Redox Reactions Question Answers: NCERT Class 11 Chemistry

(a) NaH₂PO₄

Let's assume oxidation number of P is x.

We know that,

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = -2

Then we have

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

⇒ 1 + 2 + x - 8 = 0

⇒ x - 5 = 0

⇒ x = + 5

Hence, oxidation number of P is +5

(b)  NaHSO₄

Let's assume oxidation number of S is x.

Oxidation number of Na = +1

Oxidation number of H = +1

Oxidation number of O = -2

Then we have:

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

⇒ 1 + 1 + x - 8 = 0

⇒ x-6 = 0

⇒ x = +6

Hence, oxidation number of S is +6

 

(c)  H₄P₂O₇

Let's assume oxidation number of P is x.

Oxidation number of H = +1

Oxidation number of O = -2

Then we have:

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

⇒ 4 + 2x - 14 = 0

⇒ 2x - 10 = 0

⇒ 2x = +10

⇒ x = +5

Hence, Oxidation number of P is +5

 

(d) K₂MnO₄

 

Let's assume oxidation number of Mn is x.

Oxidation number of K = +1

Oxidation number of O = -2

Then we have:

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

⇒ 2 + x - 8 = 0

⇒ x - 6 = 0

⇒ x = +6

Hence, Oxidation number of Mn is +6

 

(e) CaO₂

Let's assume oxidation number of O is x.

Oxidation number of Ca = +2

Then we have:

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

⇒ 2 + 2x  = 0

⇒ 2x = -2

⇒ x  = -1

Hence, Oxidation number of O is -1

 

(f) NaBH₄

Let's assume oxidation number of B is x.

Oxidation number of Na = +1

Oxidation number of H = -1

Then we have:

1(+1) + 1(x) + 4(-1)  = 0

⇒ 1 + x -4 = 0

⇒ x - 3 = 0

⇒ x = +3

Hence, Oxidation number of B is +3.

 

(g) H₂S₂O₇

Let's assume oxidation number of S is x.

Oxidation number of O = -2

Oxidation number of H = +1

Then we have:

2(+1) + 2(x) + 7(-2)  = 0

⇒ 2 + 2x - 14 = 0

⇒ 2x - 12 = 0

⇒ x  = +6

Hence, Oxidation number of S is +6.

 

(h) KAl(SO₄)₂.12 H₂O

Let's assume oxidation number of S is x.

Oxidation number of K = +1

Oxidation number of Al = +3

Oxidation number of O = -2

Oxidation number of H = +1

Then we have:

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

⇒ 1 + 3 + 2x -16 +24 -24 = 0

⇒ 2x - 12 = 0

⇒ 2x = +12

⇒ x = +6

Hence, Oxidation number of S is +6.

 

(a) KI₃

Let assume oxidation number of l is x.

In KI₃, the oxidation number (O.N.) of K is +1.

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

⇒ +1 +3x = 0

⇒ 3x = -1

⇒ x = -1/3

Hence, the average oxidation number of I is - 1/3

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.

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

Let assume oxidation number of S is x.

The oxidation number (O.N.) of H is +1.

The oxidation number (O.N.) of O is -2.

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

⇒ 2 + 4x - 12 = 0

⇒ 4x -10 = 0

⇒ 4x  =  +10

⇒ x  = +10/4

However, 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. 

 

(c) Fe₃O₄ 

Let assume oxidation number of Fe is x.

The oxidation number (O.N.) of O is -2.

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

⇒ 3x  - 8 = 0

⇒ 3x  = 8

⇒ x  = 8/3

However, O.N. cannot be fractional. 

Here, one of the three Fe atoms exhibits the O.N. of +2 and the other two Fe atoms exhibit the O.N. of +3. 

 

(d) CH₃CH₂OH

Let assume oxidation number of C is x.

The oxidation number (O.N.) of O is -2.

The oxidation number (O.N.) of H is +1.

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

⇒ x +3 + x +2 - 2  + 1 = 0

⇒ 2x  + 4 = 0

⇒ 2x  = -4

⇒ x  = -2

Hence, the oxidation number of C is -2.

