NCERT Solutions Class 11 Chemistry Chapter 8 Redox Reactions

Question. Assign oxidation number to the underlined elements in each of the following species:
(a) NaH₂ P O₄
(b) NaH SO₄
(c) H₄ P₂O₇
(d) K₂MnO₄
(e) CaO₂
(f) NaBH4
(g) H₂S2O₇
(h) KAl(SO₄)₂ ⋅ 12H₂O
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
NaH₂PO₄

(f) 9NaBH₄
Then, we have
1(+1) +1(x)+4(−1)=0
⇒1+x−4=0
⇒x=+3
Hence, the oxidation number of B is + 3.

(g) H₂S₂O₇
Then, we have
2(+1) +2(x) +7(−2)=0
⇒2 +2x− 14=0
⇒2x=12
⇒x=+6
Hence, the oxidation number of S is + 6.

(h ) KAl (SO₄)2 ⋅ 12H₂O
Then, we have
1(+1) 1(+3)+ 2(x) +8(−2) +24(+1) +12(−2) =0
⇒1 +3 +2x −16 +24 −24=0
⇒2x=12
⇒x=+6
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
1(+1) +1(+3) +2(x) +8(−2) =0
⇒1+3+2x−16=0
⇒2x=12
⇒x=+6
Hence, the oxidation number of S is + 6.
oxidation number of P is +5.

Question. What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?
(a) Kl₃
(b) H₂S₄O₆
(c) Fe₃O₄
(d) CH₃ CH₂ OH
(e) CH₃ COOH
Answer : (a) Kl₃
In  Kl₃,the oxidation number (O.N.) of K is +1. Hence, the average oxidation number of I is −1/3 .However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI₃to find the oxidation states.
In a Kl₃ molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule. Hence, in a  Kl₃ molecule, the O.N. of the two I atoms forming the l₂> molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is as 1.

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) Fe₃O₄
On taking the O.N. of O as 2, the O.N. of Fe is found to be +2(2/3).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) CH₃ CH₂ OH
2(x) +6(+1) +1(−2)=0
2x+4 = 0
x = −2
Hence, the O.N. of C is −2

(e) CH₃ COOH
2(x) +4(+1) +2(−2)=0
2x = 0
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 CH₃ COOH.

Question. Justify that the following reactions are redox reactions:
(a) CuO(s) + H₂(g) → Cu(s) + H₂O(g)
(b) Fe₂O_3(s) + 3cO_(g) → 2Fe_(s) + 3CO_2(g)
(c) 4BCl_3(g) + 3LiAlH_4(s)→ 2B2H_6(g) + 3LiCl_(s)+3AlCl_3(s)
(d) 2K(s) + F_2(g) → 2K + F−_(s)
(e) 4NH_3(g) + 5O_2(g) → 4NO_(g)+6H₂O_(g)
Answer : (a) CuO(s)+H_2(g) longrightarrow Cu(s)+H₂O_(g)
Let us write the oxidation number of each element involved in the given reaction as:

Here, the oxidation number of Cu decreases from +2 in CuO to 0 in Cu i.e., CuO is reduced to Cu. Also, the oxidation number of H increases from 0 in H₂ to +1 in H₂O i.e., H₂ is oxidized to H₂O .Hence, this reaction is a redox reaction.

(b) Fe₂O_3(s) + 3CO_(g)⟶ 2Fe_(s) + 3CO_2(g)
Let us write the oxidation number of each element in the given reaction as:

(e) 4NH_3(g) + 5O₂(g) ⟶4NO_(g) + 6H₂O_(g)
The oxidation number of each element in the given reaction can be represented as:
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.

Question. 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:
Answer : Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

Here, we have observed that the oxidation number of F increases from 0 in F₂ to +1 in HOF. Also, the oxidation number decreases from 0 in F₂ to –1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

Question. Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅,Cr₂O^2 −₇  and NO⁻₃. Suggest structure of these compounds. Count for the fallacy.
Answer : (i) H₂SO₅
2(+1) + 1(x) +5(−2)=0
⇒2 +x −10=0
⇒x=+8
However, the O.N. of S cannot be +8. S has six valence electrons. Therefore, the O.N. of S cannot be more than +6.
The structure of H₂SO₅ is shown as follows:

Question. Write the formula 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
Answer : (a)Mercury (II) chloride: HgCl₂
(b)Nickel (II) sulphate: NiSO₄
(c)Tin (IV) oxide: SnO₂
(d)Thallium (I) sulphate: TI₂SO₄
(e)Iron (III) sulphate: Fe₂(SO₄)₃
(f)Chromium (III) oxide: Cr₂O₃

Question. Suggest a list of the substances where carbon can exhibit oxidation states from –4 to +4 and nitrogen from –3 to +5.
Answer : The substances where carbon can exhibit oxidation states from –4 to +4 are listed in the following table.
The substances where nitrogen can exhibit oxidation states from –3 to +5 are listed in the following table.

