Buy BOOKS at Discounted Price

Redox Reactions

Class 11th Chemistry Part Ii Bihar Board Solution
Exercise
  1. NaH2PO4 Assign oxidation number to the underlined elements in each of the following…
  2. NaHSO4 Assign oxidation number to the underlined elements in each of the following…
  3. H4P2O7 Assign oxidation number to the underlined elements in each of the following…
  4. K2MnO4 Assign oxidation number to the underlined elements in each of the following…
  5. CaO2 Assign oxidation number to the underlined elements in each of the following species:…
  6. NaBH4 Assign oxidation number to the underlined elements in each of the following species:…
  7. H2S2O7 Assign oxidation number to the underlined elements in each of the following…
  8. KAl(SO4)2.12 H2O Assign oxidation number to the underlined elements in each of the…
  9. KI3 What are the oxidation number of the underlined elements in each of the following and…
  10. H2S4O6 What are the oxidation number of the underlined elements in each of the following…
  11. Fe3O4 What are the oxidation number of the underlined elements in each of the following…
  12. CH3CH2OH What are the oxidation number of the underlined elements in each of the following…
  13. CH3COOH What are the oxidation number of the underlined elements in each of the following…
  14. CuO(s) + H2(g) → Cu(s) + H2O(g) Justify that the following reactions are redox reactions:…
  15. Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) Justify that the following reactions are redox…
  16. 4BCl3(g) + 3LiAlH4(s) → 2B2H6(g) + 3LiCl(s) + 3 AlCl3 (s) Justify that the following…
  17. 2K(s) + F2 (g) → 2K + F-(s) Justify that the following reactions are redox reactions:…
  18. 4 NH3(g) + 5 O2 (g) → 4NO(g) + 6H2O(g) Justify that the following reactions are redox…
  19. Fluorine reacts with ice and results in the change: H2O(s) + F2(g) → HF(g) + HOF(g)…
  20. Calculate the oxidation number of sulphur, chromium, and nitrogen in H2SO5, Cr2O2-7 and…
  21. Mercury(II) chloride Write formulas for the following compounds:
  22. Nickel(II) sulphate Write formulas for the following compounds:
  23. Tin(IV) oxide Write formulas for the following compounds:
  24. Thallium(I) sulphate Write formulas for the following compounds:
  25. Iron(III) sulphate Write formulas for the following compounds:
  26. Chromium(III) oxide Write formulas for the following compounds:
  27. Suggest a list of the substances where carbon can exhibit oxidation states from-4 to + 4…
  28. While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing…
  29. (a) 6 CO2(g) + 6H2O(l) → C6 H12 O6(aq) + 6O2(g) (b) O3(g) + H2O2(l) → H2O(l) + 2O2(g) Why…
  30. The compound AgF2 is unstable compound. However, if formed, the compound acts as a very…
  31. Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a…
  32. Though alkaline potassium permanganate and acidic potassium permanganate both are used as…
  33. When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we…
  34. 2AgBr (s) + C6H6O2 (aq) → 2Ag(s) + 2HBr (aq) + C6H4O2(aq) Identify the substance oxidized…
  35. HCHO(l) + 2[Ag (NH3)2] + (aq) + 3OH- (aq) → 2Ag(s) + HCOO- (aq) + 4NH3(aq) + 2H2O(l)…
  36. HCHO (l) + 2 Cu2+ (aq) + 5 OH- (aq) → Cu2O(s) + HCOO-(aq) + 3H2O(l) Identify the substance…
  37. N2H4 (l) + 2H2O2 (l) → N2(g) + 4H2O(l) Identify the substance oxidized reduced, oxidizing…
  38. Pb(s) + PbO2(s) + 2H2SO4(aq) → 2PbSO4(s) + 2H2O(l) Identify the substance oxidized…
  39. Why does the same reductant, thiosulphate react differently with iodine and bromine?…
  40. Justify giving reactions that among halogens, fluorine is the best oxidant and among…
  41. Why does the following reaction occur? What conclusion about the compound Na4XeO6 (of…
  42. Consider the reactions: (a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) H3PO4(aq) + 4Ag(s) +…
  43. (in basic medium) Balance the following redox reactions by ion - electron method:…
  44. (in acidic solution) Balance the following redox reactions by ion - electron method:…
  45. Balance h2o2+fe2+=fe3++h2o in acidic medium
  46. (in acidic solution) Balance the following redox reactions by ion - electron method:…
  47. P4(s) + OH-(aq) → PH3(g) + HPO-2(aq) Balance the following equations in basic medium by…
  48. N2H4(l) + ClO-3(aq) → NO(g) + Cl-(g) Balance the following equations in basic medium by…
  49. Cl2O7(g) + H2O2(aq) → ClO-2(aq) + O2(g) + H+ Balance the following equations in basic…
  50. What sorts of informations can you draw from the following reaction? (CN)2(g) + 2OH-(aq) →…
  51. The Mn3+ ion is unstable in solution and undergoes disproportionation to give Mn2+, MnO2,…
  52. Consider the elements: Cs, Ne, I and F (a) Identify the element that exhibits only…
  53. Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of…
  54. Refer to the periodic table given in your book and now answer the following questions: (a)…
  55. In Ostwald’s process for the manufacture of nitric acid, the first step involves the…
  56. Fe3+(aq) and I-(aq) Using the standard electrode potentials given in the Table 8.1,…
  57. Ag+(aq) and Cu(s) Using the standard electrode potentials given in the Table 8.1, predict…
  58. Fe3+(aq) and Cu(s) Using the standard electrode potentials given in the Table 8.1, predict…
  59. Ag(s) and Fe3+(aq) Using the standard electrode potentials given in the Table 8.1, predict…
  60. Br2(aq) and Fe2+(aq). Using the standard electrode potentials given in the Table 8.1,…
  61. Predict the products of electrolysis in each of the following: (i) An aqueous solution of…
  62. Arrange the following metals in the order in which they displace each other from the…
  63. Given the standard electrode potentials, K+/K = -2.93V, Ag+/Ag = 0.80V, Hg2+/Hg = 0.79V…
  64. Depict the galvanic cell in which the reaction Zn(s) + 2Ag+(aq) → Zn2+(aq) +2Ag(s) takes…

