GSEB Solutions Class 11 Chemistry Chapter 8 Redox Reactions

Gujarat Board GSEB Textbook Solutions Class 11 Chemistry Chapter 8 Redox Reactions Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Chemistry Chapter 8 Redox Reactions

GSEB Class 11 Chemistry Redox Reactions Text Book Questions and Answers

Question 1.
Assign oxidation number to the underlined elements in each of the following species.
(a) NaH₂PO₄
(b) NaHSO₄
(c) H₄P₂O₇
(d) K₂MnO₄
(e) CaO₂
(f) NaBH₄
(g) H₂S₂O₇
(h) KAl(SO₄)₂ . 12H₂O
Answer:
(a) Let the oxidation number of P be x. Writing the oxidation number of each atom above its symbol we have,

sum of oxidation number of various atom in NaH₂PO₄
= 1(+1) 4- 2(+1) + 1 (x) + 4 (-2) = 0
But the sum of oxidation number of various atoms in NaH₂PO₄ is zero.
x – 5 = 0
or x = + 5
oxidation number of P in NaH₂PO₄ = +5

(b) Let the oxidation number of S be x.
+ 1 + 1 + x + 4 (- 2)
= 0 + 2 + x – 8 = 0
x = +6
oxidation number of S in NaHSO₄ is +6.

(c) Let the oxidation number of P be x.
4(+1) + 2x + 7 (- 2) = 0
+ 4 + 2x – 14 = 0
2x = 10
x = +5
oxidation number of P in H₄P₂O₇ = + 5.

(d) Let the oxidation number of Mn be x.
2(+1) + x + 4 (- 2) = 0
+ 2 + x – 8 = 0
x = +6
oxidation number of Mn in K₂MnO₄ = +6.

(e) Let the oxidation number of oxygen in CaO₂ be x.
+2 + 2x = 0
2x = – 2
x = – 1
oxidation number of O in CaO₂ = -1

(f) Let the oxidation number of B in NaBH₄ be x.
+ 1 + x + 4(-1) – 0
x – 3 = 0
x = +3
oxidation number of B in NaBH₄ = +3

(g) Let the oxidation number of S in H₂S₂O₇ be x.
2(+1) + 2x + 7(-2) = 0
+2 + 2x – 14 = 0
2x = 12
x = +6
oxidation number of S in H₂S₂O₇ = + 6

Question 2.
What are the oxidation number of the underlined elements in each of the followings & how do you rationalise your result?
(a) KI₃
(b) H₂S₄O₆
(c) Fe₃O₄
(d) CH₃CH₂OH
(e) CH₃COOH
Answer:
(a) Oxidation of I in KI₃ is
KI₃ is made up of K⁺I^– + I₂
oxidation number of I in KI is -1 whereas it is O in I₂
The oxidation number of I in KI₃ is = – 1

(b) Oxidation number of S in H₂ S₄O₆ is
structure of H₂ S₄O₆ is

The oxidation number of each of S atom linked with each other in the middle is zero while that of each of me remaining two S atom is +5.

(c) Oxidation number of Fe in Fe₃O₄

Fe has oxidation number of +2 and +3.

(d) CH₃CH₂OH
oxidation number of carbon atom can be calculated as

C₂ is attached to three H-atom (less electronegative than – c) and one – CH₂OH (more electronegative than C) oxidation number of C₂ is 3(+1) +x + 1 (-1) = 0
+ 3 + x – 1 = 0
x = – 2
[For each covalent bond between two dissimilar atom, assign an oxidation number of +1 to the less electronegative atom and – 1 to the more electronegative elements]
C₁ is attached to one OH (O. No. = – 1) one – CH₃ group (O. No. + 1), two H atom (O. No. = +1) O. No. of C₁ is
+ 1 + 2(+1) + x – 1 = 0
+ 1 + 2 + x – 1 = 0
x = – 2

(e) Let us calculate the oxidation no of two atom C₁ & C₂ in acetic acid.