The substances where nitrogen can exhibit oxidation states from –3 to +5 are listed in the following table.

Question. 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?
Answer : In sulphur dioxide (SO₂), the oxidation number (O.N.) of S is +4 and the range of the O.N. that S can have is from +6 to –2.
Therefore, SO₂ can act as an oxidising as well as a reducing agent.
In hydrogen peroxide (H₂O₂), the O.N. of O is –1 and the range of the O.N. that O can have is from 0 to –2. O can sometimes also attain the oxidation numbers +1 and +2. Hence, H₂O₂ can act as an oxidising as well as a reducing agent.
In ozone (O₃), the O.N. of O is zero and the range of the O.N. that O can have is from 0 to –2. Therefore, the O.N. of O can only decrease in this case. Hence, O₃ acts only as an oxidant.
In nitric acid (HNO₃), the O.N. of N is +5 and the range of the O.N. that N can have is from +5 to –3. Therefore, the O.N. of N can only decrease in this case. Hence, HNO₃ acts only as an oxidant.

Question. Consider the reactions:
(a) 6 CO_2(g) + 6H₂O(l) → C₆ H₁₂ O6(aq) + 6O_2(g)
(b) O₃(g) + H₂O₂(l) → H₂O(l) + 2O_2(g)
Why it is more appropriate to write these reactions as
(a) 6CO_2(g) + 12H₂O(l) → C₆ H₁₂ O_6(aq) + 6H₂O(l) + 6O_2(g)
(b) O₃(g) + H₂O₂ (l) → H₂O(l) + O_2(g) + O_2(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.
Answer : (a)The process of photosynthesis involves two steps.
Step 1:
H₂O decomposes to give H₂ and O₂.
2H₂O(l) ⟶ 2H_2(s) + O_2(s)
Step 2:
The H₂ produced in step 1 reduces CO₂, thereby producing glucose (C₆H₁₂O₆) and H₂O.
Now, the net reaction of the process is given as:

It is more appropriate to write the reaction as given above because water molecules are also produced in the process of photosynthesis.
The path of this reaction can be investigated by using radioactive H₂O₁₈ in place of H₂O.

(b) O₂ is produced from each of the two reactants O₃ and H₂O₂. For this reason, O₂ is written twice.
The given reaction involves two steps. First, O₃ decomposes to form O₂ and O. In the second step, H₂O₂ reacts with the O produced in the first step, thereby producing H₂O and O₂.

The path of this reaction can be investigated by using H₂O¹⁸₂ or O¹⁸₃.

Question. The compound AgF₂ is an unstable compound. However, if formed, the compound acts as a very strong oxidizing agent. Why?
Answer : The oxidation state of Ag in AgF₂ is +2. But, +2 is an unstable oxidation state of Ag. Therefore, whenever AgF₂ is formed, silver readily accepts an electron to form Ag⁺. This helps to bring the oxidation state of Ag down from +2 to a more stable state of +1. As a result, AgF₂ acts as a very strong oxidizing agent.

Question. 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.
Answer : 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. This can be illustrated as follows:

(i) P₄ and F₂ are reducing and oxidising agents respectively.
If an excess of P₄ is treated with F₂, then PF₃ will be produced, wherein the oxidation number (O.N.) of P is +3.

Question. 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?
Answer : (a) In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant because of the following reasons.
(i) In a neutral medium, OH– ions are produced in the reaction itself. As a result, the cost of adding an acid or a base can be reduced.
(ii) KMnO₄ and alcohol are homogeneous to each other since both are polar. Toluene and alcohol are also homogeneous to each other because both are organic compounds. Reactions can proceed at a faster rate in a homogeneous medium than in a heterogeneous medium. Hence, in alcohol, KMnO₄ and toluene can react at a faster rate.
The balanced redox equation for the reaction in a neutral medium is give as below:

(b) When conc.H₂SO₄ is added to an inorganic mixture containing bromide, initially HBr is produced. HBr, being a strong reducing agent reduces H₂SO₄ to SO₂ with the evolution of red vapour of bromine.
2NaBr + 2H₂SO₄ ⟶2NaHSO4+2HBr
2HBr + H₂SO₄ ⟶ Br₂ + SO₂ + 2H₂O
But, when conc. H₂SO₄ is added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce  H₂SO₄ to SO₂.
2NaCl+2H₂SO₄⟶2NaHSO₄+2HCl.

