.5 But all the four S atoms cannot be in the same oxidation state. So by chemical bonding method: The O.N. of each of the S-atoms linked with each other in the middle is zero while that of each of the remaining two S-atoms is +5. (c) By conventional method O.N. of Fe in Fe3O4 is calculated as: 3x + 4 (-2) = 0 or x = 8/3 By stoichiometry O.N. of Fe in Fe3O4 is +2 and +3. (d) By conventional method, O.N. of C in CH3CH2OH is calculated as: 2x + 6 (+1) +1 (-2) = 0 or x = -2 (e) By conventional method, O.N. of C in CH3COOH is calculated as: 2x + 4 - 4 = 0 or x = 0 By chemical bonding method, C2 is attached to three H-atoms (less electronegative than carbon) and one -COOH group (more electronegative than carbon). Therefore, O.N. of C2 = 3 (+1) + x + 1 (-1) = 0 or x = -2 But C1 is attached to one oxygen atom by a double bond, one OH (-1) and one CH3 (+1) group, therefore, O.N. of C1 = +1 + x+ 1 (-2) + 1 (-1) = 0 or x = +2

Q: Q 8.3 Justify that the following reactions are redox reactions:

A 8.3 (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) 4 NH3 (g) + 5 O2 → 4NO (g) + 6H2O (g) A 8.3 (a) Here, O is removed from CuO, therefore, it is reduced to Cu while O is added to H2 to form H2O, therefore, it is oxidised. Further, O.N. of Cu decreases from +2 in CuO to 0 in Cu but that of H increases from 0 in H2 to +1 in H20. Therefore, CuO is reduced to Cu but H2 is oxidised to H2O. Thus, this is a redox reaction. (b) Here O.N. of Fe decreases from +3 in Fe2O3 to 0 in Fe while that of C increases from +2 in CO to +4 in CO2. Further, oxygen is removed from Fe2O3 and added to CO, therefore, Fe2O3 is reduced while CO is oxidised. Thus, this is a redox reaction. (c) Here, O.N. of B decreases from +3 in BrCl3to -3 in B2H6 while that of H increases from -1 in LiAlH4to +1 in B2H6. Therefore, BCl3 is reduced while LiAlH4 is oxidised. Further, H is added to BCl3 but is removed from LiAlH4, therefore, BC13 is reduced while LiAlH4 is oxidised. Thus, it is a redox reaction (d) Here, each K atom as lost one electron to form K+ while F2 has gained two electrons to form two F– ions. Therefore, K is oxidised while F2 is reduced. Thus, it is a redox reaction. (e) Here, the oxidation number of N increases from –3 in NH3 to +2 in NO. On the other hand, the oxidation number of O2 decreases from 0 in O2 to –2 in NO and H2O i.e., O2 is reduced. Hence, the given reaction is a redox reaction.

Q: Q 8.4 Fluorine reacts with ice and results in the change, Justify that this reaction is a redox reaction.

H2O(s) + F2 (g) ——> HF (g) + HOF (g) A 8.4 Writing the O.N. of each atom above its symbol, we have, here, the O.N. of F decreases from 0 in F 2 to -1 in HF and increases from 0 in F 2 to +1 in HOF. Therefore, F 2 is both reduced as well as oxidised. Thus, it is a redox reaction and more specifically, it is a disproportionation reaction.

Q: Q 8.5. Calculate the oxidation number of sulphur, chromium and nitrogen in (equation) Suggest structure of these compounds. Count for the fallacy.

H2SO5, Cr2O72- and NO3- A 8.5 (a) H2SO5 by conventional method. Let x be the oxidation number of S 2(+1) + x + 5(−2) = 0 x = +8 +8 oxidation state of S is not possible as S cannot have oxidation number more than 6. The fallacy is overcome if we calculate the oxidation number from its structure HO−S(O2)−O−O−H. −1+X+2(−2)+2(−1)+1=0 x=+6 (b) Dichromate ion Let x be the oxidation number of Cr in dichromate ion 2x+7(−2)=−2 x=+6 Hence the oxidation number of Cr in dichromate ion is +6. This is correct and there is no fallacy. (c) Nitrate ion, by conventional method Let x be the oxidation number of N in nitrate ion. x+3(−2)=−1 From the structure O−−N+(O)−O− x+1(−1)+1(−2)+1(−2)=0 x=+5 Thus there is no fallacy.

Q: Q 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.

A 8.6 (a) HgCl 2, (b) NiSO 4, (c)SnO 2 (d) Tl 2SO 4, (e) Fe 2(S0 4) 3, (f) Cr 2O 3

Q: Q 8.7. Suggest a list of substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

A 8.7 Substance Oxidation number of C Substance Oxidation number of N CH2Cl2 0 N2 0 FC≡CF +1 N2O +1 HC≡CH -1 N2H2 -1 CHCl3,CO +2 NO +2 CH3Cl -2 N2H4 -2 Cl3C−CCl3 +3 N2O3 +3 H3C−CH3 -3 NH3 -3 CCl4,CO2 +4 NO2 +4 CH4 -4 N2O5 +5

Q: Q 8.8. While sulphur dioxide and hydrogen peroxide can act as an oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

A 8.8 (i) In SO2, O.N. of S is +4. In principle, S can have a minimum O.N. of -2 and maximum of +6. Therefore, S in SO2 can either decrease or increase its O.N. and hence can act both as an oxidising as well as a reducing agent. (ii) In H2O2, the O.N. of O is -1. In principle, O can have a minimum O.N. of -2 and maximum of zero (+1 is possible in O2F2 and +2 in OF2). Therefore, O in H2O2 can either decrease its O.N. from -1 to -2 or can increase its O.N. from -1 to zero. Therefore, H2O2 acts both as an oxidising as well as a reducing agent. (iii) In O3, the O.N. of O is zero. It can only decrease its O.N. from zero to -1 or -2, but cannot increase to +2. Therefore, O3 acts only as an oxidant. (iv) In HNO3, O.N. of N is +5 which is maximum. Therefore, it can only decrease its O.N. and hence it acts as an oxidant only.