C₂ is attached to three H^– atom and one COOH group.
oxidation number of C₂ is 3(+1) +x + 1 (- 1) = 0
or x = – 2
oxidation dumber of C₂ = – 2
C₁ is however attached to one O-atom by a double bond,
one OH group, (oxidation number = – 1) and one CH₃
group (oxidation no. = +1)
oxidation number of C₁ is +1 + x + (- 2) +1( – 1) = 0
x = +2
oxidation number of C₁ is = +2

Question 3.
Justify that the following reactions are redox reactions :
(a) CuO(S) + H₂(g) → Cu(S) + H₂O(g)
Answer:
CuO has been reduced to Cu whereas H₂ has been oxidised to H₂O.

(b) Fe₂O₃(S) + 3CO(g) → 2Fe(S) + 3CO₂(g)
Fe₂O₃ has been reduced to Fe whereas CO has been oxidised to CO₂

Fe₂O₃ has been reduced to Fe, because there has occured a decrease in its oxidation number whereas CO has been oxidised to CO₂, because there has occured as increase in its oxidation number.

(c) 4BCl₃(g) + 3LiAlH₄(S) → 2B₂H₆ (g) + 3LiCl(s) + AlCl₃
Here BCl₃ is reduced to B₂H₆ whereas both Li and Al have both been oxidised (addition of electronegative element chlorine).

(d) 2K(S) + F₂(g) → 2K⁺F^–(S)
Here K is oxidised to K⁺F^– (addition electronegative element F) where F₂ is reduced to K⁺F faddition of electro positive element K).

K has been oxidised (increase in oxidation number). F₂ has been reduced (decrease in oxidation number).

(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)
Here NH₃ has been oxidised to NO. (removal to Hydrogen, addition of oxygen) where as O₂ has been reduced to H₂O (addition of hydrogen).

NH₃ has been oxidised to NO (oxidation number of N has increased from – 3 to + 2) whereas O₂ has been reduced to H₂O (decrease of oxidation number).

Question 4.
Fluorine reacts with ice and result in the change.
H₂O(s) + F₂(g) → HF(g) + HOF(g)
Justify this reaction is a redox reaction.
Answer:
F undergoes reduction to form HF. Whereas F undergoes oxidation to HOF. F₂ undergoes oxidation as well as reduction. It is a disproportionation reaction.
HOF is an unstable compound and decompose on its own to form H₂O and OF₂
2HOF → H₂O + OF₂.
Fluorine as such does not undergo disproportionation as it can show only one stable oxidation state of -1.
(F₂ can undergo only reduction and not oxidation).

Question 5.
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:
Oxidation number of s in H₂SO₅ can be calculated by understanding its structure.
(i) By conventional method.

But this is wrong as the maximum oxidation number of sulphur cannot be more than +6, since, it has only six valence electrons. The structure of H₂SO₅ is

in which two oxygen atoms are joined by peroxide linkage.
The oxidation number of s can be calculated as
2( +1) + x + 3(- 2) + 2(-1) = 0
for H (for O – O)
+ 2 + x – 6 – 2 = 0
x = +6
oxidation number of s in H₂SO₅ = + 6

(ii) Oxidation number of Cr in CrO₅ By conventional method
Cr^x O^-2₅
or x + 5 (- 2) = 0
x = +10
This is not possible as Cr has five 3d electrons and one 4s electrons.
Max oxidation number of Cr cannot be more than six.
In CrO₅ out of 5 oxygen atoms, 4 oxygen atoms are linked as two peroxide bonds as shown.

oxidation number of Cr is x + 4(- 1) + 1(- 2) = 0
x – 4 – 2 – 0
x = +6
oxidation number of Cr in CrO₅ is +6.

