NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Chemistry Class 11

Created by: Mohit Rajpoot

Class 11 : NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The document NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is a part of the Class 11 Course Chemistry Class 11.
All you need of Class 11 at this link: Class 11

NCERT QUESTIONS:

( Redox Reactions )

Ques 8.1: Assign oxidation numbers 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)2.12 H2O

Ans: (a) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev NotesNCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Then, we have

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, the oxidation number of P is + 5.

(b) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes Then, we have

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, the oxidation number of S is +6.

(c) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev NotesThen, we have

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, the oxidation number of P is + 5.

(d) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes Then, we have

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, the oxidation number of Mn is + 6.

(e) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes Then, we have

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, the oxidation number of O is – 1.

(f) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes Then, we have

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, the oxidation number of B is +3.

(g) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  Then, we have

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, the oxidation number of S is +6.

(h) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  Then, we have

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, the oxidation number of S is +6.


Ques 8.2: What are the oxidation numbers of the underlined elements in each of the following and how do you rationalise your results?

(a) KI3
 (b) H2S4O6 
 (c) Fe3O4 
 (d) CH3CH2OH
 (e) CH3COOH

Ans: (a)In KI3, the oxidation number (O.N.) of K is 1. Hence, the average oxidation number of I is  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes     . However, O.N. cannot be fractional . Therefore, we will have to consider the structure of KI3 to find the oxidation states.

In a KI3 molecule, an atom of iodine forms a coordinate covalent bond with an iodine molecule.

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, in a KI3 molecule, the O.N. of the two I atoms forming the I2 molecule is 0, whereas the O.N. of the I atom forming the coordinate bond is –1.

(b) H2S4O6

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes 2 + 4x – 12 = 0

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes 4x = 10

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes x = +2½

However, O.N. cannot be fractional. Hence, S must be present in different oxidation states in the molecule.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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 & the other two Fe atoms exhibit the O.N of +3.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

(d)NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes 2(x)+ 4( +1)+ 1(-2) = 0

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes 2x +6 – 2 = 0

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes x = – 2

Hence, the O.N. of C is –2.

(e)NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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 CH3COOH.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes


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

(a) CuO(s) +H2(g) → Cu(s) +H2O(g)

(b) Fe2O3(s) +3CO(g) → 2Fe(s) +3CO2(g)

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

(d) 2K(s) +F2(g) → 2K+F- (s)

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

Ans: (a) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Let us write the oxidation number of each element involved in the given reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Let us write the oxidation number of each element in the given reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Here, the oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe i.e., Fe2O3 is reduced to Fe. On the other hand, the oxidation number of C increases from +2 in CO to +4 in COi.e., CO is oxidized to CO2. Hence, the given reaction is a redox reaction.

(c) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number of each element in the given reaction can be represented as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

In this reaction, the oxidation number of B decreases from +3 in BCl3 to –3 in B2H6. i.e., BCl3 is reduced to B2H6. Also, the oxidation number of H increases from –1 in LiAlH4 to +1 in B2H6 i.e., LiAlH4 is oxidized to B2H6. Hence, the given reaction is a redox reaction.

(d) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number of each element in the given reaction can be represented as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

In this reaction, the oxidation number of K increases from 0 in K to +1 in KF i.e., K is oxidized to KF. On the other hand, the oxidation number of F decreases from 0 in F2 to –1 in KF i.e., F2 is reduced to KF.

Hence, the above reaction is a redox reaction.

(e) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number of each element in the given reaction can be represented as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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.


Ques 8.4: Fluorine reacts with ice and results in the change:

H2O(s) +F2(g) → HF(g) +HOF(g)

Justify that this reaction is a redox reaction.

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

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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 Fto –1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.


Ques 8.5: Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes. Suggest structure of these compounds. Count for the fallacy.

Ans: (i) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Therefore, the O.N. of S is +6.

(ii)NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Here, there is no fallacy about the O.N. of Cr in NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes.

The structure of NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is shown as follows:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Here, each of the two Cr atoms exhibits the O.N. of +6.

(iii) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Here, there is no fallacy about the O.N. of N in NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes.

The structure of NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is shown as follows:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The N atom exhibits the O.N. of +5.


Ques 8.6: Write the formulae 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

Ans: (a) Mercury (II) chloride: HgCl2

(b) Nickel (II) sulphate: NiSO4

(c) Tin (IV) oxide: SnO2

(d) Thallium (I) sulphate: Tl2SO4

(e) Iron (III) sulphate: Fe2(SO4)3

(f) Chromium (III) oxide: Cr2O3


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

Ans: The substances where carbon can exhibit oxidation states from –4 to +4 are listed in the following table.

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 

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

Ans: 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, H2O2can 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.