Question. 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 (1) + 2Cu₂ +(aq) + 5OH − (aq)→ Cu₂O_(s) + HCOO − (aq)+3H₂O(l)
(d) N2H₄(l) + 2H₂O₂(l) → N_2(g) + 4H₂O(l)
(e) Pb_(s) + PbO_2(s) + 2H₂SO₄ (aq) → 2PbSO_4(s) + 2H₂O(l)
Answer : (a) Oxidised substance →C₆H₆O₂
Reduced substance → AgBr
Oxidising agent → AgBr
Reducing agent → C₆H₆O₂

(b) Oxidised substance → HCHO
Reduced substance →[Ag (NH₃ )₂]⁺
Oxidising agent → [Ag (NH₃)₂]⁺
Reducing agent → HCHO

(c) Oxidised substance → HCHO
Reduced substance → Cu²⁺
Oxidising agent → Cu²⁺
Reducing agent → HCHO

(d) Oxidised substance →N₂H₄
Reduced substance → H₂O₂
Oxidising agent → H₂O₂
Reducing agent → N₂H₄

(e) Oxidised substance → Pb
Reduced substance → PbO₂
Oxidising agent → PbO₂
Reducing agent → Pb

Question. Consider the reactions:
2S₂O₂ − 3(aq) + I_2(s) → S4O₂ − 6(aq)+2I − (aq)
S₂O₂ − 3aq) + 2Br₂(l)+5H₂O(l) → 2SO₂ − 4(aq) + 4Br − (aq) + 10H⁺
Answer : The average oxidation number (O.N.) of S in S₂O²⁻₃ is +2. Being a stronger oxidising agent than I₂, Br₂ oxidises S2O2−3 to SO2−4 , in which the O.N. of S is +6. However, I₂ is a weak oxidising agent. Therefore, it oxidises S₂O²⁻₃ to
S4O²⁻₆ , in which the average O.N. of S is only +2.5. As a result, S₂O²⁻₃ reacts differently with iodine and bromine.

Question. Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer : F₂ can oxidize Cl− to Cl₂,Br− to Br₂, and I− to I₂
On the other hand, Cl₂,Br₂, and I₂ cannot oxidize F− to F₂ .The oxidizing power of halogens increases in the order of I₂ < Br₂ < Cl₂ < F₂ . Hence, fluorine is the best oxidant among halogens.
HI and HBr can reduce H₂SO₄ to SO₂, but HCl and HF cannot. Therefore, HI and HBr are stronger reductants than HCl and HF.
2HI + H₂SO₄ ⟶ I₂ + SO₂ + 2H₂O
2HBr + H₂SO₄ ⟶ Br₂ + SO₂+2H₂O
1⁻ can reduce Cu²⁺ to Cu⁺, but Br⁻
4I− (aq) + 2Cu₂ + (aq) ⟶ Cu₂I_2(s) + I₂(aq)
Hence, hydroiodic acid is the best reductant among hydrohalic compounds.
Thus, the reducing power of hydrohalic acids increases in the order of
HF < HCl < HBr < HI.

Question. Why does the following reaction occur?
XeO₄− 6(aq) +2F − (aq) + 6H+ (aq) → XeO_3(g) + F_2(g) + 3H₂O(l)
What conclusion about the compound Na4XeO6 (of which XeO⁴⁻₆ is a part) can be drawn from the reaction.
Answer : The given reaction occurs because XeO⁴⁻₆ oxidises F− and F− reduces XeO⁴⁻₆.
XeO + 6(aq)+ 2F − 1(aq) + 6H + (aq)-> X_3(g)+0F₂(x)+3H₂O(l)
In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in
XeO⁴⁻₆ to +6 in XeO₃ and the O.N. of F increases from –1 in F⁻ to O in F2
Hence, we can conclude that Na₄ XeO₆ is a stronger oxidising agent than F⁻.

Question. consider the reactions:
(a) H₃PO_{2 (aq)} + 4AgNO_{3 (aq)} + 2H₂O(l) → H₃PO_{4 (aq)} + 4Ag_(s)+ 4HNO₃ (aq)
(b) H₃PO_2 (aq) + 2CuSO_4 (aq) + 2H₂O(I) → H₃PO_4(aq) + 2Cu_(s) + H₂SO₄(aq)
(c) C₆H₅ CHO(I) + 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.
Answer : Cu²⁺ and Cu²⁺ act as oxidising agents in reactions
(a) and
(b)respectively. In reaction
(c), Cu²⁺ oxidises C₆H₅ CHO to C₆H₅ COO⁻, but in reaction
(d), Cu²⁺ cannot oxidise C₆H₅CHO.
Hence, we can say that  Cu²⁺ is a stronger oxidising agent than Cu²⁺.