 

(e) CH₃COOH

Let assume oxidation number of C is x.

The oxidation number (O.N.) of O is -2.

The oxidation number (O.N.) of H is +1.

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

⇒ 2x + 3 - 2 -  2  +  1 = 0

⇒ 2x + 0 = 0

⇒ x = 0

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. 

 

(a) CuO_(s) + H_2(g) → Cu_(s) + H₂O_(g)

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

+2  -2        0             0             +1    -2

Cu O_(s) +  H_2(g)  →  Cu_(s)  +  H₂  O_(g)

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) Fe₂O_3(s) + 3CO_(g) → 2Fe_(s) + 3CO_2(g)

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

+3   -2          +2  -2            0          +4  -2

Fe₂ O_3(s) + 3C  O_(g) → 2Fe_(s) + 3C O_2(g)

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 CO₂ i.e., CO is oxidized to CO₂. Hence, the given reaction is a redox reaction.

 

(c) 4BCl_3(g) + 3LiAlH_4(s) → 2B₂H_6(g) + 3LiCl_(s) + 3 AlCl_{3 (s)}

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

 +3  -1            +1   +3   -1            -3   +1         +1  -1             +3   -1

4B Cl_3(g) + 3  Li   Al   H_4(s) → 2B₂ H_6(g) + 3Li   Cl_(s)   + 3 Al  Cl_{3 (s)}

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) 2K_(s) + F_2(g) → 2K⁺F^– _(s)

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

0               0                    +1   -1

2K_(s)   +   F_2(g)    →    2K⁺  F^– _(s)

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) 4 NH_3(g) + 5 O_2(g) → 4NO_(g) + 6H₂O_(g)

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

  -3   +1            0              +2  -2         +1    -2

4 N  H_3(g) + 5 O_2(g) → 4N  O_(g) + 6H₂  O_(g)

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.

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 oxidation number decreases from 0 in F₂ to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction is a redox reaction.

(a) H₂SO₅

Let assume oxidation number of S is x.

We know that,

Oxidation number of H = +1

Oxidation number of O = -2

Then we have

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

⇒ 2 + x - 10 = 0

⇒ x  =  +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.

 

(b) Cr₂O^2-₇

Let assume oxidation number of Cr is x.

We know that,

Oxidation number of O = -2

Then we have

 2(x) + 7 (-2) =  -2

⇒ 2x  -14  = -2

⇒ 2x   =  +12

x  = +6

Here, there is no fallacy about the O.N. of Cr in Cr₂O₇^2-

The structure of Cr₂O₇^2- is shown as follows: 

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

 

(c) NO₃^-

Let assume oxidation number of N is x.

We know that,

Oxidation number of O = -2

Then we have

1(x) + 3 (-2) =  -1

⇒ x  - 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. 

 

The representation of a molecule in terms of the symbols of the constituting atoms is called its chemical formula.

Rules for building the formula for a compound:

1) Write side by side the symbols of the radicals involved, with their valencies at the top, the positive radical to the left & the negative radical to the right.

2) Remove the common factor, if any, to get a simple ratio of the valencies of combining radicals.

3) Criss cross the valencies i.e, shift the valencies cross wise to the lower right of radical.

(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

The substances where carbon can exhibit oxidation states from -4 to +4 are listed below:

Substance	Formula	Oxidation State of (C)
Methan	CH4	-4
Ethane	C2H6	-3
Ethene	C2H4	-2
Ethyne	C2H2	-1
Dichloromethane	CH2Cl2	0
Chloroform	CHCl3	+2
Oxalic Acid	(COOH)2	+3
Carbon TetraChloride	CCl4	+4
Carbon dioxide	CO2	+4

 

The substances where nitrogen can exhibit oxidation states from -3 to +5 are listed below:

Substance	Formula	Oxidation State of Nitrogen
Ammonia	NH3	-3
Hydrazine	N2H4	-2
Hydride	N2H2	-1
Dinitrogen Gas	N2	0
Nitrous Oxide	N2O	+1
Nitric Oxide	NO	+2
Dinitrogen Trioxide	N2O3	+3
Dinitrogen Dioxide	NO2	+4
Nitrogen Pentroxide	N2O5	+5

 

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, H₂O₂ can 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.