Q: Q 8.9 Consider the reactions, Also suggest a technique to investigate the path of above (a) and (b) redox reactions.

(a) 6CO2(g) + 6H2O(l) →C6H12O6(s) + 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(s) + 6H2O(l) + 6O2(g) (b) O3(g) + H2O2 (l) →H2O(l) + O2(g) + O2(g) A 8.9

Q: Q 8.10. The compound AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?

A 8.10 In AgF 2, oxidation state of Ag is +2 which is very unstable. Therefore, it quickly accepts an electron to form the more stable +1 oxidation state. Ag 2+ + e – →Ag + Therefore, AgF 2, if formed, will act as a strong oxidising agent.

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This chapter has multiple topics listed below:

NCERT Chemistry 11th Some Basic Concepts of Chemistr...Structure of Atoms Classification of Elements and ...Chemical Bonding and Molecular ...States of Matter Thermodynamics Equilibrium Redox Reactions Hydrogen The S-block Elements The p -Block Elements Organic Chemistry - Some Basic ...Hydrocarbons Environmental chemistry

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Updated on Oct 9, 2023 12:12 IST

Redox Reactions are the chemical reactions that involve transfer of electrons. In Redox Reactions, electrons get transferred between the two reactants and the state of atoms are changed. Redox Reactions are the combination of two reactions, Oxidation and Reduction Reactions.

Chemistry deals with varieties of matter and change of one kind of matter into the other. A number of phenomena, both physical as well as biological, are concerned with redox reactions. These reactions find extensive use in pharmaceutical, biological, industrial, metallurgical and agricultural areas. Students can find here NCERT class 11 Chemistry for Redox Reactions chapter. These solutions have been prepared by subject experts and will help in securing good marks in the exam. Redox Reactions is important chapter from the exam point of view as several questions are asked from this particular chapter. Those preparing for entrance exams can refer NCERT Solutions Chemistry class 12.

Table of Contents

Redox Reaction Chapter Subtopics
Redox Reactions Solutions and FAQs

Redox Reaction Chapter Subtopics

Given here are the subtopics that students will be studying in the Chemistry class 11 NCERT solution for Redox Reactions.

Classical Idea of Redox Reactions – Oxidation and Reduction Reactions
Redox Reactions in Terms of Electron Transfer Reactions
- Competitive Electron Transfer Reactions
Oxidation Number
Types of Redox Reactions
- Combination Reactions
- Decomposition Reactions
- Displacement Reactions
- Disproportionation Reactions
Balancing of Redox Reactions
Redox Reactions as the Basis for Titrations
Limitations of Concept of Oxidation Number
Redox Reactions and Electrode Processes.

Redox reaction Additional Questions

8.1. Assertion: Standard hydrogen electrode is called as reversible electrode.
Reason: It can react both as anode as well as cathode in an electrochemical cell.
Answer: (a) A standard hydrogen electrode is called reversible electrode because it can react both as anode as well as cathode in an electrochemical cell.

8.2. Assertion: Na+ ion has an oxidation number of +1, Mg2+ ion, +2, Fe3+ ion, +3, Cl– ion, –1, O2– ion, –2; and so on.
Reason: For ions composed of only one atom, the oxidation number is equal to the charge on the ion.
Answer: (a)

8.3. Assertion: The highest value of oxidation number exhibited by an atom of an element generally increases across the period in the periodic table.
Reason: The oxidation number denotes theoxidation state of an element in a compound.
Answer: (b)Both are correct statements but reason is not correct for the assertion. The highest oxidation number of a representative element is the group number for the first two groups and the group number minus 10 (following the long form of periodic table) for the other groups.

8.4. Assertion: While H2O2 can only act as oxidising agent in their reactions.
Reason: In H2O2 oxidation number of O is -1 and can vary from 0 to -2
Answer: (d) In H2O2 oxidation number of O = -1 and can vary from 0 to -2 (+2 is possible in OF2). The oxidation number can decrease or increase, because of this H2O2 can act both asoxidising and reducing agent.

8.5. Assertion: ClO4– does not show disproportionation reaction.
Reason: In ClO4–, chlorine is in minimum oxidation state of +1.
Answer: (c)In ClO4–,chlorine is in maximum oxidation state of +7. So, it does not show the disproportionation reaction.

MCQs

8.1. oxidation number ofS in S2O8-2 is
(a) -2 (b) -6 (c) +2 (d) +6
Answer: (d) +6

8.2. Which element is considered to have oxidation number of +3 in all its compounds?
(a) Be (b) Li (c) Al (d) Sc
Answer: (c) Al

8.3. Choose the incorrect statement.
(a) HNO3 acts only as an oxidising agent while HNO3 can act both as reducing and oxidising agent.
(b) ClO4– does not show disproportionation reaction.
(c) Ozone acts as an oxidising agent.
(d) none of the above
Answer: (d) none of the above

8.4. Choose the incorrect statement.
(a) A redox couple consists of oxidised and reduced form of the same substance taking part in the oxidation or reduction half reaction.
(b) Oxidation involves gain of one or more electrons by a species during a reaction.
(c) half reactions that involve loss of electrons are called oxidation reactions.
(d) half reactions that involve gain of electrons are called reduction reactions.
Answer: (b)Oxidation involves gain of one or more electrons by a species during a reaction.

8.5. Best reducing agent is:
(a) He (b) Be (c) Li (d) Al
Answer: (c) Li

Questions and Answers:

8.1. Define electrochemical cell. How will you identify cathode and anode in an electrochemical cell?
Answer: Electrochemical cell is a device in which the redox reaction is carried indirectly and the decrease in free energy appears as electrical energy.
At cathode there is gain of electrons.
At anode there is loss of electrons.
In electrochemical cell anode is written on L.H.S while cathode is written on R.H.S.

8.2. What is meant by electrochemical series? What are the characteristics of electrochemical series?
Answer: Electrochemical series is the series of elements in which elements are arranged in decreasing order of their reduction potential.
Reducing power goes on increasing whereas oxidising power goes on decreasing down the series.

8.3. Name the different types of redox reactions.
Answer: (a) Combination reactions (b) Decomposition reactions (c) displacement reactions (d) Disproportionation reactions.

