NCERT Solutions For Class 11 Chemistry Chapter 8 Redox Reactions are provided here in detail. This topic is an extremely important topic of chemistry and is important for class 11, Class 12 and for competitive exams like JEE and NEET. Students must have a good knowledge of the topic in order to excel in the examination. We at BYJU’S provide the NCERT Solutions For Class 11 Chemistry Chapter 8 pdf which students can download. Practicing all the questions will be very helpful for the students as many questions are framed from the topic.

*1. Assign oxidation no. to the elements underlined:*

*(i) NaH2P−−O4*

*(ii) NaHS−−O4*

*(iii) H4P−−2O7*

*(iv) K2Mn−−−O4*

*(v) CaO−−2*

*(vi) NaB−−H4*

*(vii) H2S−−2O7*

*(viii) KAl(S−−O4)2.12H2O*

**Answer:**

(i)

Let x be the oxidation no. of P.

Oxidation no. of Na = +1

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

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

1 + 2 + x -8 = 0

x = +5

Therefore, oxidation no. of P is +5.

(ii)

Let x be the oxidation no. of S.

Oxidation no. of Na = +1

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

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

1 + 1 + x -8 = 0

x = +6

Therefore, oxidation no. of S is +6.

*(iii) H4P−−2O7*

Let x be the oxidation no. of P.

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

4(+1) + 2(x) + 7(-2) = 0

4 + 2x – 14 = 0

2x = +10

x = +5

Therefore, oxidation no. of P is +5.

*(iv) K2Mn−−−O4*

Let x be the oxidation no. of Mn.

Oxidation no. of K = +1

Oxidation no. of O = -2

Then,

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

2 + x – 8 = 0

x = +6

Therefore, oxidation no. of Mn is +6.

*(v) CaO−−2*

Let x be the oxidation no. of O.

Oxidation no. of Ca = +2

Then,

(+2) + 2(x) = 0

2 + 2x = 0

2x = -2

x = -1

Therefore, oxidation no. of O is -1.

*(vi) NaB−−H4*

Let x be the oxidation no. of B.

Oxidation no. of Na = +1

Oxidation no. of H = -1

Then,

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

1 + x -4 = 0

x = +3

Therefore, oxidation no. of B is +3.

*(vii) H2S−−2O7*

Let x be the oxidation no. of S.

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

2(+1) + 2(x) + 7(-2) = 0

2 + 2x – 14 = 0

2x = +12

x = +6

Therefore, oxidation no. of S is +6.

*(viii) KAl(S−−O4)2.12H2O*

Let x be the oxidation no. of S.

Oxidation no. of K = +1

Oxidation no. of Al = +3

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

1(+1) + 1(+3) + 2(x) + 8(-2) + 24(+1) + 12(-2) = 0

1 + 3 + 2x – 16 + 24 – 24 = 0

2x = +12

x = +6

Therefore, oxidation no. of S is +6.

OR

Ignore the water molecules because it is neutral. Then, the summation of the oxidation no. of all atoms of water molecules can be taken as 0. Hence, ignore the water molecule.

1(+1) + 1(+3) + 2(x) + 8(-2) = 0

1 + 3 + 2x -16 = 0

2x = 12

x = +6

Therefore, oxidation no. of S is +6.

*2.Assign oxidation no. to the elements underlined:*

*(i) KI–3*

*(ii) H2S−−4O6*

*(iii) Fe−−−3O4*

*(iv) C−−H3C−−H2OH*

*(v) C−−H3C−−OOH*

**Answer:**

*(i) KI–3*

Let x be the oxidation no. of I.

Oxidation no. of K = +1

Then,

1(+1) + 3(x) = 0

1 + 3x = 0

x =

Oxidation no. cannot be fractional. Hence, consider the structure of

In

Therefore, in

*(ii) H2S−−4O6*

Let x be the oxidation no. of S.

Oxidation no. of H = +1

Oxidation no. of O = -2

Then,

2(+1) + 4(x) + 6(-2) = 0

2 + 4x -12 = 0

4x = 10

x =

Oxidation no. cannot be fractional. Therefore, S would be present with different oxidation state in molecule.

The oxidation no. of two out of the four S atoms is +5 while that of other two atoms is 0.

*(iii) Fe−−−3O4*

Let x be the oxidation no. of Fe.

Oxidation no. of O = -2

Then,

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

3x -8 = 0

x =

Oxidation no. cannot be fractional.

One of the three atoms of Fe has oxidation no. +2 and other two atoms of Fe has oxidation no. +3.