Question. Balance the following redox reactions by ion-electron method:
(a) MnO − 4(aq) + I⁻(aq) → MnO_2(s) + I_2(s) ( in basic medium)
(b) MnO − 4(aq) + SO_2(g) → Mn²⁺(aq) + HSO− 4(aq)
(c) H₂O₂ (aq) + Fe²⁺ (aq) → Fe³⁺ (aq) + H₂O(l) (in acidic  solution)
Cr₂O²⁻₇ + SO_2(g) → Cr³⁺(aq) + SO²⁻₄ (aq) ( in acidic solution )
Answer : (a) Step 1: The two half reactions involved in the given reaction are:
Oxidation half reaction:

Step 2: Balancing I in the oxidation half reaction, we have:
2I−(aq)⟶I2(s)
Now, to balance the charge, we add 2 e^– to the RHS of the reaction.
2I⁻(aq)⟶I₂(x)+2e⁻
Step 3: In the reduction half reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the reaction.
MnO⁻_4(aq)+3e⁻⟶MnO_2(aq)
Now, to balance the charge, we add 4 OH^– ions to the RHS of the reaction as the reaction is taking place in a basic medium.
MnO⁻_4(oq) + 3e⁻⟶MnO_2(cq) + 4OH⁻
Step 4: In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the LHS.
MnO⁻_4(cq) + 2H₂O + 3e⁻⟶ MnO_2(ca) + 4OH⁻
Step 5: Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
6I−(aq)⟶ 3I_2(s) + 6e⁻
2MnO⁻_4(aq) + 4H₂O + 6e⁻⟶ 2MnO_2(s) + 8OH⁻(aq)
Step 6: Adding the two half reactions, we have the net balanced redox reaction as:
6I⁻(eq) + 2MnO − 4(cy) + 4H₂O(l) ⟶  3I_2(s) + 2MnO_2(s) + 8OH⁻(aq)

(b) Following the steps as in part (a), we have the oxidation half reaction as:
SO₂(x)+2H₂O(l)
And the reduction half reaction as:
MnO− 4(αq) + 8H + (aq) + 5e⁻⟶ Mn²⁺(aq) + 4H₂O(l)
Multiplying the oxidation half reaction by 5 and the reduction half reaction by 2, and then by adding them, we have the net balanced redox reaction as:
2MnO − 4_(aq) + 5SO_2(g) + 2H₂O(l) +H+(aq) ⟶ 2Mn²⁺_(aq) + 5HSO₄⁻_(aq)

(c) Following the steps as in part (a), we have the oxidation half reaction as:
Fe²⁺(αq)⟶Fe³⁺(aq)+e−
And the reduction half reaction as:
H₂O₂(aq) + 2H+(aq) + 2e⁻⟶ 2H₂O(l)
Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
H₂O₂(aq) + 2Fe²⁺(aq) + 2H+(aq) ⟶ 2Fe³⁺(aq) + 2H₂O(l)

(d) Following the steps as in part (a), we have the oxidation half reaction as:
SO₂(g)+2H₂Ol) ⟶ SO²⁻₄ + 4H⁺(aq)+2e⁻
And the reduction half reaction as:
Cr₂O²⁻₇ + 14H⁺(aq) + 6e⁻ ⟶ 2Cr³⁺(aq)+7H₂O(l)
Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:
Cr₂O²⁻₇ + 3SO_2(g) + 2H⁺(aq) ⟶ 2Cr³⁺(aq) + 3SO²⁻ ₄ + H₂O(l)

Question. Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
P_4(s) + OH−_(oq)⟶ PH_3(g) + HPO₂⁻
N₂H₄(l) + ClO⁻aq ⟶ NO_(g) + Cl⁻_(g)
Cl₂O_7(g) + H₂O_2(aq) ⟶ ClO₂⁻_(aq) + O_2(g)+ H⁺_(aq)
Answer : 
(a)

The O.N. (oxidation number) of P decreases from 0 in P₄ to –3 in PH₃ and increases from 0 inP₄ to + 2 inHPO₂⁻. Hence, P4acts both as an oxidizing agent and a reducing agent in this reaction.
Ion–electron method :
The oxidation half equation is:
P_4s → HPO₂⁻(aq)
The P atom is balanced as:
P₄→ 4HPO₂^- (aq)
The O atom is balanced by adding 8 H₂O molecules:
P₄ + 8H₂O →  4_(aq)
The H atom is balanced by adding 12 H⁺ ions:
P₄ + 8H₂O → 4HPO − 2_(aq) + 12H+
The charge is balanced by adding e^– as:
P₄ + 8H₂O → 4HPO₂⁻ _(aq) + 12H⁺+ 8e−…(i)
The reduction half equation is :
P_4(s) ⟶ PH_3(g)
The P atom is balanced as:
P₄→ 4PH_3(g)
The H is balanced by adding 12 H⁺ as:
P₄ + 12H⁺ → 4PH_3(g)
The charge is balanced by adding 12e– as:
P₄+ 12H⁺+12e^-→ 4PH_3(g) …(ii)
By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:
5P_4(s) + 24H₂O → 12HPO₂⁻– + 8PH_3(g) + 12H⁺
As, the medium is basic, add 12OH– both sides as:
5P_4(s) +12H₂O + 12OH^- →12HPO₂⁻+8PH_3(g)
This is the required balanced equation.
Oxidation number method:
Let, total no of P reduced = x
∴Total no of P oxidised = 4– x
P4(s) +OH^- →xPH₃(g) + 4-x HHPO₂⁻… (i)
Total decrease in oxidation number of P = x × 3 = 3x
Total increase in oxidation number of P = (4 – x) × 2 = 8 – 2x
∵ 3x = 8 – 2x x = 8/5 From (i),
5P_{4 (s)} +5OH^- →8PH_3(g)) +12HPO₂⁻
Since, reaction occurs in basic medium, the charge is balanced by adding 7OH– on LHS as:
5P_4(s) +12OH^- →8PH_3(g)+12HPO₂⁻
The O atoms are balanced by adding 12H₂O as:
5P_4(s) + 12H₂O+ 12OH^-→ +12HPO₂⁻+ 8PH_3(g)
This is the required balanced equation.
(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 ClO₃⁻ to – 1 in Cl⁻. Hence, in this reaction, N₂H₄ is the reducing agent and ClO₃⁻ is the oxidizing agent.
Ion–electron method:
The oxidation half reaction:

Balance atom N:
N₂H₄(l)→2NO_(g)
Add 8 electrons to balance oxidation no:
N₂H₄(l) → 2NO_(g) + 8e⁻
 Add 8OH^– to balance the charge:
N₂H₄(l) + 8OH^–_(aq) → 2NO_(g) + 8e⁻
 Add 6 H₂O to balance O atoms:
N₂H₄(l) + 8OH⁻_(aq) → 2NO_(g) + 6H₂O(l) + 8e⁻  ——— (1)
The reduction half reaction:
ClO₃^–_(aq)  →  Cl^–_(aq)
Add 6 electrons to balance oxidation no.
ClO₃⁻_(aq)  +  6e⁻→ Cl⁻_(aq)
Add 6OH^– ions to balance the charge:
ClO₃⁻_(aq) + 6e⁻ → Cl⁻_(aq) + 6OH⁻_(aq)
Add 3 H₂O to balance O atoms:
ClO₃⁻_(aq) + 3H₂O(l) + 6e⁻→ Cl⁻_(aq) + 6OH⁻_(aq) ——— (2)
Now, multiply the equation (1) by 3 and equation (2) by 4. Then, after adding them, we get the balanced redox reaction as given below:
3N₂H₄(l) + 4ClO₃^–_(aq) → 6NO_(g)+4Cl^–_(aq) + 6H₂O(l)
Oxidation number method :
Reduction in the oxidation no. of N = 2 × 4 = 8
Increment in the oxidation no. of Cl = 1 × 6 = 6
Multiply N₂H₄ by 3 and ClO₃^– by 4 to balance the reduction and increment of the oxidation no. :
3N₂H₄(l) + 4ClO₃^–_(aq) → NO_(g) + Cl^–_(aq)
Balance Cl and n atoms :
3N₂H₄(l) + 4ClO₃^–_(aq)→ 6NO_(g) + 4Cl^–_(aq)
Add 6 H₂O to balance O atoms :
3N₂H₄(l) + 4ClO₃^–_(aq) → 6NO_(g) + 4Cl^–_(aq) + 6H₂O(l)
This is the required reaction equation.
(c)  

The oxidation number of Cl decreases from + 7 in Cl₂O₇ to + 3 in ClO₂⁻ and the oxidation number of O increases from – 1 in H₂O₂ to zero in O₂ .Hence, in this reaction, Cl₂O₇ is the oxidizing agent and H₂O₂ is the reducing agent.
Ion–electron method:
The oxidation half reaction:
H₂O_2(aq) → O₂(g)

Add 2 electrons to balance oxidation no:
H₂O_2(aq)→ O_2(g) + 2e^–
Add 2OH^– to balance the charge:
H₂O_2(aq) + 2OH^–_(aq) → O_2(g)+2e^–
Add 2 H₂O to balance O atoms:
H₂O_2(aq) + 2OH^– _(aq)→ O_2(g)+ 2H₂O(l)+2e^– ——– (1)
The reduction half reaction:

Cl₂O_7(g)→ ClO₂^–_(aq)
Balance Cl atoms:
Cl₂O_7(g) → 2ClO₂^–_(aq)
Add 8 electrons to balance oxidation no.
Cl₂O_7(g) + 8e^– → 2ClO₂^–_(aq)
Add 6OH^– ions to balance the charge:
Cl₂O_7(g) + 8e^–→ 2ClO₂^–_(aq) + 6OH^–_(aq)
Add 3 H₂O to balance O atoms:
Cl₂O_7(g)+3H₂O(l) + 8e^–→ 2ClO₂^–_(aq)+ 6OH^–_(aq)
Now, multiply the equation (1) by 4. Then, adding equation (1) and (2), we get the balanced redox reaction as given below:
Cl₂O_7(g) + 4H₂O_2(aq) + 2OH^–_(aq) → 2ClO₂^–_(aq) + 4O_2(g) + 5H₂O(l)
Oxidation number method :
Reduction in the oxidation no. of Cl₂O₇ = 4× 2 = 8
Increment in the oxidation no. of H₂O₂ = 2× 1 = 2
Multiply H₂O₂ by 4 and O2 by 4 to balance the reduction and increment of the oxidation no. :
3N₂H₄(l) + 4ClO₃^– _(aq) → NO_(g) + Cl^–_(aq)
Balance Cl and n atoms:
Cl₂O_7(g) + 4H₂O_2(aq) → 2ClO – 2_(aq)+ 4O_2(g)
 Add 3 H₂O to balance O atoms:
Cl₂O_7(g)+4H₂O_2(aq) → 2ClO₂^–_(aq) + 4O_2(g) + 3H₂O(l)
 Add 2OH^– and 2H₂O to balance H atoms:
Cl₂O_7(g)+ 4H₂O_2(aq) 2OH^–_(aq)→ 2ClO₂^–_(aq) + 4O_2(g)+ 5H₂O(l)
This is the required reaction equation.

Question. What sorts of informations can you draw from the following reaction ?
(CN)_2(g) + 2OH⁻ _(aq) ⟶ CN⁻_(g) + CNO⁻_(aq) + H₂O(l)
Answer : The oxidation numbers of carbon in (CN)₂,CN⁻ and CNO⁻ are +3, +2 and +4 respectively. These are obtained as shown below:
Let the oxidation number of C be x.
(CN)₂
2(x – 3) = 0
∴x = 3
CN− x – 3 = –1
∴x = 2
CNO–x – 3 – 2 = –1
∴x = 4
The oxidation number of carbon in the various species is:

It can be easily observed that the same compound is being reduced and oxidised simultaneously in the given equation. Reactions in which the same compound is reduced and oxidised is known as disproportionation reactions. Thus, it can be said that the alkaline decomposition of cyanogen is an example of disproportionation reaction.

Question. 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.
Answer : The given reaction can be represented as:
Mn³⁺_(αq)  ⟶ Mn²⁺_(aq) + MnO_2(s) + H⁺_(aq)
The oxidation half equation is:
Mn³⁺_(aq) ⟶ +4MnO_2(s)

The oxidation number is balanced by adding one electron as:
Mn³⁺_(aq) ⟶ MnO_2(s) +e⁻
The charge is balanced by adding 4H+ ions as:
Mn³⁺_(α) ⟶ MnO_2(s) + 4H⁺_(aq) + e⁻
The O atoms and H⁺ ions are balanced by adding 2H₂O molecules as:
The reduction half equation is:
Mn³⁺_(aq) ⟶ Mn²⁺_(aq)
The oxidation number is balanced by adding one electron as:
The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

Question. 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 positive 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.
Answer : (a) F exhibits only negative oxidation state of –1.
(b) Cs exhibits positive oxidation state of +1.
(c) I exhibits both positive and negative oxidation states. It exhibits oxidation states of – 1, + 1, + 3, + 5, and + 7.
(d) The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states.

Question. 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.
Answer : The redox reaction is as given below:
Cl_2(s)+ SO_2(aq) +H₂O(l) → Cl^–_(aq) + SO₄²⁻_(aq)
The oxidation half reaction:
SO_2(aq) → SO₄²⁻_(aq)
Add 2 electrons to balance the oxidation no. :
SO_2(aq)→ SO₄²⁻_(aq)+ 2e^–
Add 4H⁺ ions to balance the charge:
SO_2(aq)→ SO₄²⁻_(aq)+ 4H+ (aq)+2e^–
Add 2 H₂O to balance O atoms and H+ ions:
SO_2(aq) + 2H₂O → SO₄²⁻_(aq) + 4H⁺_(aq)+2e^– ——— (1)
The reduction half reaction:
Cl_2(s)→ Cl⁻(aq)
Balance Cl atoms:
Cl_2(s)→ 2Cl^–_(aq)
Add 2 electrons to balance the oxidation no. :
Cl_2(s)+ 2e^–→ 2Cl^–_(aq) ——— (2)
Add equation (1) and (2) to get the balanced chemical equation:
Cl_2(s)+ SO_2(aq)+ 2H₂O(l) → 2Cl^–_(aq)+ SO₄²⁻_(aq) + 4H⁺_(aq)

Question. 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.
Answer : In disproportionation reactions, one of the reacting substances always contains an element that can exist in at least three oxidation states,
(a) P, Cl, and S can show disproportionation reactions as these elements can exist in three or more oxidation states,
(b) Mn, Cu, and Ga can show disproportionation reactions as these elements can exist in three or more oxidation states.