8.4. What are the methods of balancing redox reactions.
Answer: Two methods are used to balance chemical equations for redox processes. One of these methods is based on the change in the oxidation number of reducing agent and the oxidising agent (i.e. oxidation number method) and the other method is based on splitting the redox reaction into two half reactions — one involving oxidation and the other involving reduction (half reaction method). Both these methods are in use and the choice of their use rests with the individual using them.

8.5. What do you understand by Standard Electrode Potential? Explain.
Answer: The potential associated with each electrode is known as electrode potential. If the concentration of each species taking part in the electrode reaction is unity (if any gas appears in the electrode reaction, it is confined to 1 atmospheric pressure) and further there action is carried out at 298K, then the potential of each electrode is said to be the Standard Electrode Potential. By convention, the standard electrode potential (Eϴ) of hydrogen electrode is 0.00 volts. The electrode potential value for each electrode process is a measure of the relative tendency of the active species in the process to remainin the oxidised/reduced form. A negative E means that the redox couple is a stronger reducing agent than the H+/H2 couple. A positive Eϴ means that the redox couple is a weaker reducing agent than the H+/H2 couple.

Redox Reactions Solutions and FAQs

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

A 8.1 Let x be the oxidation number to the underlined elements in the given species:

(a) NaH₂PO₄
(+1) + 2(+1) + x + 4 (-2) = 0
x + 3 – 8 = 0
x = +5

(b) NaHSO₄
(+1) + (+1) + x + 4(-2) = 0
x – 6 = 0
x = +6
(c) H₄P₂O₇
4 (+1) + 2x + 7 (-2) = 0
2x -10 =0
x = +5

(d) K₂MnO₄
2 (+1) + x + 4 (-2) = 0
x – 6 = 0
x = +6

(e) CaO₂
2 + 2x = 0
x = -1

(f) NaBH₄
1 + x + 4 (-1) = 0 (Since H is present as hydride ion.)
x = +3

(g) H₂S₂O₇
2 (+1) + 2x + 7 (-2) = 0
x = +6

(h) KAl(SO₄)₂.12H₂O
+1 + 3 + 2x + 8 (-2) + 12 (2 x 1 - 2) = 0
x = +6

Q 8.2. What is the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

A 8.2

(a) In Kl₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K⁺[I —I <— I]^–. Here, a coordinate bond is formedbetween I₂molecule and I^– ion. The oxidation number of two iodine atoms forming the I₂ molecule is zero while that of iodine forming the coordinate bond is -1. Thus, the oxidation number of three I atoms, atoms in Kl₃ is 0, 0 and -1 respectively.
(b) By conventional method O.N. of S in H₂S₄O₆is calculated as:
2 (+1) +4x + 6) (-2) = 0
Or x = +2.5
But all the four S atoms cannot be in the same oxidation state.
So by chemical bonding method:
The O.N. of each of the S-atoms linked with each other in the middle is zero while that of each of the remaining two S-atoms is +5.

(c) By conventional method O.N. of Fe in Fe₃O₄ is calculated as:
3x + 4 (-2) = 0 or x = 8/3
By stoichiometry O.N. of Fe in Fe₃O₄is +2 and +3.

(d) By conventional method, O.N. of C in CH3CH2OH is calculated as:
2x + 6 (+1) +1 (-2) = 0 or x = -2
(e) By conventional method, O.N. of C in CH₃COOH is calculated as:
2x + 4 - 4 = 0 or x = 0
By chemical bonding method, C2 is attached to three H-atoms (less electronegative than carbon) and one -COOH group (more electronegative than carbon).
Therefore, O.N. of C₂ = 3 (+1) + x + 1 (-1) = 0 or x = -2
But C₁ is attached to one oxygen atom by a double bond, one OH (-1) and one CH3 (+1) group, therefore, O.N. of C₁ = +1 + x+ 1 (-2) + 1 (-1) = 0 or x = +2

Q 8.3 Justify that the following reactions are redox reactions:

A 8.3 (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) 4 NH₃(g) + 5 O₂ → 4NO (g) + 6H₂O (g)

A 8.3 (a) Here, O is removed from CuO, therefore, it is reduced to Cu while O is added to H₂ to form H₂O, therefore, it is oxidised. Further, O.N. of Cu decreases from +2 in CuO to 0 in Cu but that of H increases from 0 in H₂ to +1 in H₂0. Therefore, CuO is reduced to Cu but H₂ is oxidised to H₂O. Thus, this is a redox reaction.
(b) Here O.N. of Fe decreases from +3 in Fe₂O₃ to 0 in Fe while that of C increases from +2 in CO to +4 in CO₂. Further, oxygen is removed from Fe₂O₃ and added to CO, therefore, Fe₂O₃ is reduced while CO is oxidised. Thus, this is a redox reaction.
(c) Here, O.N. of B decreases from +3 in BrCl₃to -3 in B₂H₆ while that of H increases from -1 in LiAlH₄to +1 in B₂H₆. Therefore, BCl₃ is reduced while LiAlH₄ is oxidised. Further, H is added to BCl₃ but is removed from LiAlH₄, therefore, BC1₃ is reduced while LiAlH₄ is oxidised. Thus, it is a redox reaction
(d) Here, each K atom as lost one electron to form K⁺ while F₂ has gained two electrons to form two F^– ions. Therefore, K is oxidised while F₂ is reduced. Thus, it is a redox reaction.
(e) Here, the oxidation number of N increases from –3 in NH₃ to +2 in NO. On the other hand, the oxidation number of O₂ decreases from 0 in O₂ to –2 in NO and H₂O i.e., O₂ is reduced. Hence, the given reaction is a redox reaction.

Q 8.4 Fluorine reacts with ice and results in the change, Justify that this reaction is a redox reaction.

H₂O(s) + F₂ (g) ——> HF (g) + HOF (g) A 8.4 Writing the O.N. of each atom above its symbol, we have,

here, the O.N. of F decreases from 0 in F ₂ to -1 in HF and increases from 0 in F ₂ to +1 in HOF. Therefore, F ₂ is both reduced as well as oxidised. Thus, it is a redox reaction and more specifically, it is a disproportionation reaction.

