Redox Reactions - shikshahouse2019/01/08 · Redox Reactions 1) Assign oxidation numbers to the underlined elements in each of the following species: a) NaH 2PO 4 b) NaHSO 4 c) H

Redox Reactions

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)2.12H2O

Solution

a) +1 +2x -8

NaH2PO4

According to oxidation number (ON) rules, sum of all the atoms in an neutral molecule

should be zero.

Then,

1+2+x -8=0

So, x=+5

Hence, the ON of P is +5.

b) +1 +1x -8

NaHSO4


should be zero.

Then,

1+1+x -8=0

So, x=+6


c) +4 2x -14

H4P2O7


should be zero.

Then,

4+2x -14=0

So, x=+5


d) +2 x -8

K2MnO4


should be zero.

Then,

2-x -8=0

So, x=+6


e) x -2

CaO2


should be zero.

Then,

x -2=0

So, x=+2


f) + 1 x -4

NaBH4


should be zero.

Then,

x1+x-4=0

So, x=+3


g) +2 2x -14

H2S2O7


should be zero.

Then,

2+2x-14=0

So, x=+6


h) +1 +3 x -8 12x 0

KAl(SO4)2.12H2O


should be zero.

Then,

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

So, x=+6


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

Solution

a) +1 3x

K I3

1+3x =0

x= -1/3

ON of I = -1/3

The ON of iodine comes in a fraction . It is explained by the structure of I3- as

So, the average ON of I= -1/3

b) +1 4x -12

H2 S4 O6

2 + 4x - 12 =0

x= +5/2

The ON of S comes in a fraction, and it indicates that the S atoms are present in different

oxidation states.

It is explained as

i) There is no distribution of electrons between S-S bond, and so the ON of such S- atoms

is 0.

ii) The ON of the two terminal S-atoms is 5 each.

So, the average ON of S

= 5+0+0+5/4

= 2.5

The average ON of S is 2.5.

ii) c) Fe3O4 d) CH3CH2OH

c) 3x -8

Fe3 O4

3 x - 8=0

x=8/3

ON of Fe= +8/3

In fact, in Fe3O4 , the ON of the Fe atoms are +2 and +3, respectively.

+2 -2 +6 -6

Fe O . Fe2 O3

So, average

ON of Fe = 2+6/3 =+8/3

The average of ON of Fe is +8/3

d) CH3 CH2 OH

2x +6 -2

C2 H6 O

Here, 2x +6-2=0

x=-2

By convention ON of C should be -2 . But in the compound , two C –atoms have different

ON. It can be explained on basis of the structure of the compound.

So, the ON of one C-atom is -3.

The ON of other C-atom is -1

So, the average ON of C-atom= -3-1/2

= -2

The average ON of C-atom is -2.

iii) CH3COOH

Solution

2x +4 -4

C2 H4 O2

2x + 4 - 4 =0

= x=0

ON of C=0

By convention, the ON of C in compound should be 0. However, the C-atom in the

compound has different ON. It can be explained on the basis of its structure as

So, the ON of one C-atom is -3.

The ON of other C-atom is +3

So, the average ON of C-atom= -3+3/2

The average ON of C-atom is 0.

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) + 2CO2(g)

Solution

a) In terms of oxygen and hydrogen, the addition of oxygen or the removal of hydrogen is

oxidation, and the addition of hydrogen and the removal of oxygen is reduction.

b) In terms of loss and gain of electrons, and change in oxidation number, if there is an

increases in the oxidation number, it is oxidation, and if there is a decrease in the oxidation

number, it is reduction.

a) CuO(s) + H2(g) →Cu(s) +H2O(g)

+2 -2 0

Cu O→ Cu

i) Oxygen is removed from CuO, so it is reduced.

ii) When CuO is converted into Cu, there is a decrease in the oxidation number of Cu from

+2 to 0. Hence, it is reduction.

Cu from +2 to 0. Hence, it is reduction.