Exercise
Question 1.

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

NaH2PO4


Answer:

Let the oxidation number(O.N.) of P be ‘x’


We know that,


O.N. of Na = + 1


O.N. of H = + 1


O.N. of O = -2


Therefore, we have,


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


1 + 2 + x-8 = 0


X = + 5


The O.N. of P is + 5.



Question 2.

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

NaHSO4


Answer:

Let the oxidation number (O.N.) of S be ‘x’


We know that,


O.N. of Na = + 1


O.N. of H = + 1


O.N. of SO4 = -2


Therefore, we have,


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


1 + 1 + x-8 = 0


X = + 6


The O.N. of S is + 6.



Question 3.

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

H4P2O7


Answer:

Let the oxidation number(O.N.) of P be ‘x’


We know that,


O.N. of H = + 1


O.N. of O = -2


Therefore, we have,


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


4 + 2x-14 = 0


X = + 5


The O.N. of P is + 5.



Question 4.

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

K2MnO4


Answer:

Let the oxidation number(O.N.) of Mn be ‘x’


We know that,


O.N. of K = + 1


O.N. of O = -2


Therefore, we have,


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


2 + x-8 = 0


X = + 6


The O.N. of Mn is + 6.



Question 5.

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

CaO2


Answer:

Let the oxidation number(O.N.) of O be ‘x’


We know that,


O.N. of Ca = + 2


Therefore, we have,


(+2) + 2(x) = 0


X = -1


The O.N. of O is -1.


Question 6.

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

NaBH4


Answer:

Let the oxidation number(O.N.) of B be ‘x’


We know that,


O.N. of Na = + 1


O.N. of H = -1


Therefore , we have,


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


1 + x-4 = 0


X = + 3


The O.N. of B is + 3.



Question 7.

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

H2S2O7


Answer:

Let the oxidation number(O.N.) of S be ‘x’


We know that,


O.N. of H = + 1


O.N. of O = -2


Therefore, we have,


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


2x = 12


X = + 6


The O.N. of S is + 6.



Question 8.

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

KAl(SO4)2.12 H2O


Answer:

Let the oxidation number(O.N.) of S be ‘x’


We know that,


O.N. of K = + 1


O.N. of H = + 1


O.N. of O = -2


O.N. of Al = + 3


Therefore, we have,


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


1 + 3 + 2x-16 + 24-24 = 0


X = + 6


The O.N. of S is + 6.



Question 9.

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

KI3


Answer:

In KI3, O.N. of Kis + 1. Hence, the average O.N. is


However, O.N. cannot be fractional. Therefore, we will have to consider the structure of KI3 to find the oxidation states.


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



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



Question 10.

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

H2S4O6


Answer:

Let the O.N. of S be ‘x’.


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


2 + 4x-12 = 0


X = + 2


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



The O.N. of two of the four S atoms is + 5 and the O.N. of the other two S atoms is 0.



Question 11.

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

Fe3O4


Answer:

On taking the O.N. of O as -2,the O.N. of fe is found to be + 2. However, O.N. cannot be fractional. Here, one of the three Fe atoms exhibits the O.N. of + 2and the other two Fe atoms exhibit the O.N. o + 3.


i.e. in the form of FeO and Fe2O3.