(a) The process of photosynthesis involves two steps.

Step 1:

H₂O decomposes to give H₂ and O₂.

2H₂O_(l)  →  2H_2(g)  + O_2(g)

 

Step 2:

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

6CO_2(g)  +  12H_2(g)  →  C₆H₁₂O_6(s)  +  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 H₂O and O₂.

The path of this reaction can be investigated by using H₂O¹⁸ or O¹⁸₃.

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.

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.

P₄ (excess) + F₂ → P⁺³F₃

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

P₄  + F₂ (excess) → P⁺⁵F₅

 

(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.

4K (excess) + O₂ →  2K₂O^-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)  →  K₂O^-1₂

 

(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.

C (excess) + O₂ →  C⁺²O

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.

C + O₂ (excess)  →  C⁺⁴O₂

(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:

 

(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.

2NaBr +  2H₂SO₄   →    2NaHSO₄  +  2HBr

2HBr + H₂SO₄   →   Br₂  + SO2  +  2H₂O

                             (Red Vapour)

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

a) AgBr is reduced to Ag

C₆H₆O₂ is oxidised to C₆H₄O₂

AgBr is an oxidising agent

C₆H₆O₂ is a reducing agent.

 

b) [Ag(NH₃)⁺₂ reduced to Ag+

HCHO is oxidised to HCOO^-

[Ag(NH₃)⁺₂ is an oxidising agent.

 

c) HCHO is oxidised to HCOO^-

Cu²⁺ is reduced to Cu(I) state.

 

d) N₂H₄ is reduced to H₂O

H₂O₂ is an oxidising agent

N₂H₄ is reducing agent.

 

e) Pb has been oxidised to PbSO₄

PbO₂ is reduced to PbSO₄

PbO₂ is an oxidising agent

Pb is a reducing agent.

The average oxidation number (O.N.) of S in  S₂O^2– ₃ is +2. Being a stronger oxidising agent than I₂, Br₂ oxidises S₂O^2– ₃ to SO^2-₄ , in which the O.N. of S is +6. However, I₂ is a weak oxidising agent.

Therefore, it oxidises S₂O^2– ₃ to S₄O^2– ₆ , in which the average O.N. of S is only +2.5. As a result, S₂O^2– ₃ reacts differently with iodine and bromine.

F₂ can oxidize Cl^- to Cl₂, Br^-  to Br₂, and I^- to I₂ as:

F_2(aq) + S₂cl^-_(s)  →  2F^-_(aq)  +  Cl _(g)

F_2(aq) + 2Br^-_(aq)  →  2F^-_(aq)  + Br _2(l)

F_2(aq) + 2l^-_(aq)  →  2F^-_(aq)  + 1_2(s)

 

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.

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  →  l₂  +  SO₂  +  2H₂O

2HBr + H₂SO₄  →  Br₂ + SO₂  +  2H₂O

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

4l^-_(aq)  +   2Cu²⁺_(aq)  →  Cu₂l_2(s)  + l²_(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.

The given reaction occurs because XeO^4– 6 oxidises F^- and F^- reduces XeO^4-₆.

In this reaction, the oxidation number (O.N.) of Xe decreases from +8 in XeO^4– 6 to +6 in XeO₃ and the O.N. of F increases from -1 in F^- to O in F₂.

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

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⁺.

Step 1:

The two half reactions involved in the given reaction are:

                                           -1              0

Oxidation half reaction:  l _(aq)  →  l_2(s)

                                                  +7                         +4

Reduction half reaction: Mn O^-_4(aq)   →  MnO_2(aq)

 

Step 2:

Balancing I in the oxidation half reaction, we have:

2l^-_(aq)  →  l_2(s)

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

2l^-_(aq) →  l_2(s) + 2e^-

 

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^-   →MnO_2(aq)

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.

MnO^-_4(aq) + 3e^-   →MnO_2(aq)  +  4OH^-

 

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.