Q 8.5. Calculate the oxidation number of sulphur, chromium and nitrogen in (equation) Suggest structure of these compounds. Count for the fallacy.

H₂SO₅, Cr₂O₇^2-and NO₃^- A 8.5

(a) H₂SO₅ by conventional method.
Let x be the oxidation number of S
2(+1) + x + 5(−2) = 0
x = +8
+8 oxidation state of S is not possible as S cannot have oxidation number more than 6. The fallacy is overcome if we calculate the oxidation number from its structure HO−S(O₂)−O−O−H.
−1+X+2(−2)+2(−1)+1=0
x=+6
(b) Dichromate ion
Let x be the oxidation number of Cr in dichromate ion
2x+7(−2)=−2
x=+6
Hence the oxidation number of Cr in dichromate ion is +6. This is correct and there is no fallacy.
(c) Nitrate ion, by conventional method
Let x be the oxidation number of N in nitrate ion.
x+3(−2)=−1
From the structure O−−N+(O)−O−
x+1(−1)+1(−2)+1(−2)=0
x=+5
Thus there is no fallacy.

Q 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.

A 8.6 (a) HgCl ₂, (b) NiSO ₄, (c)SnO ₂ (d) Tl ₂SO ₄, (e) Fe ₂(S0 ₄) ₃, (f) Cr ₂O ₃

Q 8.7. Suggest a list of substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

A 8.7

Substance	Oxidation number of C	Substance	Oxidation number of N
CH₂Cl₂	0	N₂	0
FC≡CF	+1	N₂O	+1
HC≡CH	-1	N₂H₂	-1
CHCl₃,CO	+2	NO	+2
CH₃Cl	-2	N₂H₄	-2
Cl₃C−CCl₃	+3	N₂O₃	+3
H₃C−CH₃	-3	NH₃	-3
CCl₄,CO₂	+4	NO₂	+4
CH₄	-4	N₂O₅	+5

Q 8.8. While sulphur dioxide and hydrogen peroxide can act as an oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why?

A 8.8 (i) In SO₂, O.N. of S is +4. In principle, S can have a minimum O.N. of -2 and maximum of +6. Therefore, S in SO₂ can either decrease or increase its O.N. and hence can act both as an oxidising as well as a reducing agent.
(ii) In H₂O₂, the O.N. of O is -1. In principle, O can have a minimum O.N. of -2 and maximum of zero (+1 is possible in O₂F₂and +2 in OF₂). Therefore, O in H₂O₂ can either decrease its O.N. from -1 to -2 or can increase its O.N. from -1 to zero. Therefore, H₂O₂ acts both as an oxidising as well as a reducing agent.
(iii) In O₃, the O.N. of O is zero. It can only decrease its O.N. from zero to -1 or -2, but cannot increase to +2. Therefore, O₃ acts only as an oxidant.
(iv) In HNO₃, O.N. of N is +5 which is maximum. Therefore, it can only decrease its O.N. and hence it acts as an oxidant only.

Q 8.9 Consider the reactions, Also suggest a technique to investigate the path of above (a) and (b) redox reactions.

(a) 6CO₂(g) + 6H₂O(l) →C₆H₁₂O₆(s) + 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(s) + 6H₂O(l) + 6O₂(g)
(b) O₃(g) + H₂O₂ (l) →H₂O(l) + O₂(g) + O₂(g) A 8.9

Q 8.10. The compound AgF2 is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why?

A 8.10 In AgF ₂, oxidation state of Ag is +2 which is very unstable. Therefore, it quickly accepts an electron to form the more stable +1 oxidation state.
Ag ²⁺ + e ^– →Ag ⁺
Therefore, AgF ₂, if formed, will act as a strong oxidising agent.

Q 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 oxidising agent is in excess. Justify this statement giving three illustrations.

A 8.11 The three examples are:

(i) When excess P₄ (reducing agent) reacts with F₂(oxidizing agent), PF₃ is produced in which P has +3 oxidation number
P₄ (excess) + F₂ → PF₃
But if fluorine is in excess, PF₅ is formed in which P has oxidation number of +5.
P₄ + F₂ (excess) → PF₅
(ii) Oxidizing agent is oxygen and reducing agent is K. When excess K reacts with oxygen, K₂O is formed in which oxygen has oxidation number of -2.
4K (excess) + O₂ → 2K₂O
But if oxygen is in excess, then K₂O₂ is formed in which O has oxidation number of -1.
2K + O₂ (excess) → K₂O₂
(iii) The oxidizing agent is oxygen and the reducing agent is C. When an excess of C reacts with oxygen, CO is formed in which C has +2 oxidation number.
C (excess) + O₂ → CO
When excess of oxygen is used, CO₂ is formed in which C has +4 oxidation number.
C + O₂ (excess) → CO₂

Q 8.12 How do you account 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?

A 8.12 (a) Toluene can be oxidised to benzoic acid in acidic, basic and neutral media according to the following redox equations:

In the laboratory, benzoic acid is usually prepared by alcoholic KMnO₄ oxidation of toluene. However, in industry alcoholic KMnO₄ is preferred over acidic or alkaline KMnO₄ because of the following reasons:
(i) The cost of adding an acid or the base is avoided because in the neutral medium, the base (OH^- ions) are produced in the reaction itself.
(ii) Since reactions occur faster in homogeneous medium than in heterogeneous medium, therefore, alcohol helps in mixing the two reactants, i.e., KMnO₄ (due to its polar nature) and toluene (because of its being an organic compound).
(b) KBr + H₂SO₄ → KHSO₄ + HBr
HBr is a strong reducing agent which further reduces H₂SO₄ as follows
2HBr + H₂SO₄ → Br₂ + SO₂ + H₂O
But in case of mixture containing Chloride the reaction ends up on forming HCl as HCl is a weak reducing agent which can't further reducethe sulphuric acid further.
KCl+H₂SO₄→KHSO₄+HCl (colourless pungent smell gas)

Q 8.13. Identify the substance oxidised, reduced, oxidising agent and reducing agent for each of the following reactions.