0 +2 -2

H2 → H2 O

i) H2 is oxidised due to the addition of oxygen.

ii) When H2 is converted into H2O, there is an increase in oxidation number of H from 0 to

+2 . Hence, it is oxidation.

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

+3 -6

Fe2 O3 → Fe

ON of Fe = +3 ON of Fe = 0

i) Oxygen is removed from Fe2O3. Therefore, it is reduction.

ii) A decrease in the oxidation number of Fe from +3 to 0. So, the conversion of Fe2O3 into

Fe is reduction.

+2 -2 +4 -4

C O → C O2

i) CO2 is formed by the addition of oxygen to CO. Therefore, it is oxidation.

ii) In the conversion of CO to CO2 there is an increase in the oxidation number of C from

+2 to +4. Hence, it is oxidation.

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

d) 2K (s) + F2(g) → 2K+F

-(s)

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

Solution

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

+3 -3 -6 +6

B Cl2 → B2 H6

(-3)

i) BCl3 is converted into B2H6 by the addition of H. So, it is reduction.

ii) When BCl3 is converted into B2H6 , there is a decrease in the oxidation number of boron

from +3 to -3. Therefore, it is reduction.

+1 +3 -4

Li Al H4 → LiCl +AlCl3

a) Removal of hydrogen from LiAlH4 . Therefore , it is oxidation. The oxidation number

of hydrogen increases from -1 in LiAlH4 to +1 in B2H6.

So, it is oxidation.

d) 2K (s) + F2 → 2K+F

-(g)

i) 0 +1

K → KF

There is an increase in the oxidation number of K from 0 to +1. So, the oxidation of K to

KF.

ii) F2 → K+F

-

a) Decrease in the ON of F from 0 to -1.

b) So, reduction of F2 to KF.

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

i) -3 +3 -2 +2

N H3 → N O

a) Removal of H from NH3.

b) Increase in the ON of N from -3 to +2.

So, oxidation of NH3 to NO.

ii) 0 -2

O2 → H2O

a) The addition of hydrogen to oxygen.

b) A decreases in the ON of oxygen from 0 to -2.

So, reduction of O2 to H2O.

Thus , all the above reactions are redox reactions.

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.

Solution

+2 -2 0 +1 -1 +1 0 -1

H2 O + F2 → H F + H O F

Here, the oxidation number of oxygen increases from -2 to 0. Hence, H2O is oxidised to

HOF. The oxidation number of F2 decreases from 0 to -1, so F2 is reduced to HF and HOF.

Hence, the reaction is a redox reaction.

5) Calculate the oxidation number of sulphur, chromium and nitrogen in H2SO5, Cr2O7-2

and NO3- .Suggest structure of these compounds. Count for the fallacy.

Solution

+2 x x

H2 S O5

2 + x – 10 = 0

x =+8

Oxidation number of S= +8

But the oxidation number of sulphur in any of its compounds does not exceed 6 . So, the

correct value of the oxidation number of S is obtained from its structure, in which it has a

peroxide bond and the two peroxide oxygen atoms have the oxidation number of -1 each

as

Oxidation number of S= 1-1-1 +x -6+1=0

∴ x=6

So, the oxidation number of S in H2SO5 in +6.

ii)

x -10

Cr O5

x - 10 =5

∴ x=10

oxidation number of Cr =+10

But the maximum oxidation number of Cr is +6, so it cannot be +10 . The oxidation

number of Cr in CrO5 is calculated from its structure which is observed to have two

peroxide bonds as

Then, x-4-2 =0

Oxidation number of Cr, x= +6

So, the oxidation number of Cr is +6.

iii)

x -6

N O3

x - 6 = -1

x= +5

Oxidation number of N= +5

There is no fallacy in the oxidation number of N in NO3- ion. Its structure is as

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

Solution

a) HgCl2

b) NiSO4

c) SnO2

d) Tl2SO4

e) Fe2(SO4)3

f) Cr2O3

7) Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4,

and nitrogen from -3 to +5.

Solution

Compounds of carbon

Compounds of Nitrogen

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 oxidant. Why?