MnO^-_4(aq) + 2H₂O + 3e^-   →MnO_2(aq)  +  4OH^-

Step 5:

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

6l^-_(aq) →  3l_2(s) + 2e^-

2MnO^-_4(aq) + 4H₂O + 6e^-   →  2MnO_2(s)  +  8OH^-_(aq)

Step 6:

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

6l^-_(aq)  +   2MnO^-_4(aq) + 4H₂O_(l)  →   3l2(s) + 2MnO_2(s)  +  8OH^-_(aq)

 

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

SO_2(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_(l)

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:

2MnO^-_4(aq)  + 5SO_2(g) + 2H₂O_(l) +  H⁺_(aq)  → Mn²⁺_(aq)  +  HSO^-_4(aq)

 

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

Fe²⁺_(aq)  →   Fe³⁺_(aq)  +  e^-

And the reduction half reaction as:

H₂O_2(aq)  +  2H⁺_(aq) +  2e^-  →  2H₂O_(l)

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

H₂O_2(aq)  +  2Fe²⁺_(aq)  +  2H⁺_(aq)    →   2Fe³⁺_(aq) +  2H₂O_(l)

 

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

SO_2(g)  +  2H₂O_(l)    →  SO^2-_4(aq)  +   4H⁺ _(aq)   +  2e^-

And the reduction half reaction as:

Cr₂O^2-_7(aq) + 14H⁺_(aq)  +  6e^-   →  2Cr³⁺_(aq)  + 3SO^2-_4(aq)  +  H₂O_(l)

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

Cr₂O^2-_7(aq) +  3SO_2(g)  +  2H⁺_(aq)   →  2Cr³⁺_(aq)  + 3SO^2-_4(aq) +  H₂O_(l)

 

(a) 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 HPO^-₂. Hence, P₄ acts both as an oxidizing agent as well as a reducing agent in this reaction.

Ion-electron method:

The oxidation half equation is:

P_4(s) →  H₂PO^-_(aq)

The P atom is balanced as:

P⁰ 4_(s)  →  4H₂P⁺¹O^-_(aq)

The O.N. is balanced by adding 4 electrons as:

P_4(s)    →   4H₂PO^-_(aq)    +   4e^-

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

P_4(s)    +  8OH ^- _(aq)  →   4H₂PO^-_2(aq)

The O and H atoms are already balanced. The reduction half equation is:

P_4(s)    →    PH_3(g)

The P atom is balanced as

P⁰_4(s)    →   4 P^-3H_3(g)

The O.N. is balanced by adding 12 electrons as:

P_4(s)  +  12e^-   →    4 PH_3(g)

The charge is balanced by adding 12OH- as:

P_4(s)  +  12e^-   →    4 PH_3(g)   +  12OH^-_(aq)  .....(i)

The O and H atoms are balanced by adding 12H₂O as:

P_4(s) + 12H₂O_(l)  +  12e^-   →    4 PH_3(g)   +  12OH^-_(aq)  -- (ii)

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

P_4(s) + 3OH^-_(aq) + 3H₂O  →  PH₃   +   3H₂PO^-_2(aq)

 

(b)

The oxidation number of N increases from -2 in N₂H₄ to +2 in NO and the oxidation number of Cl decreases from + 5 in CIO^-₃ to -1 in Cl-. Hence, in this reaction, N₂H₄ is the reducing agent and CIO^-₃ is the oxidizing agent. Ion-electron method:

The oxidation half equation is:

N^-2₂ H_4(l)  →  N⁺² O_(g)

The N atoms are balanced as:

N₂H_4(l)   →   2NO_(g)

The oxidation number is balanced by adding 8 electrons as:

N₂H_4(l)   →    2NO_(g)  +  8e^-

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

N₂H_4(l)  + 8OH^-_(aq)  →  2NO_(g)  +  8e^-

The O atoms are balanced by adding 6H₂O as:

N₂H_4(l)  + 8OH^-_(aq)  → 2NO_(g)  + 6H₂O_(l) +  8e^- .... (i)

The reduction half equation is:

C⁺⁵IO^-_3(aq)  →  C^-1l^-_(aq)