Substance oxidised	Substance reduced	Oxidising agent	Reducing agent
(a) C₆H₁₂O₆	AgBr	AgBr	C₆H₁₂O₆
(b) HCHO	[Ag(NH₃)₂]⁺	[Ag(NH₃)₂]⁺	HCHO
(c) HCHO	Cu²⁺	Cu²⁺	HCHO
(d) N₂H₄	H₂O₂	H₂O₂	N₂H₄
(e) Pb	PbO₂	PbO₂	Pb

Q 8.14. Consider the reactions,why does the same reductant, thiosulphate react differently with iodine and bromine?

A 8.14 The average O.N. of S in S ₂O _3^2- is +2 while in S ₄O _6^2- it is + 2.5. The O.N. of S in SO _4^2- is +6. Since Br ₂ is a stronger oxidising agent than I ₂, it oxidises S of S ₂O _3^2- to a higher oxidation state of +6 and hence forms SO ₄ _^2- ion. I ₂, however, being weaker oxidising agent oxidises S of S ₂O ₃ _^2- ion to a lower oxidation of +2.5 in S ₄O _6^2- ion. It is because of this reason that thiosulphate reacts differently with Br ₂ and I ₂.

Q 8.15. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.

A 8.15

Fluorine oxidizes chloride ion to chlorine, bromide ion to bromine and iodide ion to iodine respectively.
F₂ + 2Cl⁻ → 2F⁻ + Cl₂
F₂ + 2Br⁻ → 2F⁻ + Br₂
F₂ + 2I⁻ → 2F⁻ + I₂.
Chlorine oxidizes bromide ion to bromine and iodide ion to iodine.
Cl₂ + Br⁻ → 2Cl⁻ + Br₂
Cl₂ + I⁻ → 2Cl⁻ + I₂
Bromine oxidizes iodide ion to iodine.
Br₂ + I⁻ → 2Br⁻ + I₂
But bromine and chlorine cannot oxidize fluoride to fluorine. Hence, fluorine is the best oxidizing agent amongst the halogens. The decreasing order of the oxidizing power of halogens is F₂>Cl₂>Br₂>I₂.
HI and HBr can reduce sulphuric acid to sulphur dioxide but HCl and HF cannot. Thus, HI and HBr are stronger reducing agents than HCl and HF.
2HI+H₂SO₄→I₂+SO₂+2H₂O
2HBr+H₂SO₄→Br₂+SO₂+2H₂O
Iodide ion can reduce Cu(II) to Cu(I) but bromide cannot.
4I⁻+2Cu²⁺→Cu₂I₂+I₂
Hence, among the hydrohalic compounds, hydroiodic acid is the best reductant. The reducing power of hydrohalic acids is HF

Q 8.16. Why does the following reaction occur? What conclusion about the compound Na4XeO6 (of which XeO64– is a part) can be drawn from the reaction?

XeO₆^4– (aq) + 2F^– (aq) + 6H⁺(aq) → XeO₃(g)+ F₂(g) + 3H₂O(l) A 8.16 XeO ₆ ⁴⁻oxidizes F ⁻ and F ⁻ reduces XeO ₆ ⁴⁻.
Hence, the given reaction occurs.
The oxidation number of Xe decreases from +8 to +6. The oxidation number of F increases from -1 to 0.
Thus, Na ₄XeO ₆is a stronger oxidising agent than F ⁻.

Q 8.17. Consider the reactions, What inference do you draw about the behaviour of Ag+ and Cu2+ from these 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
A 8.17 Reactions (a) and (b) indicate that H ₃PO ₂ (hypophosphorous acid) is a reducing agent and thus reduces both AgNO ₃ and CuSO ₄ to Ag and Cu respectively. Conversely, both AgNO ₃ and CuSO ₄ act as oxidising agent and thus oxidise H ₃PO ₂to H ₃PO ₄ (orthophosphoric acid) Reaction (c) suggests that [Ag(NH ₃) ₂] ⁺ oxidises C ₆H ₅CHO (benzaldehyde) to C ₆H ₅COO ^– (benzoate ion) but reaction (d) indicates that Cu ²⁺ ions cannot oxidise C ₆H ₅CHO to C ₆H ₅COO ^–. Therefore, from the above reactions, we conclude that Ag ⁺ ion is a strong deoxidising agent than Cu ²⁺ ion.

Q 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) +H₂SO₄^–(in acidic solution)
(c) H₂O₂(aq) + Fe²⁺(aq) → Fe³⁺(aq) + H₂O(l) (in acidic solution)
(d) Cr₂O₇^2-(aq) + SO₂ (g) → Cr³⁺(aq) + SO₄^2-(aq) (in acidic solution)
A 8.18 (a) The balanced half reaction equations are:
Oxidation half equation:

I⁻(aq) → I₂(s) -------------------------------- (i)
Reduction half reaction equation:
MnO₄⁻(aq) → MnO₂(aq) --------------------- (ii)
Balance I atoms and charges in the oxidation half reaction.
2I⁻(aq) → I₂(s) + 2e⁻
In the reduction half reaction, the oxidation number of Mn changes from +7 to +4. Hence, add 3 electrons to reactant side of the reaction.
MnO₄⁻(aq) + 3e⁻→ MnO₂(aq)
Balance charge in the reduction half reaction by adding 4 hydroxide ions to product side.
MnO4⁻(aq) + 3e⁻→ MnO₂(aq)+4OH⁻
To balance O atoms, add 2 water molecules to reactant side.
MnO4⁻ (aq)+ 3e⁻+2H₂O→MnO₂(aq) + 4OH⁻
To equalize the number of electrons, multiply the oxidation half reaction by 3 and multiply the reduction half reaction by 2.
6I⁻ (aq) → 3I₂(s)+6e⁻
2MnO₄⁻ (aq) + 6e⁻+4H₂O → 2MnO₂(aq)+8OH⁻
Add two half-cell reactions to obtain the balanced equation.
2MnO₄⁻(aq) + 6I⁻(aq) + 4H₂O₂ (l) → 2MnO₂(s) + 3I₂ (s) + 8OH⁻
(b) The balanced half reaction equations are:
Oxidation half equation:
SO₂(g) + 2H₂O(l) → HS0₄^– (aq) + 3H⁺(aq) +2e^–…(i)
Reduction half equation:
MnO₄^–(aq) + 8H⁺(aq) + 5e^– → Mn²⁺(aq) + 4H₂O(l) ………..(ii)
Multiply Eq. (i) by 3 and Eq. (ii) by 2 and add, we have,
2MnO₄^–(aq) + 5S0₂(g) + 2H₂0(l) + H⁺(aq) → 2Mn²⁺(aq) + 5HSO₄^–(aq)
(c) Oxidation half equation: Fe²⁺(aq) → Fe³⁺(aq) + e^–…(i)
Reduction half equation: H₂O₂(aq) + 2H⁺(aq) + 2e^–→ 2H₂O(l) …(ii)
Multiply Eq. (i) by 2 and add it to Eq. (ii), we have,
H₂O₂(aq) +2Fe²⁺(aq) +2H⁺(aq) → 2Fe³⁺(aq) + 2H₂O(l)
(d)Oxidation half equation:
SO₂(g) + 2H₂O(l) → SO₄^2-(aq) + 4H⁺(aq) + 2 e^– …(i)
Reduction half equation:
Cr₂O₇^2–(aq) + 14H⁺(aq) + 6e^– → 2Cr³⁺(aq) + 7H₂O(l) …(ii)
Multiply Eq. (i) by 3 and add it to Eq. (ii), we have,
Cr₂O₇^2–(aq) + 3SO₂(q) + 2H⁺(aq) → 2Cr³⁺(aq) + 3SO₄^2-(aq) + H₂O(l)

Q 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) + H₂PO₂^–(aq)
(b) N₂H₄(l) + ClO^–(aq) → NO(g) + Cl^-(aq)
(c) Cl₂O₇(g) + H₂O₂(aq) → ClO₂^–(aq) + O₂(g) + H⁺ A 8.19

(a) Oxidation number method:
The oxidation number of P decreases from 0 to -3 and increases from 0 to +2. Hence, P₄ is oxidizing as well as reducing agent.
During reduction, the total decrease in the oxidation number for 4 P atoms is 12.
During oxidation, total increase in the oxidation number for 4 P atoms is 4.
The increase in the oxidation number is balanced with decrease in the oxidation number by multiplying H₂PO₂⁻ with 3.
P₄(s) + OH⁻(aq) → PH₃(g) + 3H₂PO₂⁻(aq)
To balance O atoms, multiply OH⁻ ions by 6.
P₄(s) + 6OH⁻(aq) → PH₃(g) + 3H₂PO₂⁻(aq)
To balance H atoms, 3 water molecules are added to L.H.S and 3 hydroxide ions on R.H.S.
P₄(s) + 6OH⁻(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂⁻(aq) + 3OH⁻(aq)
Subtract 3 hydroxide ions from both sides
P₄(s) + 3OH⁻(aq) + 3H₂O(l) → PH₃(g) + 3H₂PO₂⁻(aq)