Solution

In SO2 , the oxidation number of S is +4, which lies between the range of its minimum

oxidation number of -2 and maximum oxidation number of +6. In reactions of SO2 , the

oxidation number of S may decrease or increase from +4 . So, SO2 may act both as an

oxidising as well as a reducing agent.

In H2O2 , the oxidation number of O is -1, which lies in the range of its minimum

oxidation number of -2 and maximum oxidation number of 0. In reactions of H2O2 , the

oxidation number of oxygen may decrease or increase from -1. So, H2O2 may act both as

an oxidising and a reducing agent.

In O3 , the oxidation number of oxygen is zero. Its oxidation number may decrease, so it

acts as an oxidant. In HNO3 , the oxidation number of nitrogen is +5, which is its

maximum oxidation number. So, in its reaction, the oxidation number of N must decrease

from +5. Hence, HNO3 always acts only as an oxidant.

9) Consider the reactions:

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

b) O3 (g) + H2O2 (l) → H2O (l) + 6O2(g)

Why it is more appropriate to write these reactions as:

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

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.

Solution

a) The photosynthetic reaction in plants proceeds as follows:

12H2O→ 12H2 +6O2

6CO2 + 12H2 → C6H12O6 +6 H2O

So , the overall reaction is

C6H12O6 +6 H2O → C6H12O6 + 6O2 +6 H2O

b) The reaction proceeds in two steps as

O3 →O +O2

H2O2 + O → H2O + O2

So, the overall reaction is

O3 + H2O2 → H2O + O2 + O2

Hence the reaction.

The path of reaction (a) can be determined by labelling the reaction by isotopic oxygen

(O18

) and H2O18

as

12 H2O18 → 12 H2 + 6O2

18

6 CO2 + 12 H2 → C6H12O6 +6 H2O

From the above reaction, it can be concluded that in the formation of glucose, all the 12

molecules of H2O18

are involved in the reaction, so it is appropriate to write the reaction

as

6 CO2 + 12 H2O → C6H12O6 +6 H2O+6O2

The path of reaction (b) can be determined by labelling either

H2O2 as H2O218

or O3 as O318

O318 → O2

18+ O

18

H2O2 + O18 → H2O

18 + O2

By labelling O3 to O318

, it is clear that O2 is being obtained from each of the two reactants.

10) The compound AgF2 is an unstable compound. However, if formed, the compound

acts as a very strong oxidizing agent. Why?

Solution

In AgF2 , the oxidation number of Ag is +2, which is very unstable (4d9 system). The Ag

+2

ion is AgF2 readily accepts one electron to form the more stable Ag+

ion (3d10

).

Ag+2

+ e− → Ag

+

(3d9) (3d

10)

AgF2 is readily reduced, and hence, it acts as a strong oxidising agent.

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.

Solution

+2 -2

a) Fe + Cl2 → Fe Cl2

(excess reductant) Oxidant

+3 -3

Fe + 3Cl2 → 2Fe Cl3

Oxidant (excess)

+2 -2

b) 2Fe + O2 → 2 Fe O2


+3

4Fe + 3O2 → 2Fe O3

(excess)

Reductant

+4

c) Xe + F 2 → XeF2


+4 -4

Fe + 2F 2 → Xe F4

(excess reductant)

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?

Solution

a) In manufacturing benzoic acid, alcoholic potassium permanganate is used as oxidant

due to the following facts:

i) In an alcoholic medium, the solution is homogeneous, and hence, the reaction becomes

faster.

ii) In an alkaline medium , the OH- ions present may react with the benzoic acid formed.

iii) In an acidic medium (like in the presence of conc.H2SO4), toluene may undergo

sulponation along with oxidation.

The following reaction takes place in the oxidation of toluene to benzoic acid:

[MnO4- + 8H

+ + 5 e

− → Mn

2+ + 4H2O] x 6

b) When an organic compound containing chloride is heated with concentrated H2SO4 ,

pungent smelling gas HCl is produced.