Question 12.

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

CH3CH2OH


Answer:

C2H6O


Let the oxidation number(O.N.) of C be ‘x’


We know that,


O.N. of H = + 1


O.N. of O = -2


Therefore , we have,


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


2x + 6-2 = 0


X = -2


The O.N. of C is -2.



Question 13.

What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

CH3COOH


Answer:

C2H4O2


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


2x + 4-4 = 0


X = 0


However,0 is average O.N. of C. The two carbon atoms present in this molecule are present in the different environments. Hence, they cannot have the same oxidation number.


Thus, C exhibits the oxidation states of + 2 and -2 in CH3COOH.




Question 14.

Justify that the following reactions are redox reactions:

CuO(s) + H2(g) → Cu(s) + H2O(g)


Answer:

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


In CuO, Cu = + 2 and O = -2


In H2, H = 0


In H2O, H = + 1 and O = -2


In Cu, Cu = 0


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 0in H2 to + 1in H2O i.e.,H2is oxidised to H2O. Hence, this reaction is a redox reaction.



Question 15.

Justify that the following reactions are redox reactions:

Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)


Answer:

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


In Fe2O3, Fe = + 3 and O = -2


In CO, C = + 2 and O = -2


In CO2, C = + 4 and O = -2


In Fe, Fe = 0


Here, the O.N. of Fe decreases from + 3 in Fe2O3to 0 in Fe i.e. Fe2O3 is reduced to Fe. On the other hand, the O.N. of C increases from + 2 in CO to + 4 in CO2. Hence, the given reaction is a redox reaction.



Question 16.

Justify that the following reactions are redox reactions:

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


Answer:

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


In BCl3, B = + 3 and Cl = -1


In LiAlH4, Li = + 1, Al = + 3 and H = -1


In B2H6,B = -3 and H = + 1


In LiCl, Li = + 1 and Cl = -1


In AlCl3, Al = + 3 and Cl = -1


In this reaction, O.N. of B decreases from + 3 in BCl3 to -3 in B2H6 i.e. BCl3 is reduced to B2H6. Also, the O.N. of H increases from -1 in LiAlH4 to + 1 in B2H6 i.e. is LiAlH4 oxidised to B2H6.


Hence, the given reaction is a redox reaction.



Question 17.

Justify that the following reactions are redox reactions:

2K(s) + F2 (g) → 2K + F(s)


Answer:

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


K = 0 & + 1


F = 0 &-1


Oxidation number of K increases from 0 in K to + 1 in KF i.e. K is oxidised to KF. on the other hand, the O.N. of F decreases from 0 in F2 to -1 in KF i.e. F2 is being reduced to KF.


Hence, the given reaction is a redox reaction.



Question 18.

Justify that the following reactions are redox reactions:

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


Answer:

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


In NH3, N = -3 & H = + 1


In NO, N = + 2 & O = -2


In H2O, H = + 1 &O = -2


In O2, O = 0


Here, the oxidation number of N increases from -3 in NH3 to + 2 in NO. on the other hand, the O.N. of O2 decreases from 0 to -2 i.e. O2 is reduced.


Hence, the given reaction is a redox reaction.



Question 19.

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 O.N. dereases from 0 in F2 to -1 in HF. Thus, in the above reaction, F is both oxidised and reduced. Hence, the given reaction is a redox reaction.



Question 20.

Calculate the oxidation number of sulphur, chromium, and nitrogen in H2SO5, Cr2O2–7 and NO3. Suggest structure of these compounds. Count for the fallacy.


Answer:

(i) H2SO5


Let the oxidation number(O.N.) of S be ‘x’


We know that,


O.N. of H = + 1


O.N. of O = -2


Therefore, we have,


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


2 + x-10 = 0


X = + 8


The O.N. of S is + 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:



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


2 + x-6-2 = 0


X = + 6


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


(ii) Cr2O72-


Let the oxidation number (O.N.) of Cr be ‘x’


O.N. of O = -2


Therefore, we have,


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


2x-14 = -2


X = + 6


The O.N. of Cr is + 6.


Here, there is no fallacy about the O.N. of Cr in Cr2O72-


The structure of Cr2O72- is shown as follows



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


(iii) NO3-


Let the oxidation number (O.N.) of N be ‘x’


O.N. of O = -2


Therefore, we have,


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


1x-6 = -1


X = + 5


Here, there is no fallacy about the O.N. of N in NO3-.


The structure of NO3- is shown as follows:



The N atom exhibits the O.N. of + 5.



Question 21.