The oxidation number is balanced by adding 6 electrons as:

CIO^-_3(aq)  +  6e^-  → Cl^-_(aq)

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

CIO^-_3(aq)  +  6e^-  →    Cl^-_(aq) + 6OH^-_(aq) 

The O atoms are balanced by adding 3H₂O as:

CIO^-_3(aq)  + 3H₂O(l)  + 6e^-  →  Cl^-_(aq) + 6OH^-_(aq) .... (ii)

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

3N₂H_4(l)  +  4CIO^-_3(aq)   →  6NO_(g)  +  4Cl^-_(aq) + 6H₂O(l) 

 

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 CIO^-₃ with 4 to balance the increase and decrease in O.N., we get:

3N₂H_4(l)  + 4CIO^-_3(aq)  → NO_(g)  + Cl^-_(aq)

The N and Cl atoms are balanced as:

3N₂H_4(l)  + 4CIO^-_3(aq)  → 6NO_(g)  + 4Cl^-_(aq)

The O atoms are balanced by adding 6H₂O as:

3N₂H_4(l)  + 4CIO^-_3(aq)  → 6NO_(g)  + 4Cl^-_(aq) + 6H₂O_(l)

This is the required balanced equation.

 

(c)

The oxidation number of Cl decreases from + 7 in Cl₂O₇ to + 3 in CIO^-₂and 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:

H₂O^-1_2(aq)  →  O⁰_2(g)

The oxidation number is balanced by adding 2 electrons as:

H₂O_2(aq)  →  O_2(g)  + 2e^-

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

H₂O_2(aq)  +  2OH-(aq)  →   O_2(g)  + 2e^-

The oxygen atoms are balanced by adding 2H₂O as:

H₂O_2(aq)  +  2OH-(aq)  →   O_2(g)  + 2H₂O_(l) + 2e^- ... (i)

The reduction half equation is:

C⁺⁷l₂O_7(g)   →  C⁺³lO^-_2(g)

The Cl atoms are balanced as:

Cl₂O_7(g)   →   2ClO^-_2(g)

The oxidation number is balanced by adding 8 electrons as:

Cl₂O_7(g)    +   8e^- →   2ClO^-_2(g)

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

Cl₂O_7(g)    +   8e^- →   2ClO^-_2(g)  +  6OH^- _(aq)

The oxygen atoms are balanced by adding 3H₂O as:

Cl₂O_7(g)   +    3H₂O_(l) +  8e^- →   2ClO^-_2(g)  +  6OH^- _(aq)  .... (ii)

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

Cl₂O_7(g)   +    4H₂O_2(aq) + 2OH^- _(aq)    →  2ClO^-_2(aq)  +  4O_2(g)  + 5H₂O_(l)

 

Oxidation number method:

Total decrease in oxidation number of Cl₂O₇ = 4 × 2 = 8

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

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

Cl₂O_7(g)   +    4H₂O_2(aq)  →   CIO^-_2(aq)  +  4O_2(g)

The Cl atoms are balanced as:

Cl₂O_7(g)   +    4H₂O_2(aq)  →   2CIO^-_2(aq)  +  4O_2(g)

The O atoms are balanced by adding 3H₂O as:

Cl₂O_7(g)   +    4H₂O_2(aq)  →   2CIO^-_2(aq)  +  4O_2(g)  +  3H₂O_(l)

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

Cl₂O_7(g)   +   4H₂O_2(aq)   +   2OH^-_(aq) →   2CIO^-_2(aq)  +  4O_2(g)  +  5H₂O_(l)

This is the required balanced equation.

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

∴x = 3

CN^-

x - 3 = -1

∴x = 2

CNO^-

x -3 - 2 =-1

∴x = 4

The oxidation number of carbon in the various species is:

(C⁺³N)_2(g)  +  2OH^-_(aq)  →  C⁺²N^-_(aq) + C⁺⁴NO^-_(aq) + H₂O_(l)

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.