*(iv) C−−H3C−−H2OH*

Let x be the oxidation no. of C.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

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

2x + 4 -2 = 0

x = -2

Therefore, oxidation no. of C is -2.

*(v) C−−H3C−−OOH*

Let x be the oxidation no. of C.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

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

2x + 4 -4 = 0

x = 0

Therefore, average oxidation no. of C is 0. Both the carbon atoms are present in different environments so they cannot have same oxidation no. Therefore, carbon has oxidation no. of +2 and _2 in

*3. The reactions given below are redox reactions. Justify the reactions.*

*(i) CuO(s)+H2(g)→Cu(s)+H2O(g)*

*(ii) Fe2O3(s)+3CO(g)→2Fe(s)+3CO2(g)*

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

*(iv) 2K(s)+F2(g)→2K+F(s)
*

*(v) 4NH3(g)+5O2(g)→4NO(g)+6H2O(g)
*

**Answer:**

**(i)**

Oxidation no. of Cu and O in

Oxidation no. of

Oxidation no. of Cu is 0.

Oxidation no. of H and O in

The oxidation no. of Cu decreased from +2 in

The oxidation no. of H increased from 0 to +1 in

Therefore, the reaction is redox reaction.

**(ii)**** **

In the above reaction,

Oxidation no. of Fe and O in

Oxidation no. of C and O in CO is +2 and -2 respectively.

Oxidation no. of Fe is 0.

Oxidation no. of C and O in

The oxidation no. of Fe decreased from +3 in

The oxidation no. of C increased from 0 to +2 in CO to +4 in

Therefore, the reaction is redox reaction.

**(iii)**

the above reaction,

Oxidation no. of B and Cl in

Oxidation no. of Li, Al and H in

Oxidation no. of B and H in

Oxidation no. of Li and Cl in LiCl is +1 and -1 respectively.

Oxidation no. of Al and Cl in

The oxidation no. of B decreased from +3 in

The oxidation no. of H increased from -1 in

Therefore, the reaction is redox reaction.

**(iv)**

In the above reaction,

Oxidation no. of K is 0.

Oxidation no. of F is 0.

Oxidation no. of K and F in KF is +1 and -1 respectively.

The oxidation no. of K increased from 0 in K to +1 in KF. That is K is oxidized to KF.

The oxidation no. of F decreased from 0 in

Therefore, the reaction is a redox reaction.

**(v)**** **

In the above reaction,

Oxidation no. of N and H in

Oxidation no. of

Oxidation no. of N and O in NO is +2 and -2 respectively.

Oxidation no. of H and O in

The oxidation no. of N increased from -3 in

The oxidation no. of

Therefore, the reaction is a redox reaction.

*4. Give the reaction of fluorine when reacts with ice:*

*Give reason that the above reaction is redox reaction.*

**Answer:**

In the above reaction,

Oxidation no. of H and O in

Oxidation no. of

Oxidation no. of H and F in HF is +1 and -1 respectively.

Oxidation no. of H, O and F in HOF is +1, -2 and +1 respectively.

The oxidation no. of F increased from 0 in

The oxidation no. of F decreased from 0 in

Therefore, F is both reduced as well as oxidized. So, it is redox reaction.

*5. Calculate the oxidation no. of chromium, sulphur and nitrogen in H2SO5, Cr2O2−7 and NO–3. Give the structure for the compounds. Count for the fallacy.*

**Answer:**

(i)

Let x be the oxidation no. of S.

Oxidation no. of O= -2

Oxidation no. of H = +1

Then,

2(+1) + 1(x) + 5(-2) = 0

2 + x – 10 = 0

x = +8

But the oxidation no. of S cannot be +8 as S has 6 valence electrons. Therefore, the oxidation no. of S cannot be more than +6.

The structure of

Now,

2(+1) + 1(x) + 3(-2) + 2(-1) = 0

2 + x – 6 – 2 = 0

x = +6

Therefore, the oxidation no. of S is +6.

(ii)

Let x be the oxidation no. of Cr.

Oxidation no. of O= -2

Then,

2(x) + 7(-2) = -2

2x -14 = -2

x = +6

There is no fallacy about the oxidation no. of Cr in

The structure of

Each of the two Cr atoms has the oxidation no. of +6.

(iii)

Let x be the oxidation no. of N.

Oxidation no. of O= -2

Then,

1(x) + 3(-2) = -1

x – 6 = -1

x = +5

There is no fallacy about the oxidation no. of N in

The structure of

Nitrogen atom has the oxidation no. of +5.