Ion electron method:
The oxidation half reaction is
P₄(s) → H₂PO₂⁻(aq).
The P atom is balanced.
P₄(s) → 4H₂PO₂⁻(aq)
The oxidation number is balanced by adding 4 electrons on RHS.
P₄(s) → 4H₂PO₂⁻(aq) + 4e⁻
The charge is balanced by adding 8 hydroxide ions on LHS.
P₄(s)+8OH⁻(aq)→4H₂PO₂⁻(aq)
The O and H atoms are balanced.
The reduction half reaction is P₄(s)→PH₃(g).
The oxidation number is balanced by adding 12 electrons on LHS.
P₄(s)+12e⁻→PH₃(g)
The charge is balanced by adding 12 hydroxide ions on RHS.
P₄(s)+12e⁻→PH₃(g)+12OH⁻
The oxidation half reaction is multiplied by 3 and the reduction half reaction is multiplied by 2.
The half reactions are then added to obtain balanced chemical equation.
P₄(s)+3OH⁻(aq)+3H₂O(l)→PH₃(g)+3H₂PO₂⁻(aq)
(b) The oxidation number of N increases from – 2 in N₂H₄ to + 2 in NO and the oxidation number of Cl decreases from + 5 in CIO⁻₃ to – 1 in Cl^–. Hence, in this reaction, N₂H₄ is the reducing agent and CIO⁻₃ is the oxidizing agent.
Ion - electron method:
The oxidation half equation is:
−2 +2
N₂H₄(l)⟶NO(g)
The N atoms are balanced as:
N₂H₄(l)⟶2NO(g)+8e⁻
The charge is balanced by adding 8 OH^– ions as:
N₂H₄(I)+8OH⁻(aq)⟶2NO(g)+8e⁻
The O atoms are balanced by adding 6H2O as:
N₂H₄(I)+8OH⁻(aq)⟶2NO(g)+6H₂O(I)+8e⁻......(i)
The reduction half equation is:
+5 −1
CIO⁻₃(aq)⟶CI⁻(aq)
The oxidation number is balanced by adding 6 electrons as:
CIO⁻₃(aq)+6e⁻⟶Cl⁻(aq)
The charge is balanced by adding 6OH– ions as:
ClO⁻₃(aq)+6e⁻⟶Cl⁻(aq)+6OH⁻(aq)
The O atoms are balanced by adding 3H2O as:
CIO⁻₃(aq)+3H₂O(I)+6e⁻⟶Cl⁻(aq)+6OH⁻(aq).......(ii)
The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:
3N₂H₄(I)+4ClO⁻₃(aq)⟶6NO(g)+4Cl⁻(aq)+6H₂O(I)
Oxidation number method:
Total decrease in oxidation number of N=2×4=8
Total increase in oxidation number of Cl=1×6=6
On multiplying N₂H₄ with 3 and CIO⁻₃ with 4 to balance the increase and decrease in O.N.,
we get:
3N₂H₄(I)+4CIO⁻₃(aq)⟶NO(g)+CI⁻(aq)
The N and Cl atoms are balanced as:
3N₂H₄(I)+4CIO⁻₃(aq)⟶6NO(g)+4CI⁻(aq)
The O atoms are balanced by adding 6H₂O as:
3N₂H₄(I)+4CIO⁻₃(aq)⟶6NO(g)+4CI⁻(aq)+6H₂O(I)
(c) This is the required balanced equation.
+7 -1 +3 0
Cl₂O₇(g)+H₂O₂(aq) ⟶CIO⁻₂(g)+O₂(g) + H⁺
The oxidation number of Cl decreases from + 7 in Cl₂O₇ to + 3 in CIO⁻₂ and the oxidation number of O increases from -1 inH₂O₂ to zero in O₂. Hence, in this reaction, Cl₂O₇ is theoxidizing agent and H₂O₂ is the reducing agent.
Ion–electron method:
The oxidation half equation is:
−1 0
H₂O₂(aq)⟶O₂(g)
The oxidation number is balanced by adding 2 electrons as:
H₂O₂(aq)+⟶O₂(g)+2e⁻
The charge is balanced by adding 2OH⁻ ions as:
H₂O₂(aq)+2OH⁻(aq)⟶O₂(g)+2e⁻
The oxygen atoms are balanced by adding 2H₂O as:
H₂O₂(aq)+2OH⁻(aq)⟶O₂(g)+2H₂O(I)+2e⁻(i)
The reduction half equation is:
+7 +3
Cl₂O₇(g)⟶CIO⁻₂(aq)
The Cl atoms are balanced as:
Cl₂O₂(g)⟶2CIO⁻₂(aq)
The oxidation number is balanced by adding 8 electrons as:
Cl₂O₇(g)+8e⁻⟶2CIO⁻₂(aq)
The charge is balanced by adding 6OH⁻ as:
Cl₂O₇(g)+8e⁻⟶2CIO⁻₂(aq)+6OH⁻(aq)
The oxygen atoms are balanced by adding 3H2O as:
Cl₂O₇(g)+3H₂O + 8e⁻⟶2CIO⁻₂(aq)+6OH⁻(aq) (ii)
The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as:
Cl₂O₇(g)+4 H₂O₂(aq) +2OH⁻(aq)⟶2CIO⁻₂(aq)+4O₂(aq)+5H₂O(I)
Oxidation number method:
Total decrease in oxidation number of Cl₂O₇=4×2=8
Total increase in oxidation number of H₂O₂=2×1=2
By multiplying H₂O₂ and O2 with 4 to balance the increase and decrease in the oxidation number, we get:
Cl₂O₇(g)+4H₂O₂(aq)⟶ CIO⁻₂(aq)+4O₂(g)
The Cl atoms are balanced as:
Cl₂O₇(g)+4H₂O₂(aq)⟶2CIO⁻₂(aq)+4O₂(g)
The O atoms are balanced by adding 3 H₂O as:
Cl₂O₇(g)+4H₂O₂(aq)⟶2CIO⁻₂(aq)+4O₂(g)+3H₂O(I)
The H atoms are balanced by adding 2OH⁻ and 2 H₂O as:
Cl₂O₇(g)+4H₂O₂(aq)+2OH⁻(aq)⟶2CIO⁻₂(g)+4O₂(g)+5 H₂O (I)
This is the required balanced equation.

Q 8.20. What sorts of information can you draw from the following reaction?

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

A 8.20 Let x be the O.N. of C.
O.N. of C in cyanogen, (CN)₂ = 2 (x – 3) = 0 or x = +3
O.N. of C in cyanide ion, CN^- = x – 3 = -1 or x = +2
O.N. of C in cyanate ion, CNO =x-3-2 = -l or x: = +4
The four information about the reaction are:
(i) The reaction involves decomposition of cyanogen, (CN)₂ in the alkaline medium to cyanide ion, CN and cyanate ion,CNO^–.
(ii) The O.N. of C decreases from +3 in (CN)₂ to +2 in CN^–ion and increases from +3 in(CN)₂ to +4 in CNO^– ion. Thus, cyanogen is simultaneously reduced to cyanide ion and oxidised to cyanate ion.
(iii) It is an example of a redox reaction in general and a disproportionation reaction in particular.
(iv) Cyanogen is a pseudohalogen (behaves like halogens) while cyanide ion is a pseudohalide ion (behaves like halide ion).

Q 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.

A 8.21

The unbalanced chemical reaction is:
Mn³⁺(aq) → Mn²⁺(aq) + MnO₂(s) + H⁺(aq)
The oxidation half reaction is,
Mn³⁺(aq) → MnO₂(s).
To balance oxidation number, one electron is added on R.H.S.
Mn³⁺(aq) → MnO₂(s) + e⁻
4 protons are added to balance the charge.
Mn³⁺(aq) → MnO₂(s) + 4H⁺(aq) + e⁻
2 water molecules are added to balance O atoms.
The reduction half reaction is Mn³⁺(aq) → Mn²⁺(aq).
An electron is added to balance oxidation number.
Mn³⁺(aq) + e⁻ → Mn²⁺(aq)
Two half-cell reactions are added to obtain balanced chemical equation.
2Mn³⁺(aq) + 2H₂O(l) → Mn2+(aq) + MnO₂(s) + 4H⁺(aq)

Q 8.22. Consider the elements: Cs, Ne, I, F (a) Identify the element that exhibits negative oxidation state. (b) Identify the element that exhibits positive oxidation state. (c) Identify the element that exhibits both positive and negative oxidation states. (d) Identify the element which neither exhibits negative nor does the positive oxidation state.

A 8.22

(a) F. Fluorine being the most electronegative element shows only a -ve oxidation state of -1.
(b) Cs. Alkali metals because of the presence of a single electron in the valence shell, exhibit an oxidation state of +1.
(c) I. Because of the presence of seven electrons in the valence shell, I shows an oxidation state of -1 (in compounds of I with more electropositive elements such as H, Na, K, Ca, etc.) or an oxidation state of +1 compounds of I with more electronegative elements, i.e., O, F, etc.) and because of the presence of d-orbitals it also exhibits +ve oxidation states of +3, +5 and +7.
(d) Ne. It is an inert gas (with high ionization enthalpy and high positive electron gain enthalpy) and hence it neither exhibits -ve nor +ve oxidation states.