Ques 8.9: Consider the reactions:

(a) 6 CO2(g) +6H2O(l) → C6 H12 O6(aq)+ 6O2(g)

(b) O3(g) +H2O2(l) → H2O(l) +2O2(g)

Why it is more appropriate to write these reactions as:

(a) 6CO2(g) +12H2O(l) → C6 H12 O6(aq)+ 6H2O(l) +6O2(g)

(b) O3(g) +H2O2 (l) → H2O(l) +O2(g)+ O2(g)

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

Ans:  (a)The process of photosynthesis involves two steps.

Step 1:

H2O decomposes to give H2 and O2.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Step 2:

The H2 produced in step 1 reduces CO2, thereby producing glucose (C6H12O6) and H2O.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Now, the net reaction of the process is given as:

  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The path of this reaction can be investigated by using NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes or NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  .

 

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

Ans: 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.


Ques 8.11: Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.

Ans: 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.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

However, if P4 is treated with an excess of F2, then PF5 will be produced, wherein the O.N. of P is +5.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

(ii)K acts as a reducing agent, whereas O2 is an oxidising agent.

If an excess of K reacts with O2, then K2O will be formed, wherein the O.N. of O is –2.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

However, if K reacts with an excess of O2, then K2O2 will be formed, wherein the O.N. of O is –1.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

(iii)C is a reducing agent, while O2 acts as an oxidising agent.

If an excess of C is burnt in the presence of insufficient amount of O2, then CO will be produced, wherein the O.N. of C is +2.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

On the other hand, if C is burnt in an excess of O2, then CO2 will be produced, wherein the O.N. of C is 4.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes


Ques 8.12: How do you count for the following observations?

(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why? Write a balanced redox equation for the reaction.

(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why?

Ans:  (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:

  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

(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.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

But, when conc. H2SOis added to an inorganic mixture containing chloride, a pungent smelling gas (HCl) is evolved. HCl, being a weak reducing agent, cannot reduce H2SOto SO2.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes


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

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

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

(c) HCHO (l) +2Cu2 (aq) +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)

Ans:  (a) Oxidised substance → C6H6O2

Reduced substance → AgBr

Oxidising agent → AgBr

Reducing agent → C6H6O2

(b)Oxidised substance → HCHO

Reduced substance → NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Oxidising agent → NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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


Ques 8.14: Consider the reactions:

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes(aq)+ I2(s) → NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes (aq)+ 2I(aq)

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes(aq)+ 2Br2(l)+ 5 H2O(l) →  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes(aq) +4Br(aq) + 10H+(aq) 

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

Ans: The average oxidation number (O.N.) of S in NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is +2. Being a stronger oxidising agent than I2, Br2 oxidises NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  to  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes, in which the O.N. of S is +6. However, I2 is a weak oxidising agent. Therefore, it oxidises NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes to NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes , in which the average O.N. of S is only +2.5. As a result, NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes reacts differently with iodine and bromine.


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

Ans: F2 can oxidize Cl– to Cl2, Br to Br2, and I to I2 as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Again, I can reduce Cu2+    to Cu , but Br cannot.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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.


Ques 8.16: Why does the following reaction occur?

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes(aq) +2F (aq)+ 6 H+(aq) → XeO3(g)+ F2(g)+ 3H2O(l)

What conclusion about the compound Na4XeO6 (of which NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  is a part) can be drawn from the reaction.

Ans: The given reaction occurs because NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes oxidises NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  and  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notesreduces NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes to +6 in XeO3 and the O.N. of F increases from –1 in F– to O in F2.

Hence, we can conclude that NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is a stronger oxidising agent than F.

Ques 8.17: 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?

Ans:  Ag+ and Cu2+   act as oxidising agents in reactions (a) and (b) respectively.

In reaction (c), Ag+ oxidises C6H5CHO to C6H5COO, but in reaction (d), Cu2+   cannot oxidise C6H5CHO.

Hence, we can say that Ag+  is a stronger oxidising agent than Cu2+   .


Ques 8.18: Balance the following redox reactions by ion-electron method:

(a)  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes (aq)+ I (aq) → MnO2 (s) +I2(s) (in basic medium)

(b)   NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes(aq) +SO2 (g) → Mn2+   (aq) +NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes   (aq) (in acidic solution)

(c) H2O2 (aq)+ Fe 2+   (aq) →  Fe3+ (aq)+  H2O (l) (in acidic solution)

(d) NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes+  SO2(g) → Cr3 +   (aq)+NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes(aq) (in acidic solution)

Ans:  (a) Step 1: The two half reactions involved in the given reaction are:

Oxidation half reaction:NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Reduction half reaction:NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Step 2:

Balancing I in the oxidation half reaction, we have:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Now, to balance the charge, we add 2 e to the RHS of the reaction.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Step 5:

Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Step 6:

Adding the two half reactions, we have the net balanced redox reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

(b)Following the steps as in part (a), we have the oxidation half reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

And the reduction half reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

(c) Following the steps as in part (a), we have the oxidation half reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

And the reduction half reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Multiplying the oxidation half reaction by 2 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

(d) Following the steps as in part (a), we have the oxidation half reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

And the reduction half reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction, we have the net balanced redox reaction as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes


Ques 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.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Ans: (a)

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The O.N. (oxidation number) of P decreases from 0 in P4 to – 3 in PH3 and increases from 0 in P4 to +2 in NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes. Hence, P4 acts both as an oxidizing agent and a reducing agent in this reaction.