(iii) Oxidation number of N in NO^–₃
Let oxidation number of N in NO^–₃ = x
or x + 3(- 2) = – 1
x = +5
∴ oxidation number of N in NO^–₃ = +5
Structure of NO^–₃ ions is

Question 6.
Write formulas for the following compounds :
Answer:
(a) Mercury (II) chloride : Hg (II) Cl₂
(b) Nickel (IV) sulphate : Ni (II) SO₄
(c) Tin (IV) oxide : Sn(IV) O₂
(d) Thallium (I) Sulphate : Tl₂(l) SO₄
(e) Iron (III) Sulphate : Fe₂(III) [SO₄]₃
(f) Chromium (III) Oxide : Cr₂(III) O₃

Question 7.
Suggest a list of the substance where. Carbon can exhibit oxidation state from – 4 to +4 and nitrogen from – 3 to +5
Answer:

substance containing N showing oxidation states from -3 to +5.

Question 8.
While sulphur dioxide & hydrogen peroxide can act as oxidizing as well as reducing agent both in their reactions, ozone, and nitric acid acts only as oxidants. Why?
Answer:
In both SO₂ & H₂O₂, S & O can increase as well as decrease this oxidation number therefore they can act both as reducing and oxidizing agent. Ozone can only decrease its oxidation state to -1 or -2. so it acts only oxidant.

Oxidation state exhibited by N is +5. In HNO₃ nitrogen is present in +5 oxidation state. It can only decrease and not increase. Hence HNO₃ can acts as an oxidising agent.

Question 9.
Consider the reaction :
(a) 6CO₂ (g) + 6H₂O (l) → C₆H₁₂O₆(aq) + 6O₂ (g)
(b) O₃ (g) + H₂O₂ (l) → H₂0(l) + 2O₂ (g)
Why it is more appropriate to write these reactions as :
(a) 6CO₂ (g) + 12H₂O (l) → C₆H₁₂O (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) & (b) reactions.
Answer:
(a) It is more appropriate to write down the reaction.
6CO₂ (g) + 12H₂O (l) → C₆H₁₂O₆(S) + 6H₂O (l) + 6O₂(g)
atmospheric CO₂, is reduced to C₆H₁₂O₆ and there are 6 carbon atom reduced. Similarly out of 12 H₂O molecules oxidation number of 6O₂ atom from 6H₂O increase from -2 to 0 in 6O₂ molecules whereas in remaining 6H₂O molecules its oxidation number remain the same.

(b) It is more appropriate to write down the reaction.
O₃ (g) + H₂O₂ (l) → H₂O(l) + 2O₂ (g)
because here O₃ is reduced to O₂ by the reducing action of H₂O₂.

Question 10.
The compound AgF₂ is unstable compound. However if formed, the compound acts as a here strong oxidizing agent. Why?
Answer:
AgF₂ → Ag + F₂
Due to the release of F₂(g), AgF₂ will acts as a very powerful oxidizing agent.

Question 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.
Answer:
Whenever a reaction between an oxidising agent and 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. It is clear from the following illustration.

Question 12.
How do you count for me 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) Toluene, being a non-polar liquid will not dissolve in aqueous, alkaline or acidic KMnO₄. Therefore alcoholic KMnO₄ is preferred in the manufacture of benzoic acid from toluene.

(b) When cone. H₂SO₄ is added to an inorganic mixture containing chloride, it gives colourless pungent smelling HCl gas.
MCI + H₂SO₄ → MSO₄ + HCl
But when cone. H₂SO₄ is added to inorganic mixture containing bromide we get red vapours of bromine.
MBr + H₂SO₄ → MHSO₄ + HBr
2HBr + O → H₂O + Br₂

Question 13.
Identify the substance oxidised reduced, oxidising agent an reducing agent for each of the following reactions :
(a) 2AgBr (s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr (aq) + C₆H₄O₂(aq)

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

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

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

(e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)
Answer:
(a) 2AgBr (s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr (aq) + C₆H₄O₂ (aq)
AgBr is reduced to Ag
C₆H₆O₂ is oxidised to C₆H₄O₂
AgBr is an oxidizing agent
C₆H₆O₂ is a reducing agent.