Write formulas for the following compounds:

Mercury(II) chloride


Answer:

HgCl2


The symbol for mercury is ‘Hg’ and for chloride is ‘Cl’ and since the mercury is present in + 2 oxidation state i.e. the formula is HgCl2.



Question 22.

Write formulas for the following compounds:

Nickel(II) sulphate


Answer:

NiSO4


The symbol for Nickel is ‘Ni’ and for sulphate is ‘SO4’ and since the nickel and sulphate both are present in + 2 oxidation state, therefore the oxidation state of both of them gets reduced to their simplest form i.e. the formula is NiSO4.



Question 23.

Write formulas for the following compounds:

Tin(IV) oxide


Answer:

SnO2


The symbol for tin is ‘Sn’ and for oxide is ‘O’ and since the Tin is present in + 4 oxidation state and oxide is present in + 2 oxidation state, therefore when both of them are reduced in their simplest form the formula becomes SnO2.



Question 24.

Write formulas for the following compounds:

Thallium(I) sulphate


Answer:

Tl2SO4


The symbol for Thallium is ‘Tl’ and for sulphate is ‘SO4’ and since the thallium is present in + 1 oxidation state and sulphate has -2 charge on it so the formula is Tl2SO4.



Question 25.

Write formulas for the following compounds:

Iron(III) sulphate


Answer:

Fe2(SO4)3


The symbol for Iron is ‘Fe’ and for Sulphate is ‘SO4’ and since the iron is present in + 3 oxidation state and sulphate carries -2 charge so the formula is Fe2(SO4)3.



Question 26.

Write formulas for the following compounds:

Chromium(III) oxide


Answer:

Cr2O3


The symbol for Chromium is ‘Cr’ and for oxide is ‘O’ and since the Chromium is present in + 3 oxidation state and oxide has -2 therefore when the charge gets cross-multiplied, the formula becomes Cr2O3.



Question 27.

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 following:-


Let the O.N. of C be ‘x’



The substances where Nitrogen can exhibit oxidation states from -3 to + 5 are following:-


Let the O.N. of N be ‘y’




Question 28.

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 oxidizing as well as a reducing agent.


Let the oxidation number of S in SO2 be x.


X + 2(-2) = 0 {O.N. of O = -2}


X = + 4


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 oxidizing as well as a reducing agent.


Let the oxidation number of O in H2O2 be x.


2(+ 1) + 2x = 0 {O.N. of H = + 1}


X = -1


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 can act only as an oxidant.


In nitric acid(HNO3), the O.N. of N is + 5 and the range of the O.N. that N cam 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.


Let the O.N. of N in HNO3 be ‘x’.


1 + x + 3(-2) = 0 {O.N. of H = + 1 and of O = -2}


X = + 5



Question 29.

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) + 2O2(g) + O2(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.


2 H2O(l)→ 2H2(g) + O2(g)


Step 2:


The H2 produced in step 1 reduces CO2, thereby producing glucose (C6 H12 O6) and H2O.


6CO2(g) + 12H2(g) → C6 H12 O6(aq) + 6H2O(l)


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 H2Oand O2.



The path of this reaction can be investigated by using H2O218 or O318.



Question 30.

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


Answer:

The oxidation state of 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.


Oxidation state of AgF2 :-


Let the oxidation state of Ag be x.


x + 2(-1) = 0 {oxidation state of F = -1}


x = 1



Question 31.

Whenever a reaction between an oxidizing 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 oxidizing agent is in excess. Justify this statement giving three illustrations.


Answer:

Whenever a reaction between an oxidizing 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 oxidizing agent is in excess. This can be illustrated as follows:


(a) C is a reducing agent, while O2acts as an oxidizing agent.


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


C(excess) + O2→ CO


Let the oxidation number of C be x in CO.


X + 1(-2) = 0 {oxidation number of O = -2}


X = 2


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


C + O2(excess) → CO2


Let the oxidation number of C be x in CO2.


X + 2(-2) = 0 {oxidation number of O = -2}


X = 4


(b) P4 and F2 are reducing and oxidizing agents respectively.


If an excess of P4 is treated with F2, then PF3 will be produced, wherein the O.N. of is + 3.


P4(excess) + F2→ PF3


Let the oxidation number of P be x in PF3.


X + 3(-1) = 0 {oxidation number of F = -1}


X = 3


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


P4 + F2(excess) → PF5


Let the oxidation number of P be x in PF5.


X + 5(-1) = 0 {oxidation number of F = -1}


X = 5


(c) K acts as a reducing agent, whereas O2 is an oxidizing agent.


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


4K(excess) + O2→ 2K2O


Let the oxidation number of O be x in K2O.


2(+ 1) + x = 0 {oxidation number of K = + 1}


X = -2


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


2K + O2(excess) → K2O2


let the oxidation number of O be x in K2O2.