Ion–electron method:

The oxidation half equation is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The P atom is balanced as:

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The reduction half equation is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The P atom is balanced as

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  

By multiplying equation (i) with 3 and (ii) with 2 and then adding them, the balanced chemical equation can be obtained as:

5P4 (s) + 24H2O → 12HPO2+ 8PH3(g) + 12H+

As, the medium is basic, add 12OH both sides as:

5P4 (s) + 12H2O + 12OH− →12HPO2-  + 8PH3(g)

This is the required balanced equation.

Oxidation number method:

Let, total no of P reduced = x
 ∴ 
Total no of P oxidised = 4 x

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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)

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

This is the required balanced equation.

 

(b)

  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number of N increases from – 2 in N2H4 to +2 in NO and the oxidation number of Cl decreases from +5 in  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes to – 1 in Cl. Hence, in this reaction, N2H4 is the reducing agent and  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is the oxidizing agent.

Ion–electron method:

The oxidation half equation is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The N atoms are balanced as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number is balanced by adding 8 electrons as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The charge is balanced by adding 8 OHions as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The O atoms are balanced by adding 6H2O as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The reduction half equation is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number is balanced by adding 6 electrons as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The charge is balanced by adding 6OH– ions as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The O atoms are balanced by adding 3H2O as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The balanced equation can be obtained by multiplying equation (i) with 3 and equation (ii) with 4 and then adding them as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  with 4 to balance the increase and decrease in O.N., we get:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The N and Cl atoms are balanced as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The O atoms are balanced by adding 6H2O as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

This is the required balanced equation.

(c)

  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number of Cl decreases from +7 in Cl2O7 to +3 in NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes 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 equation is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number is balanced by adding 2 electrons as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The charge is balanced by adding 2OHions as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxygen atoms are balanced by adding 2H2O as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The reduction half equation is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The Cl atoms are balanced as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number is balanced by adding 8 electrons as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The charge is balanced by adding 6OH as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxygen atoms are balanced by adding 3H2O as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The balanced equation can be obtained by multiplying equation (i) with 4 and adding equation (ii) to it as: 

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Oxidation number method:

Total decrease in oxidation number of Cl2O= 4 × 2 = 8

Total increase in oxidation number of H2O2 = 2 × 1 = 2

By multiplying H2O2 & O2 with 4 to balance the increase & decrease in the oxidation number, we get:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The Cl atoms are balanced as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The O atoms are balanced by adding 3H2O as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The H atoms are balanced by adding 2OH and 2H2O as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

This is the required balanced equation.


Ques 8.20: What sorts of information can you draw from the following reaction ?

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Ans: 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

= 3
CN

x – 3 = –1

= 2
CNO
x – 3 – 2 = –1

= 4
The oxidation number of carbon in the various species is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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.


Ques 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.

Ans: The given reaction can be represented as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation half equation is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number is balanced by adding one electron as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The charge is balanced by adding 4H +   ions as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The O atoms and H+  ions are balanced by adding 2H2O molecules as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The reduction half equation is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number is balanced by adding one electron as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes


Ques 8.22: Consider the elements: Cs, Ne, I and F

(a) Identify the element that exhibits only negative oxidation state.

(b) Identify the element that exhibits only postive oxidation state.

(c) Identify the element that exhibits both positive and negative oxidation states.

(d) Identify the element which exhibits neither the negative nor does the positive oxidation state.

Ans:  (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.


Ques 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.

Ans: The given redox reaction can be represented as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation half reaction is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number is balanced by adding two electrons as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The charge is balanced by adding 4H  ions as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The O atoms and H +   ions are balanced by adding 2H2O molecules as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The reduction half reaction is:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The chlorine atoms are balanced as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The oxidation number is balanced by adding electrons

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

The balanced chemical equation can be obtained by adding equation (i) and (ii) as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 

Ques 8.24: Refer to the periodic table given in your book and now answer the following questions:

(a) Select the possible non metals that can show disproportionation reaction.

(b) Select three metals that can show disproportionation reaction.

Ans: 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.