(b) HCHO(l) + 2[Ag(NH₃)₂]⁺ + 3OH^– → 2Ag⁺ + HCOO^– + 4NH₃ + 2H₂O (l)
Here HCHO is oxidised to HCOO^–
[Ag(NH₃)₂]⁺ reduced to Ag⁺
Ag (NH₃)₂ is an oxidising agent.

(c) HCHO + 2Cu²⁺ + 5OH^– → Cu₂O + HCOO^– + 3H₂O
HCHO is oxidised to HCOO^–
Cu²⁺ is reduced to Cu(l) state.

(d) N₂H₄ + 2H₂O₂ → N₂ + 4H₂O
N₂ H₄ is reduced to H₂O
H₂O₂ is an oxidising agent.
N₂H₄ is a reducing agent.

(e) Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O
Pb has been oxidised to PbSO₄
PbO₂ is reduced to PbSO₄
PbO₂ is an oxidising agent.
Pb is a reducing agent.

Question 14.
Consider the reactions :
2 S₂O^2-₃(aq) + I₂(s) → S₄O^2-₆ (aq) + 2I^– (aq) S₂O^2-₃ (aq) + 2Br₂(l) + 5 H₂O(l) → 2SO^2-₄ (aq) + 4Br^– (aq) + 10H⁺ (aq)
Why does the same reductant, thiosulphate react differently with iodine and bromine?
Answer:
In the reactions 2S₂ O^2-₃ (aq) + I₂(s) → S₄O^2-₆ + 2I^– S₂ O^2-₃ (aq) + 2Br2(l) + 5H₂0(l) → 2SO^2-₄ + 4Br^– + 10H⁺
The reductant i.e. S₂O^2-₃ (aq)
happens to be same but among the halogens Br₂ happens to a stronger oxidising agent as compared to I₂ which is a much weaker oxidising agent.
The oxidation state of S in S O^2-₄ is higher than in S₄O^2-₆.

Question 15.
Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
Answer:
Among the halogens best oxidant is F₂ and oxidizing character decrease in the order F₂ > Cl₂ > Br₂ > I₂
∵ F₂ + 2e^– → 2F^–(aq) . E⁰_red + 2.87
is maximum among the halogens so much so that F₂ can displace Cl^– > Br^– > I^– from their salt solution to Cl₂, Br₂, I₂ respectively.
F₂ + 2NaCl(aq) → 2NaF(aq) + Cl₂
F₂ + 2NaBr(aq) → 2NaF + Br₂
F₂ + 2NaI(aq) → 2NaF + I₂
Cl₂ + 2NaF → No reaction.
i. e. F₂ can be reduced most readily.
Hence, F₂ is the strongest oxidizing agent.
But among the hydrohalic acid HI is the strongest reducing agent.
The reducing power of halogen acids depends upon the ease with which they decompose to give H₂ & X₂ . This in turn depends upon the bond dissociation energy. Since the bond dissociation energy of the halogen acid increase in the order.
HI < HBr < HCI < HF
∴ reducing power of these acids increase in the reverse direction.
HF < HCI < HBr < HI
2HI ⇌ H₂ + I₂

Question 16.
Why does the following reaction occur?
XeO^4-₆ (aq) + 2F^– (aq) + 6H⁺ (aq) → XeO₃(g) + F₂ (g) + 3H₂O(l)
What conclusion about the compound Na₄XeO₆ (of which Xe O^4-₆ is a part) can be drawn from the reaction.
Answer:
XeO^4-₆ (aq) + 2F^– (aq) + 6H⁺ (aq) → XeO₃(g) + F₂ (g) + 3H₂O(l)
It is a redox reaction in which Xenon from XeO^4-₆ is reduce to XeO₃.
F is oxidised to F₂.
Sodium xenate (Na₆ XeO₆) is an ionic compound in which oxidation number of sodium does not undergo a change and it is Xenon whose oxidation number change from +8 to +6. It is a powerful oxidising agent which oxidises even F to F₂.