2Cl-

+ H2SO4 → 2HCl(g) + SO42-

Similarly on treatment of conc. H2SO4 with a mixture containing bromide , HBr gas is

evolved. HBr is a stronger reducing agent. Hence, it is readily oxidised by conc. H2SO4

into elemental bromine and red vapours of bromine are obtained.

2Br-

+ H2SO4 → 2HBr + SO42-

2HBr + H2SO4 → Br2 + SO2 + H2O

However, HCl being a weak reducing agent is not oxidised by conc. H2SO4 .

13) Identify the substance oxidised , reduced , oxidising agent and reducing agent for each

of the following reactions:

Solution:

• If one of the atoms of a compound or ion undergoes an increase in the oxidation number,

then the compound is said to be oxidised.

• If one of the atoms of a compound or ion undergoes a decrease in the oxidation number ,

then the compound or ion is said to be oxidised.

• A compound undergoing oxidation act as a reluctant or reducing agent.

• A compound undergoing reduction acts as an oxidant or oxidising agent.

Now applying the above in the given examples:

Species oxidised= C6H6O2

Species reduced=AgBr

Oxidant = AgBr

Reductant = C6H6O2

Species oxidised= HCHO

Species reduced=[Ag(NH3)2]+

Oxidant = =[Ag(NH3)2]+

Reductant = HCHO

Species oxidised= HCHO

Species reduced=Cu2+

Oxidant = HCHO

Reductant = Cu2+

Species oxidised= N2H4

Species reduced=H2O2

Oxidant = H2O2

Reductant = N2H4

Species oxidised= Pb

Species reduced= PbO2

Oxidant = PbO2

Reductant = Pb


2S2O32-

(aq) + I2(s) → S4O62-

(aq) + 2I- (AQ)

S2O32-

(aq) + 2Br(l) + 5H2O(l)→2SO42-

(aq)+4Br-(aq)+10 H

+(aq)

Solution:

In both the cases, there is oxidation of S2O32 ion by I2 and Br2 . But the oxidising power of

both I2 and Br2 are different . Since Br2 is a stronger oxidising agent, it oxidises S2O32 to a

higher oxidation state in SO42-

(+2 + to +6) , while I2 , being a weaker oxidising agent,

oxidises S2O32-

to a lower oxidation state in S S4O62-

(+2 to +2.5). Hence, different products

are obtained by the action of I2 and Br2 on S2O32-

ion.

15) Justify giving reactions that among halogens, fluorine is the best oxidant, and that

among hydrohalic compounds, hydroiodic acid is the best reductant.

Solution:

The relative oxidising powers of elements depend upon their standard reduction potential

(E°) values.

The larger of E° values of an element, the stronger oxidising agent is the element. Among

halogens, fluorine has the highest E° value, while I2 has the least E° value , so the relative

oxidising power of halogens is:

I2 > Br2 > Cl2 > F2

E° (V) 0.54 1.08 1.36 2.87

For a non-metallic strong oxidising agent, its corresponding anion is a weak reducing

agent. So , HF should be the weakest reducing agent among all hydrohalic compounds.

Hence, among those, hydriodic acid is the strongest reducing agent.

16) Why does the following reaction occur?

XeO64-

(aq) + 2 F-(aq) + 6H

+ → XeO3(g) + F2(g) + 3H2O(l)

What conclusion about the compound Na4XeO6 (of which XeO64-

is a part) can be drawn

from the reaction?

Solution:

The given reaction occurs as XeO64-

ion is able to oxidise F- ion into F2.

XeO64-

(aq) + 2 F-(aq) + 6H

+ → XeO3(g) + F2(g) + 3H2O(l)

(Xe= +8) (-1) (Xe=+6) (0)

From the reaction , it can be concluded that the XeO64-

ion is a stronger oxidising agent

than F2 .


What inference do you draw about the behaviour of Ag+

and Cu2+

from these reactions?