2(+ 1) + 2x = 0 {oxidation number of K = + 1}


X = -1



Question 32.

How do you count for the following observations?

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.


Answer:

In the manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used as an oxidant because of the following reasons.


(i) KMnO4 and alcohol are homogeneous to each other since both are polar. Similarly, toluene and alcohol are also homogeneous to each other since both are organic compounds. Reactions a 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 given below:



(ii) 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.



Question 33.

How do you count for the following observations?

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:

When conc. H2SO4 is added to an inorganic mixture containing bromide, initially HBr is produced. HBr, being strong reducing agent reduces H2SO4 to SO2with the evolution of red vapour of bromine.



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 34.

Identify the substance oxidized reduced, oxidizing agent and a reducing agent for each of the following reactions:

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


Answer:

Oxidised substance:- C6H6O2


Reduced Substance:-AgBr


Reducing agent:- C6H6O2


Oxidising agent:-AgBr


Explanation:- Oxidising agent is a reagent which can increase the O.N. of an element in a given substance.


Here, AgBr helps carbon in C6H6O2 to increase its oxidation number.


The reduced substance is a reagent whose oxidation state has been decreased.


Here, in this case, Ag in AgBr has + 1 oxidation state and the O.N. is reduced to zero as Ag in the product side is present in zero O.N. (its elemental form)


The oxidised substance is a reagent whose oxidation state has been decreased.


C in C6H6O2 has O.N. of -0.33 which is being increased to zero in C6H4O2


Reducing Agent is a reagent which lowers the oxidation number of an element in a given substance.


C6H6O2 helps Ag of AgBr to decrease its oxidation state.



Question 35.

Identify the substance oxidized reduced, oxidizing agent and a reducing agent for each of the following reactions:

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


Answer:

Oxidised substance:-HCHO


Reduced Substance:- [Ag (NH3)2] +


Reducing agent:-HCHO


Oxidising agent:- [Ag (NH3)2] +


Explanation:- Oxidising agent is a reagent which can increase the O.N. of an element in a given substance.


Here, [Ag (NH3)2] + helps HCHO to increase its oxidation state.


The reduced substance is a reagent whose oxidation state has been decreased.


Here, in this case, Ag in [Ag (NH3)2] + has + 1 oxidation state and the O.N. is reduced to zero as Ag in the product side is present in zero O.N. (its elemental form)


The oxidised substance is a reagent whose oxidation state has been decreased.


Carbon in HCHO is present in zero oxidation state which is increased to + 2 in HCOO-.


Reducing Agent is a reagent which lowers the oxidation number of an element in a given substance.


HCHO helps [Ag (NH3)2]+ to decrease its oxidation state.



Question 36.

Identify the substance oxidized reduced, oxidizing agent and a reducing agent for each of the following reactions:

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


Answer:

Oxidised substance:- HCHO


Reduced Substance:- Cu2 +


Reducing agent:- HCHO


Oxidising agent:- Cu2+


Explanation:- Oxidising agent is a reagent which can increase the O.N. of an element in a given substance.


Cu+ 2 helps HCHO to increase its oxidation state.


The reduced substance is a reagent whose oxidation state has been decreased.


Here, in this case, Cu+2 has + 2 oxidation state and the O.N. is reduced to + 1 as Cu+1 in the product side is present in + 1 O.N.


The oxidised substance is a reagent whose oxidation state has been decreased.


Carbon in HCHO is present in zero oxidation state which is increased to + 2 in HCOO-.


Reducing Agent is a reagent which lowers the oxidation number of an element in a given substance.


HCHO helps Cu+2 to decrease its oxidation state.



Question 37.

Identify the substance oxidized reduced, oxidizing agent and a reducing agent for each of the following reactions:

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


Answer:

Oxidised substance:- N2H4


Reduced Substance:- H2O2


Reducing agent:- N2H4


Oxidising agent:- H2O2


Explanation:- Oxidising agent is a reagent which can increase the O.N. of an element in a given substance.


H2O2 helps nitrogen to increase its Oxidation state.


The reduced substance is a reagent whose oxidation state has been decreased.


Here in this case, O in H2O2 has -1 oxidation state and the O.N. is reduced to -2 as oxygen in the product side is present in -2 O.N.


The oxidised substance is a reagent whose oxidation state has been decreased.


N in N2H4 is present in -2 O.N which is increased to zero in the product side where N is present as N2 (its elemental form)


Reducing Agent is a reagent which lowers the oxidation number of an element in a given substance.


N2H4 helps oxygen to decrease its O.N.



Question 38.