Question 17.
Consider the reactions;
(a) H₃PO₂(aq) + 4 AgNO₃(aq) + 2 H₂O(l) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)

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

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

(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^–(aq) → No change observed.
What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions?
Answer:
(a) H₃PO₂(aq) + 4AgNO₃(aq) + 2H₂O(l) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)
In this reaction, Ag⁺ ions have been reduced to Ag. It is oxidizing.
H₃PO₂ to H₃PO₄.

(b) H₃PO₂(aq) + 2CuSO₄ + 2H₂O(l) → H₃PO₄(aq) + 2Cu(s) + H₂SO₄
In this reaction Cu⁺⁺ ions are reduced to Cu(s)
Thus, it acts as an oxidising agent and oxidises H₃PO₂ to H₃PO₄

(c) C₆H₅CHO(l) + 2[Ag(NH₃)₂]⁺(aq) + 30H^– (aq) → C₆H₅COO^– + 2Ag + 4NH₃ + 2H₂O
In this reaction Ag⁺ present in the complex [Ag(NH₃)₂]⁺ is reduced to Ag(s). It oxidizes C₆H₅CHO to C₆H₅COO

(d) C₆H₅CHO(l) + 2Cu⁺ (aq) + 5OH^– (aq) → No change observed.
Here Cu²⁺ ions are not in a position to oxidise C₆H₅CHO.

Question 18.
Balance the following redox reactions by ion – electron method :

Answer:
(a) MnO^–₄(aq) + I (aq) → MnO₂(s) + I₂(s) in basic medium.
oxidation half reaction is
(i) I^–(aq) → I₂(s)
(ii) 2I^– (aq) → I₂(s)
(iii) 2I^– (aq) → I₂(s) + 2e^– … (i)

Reduction half reaction is :
MnO^–₄ → MnO₂(s)
(i) MnO^–₄ → MnO₂ + 2H₂O (To balance O atom).

(ii) MnO^–₄ + 4H₂O → MnO₂(s) + 2H₂O + 4OH^– (To balance H atom)

(iii) MnO^– (aq) + 2H₂O → MnO₂ + 4OH (aq) (Remove duplication)

(iv) MnO^–₄ (aq) + 2H₂O + 3e^– → MnO₂ + 40H^– (aq) … (2)
To equate number of electrons in both half reactions multiply eq. (i) by (3) & equation no. (2) by 2 and add the two half reaction to cancel out electrons.

(b) MnO^–₄ + SO₂(aq) → Mn²⁺ + HSO^–₄ (aq) (in acidic solution)

oxidation half reaction is :
(i) SO₂(g) → HS O^–₄ (aq)
There has occurred an increase in oxidation number
from +4 in SO₂ to 6 in HS O^–₄.

(ii)

Reduction half – reaction is:
(i) MnO^–₄(aq) → Mn²⁺ (aq)
There has occurred a decrease in oxidation number of Mn from +7 to +2.

(ii)

Multiply equation (1) by (5) and equation number (2) by 2 to equalise to the number of electrons and add the two half reaction cancelling electrons.
5SO₂(g) + 10H₂O^–(l) → 5HS O^–₄ + 15H⁺ + 10e^–
2MnO^–₄(aq) + 16H⁺ (aq) + 10e^– → -2Mn²⁺ + 8H₂O
2MnO^–₄(aq) + 5SO₂(g) + 2H₂0 + H⁺ → 2Mn²⁺ + 5HSO^–₄
To balance atoms and charges
2(-1) + 1 = 2(+ 2) + 5( – 1)
– 1 = – 1

(c) H₂O₂ (aq) + Fe²⁺ (aq) → Fe³⁺+(aq) + H₂O(l) (In acidic medium)
(i) The oxidation half reaction is
Fe²⁺ (aq) → Fe³⁺ (aq)

(ii) Fe²⁺ (aq) → Fe³⁺(aq) + e^– … (1)
The reduction half reaction is
H₂O₂ (aq) → H₂0(l)
(Oxidation number of 0 decrease from – 1 in H₂O₂ to -2 in H₂O) (To balance oxygen atoms)