*6.Give the formula for the given compounds:*

*(i) Mercury*

* (II) chloride*

*(ii) Nickel (II) sulphate*

*(iii) Tin (IV) oxide*

*(iv) Thallium (I) sulphate*

*(v) Iron (III) sulphate*

*(vi) Chromium (III) oxide*

**Answer:**

(i) Mercury (II) chloride

(ii) Nickel (II) sulphate

(iii) Tin (IV) oxide

(iv) Thallium (I) sulphate

(v) Iron (III) sulphate

(vi) Chromium (III) oxide

**7.***Give a list of the compounds where carbon has oxidation no. from -4 to +4 and nitrogen from -3 to +5.*

**Answer:**

The compound where carbon has oxidation no. from -4 to +4 is as given below in the table:

Compounds | Oxidation no. of carbon |

0 | |

-1 | |

+1 | |

-2 | |

+2 | |

-3 | |

+3 | |

-4 | |

+4 |

Compounds | Oxidation no. of nitrogen |

0 | |

-1 | |

+1 | |

-2 | |

+2 | |

-3 | |

+3 | |

+4 | |

+5 |

*Hydrogen peroxide and sulphur dioxide can act as oxidizing and reducing agents in the reactions, nitric acid and ozone act only as oxidants. Why?*

**Answer:**

In sulphur dioxide (

Hence,

In hydrogen peroxide (

Therefore,

In ozone (

Therefore,

In nitric acid (

Therefore,

*9.Consider the following reactions:*

*(i) 6CO2(g)+6H2O(l)→C6H12O6(aq)+6O2(g)*

*(ii) O3(g)+H2O2(l)→H2O(l)+2O2(g)*

*It is suitable to write the above equations as given below. Why?*

*(i) 6CO2(g)+12H2O(l)→C6H12O6(aq)+6H2O(l)+6O2(g)*

*(ii) O3(g)+H2O2(l)→H2O(l)+O2(g)+O2(g)*

*Give a technique to find the path of above (i) and (ii) redox reactions.*

**Answer:**

(i)

Step 1 :

Step 2 :

The

The net reaction is as given below:

————————————————————————————————————————–

This is the suitable way to write the reaction as the reaction also produce water molecules in the photosynthesis process.

The path can be found with the help of radioactive

(ii)

Step 1 :

Step 2 :

———————————————————————————————-

The path can be found with the help of

*10. AgF2 is a compound which is unstable. Even if formed, the compound would act as a very strong oxidizing agent. Why?*

**Answer:**

The oxidation no. of Ag in

*11. When there is a reaction between reducing agent and oxidizing agent, a compound is formed which has lower oxidation number if the reducing agent is in excess and a compound is formed which has higher oxidation number if the oxidizing agent is in excess.*

*Justify the above statement with three examples.*

**Answer:**

When there is a reaction between reducing agent and oxidizing agent, a compound is formed which has lower oxidation number if the reducing agent is in excess and a compound is formed which has higher oxidation number if the oxidizing agent is in excess.

(i)

In an excess amount of

If

(ii) K and

If an excess of K reacts with

If K reacts with an excess of

(iii) C and

If an excess amount of C is reacted with insufficient amount of

If C is burnt in excess amount of

*12.How do you count for the following observations?*

*(i) Acidic potassium permanganate and alkaline 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.*

*(ii) 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:**

(i) While manufacturing benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant due to the given reasons.

(a) In a neutral medium,

(b)

The redox reaction is as given below:

(ii) When concentrated

When concentrated

*13. Find the oxidizing agent, reducing agent, the substance oxidized and reduced for the given reactions:*

*(i) 2AgBr(s)+C6H6O2(aq)→2Ag(s)+2HBr(aq)+C6H4O2(aq)*

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

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

*(iv) N2H4(l)+2H2O2(l)→N2(g)+4H2O(l)*

*(v) Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(aq)+2H2O(l)*

**Answer:**

(i)

AgBr => Reduced substance

AgBr =>Oxidizing agent

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

HCHO => Oxidized substance

HCHO=> Reducing agent

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

HCHO => Oxidized substance

HCHO => Reducing agent

*(iv) N2H4(l)+2H2O2(l)→N2(g)+4H2O(l)*

*(v) Pb(s)+PbO2(s)+2H2SO4(aq)→2PbSO4(aq)+2H2O(l)*

Pb=> Oxidized substance

Pb => Reducing agent

*14. Consider the given reactions:*

*Thiosulphate, the reductant, react differently with bromine and iodine. Why?*

**Answer:**

The average oxidation no. of S in

The average oxidation no. of S in

The oxidation no. of S in

The oxidation no. of S in

As

As

Thus, thiosulphate react differently with

*15.Among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.Justift this statement with the help of examples*

**Answer:**