Identify the substance oxidized reduced, oxidizing agent and a reducing agent for each of the following reactions:

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


Answer:

Oxidised substance:-Pb


Reduced Substance:- PbO2


Reducing agent:-Pb


Oxidising agent:- PbO2


Explanation:- Oxidising agent is a reagent which can increase the O.N. of an element in a given substance.


Here, the Pb of PbO2 helps Pb to increase its oxidation number.


The reduced substance is a reagent whose oxidation state has been decreased.


Here, in this case, Pb in PbO2 has + 4 oxidation state and the O.N. is reduced to + 2 as Pb in PbSO4 in the product side is present in + 2 O.N.


The oxidised substance is a reagent whose oxidation state has been decreased.


Here, in this case, Pb has zero oxidation state and the O.N. is increased to + 2 as Pb in PbSO4 in the product side is present in + 2 O.N.


Reducing Agent is a reagent which lowers the oxidation number of an element in a given substance.


Here, Pb helps itself to decrease its oxidation number.



Question 39.

Consider the reactions:



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


Answer:

The average O.N. of S in S2O32- is + 2 while in S4O62- it is + 2.5.The O.N. of S in S4O62- is + 6. Since Br2 is a stronger oxidizing agent than I2, it oxidizes S of S2O32- to a higher oxidation state of + 6 and hence forms SO42- ion. I2, however, being weaker oxidizing agent oxidizes S of S2O32- ion to a lower oxidation state of + 2.5 in S4O62- ion. It is because of this reason that thiosulphate reacts differently with Br2 and I2.



Question 40.

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


Answer:

F2 can oxidise Cl- to Cl2, Br- to Br2and I- to I2 as:


F2(aq) + 2Cl-(s)→ 2F-(aq) + Cl2(g)


F2(aq) + 2Br-(aq)→ 2F-(aq) + Br2(l)


F2(aq) + 2I-(aq)→ 2F-(aq) + I2(s)


On the other hand, Cl2, Br2and I2 cannot oxidise F- to F2. The oxidising 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


Again, I- can reduce Cu + 2 to Cu+, but Br-cannot.


4I-(aq) + 2Cu2+(aq) → Cu2I2(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 41.

Why does the following reaction occur?



What conclusion about the compound Na4XeO6 (of which is a part) can be drawn from the reaction.


Answer:

The given reaction occurs because XeO64- oxidises fluorine from -1 oxidation state to 0 oxidation state and reduces XeO64- from +8 oxidation state to +6 oxidation state as shown in the equation below :

+8XeO4- (aq) + 2F-1(aq) + 6H+1(aq) → +6XeO3(g) + F2 (g) + 3H2O (l)


Hence, we can easily conclude that XeO64- is a stronger oxidising agent than F-.



Question 42.

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:

In the above given reactions we can clearly see that:

(a) H3PO2(aq) + 4 AgNO3(aq) + 2 H2O(l) → H3PO4(aq) + 4Ag(s) + 4HNO3(aq)
Inference: Ag is getting reduced and P is getting oxidised.

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

Inference: Cu is getting reduced and P is oxidised.

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

Inference: Ag is getting reduced and C6H5CHO is getting oxidised.

(d) C6H5CHO(l) + 2Cu2+(aq) + 5OH(aq) → No change observed.

Inference: No reaction as Cu cannot oxidize C6H5CHO.

Hence, as a whole 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+.


Question 43.

Balance the following redox reactions by ion – electron method:

(in basic medium)


Answer:

Step 1: The two half reactions involved in the given reaction are:


The oxidation half reaction is as below :


I-(aq)→ I2(s)


The reduction half reaction is as below :


MnO-4(aq) → MnO2(aq)


Step 2: Balancing I in the oxidation half reaction, we have:


2I-(aq)→ I2(aq)


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


2I-(aq)→ I2(aq) + 2 e-


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-→ MnO2(aq)


Step 4:To balance reduction half reaction first balance oxidation number by writing required number of electrons to LHS and then balance charge by adding OH- ions because reaction occurs in basic medium.


MnO-4(aq) + 3e-→ MnO2(aq) + 4OH-


Step 5: Balance oxygen atoms by adding H2O to get :-


MnO-4(aq) + 3e- + 2H2O → MnO2(aq) + 4OH-


Step 6: Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have the following equations:-


6I-(aq)→ 3I2(aq) + 6 e-


2MnO-4(aq) + 6e- + 4H2O → 2MnO2(aq) + 8OH-


Step 7: Adding the two half reactions, we have the net balanced redox reaction as:-


6I-(aq) + 2MnO-4(aq) + 4H2O → 3I2(aq) + 2MnO2(aq) + 8OH-



Question 44.