(iii) H₂O₂(aq) + 2H⁺(aq) → 2H₂O(l) (To balance H atoms in acidic medium)

(iv) H₂O₂(aq) + 2H⁺ + 2e^– → 2H₂O(l) … (2) (To balance charge)

(v) Multiply eq. number (1) by 2 to balance the number of electrons in both the half reaction and add them.
2Fe²⁺ (aq) + H₂O₂ + 2H⁺ (aq) → 2Fe³⁺(aq) + 2H₂O(l)
To balance atoms & charge
2(+2) + 2(+1) = 2(+ 3)
+4+2 = +6 + 6 = +6

(d) Cr₂O^2-₇(aq) + SO₂(aq) → Cr³⁺(aq) + SO^2-₄(aq) (in acidic solution)

Divides the equation in two parts.

Oxidation half reaction –

Reduction half reaction –

(vi) Multiply equation number (1) by 3 to balance electrons in both half reactions & add them to cancel electrons.
Cr₂ O^2-₇(aq) + 14H³⁺ (aq) + 6e^– → 2Cr³⁺ + 7H₂O(l)
3SO₂(aq) + 6H₂O (l) → 3SO^2-₄ (aq) + 12H⁺(aq) + 6e^–
Cr₂ O^2-₇ (aq) + 3SO₂(aq) + 2H⁺(aq) → 2Cr³⁺ + 3SO^2-₄ + H₂O(l)
– 2 + 2 = + 6 – 6
0 = 0

Question 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) + HPO^–₂ (aq)
(b) N₂H₄(l) + ClO^–₃ (aq) → NO(g) + Cl^– (g)
(c) Cl₂O₇ (g) + H₂O₂(aq) → ClO^–₂ (aq) + O₂(g) + H⁺
Answer:
Balance the equation by Ion electron method.

oxidation half reaction
P → H₂PO^–2
P + 2OH^– → H₂PO^–2
P + 2OH^– → H₂PO^–2 + e^– … (1)
Reduction half reaction
P → PH₃
P + 3H₂O → PH₃ + 3OH^–
P + 3H₂O + 3e^– → PH₃ + 3OH^– … (2)
multiply equation (1) by 3 to equalise the number of electrons in both half reaction.
Add the two half reactions
3P + 6OH^– → 3H₂PO^–₂ + 3e^–
P + 3H₂O + 3e^– → PH₃ + 3OH^–
4P + 3H₂O + 3OH^– → 3H₂PO^–₂ + PH₃(g)
or P₄ + 3H₂O(l) + 3OH^– (aq) → 3H₂PO₂ + PH₃(g)
Balance the atoms and charge on both sides
3(- 1) = +3( -1)
-3 = -3
It is a disproportionation reaction.
Here phosphorus undergoes oxidation as well as reduction. Therefore P is both an oxidant & as a reductant.

N ↑ 8 x 3
Cl ↑ 6 x 4
3N₂H₄(l) + 4ClO^–₃ (aq) → 6NO + 4Cl^–
Now balance H and O atom
3N₂H₄(1) + 4ClO^–₃ (aq) → 6NO(g) + 4Cl^– + 6H₂O
4(- 1) = 4(- 1)
– 4 = – 4
Here Cl O^–₃ ions is an oxidant and N₂H₄ is a reductant.

(c)

Reduction half reaction
Cl₂O₇(g) → ClO^–₂
Cl₂O₇(g) → 2ClO^–₂
Cl₂O₇(g) → 2ClO^–₂ + 3H₂O
Cl₂O₇(g) + 6H₂O(l) → 2ClO^–₂ + 3H₂O + 6OH^–
Cl₂O₇(g) + 3H₂O(l) → 2ClO^–₂ + 6OH^–
Cl₂O₇(g) + 3H₂0 + 8e → 2ClO^–₇ + 6OH^–
To equalise electrons in both half reactions multiply equation number (1) by 4. Add both the half reactions to cancel out electrons.
4H₂O^–₂(aq) + 80H^–(aq) → 4O₂ + 8H₂O(l) + 8e^–
Cl₂O₇ + 3H₂O(l) + 8e^–
Cl₂O₇(g) + 4H₂O (aq) + 2OH^– (aq) → 2ClO^–₂ (aq) + 5H₂O + 4O₂
To balance atoms and charges
2(- 1) = 2( -1)
– 2 = – 2
In this reaction, Cl₂O₇ is an oxidising agent and H₂O₂ acts as a reducing agent.