Q 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.

A 8.23 The balanced chemical reaction for the redox reaction between chlorine and sulphur dioxide is:
Cl ₂ + SO ₂ + 2H ₂O → 2Cl ⁻ + SO ₄ ²⁻+ 4H ⁺

Q 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 show disproportionation reaction.

A 8.24

a) Phosphorous, chlorine and sulphur are the non metals which can show disproportionation reaction. The reactions shown are:
P₄ + 3OH⁻ + 3H₂O → PH₃ + 3H₂PO₂⁻
Cl₂ + 2OH⁻ → Cl⁻ + ClO⁻ + H₂O
S₈+12OH⁻→4S²⁻+2S₂O₃²⁻+6H₂O
(b) Copper, gallium and indium are the metals that show disproportionation reaction.
The reactions are shown below.
2Cu⁺ → Cu²⁺ + Cu
3Ga⁺ → Ga³⁺ + 2Ga
3In⁺ → In³⁺ + 2In

Q 8.25. In Ostwald’s process for the manufacture of nitric add, 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 10.0 g of ammonia and 20.0 g of oxygen?

A 8.25

The balanced equation for the reaction is:
4NH₃(g)+5O₂(g)→4NO(g)+6H₂O(g)
The molar masses of ammonia and oxygen are 17 g/mol and 32 g/mol respectively.
68 g of NH₃ will react with O₂ = 160 g.
Therefore, 10 g of NH₃ will react with O₂ = 160/68 x 10 g = 23.6 g
But the amount of O₂ which is actually available is 20.0 g which is less than the amount which is needed. Therefore, O₂ is the limiting reagent and hence calculations must be based upon the amount of O₂ taken and not on the amount of NH₃ taken. From the equation,
160 g of O₂ produce NO = 120 g
Therefore, 20 g of O₂ will produce NO =120/160 x 20 = 15 g

Q 8.26. Using the standard electrode potentials given in 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 Fe³⁺(aq). (Advance) A 8.26 (i) When emf is positive, the reaction is feasible.

E^o_cell = E^o_Fe−E^o_I2=0.77−0.54=0.23V
Reaction is feasible.
(ii) E^ocell=E^o_Ag−E^o_Cu=0.80−0.34=0.46V
Reaction is feasible.
(iii) E^ocell=E^o_Fe−E^o_Cu=0.77−0.34 =0.43 V
Reaction is feasible.
(iv) E^ocell=E^o_Fe−E^o_Ag=0.77−0.80=−0.03 V
Reaction is not feasible.
(v) E^ocell=E^o_Br2−E_Fe=1.09−0.77=0.32 V
Reaction is feasible.

Q 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 of silver nitrate with platinum electrodes. (iii) A dilute solution of H2SO4with platinum electrodes. (iv) An aqueous solution of CuCl2 with platinum electrodes.

A 8.27

(i) An aqueous solution of AgNO₃ with silver electrodes.
At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.
At anode: Silver anode will dissolve to form silver ions in the solution.
Ag → Ag⁺ + e⁻
(ii) An aqueous solution of AgNO₃ with platinum electrodes.
At cathode: Silver ions have lower discharge potential than hydrogen ions. Hence, silver ions will be deposited in preference to hydrogen ions.
At anode: Hydroxide ions having lower discharge potential will be discharged in preference to nitrate ions. Hydroxide ions will decompose to give oxygen.
4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
(iii) A dilute solution of H₂SO₄ with platinum electrodes.
At cathode: 2H⁺ + 2e⁻ → H₂(g)
At anode: Hydroxide ions having lower discharge potential will be discharged in preference to sulphate ions. Hydroxide ions will decompose to give oxygen.
4OH⁻(aq) → 2H₂O(l) + O₂(g) + 4e⁻
(iv) An aqueous solution of CuCl₂ with platinum electrodes
At cathode: Cupric ions will be reduced in preference to protons
Cu²⁺ + 2e⁻ → Cu
At anode: Chloride ions will be oxidized in preference to hydroxide ions
2Cl⁻ → Cl₂ + 2e⁻

Q 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.

A 8.28 Based on the relative positions of these metals in the activity series, the correct order is Mg, Al, Zn, Fe, Cu.

Q 8.29. Given the standard electrode potentials, Arrange these metals in increasing order of their reducing power.

K⁺/K = -2.93 V, Ag⁺/Ag = 0.80 V, Hg²⁺/Hg = 0.79 V, Mg²⁺/Mg = -2.37 V, Cr³⁺/Cr = -0.74 V.
A 8.29 Lower the electrode potential, better is the reducing agent. Since the electrode potentials increase in the order: K ⁺/K (-2.93 V), Mg ²⁺/Mg (-2.37 V), Cr ³⁺/Cr (-0.74 V), Hg ²⁺/Hg (0.79 V), Ag ⁺/Ag (0.80 V), therefore, reducing power of metals decreases in the same order, i.e., K, Mg, Cr, Hg, Ag.

Q 8.30. Depict the galvanic cell in which the reaction, Zn(s) + 2Ag+(aq) → Zn^2+(aq) + 2Ag(s) takes place. Further show: (i) which of the electrode is negatively charged. (ii) the carriers of current in the cell and (iii) individual reaction at each electrode. (Advance)

A 8.30

The given redox reaction can be depicted as
Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)
Since Zn gets oxidised to Zn²⁺ ions, and Ag⁺ gets reduced to Ag metal, therefore,
Oxidation occurs at the zinc electrode and reduction occurs at the silver electrode. Thus, galvanic cell corresponding to the above redox reaction may be depicted as:
Zn|Zn²⁺ (aq) || Ag⁺(aq) | Ag
Zinc electrode is negatively charged because oxidation occurs at the zinc electrode (i.e. electrons accumlulate on the zinc electrode)
The ions carry current. The electrons flow from Zn to Ag electrode while the current flows from Ag to Zn electrode.
Reaction occurring at each electrode are:
Zn(s) → Zn²⁺ (aq) + 2e^-
Ag⁺ (aq) + e^- → Ag(s)