Solution:

a) In reaction (a), the Ag+

ion is oxidising H3PO2 and H3PO4

(oxidation number of P changes +1 to +5) and is itself reduced to Ag. So, the Ag+

ion acts

as an oxidant.

b) In reaction (c), the [Ag(NH3)2]+

ion is oxidising C6H5CHO to C6H5COO - .So , the and

H3PO4 Ag+

ion acts as an oxidant.

c) In reaction (b), the Cu2+

ion is oxidising H3PO2 to H3PO4 (oxidation number of P

changing from +1 to +5). Hence, the ion acts as an oxidant.

d) In reaction (a), the Cu2+

is not oxidising C6H5CHO .

So, it can be said that the Ag+

ion is a stronger oxidising agent than the Cu2+

ion.

18) Balance the following redox reactions by ion-electron method:

a) MnO4- (aq) + I

- (aq) → MnO2

(s) + I2 (s) (in basic medium)

b) MnO4- (aq) + SO2(g) → Mn

2+ (aq) + HSO4

- (in acidic solution)

c) H2O2(aq) + Fe2+

(aq) → Fe3+

(aq) + H2O (l) (in acidic solution)

d) Cr2O32-

(aq) + SO2(g) → Cr3+

(aq) + SO42-

(aq) (in acidic solution)

Solution:

a) MnO4- (aq) + I

- (aq) → MnO2

(s) + I2 (s) (in basic medium)

Oxidation half reaction

2 I-→ I2 +2 e

− --------1

Reduction half reaction

MnO-4 → MnO2

MnO-4 + 2H2O + 2 e

− → MnO2 + 4OH

--------2

Now,

3 x equation (1) + 2 x equation 2= Balanced equation

6I- → 3 I2 + 6 e

−

b) MnO4- (aq) + SO2(g) → Mn

2+ (aq) + HSO4

- (in acidic solution)

Oxidised half equation

SO2 → HSO4-

SO2 + 2H2O → HSO4- + 3H

+ + 2e

−--------1

Reduced half equation

MnO4- → Mn

2+

MnO4- + 8H

+ + 5e

− → Mn

2+ + 4H2O------2

Now,

5 x equation 1+2 x equation 2= Balanced equations

c) H2O2 (aq) + Fe2+

(aq) → Fe3+

(aq) + H2O (l) (in acidic solution)


Fe2+ → Fe

3+ + e

− --------1


H2O2 → H2O

H2O2 + 2H+

+ 2e− → H2O + H2O --------2

Now. 2 x equation 1 + equation 2 = Balanced equation

2Fe2+ → Fe

3+ + 2e

−

H2O2 + 2H+

+ 2e− → H2O

d) Cr2O32-

(aq) + SO2(g) → Cr3+

(aq) + SO42-

(aq) (in acidic solution)


SO2 → SO42-

SO2 + 2 H2O → SO42-

+ 4H+

+ 2e−

-------1


Cr2O72-

→ Cr3+

Cr2O72-

+ 14H+

+ 6e−→

2 Cr

3+ + 7H2O-------2

Now,

3 x equation 1 + equation 2 = Balance equation

3SO2 + 6H2O → 3SO42-

+ 12H+

+ 6e−

19) Balance the following equations in basic medium by ion-electron method and

oxidation number methods, and identify the oxidising agent and the reducing agent.

a) P4 (s) + OH- (aq) → PH3 (g) + HPO2

- (s)

Solution

a) P4 (s) + OH- (aq) → PH3 (g) + HPO2

- (s)

i) Oxidation half reaction:

P4 → H 2PO2-

--------1

Balancing of oxygen:

P4 → 8H 2 O → 2H 2PO2-

Balancing of hydrogen

P4 + 8 OH- + 4H 2PO→ 2H 2PO2

- + 8H 2 O

Cancelling the common terms on either side:

P4 + 8 OH- → 4H 2PO2

-

Balancing of charge

P4 + 8 OH- → 4H 2PO2

- + 4e

−

Reduction half equation:

P4 → PH3 -------2

Balancing of P

P4 → 4PH3

Balancing of hydrogen

P4 + 12H2O + 12 e− → 4PH3 +12 OH

-

Now,

3 x equation (i) + equation (ii) = balance equation

P4 + 24OH- → 12H 2PO2

- + 12e

−

P4 + 12H2O + 12 e− → 4PH3 +12 OH

-

Or

P4 + 3OH- + 3H2O

→ PH3 + 3H 2PO2

-

ii)

Total decrease in oxidation number of P4 in PH3 = 3 x 4 =12

Total increase in oxidation number of P4 in H 2PO2- = 1 x 4 = 4

So, to make the change in the oxidation number equal , multiply PH3 by 1 and H2PO2- by 3.