Question 20.
What sorts of informations can you draw from the following reaction?
(CN)₂(g) + 2OH^–(aq) → CN^–(aq) + CNO^–(aq) + H₂O(l)
Answer:
(CN)₂ (g) + 2OH^–(aq) → CN^–(aq) + CNO^– + H₂O(l)
(i) In the above reaction (CN)₂ is called a pseudohalogen while CN^–, CNO^– ions are called Pseudohalides ions.

(ii) Ions which consist of two or more electronegative atoms of which atleast one is nitrogen and have properties similar to those of halides ions are called pseudohalides ions.

(iii) Just as halide ions (Cl^–) have corresponding diatomic molecules (Cl₂) pseudohalides ions (CN^–, CNO^–) have also corresponding dimeric molecules (CN)₂, (CNO)₂. These are called pseudohalogens and show properties similar to those of halogens. (CN)₂ is cyanogen and similar to Cl₂.

(iv) They undergoes reactions similar to halogens as Cl₂(g) disproportionate to give Cl^– ions and ClO^– ions in basic medium. So does (CN) does above.

Question 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.
Answer:
The skeletal equation is :
Mn³⁺ → Mn²⁺ + MnO₂ + H⁺ (aq)
Oxidation half reaction

Reduction half reaction

It is the net balanced ionic equation
2(+ 3) = +2 + 4(+1)
+6 = +6

Question 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 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 exhibit only -ve oxidation state (F^–).
(b) Cs exhibit only +ve oxidation state (CS⁺).
(c) I exhibit both -ve & +ve oxidation state I^– & I⁺
(d) Ne neither shows +ve oxidation state nor does it shows -ve oxidation state.

Question 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.
Answer:
Cl₂(g) + SO₂(aq) + 2H₂O(l) → 2HCl(aq) + H₂SO₄(aq)
Here Cl₂(g) undergoes reduction to form HCl whereas SO₂ undergoes oxidation to form H₂SO₄.

Question 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.
Answer:
(a) Oxygen, chlorine, bromine, phosphorus etc.
(b) Copper, manganese.

Question 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 ?
Answer:

not of NH₃ given = 10.00 gm
not of O₂ given = 20 gm.
Let us arbitrarily choose NH₃ which will react with 20 gm of O₂.
Mass of NH₃ required = \(\frac { 60 }{ 160 }\) x 20 = 8.5 gm
It means 20 gm of O₂ will require 8.5 gm of NH₃.
∴ Oxygen must be the limiting reactant and NH₃ is present in excess i.e. 10 gm.
Now if the entire mass of O₂ react how much NO is formed.
Mass of NO produce from 20 gm of O₂
\(\frac { 120 }{ 160 }\) x 20 = 15 gm of NO.
Maximum wt. of NO produced = 15 gm.

Question 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) Ag³⁺ (aq) and Cu(s)
(c) Fe³⁺ (aq) and Cu(s)
(d) Ag(s) and Fe³⁺ (aq)
(e) Br^– (aq) and Fe²⁺ (aq).
Answer:
(a) Fe³⁺ (aq) and I^– (aq)

since the EMF is positive, the reaction is feasible.

(b) Here Cu(s) loses electrons and Ag⁺(aq) gains electrons. Thus,
oxidation half cell reaction is
Cu(s) → Cu²⁺ + 2e^–] E° = – 0.34V
Reduction Ag⁺ + e^– → Ag(s) x 2 E° = – 0.80 V
overall reaction is :
Cu(s) + 2Ag⁺ → Cu²⁺ (aq) + 2Ag(s)
E°_cell = – 0.46V
Since E°cell is positive the reaction is feasible.