Question. 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?
Answer : The balanced chemical equation for the given reaction is given as:
Thus, 68 g of NH₃ reacts with 160 g of O₂
Therefore, 10g of NH₃ reacts with
160 x 10 / 68
g of O₂, or 23.53 g of O₂.
But the available amount of O₂ is 20 g.
Therefore, O₂ is the limiting reagent (we have considered the amount of O2 to calculate the weight of nitric oxide obtained in the reaction).
Now, 160 g of O₂ gives 120g of NO.
Therefore, 20 g of O₂ gives
120 x 20 / 160
g of N, or 15 g of NO.
Hence, a maximum of 15 g of nitric oxide can be obtained.

Question. 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) Br_2(aq) and Fe²⁺_(aq)
Answer : (i) Fe³⁺_(aq) and I^–_(aq)
2Fe³⁺_(aq) + 2I^–_(aq) → 2Fe²⁺_(aq)+I_2(s)
Oxidation half reaction: 2I^–_(aq) → I_2(s)+2e^– ;E∘=−0.54V
Reduction half reaction: [Fe³⁺_(aq) + e^– → Fe²⁺_(aq)]×2 ; E∘=+0.77V
------- ----------- -------- ---------- ------ ---------- ------ --------- ----------- -----
2Fe³⁺_(aq) + 2I^–→ 2Fe²⁺_(aq)+ I_2(s) ; E∘=+0.23V
E^∘for the overall reaction is positive. Therefore, the reaction between Fe³⁺_(aq) and I^–_(aq) is feasible.

(ii) Ag+(aq) and Cu_(s)
2Ag+(aq)+Cu_(s) → 2Ag_(s) + Cu²⁺_(aq)
Oxidation half reaction: Cu_(s) → Cu²⁺_(aq)+ 2e^–;  E^∘=−0.34V
Reduction half reaction: [Ag+(aq)+e^– → Ag_(s)]×2; E^∘= +0.80V
--------- ----------- ------------ -------------- ----------- ---------- -------------- ----
2Ag + (aq)+ Cu_(s)→2Ag(s)+Cu²⁺; E∘=+0.46V
E^∘ for the overall reaction is positive. Therefore, the reaction between Ag⁺_{(aq) and Cu(s) is feasible}

(iii) Fe³⁺_(aq) and Cu_(s)
2Fe³⁺(aq)+ Cu_(s)→2Fe²⁺_(s) + Cu²⁺_(aq)
Oxidation half reaction: Cu(s)→ Cu²⁺_(aq)+ 2e^–;   E∘=−0.34V
Reduction half reaction: [Fe³⁺_(aq) + e^–→ Fe²⁺_(s)] ×2;   E∘=+0.77V
---- --------- ---------- ----------- ---------- --------- --------- ------- ------------
2Fe³⁺_(aq)+ Cu_(s)→ 2Fe²⁺_(s) + Cu²⁺_(aq);             E^∘=+0.43V
E^∘ for the overall reaction is positive. Therefore, the reaction between Fe3+(aq)and Cu(s) is feasible.

(iv) Ag_(s) and Fe³⁺(aq)
Ag_(s)+2Fe³⁺_(aq)→ Ag⁺_(aq)+ Fe²⁺_(aq)
Oxidation half reaction: Ag(s)→ Ag+(aq)+e^–;   E∘=−0.80V
Reduction half reaction: Fe³⁺_(aq) + e^–→Fe²⁺_(aq); E∘=+0.77V
------- ----------- -------- ------- ----------- -------- ------- ------- ------- ---------
Ag_(s)+Fe³⁺_(aq)→ Ag⁺_(aq) + Fe²⁺_(aq);             E^∘=−0.03V
E^∘ for the overall reaction is positive. Therefore, the reaction between Ag(s) and Fe3+(aq) is feasible.

(v) Br_2(aq) and Fe²⁺_(aq)
Br_2(s)+ 2Fe²⁺_(aq) → 2Br^–_(aq) + 2Fe³⁺_(aq)
Oxidation half reaction: [Fe²⁺_(aq) → Fe³⁺_(aq) +e^–] ×2;            E^∘= −0.77V
Reduction half reaction: Br_2(aq)+2e^–→2Br^–_(aq);                     E^∘= +1.09V
---------------------------------------------------------------------------------------------------
Br_2(s)+ 2Fe2+(aq) → 2Br^–_(aq) + 2Fe³⁺_(aq);                  E^∘=−0.32V
E^∘ for the overall reaction is positive. Therefore, the reaction between Br_2(aq) and Fe²⁺_(aq) is feasible.