Balance the following redox reactions by ion – electron method:

(in acidic solution)


Answer:

Following the steps as in part (A), we have the oxidation half reaction as:-


SO2(g) + 2H2O(l)→ HSO-4(aq) + 3H+(aq) + 2e-(aq)


And the reduction half reaction as:


MnO-4(aq) + 8H+(aq) + 5e-→ Mn2+(aq) + 4H2O


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 equation as:


2MnO-4(aq) + 5SO2(g) + 2H2O + H+(aq)→ 2Mn2+(aq) + 5HSO-4(aq)


Question 45.

Balance the following redox reactions by ion – electron method:

H2O2 (aq) + Fe2+ (aq) → Fe3+ (aq) + H2O (l) (in acidic solution)


Answer:

Following the steps as in part (A), we have the oxidation half reaction equation as:-


Fe2+(aq)→ Fe3+(aq) + e-


And the reduction half reaction as:-


H2O2(aq) + 2H+(aq) + 2e-→ 2H2O


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)→ 2 Fe3+(aq) + 2H2O



Question 46.

Balance the following redox reactions by ion – electron method:

(in acidic solution)


Answer:

Following the steps as in part (A), we have the oxidation half reaction equation as:-


SO2(g) + 2H2O → SO42-(aq) + 4H+(aq) + 2e-


And the reduction half reaction as:-


Cr2O72-(aq) + 14H+(aq) + 6e-→ 2Cr3+(aq) + 7H2O


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


Cr2O72-(aq) + 3 SO2(g) + 2H+(aq)→ 2Cr3+(aq) + 3SO42-(aq) + H2O



Question 47.

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

P4(s) + OH(aq) → PH3(g) + HPO2(aq)


Answer:

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 HPO2-.


Hence, P4 clearly acts both as an oxidizing agent and a reducing agent in this reaction.


Now, balancing the equation in basic medium by ion-electron method for the reduction half reaction: -


0P4 (s) → -3PH3(g)


Now, balancing P atoms :-


P4 (s) → 4PH3(g)


Balancing oxidation number by adding electrons: -


P4 (s) +12 e-→ 4PH3(g)


Balancing charge by adding OH- ions : -


P4 (s) +12 e-→ 4PH3(g) +12OH-1(aq)


Balancing 'O' atoms by adding H2O : -



Balancing the equation in basic medium by ion-electron method for the oxidation half reaction: -



Now, balancing P atoms : -



Balance oxidation number by adding electrons : -



Balance charge by adding OH- ions we get :-



Oxygen and hydrogen are balanced automatically.


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



Oxidation number method


The change in oxidation number are as follows : -



In a balanced chemical reaction loss of electrons = gain of electrons so we multiply H2PO2- by 3 and get : -



We multiply OH- by 3 to balance the charge and get : -



Balancing H by adding 3H2O to LHS we get the final balanced equation as: -




Question 48.

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

N2H4(l) + ClO3(aq) → NO(g) + Cl(g)


Answer:

The changes in the oxidation state taking place in the reaction are as shown below : -



Hence, we can infer easily that 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.


Solving by Ion–electron method:


The oxidation half equation is:-



The N atoms are balanced as:-


N2H4(l)→ 2NO(g)


The oxidation number is balanced by adding 8 electrons as:-


N2H4(l)→ 2NO(g) + 8e


The charge is balanced by adding 8 OH- ions as:-



The O atoms are balanced by adding 6H2O as:-



The reduction half equation is:-



The oxidation number is balanced by adding 6 electrons as:-



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



The O atoms are balanced by adding 3H2O as:-



Multiplying the oxidation half reaction by 3 and then adding it to the reduction half reaction multiplied by 4, to get the net balanced reaction equation as:-



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 ClO3- with 4 to balance the increase and decrease in O.N., we get:



The N and Cl atoms are balanced as:-



The O atoms are balanced by adding 6H2O to the required balanced equation:-




Question 49.

Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.

Cl2O7(g) + H2O2(aq) → ClO2(aq) + O2(g) + H+


Answer:

The oxidation number of Cl decreases from + 7 in Cl2O7 to + 3 in and the oxidation number of ClO2- increases from – 1 in H2O2 to zero in O2. Hence, in this reaction, Cl2O7 is the oxidizing agent and H2O2 is the reducing agent as could be clearly seen in the figure below: -



Solving by ion–electron method:


The oxidation half equation is:-



The oxidation number is balanced by adding 2 electrons as:-



The charge is balanced by adding 2OH- ions as:-



The oxygen atoms are balanced by adding 2H2O as:-



The reduction half equation is:-



The Cl atoms are balanced as:-



The oxidation number is balanced by adding 8 electrons as:-



The charge is balanced by adding 6OH- as:-



The oxygen atoms are balanced by adding 3H2O as:-



Multiplying the oxidation half reaction by 4 and then adding it to the reduction half reaction, we get the net balanced reaction equation as:-



Solving by oxidation number method:


Total decrease in oxidation number of Cl2O7= 4 × 2 = 8 Total increase in oxidation number of H2O2 = 2 × 1 = 2 .Now multiplying H2O2 and O2 with 4 to balance the increase and decrease in the oxidation number, we get:



The Cl atoms are balanced as:-



The O atoms are balanced by adding 3H2O as:-



The H atoms are balanced by adding 2OH- and 2H2O to get the desired balance equation :-




Question 50.