Thus , the equation is

P4 + OH- → PH3 + H 2PO2

-

Balance O atom by OH-

P4 + 6OH- → PH3 + 3H 2PO2

-

Balance H atom as

P4 + OH- + 3H2O

→ PH3 + H 2PO2

- + 3OH

-

Cancelling the common terms from either side:

The balanced equation is :

P4 + 3OH- + 3H2O

→ PH3 + 3H 2PO2

-

b) N4H4 (l) + ClO2- (aq) → NO (g) + Cl

- (g)

i) Oxidation half reaction:

N2H4→ NO

Balancing of N

N4H4 →2 NO

Balancing of O

N2H4 + 4OH- → 2NO + 4H2O

Balancing of charge

N2H4 + 4OH- + 2H2O →2NO + 4H2O +12e

−--------(1)

Reduction half reaction:

ClO3- → Cl

-

Balancing of both hydrogen and oxygen:

ClO3- + 6H2O → Cl

- + 3H2O +6OH

-

Balancing of charge

ClO3- + 3H2O + 12e

− → Cl

- + 6OH

- -------2

Now,

3 x equation (i) + 4 x equation (ii) = Balanced equation

N2H4 + 24OH- →6NO + 18H2O +24e

−

4ClO3- + 12H2O + 24e

− → 4Cl

- + 24OH

-

The balanced chemical equation is

2 N2H4 + 4ClO3- → 6NO + 4Cl

- + 6H2O

ii) Oxidation number method

Balancing of O atom by H2O molecules:

2 N2H4 + 4ClO3- → 6NO + 4Cl

- + 6H2O

Hydrogen gets itself balanced in the above step. Hence, the above equation is balanced.

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

+

i) Ion electron method:

Oxidation half equation:

H2O2 → O2

Balancing of hydrogen and oxygen

H2O2 + 2OH- → O2 + 2 H2O

Balancing of charge

H2O2 + 2OH- → O2 + 2 H2O + 2e

− --------1

Reduction half equation:

Cl2O7 → ClO2-

Balancing of Cl

Cl2O7 → 2ClO2-

Balancing of hydrogen and oxygen

Cl2O7 + 2 H2O → ClO2-

+ 6OH-

Now, 4x equation (i) + equation (ii) = Balanced equation

4H2O2 + 8OH- →4 O2 + 8H2O + 8e

−

Cl2O7 + 3H2O → ClO2-

+ 6OH-

The balanced chemical equation is

2H2O2 + Cl2O7 + 6OH-→4 O2 + 2ClO2

- + 5 H2O.

ii) Oxidation number method:

So,

Cl2O7 + 4H2O2 → 4 O2 + 2ClO2-

Balancing of oxygen:

Cl2O7 + 4H2O2 + 2OH-→4 O2 + 2ClO2

- + 3 H2O+2H2O

Cl2O7 + 4H2O2 + 2OH-→4 O2 + 2ClO2

- + 5 H2O

Hence , the equation is balanced.

20) What sorts of information can you draw from the following reaction?

(CN)2 (g) + 2HO- (aq) → CN

- (aq) + CNO

- (aq) + H2O(l)

Solution

(CN)2 (g) + 2HO- (aq) → CN

- (aq) + CNO

- (aq) + H2O(l)

ON of C=+3 ON of C= +2 ON of C= +4

In (CN)2 to CN- , the oxidation number of carbon decreases from +3 to +2, while in (CN)2

to CNO-

, the oxidation number of carbon increases from +3 to +4. Thus, (CN)2 is

undergoing a disproportionation reaction.