 

Ques 8.25: In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen?

Ans: The balanced chemical equation for the given reaction is given as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Thus, 68 g of NH3 reacts with 160 g of O2.

Therefore, 10g of NH3 reacts with NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes 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 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notesg of N, or 15 g of NO.

Hence, a maximum of 15 g of nitric oxide can be obtained.

 

Ques 8.26: Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible:

(a)  Fe3+(aq) and I(aq)
(b)  Ag+(aq) and Cu(s)
(c)  Fe3+ (aq) and Cu(s)
(d)  Ag(s) and Fe3+(aq)
(e) Br2(aq) and Fe2+(aq)

Ans:  (a) The possible reaction between NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is given by,

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

E° for the overall reaction is positive. Thus, the reaction between NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  and NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  is feasible.

(b) The possible reaction between NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is given by,

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

E° positive for the overall reaction is positive. Hence, the reaction between  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes &  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is feasible.

(c) The possible reaction between NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes and NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is given by,

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

E° positive for the overall reaction is positive. Hence, the reaction between  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes and NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  is feasible.

(d) The possible reaction between NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes and NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is given by,

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Here, E° for the overall reaction is negative. Hence, the reaction between NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes & NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes  is not feasible.

(e) The possible reaction between NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes and NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is given by,

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Here, E° for the overall reaction is positive. Hence, the reaction between NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes and  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes is feasible.


Ques 8.27: Predict the products of electrolysis in each of the following:

(i) An aqueous solution of AgNO3 with silver electrodes

(ii) An aqueous solution AgNO3 with platinum electrodes

(iii) A dilute solution of H2SO4 with platinum electrodes

(iv) An aqueous solution of CuClwith platinum electrodes.

Ans:  (i) AgNO3 ionizes in aqueous solutions to form Ag +   and  NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes ions.

On electrolysis, either Ag +   ions or H2O molecules can be reduced at the cathode. But the reduction potential of Ag +   ions is higher than that of H2O.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, Ag+    ions are reduced at the cathode. Similarly, Ag metal or H2O molecules can be oxidized at the anode. But the oxidation potential of Ag is higher than that of H2O molecules.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Therefore, Ag metal gets oxidized at the anode.

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

(iii) H2SO4 ionizes in aqueous solutions to give H +   and NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes ions.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

On electrolysis, either of H +   ions or H2O molecules can get reduced at the cathode. But the reduction potential of H +   ions is higher than that of H2O molecules.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, at the cathode, H +   ions are reduced to liberate H2 gas.

On the other hand, at the anode, either of NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes ions or H2O molecules can get oxidized. But the oxidation of NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes involves breaking of more bonds than that of H2O molecules. Hence, NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes ions have a lower oxidation potential than H2O. Thus, H2O is oxidized at the anode to liberate O2 molecules.

(iv) In aqueous solutions, CuCl2 ionizes to give Cu2 +   and Cl ions as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

On electrolysis, either of Cu2 +   ions or H2O molecules can get reduced at the cathode. But the reduction potential of Cu2 +   is more than that of H2O molecules.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

Hence, Cu2 +   ions are reduced at the cathode and get deposited.

Similarly, at the anode, either of Cl or H2O is oxidized. The oxidation potential of H2O is higher than that of Cl.

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

But oxidation of H2O molecules occurs at a lower electrode potential than 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.


Ques 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.

Ans: A metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.

The order of the increasing reducing power of the given metals is Cu < Fe < Zn < Al < Mg.

Hence, we can say that 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 given below:

Mg>Al> Zn> Fe,>Cu


Ques 8.29: Given the standard electrode potentials,

K +  /K = –2.93V, Ag /Ag = 0.80V,

Hg2 +  /Hg = 0.79V

Mg2 +  /Mg = –2.37V. 

Cr3 +  /Cr = –0.74V

Arrange these metals in their increasing order of reducing power.

Ans: The lower the electrode potential, the stronger is the reducing agent. Therefore, the increasing order of the reducing power of the given metals is Ag < Hg < Cr < Mg < K.


Ques 8.30: Depict the galvanic cell in which the reaction 

Zn(s)+ 2Ag +  (aq) → Zn2 +  (aq)+ 2Ag(s) takes place, further show:

(i) which of the electrode is negatively charged,

(ii) the carriers of the current in the cell, and

(iii) individual reaction at each electrode.

Ans: The galvanic cell corresponding to the given redox reaction can be represented as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2and the leaving electrons accumulate on this electrode.

(ii) Ions are the carriers of current in the cell.

(iii) The reaction taking place at Zn electrode can be represented as:

 NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

And the reaction taking place at Ag electrode can be represented as:

 . NCERT Solutions - Redox Reactions, Class 11, Chemistry | EduRev Notes

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