Question. 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.
Answer : (i) AgNO₃ ionizes in aqueous solution to form Ag⁺ and NO₃⁻ ions.
On electrolysis, either Ag⁺ ion or H₂O molecule can be decreased at cathode. But the reduction potential of Ag⁺ ions is higher than that of H₂O.
Ag+(aq) + e⁻→Ag_(s);  E^∘=+0.80V
2H₂O(l)+2e⁻→ H2(g)+2OH⁻_(aq); E∘=−0.83V
Therefore, Ag⁺ ions are decreased at cathode. Same way, Ag metal or H2O molecules can be oxidized at anode. But the oxidation potential of Ag is greater than that of H₂O molecules.
Ag_(s)→ Ag⁺_(aq) + e⁻; E^∘=−0.80V
2H₂O(l) → O_2(g)+ 4H + (aq) + 4e⁻; E∘=−1.23V
Hence, Ag metal gets oxidized at anode.

(ii) Pt cannot be oxidized very easily. Therefore, at anode, oxidation of water occurs to liberate O₂. At the cathode, Ag⁺ ions are decreased and get deposited.

(iii) H₂SO₄ ionizes in aqueous solutions to give H⁺ and SO₄²⁻ ions.
H₂SO_4 (aq)  → 2H⁺_(aq) + SO₄²⁻_(aq)
On electrolysis, either of H₂O molecules or H⁺ ions can get decreased at cathode. But the decreased potential of H⁺ ions is higher than that of H₂O molecules.
2H⁺_(aq) + 2e⁻→ H_2(g); E^∘=0.0V
2H₂O(aq) + 2e⁻→ H_2(g)+2OH⁻_(aq); E^∘=−0.83V
Therefore, at cathode, H⁺ ions are decreased to free H₂ gas.
On the other hand, at anode, either of H₂O molecules or SO₄²⁻ ions can be oxidized. But the oxidation of SO₄²⁻ involves breaking of more bonds than that of H₂O molecules. Therefore, SO₄²⁻ ions have lower oxidation potential than H₂O. Hence, H₂O is oxidized at anode to free O₂ molecules.

(iv)  In aqueous solutions, CuCl2 ionizes to give Cu²⁺ and Cl⁻ ions as:
CuCl_2(aq)→ Cu²⁺_(aq) + 2Cl⁻_(aq)
CuCl_2(aq)→Cu²⁺_(aq)+2Cl⁻_(aq)
On electrolysis, either of Cu²⁺ ions or H₂O molecules can get decreased at cathode. But the decreased potential ofCu²⁺ is more than that of H₂O molecules.
Cu ²⁺ _(aq)+2e⁻ → Cu_(aq); E∘=+0.34V
H₂O(l) + 2e⁻→H_2(g)+ 2OH⁻; E^∘=−0.83V
Therefore, Cu²⁺ ions are decreased at cathode and get deposited. In the same way, at anode, either of Cl⁻ orH₂O is oxidized. The oxidation potential of H₂O is higher than that of Cl⁻.
2Cl⁻_(aq)→ Cl_2(g) + 2e⁻; E^∘=+0.34V
2H₂O(l) → O_2(g) + 4H⁺_(aq)+ 4e⁻; E^∘=−1.23V
But oxidation of H₂O molecules occurs at a lower electrode potential compared to that of Cl⁻ ions because of over-voltage (extra voltage required to liberate gas). As a result, Cl− ions are oxidized at the anode to liberate Cl₂ gas.

Question. Arrange the given metals in the order in which they displace each other from the solution of their salts.
(i) Al
(ii) Fe
(iii) Cu
(iv) Zn
(v) Mg
Answer : A metal with stronger reducing power displaces another metal with weaker reducing power from its solution of salt.
The order of the increasing reducing power of the given metals is as given below:
Cu < Fe < Zn < Al < Mg
Therefore, Mg can displace Al from its salt solution, but Al cannot displace Mg. Thus, the order in which the given metals displace each other from the solution of their salts is as given below: Mg >Al>Zn> Fe >Cu

Question. The standard electrode potentials are given of the following elements:
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.
Answer : The reducing agent is stronger as the electrode potential decreases. Hence, the increasing order of the reducing power of the given metals is as given below:
Ag < Hg < Cr < Mg < K

Question. Depict the galvanic cell in which the reaction is:
Zn_(s) + 2Ag+(aq)→ Zn²⁺_(aq) + 2Ag_(s)
Show the following:
(i) Which of the electrode is negatively charged?
(ii) Name the carriers of the current in the cell.
(iii) Write the individual reaction at each electrode.
Answer : The galvanic cell corresponding to the given redox reaction can be shown as:
Zn|Zn²⁺_(aq)|| Ag⁺_(aq)|Ag
(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2+ and the leaving electrons accumulate on this electrode.
(ii) The carriers of current are ions in the cell.
(iii) Reaction at Zn electrode is shown as:
Zn_(s)→ Zn²⁺(aq) + 2e⁻
Reaction at Ag electrode is shown as:
Ag⁺_(aq) + e⁻→ Ag_(s)