(c) Oxidation Cu(s) → Cu²⁺(aq) + 2e^–, E° = – 0.34V
Reduction Fe₃4 + e → Fe²⁺ E° = +0.77V
Probable cell
Cu(s), Cu²⁺ : : Fe³⁺, Fe²⁺
E°_cell = E°_red(R.H.S) – E°_ox^x L.H.S.
= [+0.77 – 0.34] V = 0.43V.
since E°_cell is positive, reaction is feasible.

(d) Ag (s) and Fe₃+ (aq)
Oxidation Ag(s) → Ag⁺ + e^– E° = – 0.80V
Reduction Fe³⁺ + e^– → Fe²⁺, E° = + 0.77V
E°_cell = – 0.03V
since E°_cell is negative, reaction is not feasible.

(e) Br₂ and Fe²⁺
Here Fe²⁺ lose electrons and Br₂ gains them.
Oxidation Fe²⁺ → Fe³⁺ + e] x 2 [E° = 0.77V
Reduction Br₂ + 2e^– → – 2Br^–] (E° = + 1.08V)
E°_cell = +0.31V
Reaction is 2Fe²⁺ + Br₂ → 2Fe³⁺ + 2Br^–
since E°_cell is positive i.e. reaction is feasible.

Question 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 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 ⇌ Ag⁺ + NO^–₃
At cathode Ag⁺ + e^– → Ag silver is deposited
At anode = NO^–₃ ions move, Dissolve, equivalent amount of Ag concentration AgNO₃ remains unchanged.

(ii) At cathode : Ag will be liberated.
At anode : O₂ gas will be liberated.

(iii) At anode : O₂ gas will be liberated
At cathode : H₂ gas will be liberated.

(iv) At cathode : Cu will be liberated.
At anode : Cl₂ gas will be liberated.

Question 28.
Arrange the following metals in the order in which hey displace each other from the solution of their salts.
Al, Cu, Fe, Mg and Zn.
Answer:
Any metal will displace another metal from its salt solution is dependent on the values of their reduction potential values.
The order of displacement is thus :
Mg = – 2.37
Al = – 1.66
Zn = – 0.76
Fe = + 0.44
Cu = + 0.34
Thus, the decreasing order is :
Mg < Al < Zn < Fe < Cu

Question 29.
Given the standard electrode potentials,
(i) K⁺/K = – 2.93V
(ii) Mg²⁺/Mg = – 2.37V
(iii) Ag⁺/Ag = 0.80V
(iv) Hg²⁺ /Hg = 0.79V
(v) Mg²⁺/Mg = – 2.37V. Cr³⁺/Cr = – 0.74V
arrange these metals in their increasing order of reducing power.
Answer:
(i) K⁺/K = – 2.93V
(ii) Mg²⁺/Mg = – 2.37V
(iii) Cr³⁺/Cr = – 0.74V
(iv) Hg²⁺/Hg = 0.79V
(v) Ag⁺/Ag = 0.80V
The Lower the reduction potential more the tendency to get oxidised and thus more will be reducing power.
Hence increasing order of reducing power is.:
Ag > Hg > Cr > Mg > K

Question 30.
Depict the galvanic cell in which the reaction
Zn(s) + 2Ag⁺(aq) → Zn⁺(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:
The galvanic cell will be
Zn(s), Zn²⁺(aq) // Ag⁺(aq), Ag(s)
(i) The Zn electrode will be negatively charged.
(ii) Current will flow from silver electrode to zinc electrode. Electrons flow from Zn to Cu outside ions carry current in the cell.
(iii) Individual reaction at each electrode will be
At anode : oxidation half reaction
Zn(s) → Zn²⁺ + 2e^–
At cathode : reduction half reaction
2Ag⁺ (aq) + 2e^– → 2Ag(s)