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.

Solution

The skeleton reaction may be given as :

Mn3+ → Mn

2+ + MnO2 + H

+

Now, oxidised half equation:

Mn3+ → MnO2

Mn3+

+ 2H2O→ MnO2 + 4H+

+ e−

Reduced half reaction:

Mn3+ → Mn

2+

Mn3+

+ e− → Mn

2+

Now adding the two half equations:

2Mn3+

+ 2H2O→ MnO2 + 4H+

+ Mn2+

This is the balanced reaction.

22) Consider the elements:

Cs, Ne, I and F

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

b) Identify the element that exhibits only positive oxidation state.

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

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

state.

Solution

a. Fluorine (F) being the most electronegative of all elements exhibits only negative

oxidation state of -1.

b. Cs being the electropositive element exhibits only positive oxidation state.

c. Iodine (I) shows both positive and negative oxidation states ranging from -1 to +7.

d. Ne being an inert gas does not form any compound , and hence, does not exhibit either

positive or negative oxidation states.

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.

Solution

The skeleton reaction can be written as:

Cl2 + SO2 + H2O →Cl- + SO4

2-

Now, oxidised half equation:

SO2 → SO42-

SO2 + 2 H2O → SO42-

+ 4H+

+ 2 e− -------1

Reduces half equation:

Cl2 + 2 e− → 2Cl

- --------2

By adding (i) and (ii):

Cl2 + SO2 + 2 H2O → 2Cl- + SO4

2- + 4H

+

This is the balanced equation.

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.

Solution

a. Non-metals that show disproportionation reaction are:

P4 , Cl2 , Br2 , I2 and S8 .

Example :

Cl2 + NaOH → NaCl + NaOCl + H2O

P4 + KOH → PH3 + KHPO2

b. Three metals that show disproportionation reaction are:

Cu, Mn, Ga

Cu2O → Cu + CuO

Mn3+

+ 2H2O → Mn2+

+ MnO2 +4H+

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 acid that can be obtained starting only with 10 g of ammonia

and 20 g of oxygen?

Solution

The reaction is:

4NH3 + 5O2 → 4NO + 6H2O

4 x 17g 5 x 32 g 4 x 30 g

∴ 68 g of NH3 gives 120 g of NO

∴ 10 g of NH3 gives 120 x 10/68 g of NO

= 17.65 g of NO

Again,

∴ 160 g of O2 gives 120 g of NO

∴ 20 g of O2 gives 120 x 20/160 g of NO

= 15 g of NO

As oxygen is the limiting agent, the amount of NO formed will be 15 g even if 10 g of

ammonia is taken.

26) Using the standard electrode potentials given in the table 8.1 , predict if the reaction

between the following is feasible:

i) Fe3+

(aq) and I- (aq)

ii) Ag+

(aq) and Cu (s)

Solution

i) The possible reaction is

Fe3+

(aq) + 2I- (aq) → Fe

2+ (aq) + I2

(s)

The cell of the reaction is

= 0.77 - 0.54

E° cell = 0.23 V

Since E° cell of the reaction is positive, the reaction is feasible.

ii) The possible reaction is

Cu(s) + 2Ag+

(aq) → Cu2+

(aq) + 2Ag(s)

The cell of the reaction is given as

= 0.80 -0.34

E° cell = 0.46 V

Since, E° cell of the reaction is positive , the reaction is feasible.

iii) The reaction is

Fe3+

(aq) + 2Br- (aq) → Fe

2+ (aq) + Br2(l)


Then,

= 0.77 – 1.09

E° cell = -0.32 V

Since, E° cell of the reaction is negative, the reaction is not feasible.

iv) The possible reaction is

Ag(s) + Fe3+

(aq) →Ag+

(aq) + Fe2+

(aq)


Then ,

= 0.77 -0.80

E° cell = -0.03 V

Since, E° cell of the reaction is negative , the reaction is not feasible.

v) The possible reaction is

Fe2+

(aq) + ½ Br2 (l) → Fe3+

(aq) + Br- (aq)


Then,

= 1.09 – 0.77

E° cell = 0.32 V

Since, E° cell of the reaction is positive, the reaction is feasible.