What sorts of informations can you draw from the following reaction?

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


Answer:

The oxidation numbers of carbon in (CN)2, CN- and CNO- are +3, +2 and +4 respectively as shown below : -


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 concluded that the alkaline decomposition of cyanogen is an example of disproportionation reaction.



Question 51.

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 follows:


The oxidation half equation is:-



The oxidation number is balanced by adding one electron as:



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



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



The reduction half equation is:



The oxidation number is balanced by adding one electron as:



The balanced chemical equation can be obtained by adding oxidation and reduction half equations to get the desired equation as: -




Question 52.

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 is the element that exhibits only negative oxidation state of –1.


b) Cs is the element that exhibits positive oxidation state of +1.


c) I is the element that 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. Hence, Ne is the element that exhibits neither the negative nor does the positive oxidation state.



Question 53.

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 given redox reaction can be represented as follows:-



The oxidation half reaction is:-



The oxidation number is balanced by adding two electrons as:-



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



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



The reduction half reaction is:-



The chlorine atoms are balanced as:-



The oxidation number is balanced by adding electrons as:-



The balanced chemical equation can be obtained by adding oxidation and reduction half equations to get the desired equation as: -




Question 54.

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 because these elements can exist in three or more oxidation states.


b) Mn, Cu, and Ga can show disproportionation reactions because these elements can exist in three or more oxidation states.



Question 55.

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 follows with the amount of substance reacting :-


Hence, we can clearly see that 68 g of NH3 reacts with 160 g of O2.


Therefore, 10g of NH3 will react with = 160 x 10 / 68g = 23.53 g of O2.


But the available amount of O2 is 20 g.


Therefore, O2 is the limiting reagent.


Now, 160 g of O2 gives 120g of NO.


Hence, 20 g of O2 will give = 120 X 20 /160 = 15 g of NO.


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



Question 56.

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

Fe3+(aq) and I(aq)


Answer:

The possible reaction between Fe3+(aq) and I(aq) can be represented by the following equation: –




E° for the overall reaction is positive. Thus, the reaction between Fe3+(aq) and I(aq) is feasible.



Question 57.

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

Ag+(aq) and Cu(s)


Answer:

The possible reaction between Ag+(aq) and Cu(s) can be represented by the following equation:–




E° for the overall reaction is positive. Thus, the reaction between Ag+(aq) and Cu(s) is feasible.



Question 58.

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

Fe3+(aq) and Cu(s)


Answer:

The possible reaction between Fe3+(aq) and Cu(s) can be represented by the following equation:–




E° for the overall reaction is positive. Thus, the reaction between Fe3+(aq) and Cu(s) is feasible.



Question 59.

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

Ag(s) and Fe3+(aq)


Answer:

The possible reaction between Ag(s) and Fe3+(aq) can be represented by the following equation: -




E° for the overall reaction is negative. Thus, the reaction between Ag(s) and Fe3+(aq) is not feasible.



Question 60.

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

Br2(aq) and Fe2+(aq).


Answer:

The possible reaction between Br2(aq) and Fe2+(aq) can be represented by the following equation:-




E° for the overall reaction is negative. Thus, the reaction between Br2(aq) and Fe2+(aq) is not feasible.



Question 61.

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 solutions to form Ag+ and NO3- ions.


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




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.




Therefore, Ag metal gets oxidized at the anode.


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


(iii) H2SO4 ionizes in aqueous solutions to give H+ and SO42- ions as follows –



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




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


On the other hand, at the anode, either of ions or H2O molecules can get oxidized. But the oxidation of involves breaking of more bonds than that of H2O molecules. Hence, 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 shown below:



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




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


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




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.



Question 62.

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.


Answer:

According to principle a metal of stronger reducing power displaces another metal of weaker reducing power from its solution of salt.

Hence, 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



Question 63.

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.


Answer:

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



Question 64.

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.


Answer:

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



(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn2+ and the electrons liberated accumulate on this electrode.


(ii) The carriers of current in the cell are the ions.


(iii) The reaction taking place at Zn electrode can be represented as the following equation:-



And the reaction which is taking place at the Ag electrode can be represented as in the following equation:-