27) Predict the products of electrolysis in each of the following:

i) An aqueous solution of AgNO3 with silver electrodes.

ii) An aqueous solution of AgNO3 with platinum electrodes.

iii) A dilute solution of H2SO4 with platinum electrodes.

iv) An aqueous solution of CuCl2 with platinum electrodes.

Solution

i) The reactions are

AgNO3(aq) → Ag+

(aq) + NO3- (aq)

H2O(l) →H+

(aq) + OH-(aq)

At CATHODE : Since the discharge potential of Ag+

ion is preferably reduced at cathode,

Ag+

(aq) + e− → Ag(s)

At anode: Since Ag anode is attacked by NO3- ion, Ag of the anode will be oxidised and

passed into the solution as Ag+

ion.

Ag(s) →Ag+

(aq) + e−

ii) The reactions are

AgNO3(aq) → Ag+

(aq) + NO3- (aq)

H2O(l) →H+

(aq) + OH-(aq)

At cathode: Since the discharge potential of Ag+

ion is less than that of H+

ion, Ag+

ion is

reduced to Ag in preference to H+

ion.

Ag+

(aq) + e− → Ag(s)

At anode: Since the anode is not attackable , out of NO3- and OH

- ion, OH

- ion having less

discharge potential is oxidised at the anode to liberate O2 gas.

2OH- (aq) → H2O(l) + ½ O2 (g) + 2 e

−

iii) The reactions are

H2SO4(aq) → 2H+

(aq) + SO2-

4(aq)

H2O(l) → H+

(aq) +OH- (aq)

At anode : Since the discharge potential of OH- ion is less than SO

2-4 , OH

- ion is oxidised

at the anode to liberate O2 gas.

2OH- (aq) → H2O(l) + ½ O2 (g) + 2 e

−

At cathode: 2H+

(aq) + 2 e− → H2(g)

iv) The reactions are

CuSO4 (aq) →Cu2+

(aq) + SO2-

4(aq)

H2O(l) → H+

(aq) +OH- (aq)

At anode: Since the discharge potential of Cu2+

ion is less than that of H+

ion, Cu2+

ion is

reduced as Cu at the cathode in preference to H+

ion.

Cu2+

(aq) + 2 e− → Cu(s)

At anode: Since the discharge potential of Cl- ion is less than that of OH

- ion, Cl

- ion is

oxidised as Cl2 gas at the anode.

2Cl- (aq) → Cl2 (g) +2 e

−

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.

Solution

A metal with less standard reduction potential E°red displaces a metal with higher E°red

from a solution of its salt. The values of E°red of the given metals are as follows:

Hence, the order in which a metal can displace others from solutions of their salts is Mg,

Al , Zn , Fe, Cu.

29)Given the standard electrode potentials,

K+

/K = -2.93 V , Ag+

/ Ag= 0.80 V, Hg2+

/ Hg=0.79 V,

Mg2+

/ Mg=-2.37 V, Cr3+

/Cr= -0.74 V

Arrange these metals in their increasing order of reducing power.

Solution

A metal with the least E°red value will have the greatest ability to get oxidised , and hence,

will act as the powerful reducing agent.

Hence, the arrangement of metals in their increasing order of reducing power is Ag< Hg<

Cr< Mg< K.

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.

Solution

i) Anode, the left half cell, is the negatively charged electrode of the cell.

ii) The carriers of the electric current in the cell are electrons to the cathodic half cell.

iii) The reaction taking place at the anode is oxidation.

The reaction taking place at the cathode is reduction.

Documents

Redox Reactions - shikshahouse2019/01/08 · Redox Reactions 1) Assign oxidation numbers to the underlined elements in each of the following species: a) NaH 2PO 4 b) NaHSO 4 c) H