The given reaction can be represented as:

Mn3+(aq)  →  Mn2+(aq)  + MnO2(s)  +  H+(aq)

The oxidation half equation is:

M+3 n3+(aq)  →  M+4 nO2(s)

The oxidation number is balanced by adding one electron as:

Mn³⁺_(aq)     →  M nO_2(s) + e^-

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

Mn³⁺_(aq)     →  M nO_2(s)  +  4H⁺_(aq)   +    e^-

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

Mn³⁺_(aq)   +  2H2O(l)   →    MnO_2(s)  + 4H⁺_(aq)   +    e^-  ..... (i)

The reduction half equation is:

Mn³⁺_(aq)     →   Mn²⁺_(aq)  

The oxidation number is balanced by adding one electron as:

Mn³⁺_(aq)   +    e^-    →   Mn²⁺_(aq)   ....... (ii)

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

2Mn³⁺_(aq)   +  2H₂O_(l)   →   MnO_2(s)  + 2Mn²⁺_(aq) + 4H⁺_(aq)

(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.

The given redox reaction can be represented as:

Cl_2(s)  + SO_2(aq)  +  H₂O_(l)   →  Cl^-_(aq)  +   SO^2-_4(aq)

The oxidation half reaction is:

S⁺⁴ O_2(aq)  →   S⁺⁶O^2-_4(aq)

The oxidation number is balanced by adding two electrons as:

SO_2(aq)  →   SO^2-_4(aq)  + 2e^-

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

SO_2(aq)   →   SO^2-_4(aq)  +   4H⁺_(aq) + 2e^-

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

SO_2(aq)  + 2 H₂O_(l)   → SO^2-_4(aq)  + 4H⁺_(aq) + 2e^-      ..........(i)

The reduction half reaction is: The chlorine atoms are balanced as:

Cl_2(s)    →  Cl^-_(aq) 

The oxidation number is balanced by adding electrons

Cl_2(s)   + 2e^-    →   2 Cl^-_(aq) ............(ii)

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

Cl_2(s)   + SO_2(aq)  + 2 H₂O_(l)   →   2 Cl^-_(aq)  + SO^2-_4(aq)  +  4H⁺_(aq)

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.

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

4NH_3(g)   + 5O_2(g)  →  4NO_(g) + 6 H₂O_(g)

4x17g     2x32g              4x30g        6x18g

= 68g        =160g             =120g       108g

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

Therefore, 10g of NH3 reacts with 160x10 / 68g 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 120_g of NO.

Therefore, 20 g of O₂ gives120x20 / 160 g of N, or 15 g of NO.

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

(a) The possible reaction between is given by, Fe³⁺_(aq) +  I^-_(aq)

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

 

(b) The possible reaction between is given by, Ag⁺(aq) + Cu_(s)

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

 

(c) The possible reaction between and is given by, Fe³⁺ (aq) +  Cu_(s)

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

 

(d) The possible reaction betweenand is given by, Ag(s) + Fe³⁺(aq)

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

 

(e) The possible reaction betweenand is given by, Br²(aq)  +  Fe²⁺(aq)

Here, E° for the overall reaction is positive. Hence, the reaction between Br²(aq) and Fe²⁺(aq) is feasible.

(i) AgNO₃ ionizes in aqueous solutions to form Ag⁺ and NO^-₃ 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.

Ag⁺_(aq)  +  e^-  →  Ag_(s)  ;  E° = +0.80V

2H₂O_(l)  +  2e^- →  H_2(g) + 2OH^-_(aq)  ; E° = -0.83V

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.

Ag_(s) →  Ag⁺_(aq)  +  e^-  ;   E° = -0.80V

2H₂O_(l)  →  O_2(g) + 4 H⁺_(aq) + 4e^-   ; E° = -1.23V

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 SO^2-₄ ions.

H₂SO_4(aq)  →  2H⁺_(aq) + SO^2-_4(aq)

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.

2H⁺_(aq) + 2e^- →  H_2(g) ; E° = 0.0V

2H₂O_(aq) + 2e^- →  H₂g + 2OH^-_(aq)  ; E° = -0.83V

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

On the other hand, at the anode, either of SO^2-₄ ions or H₂O molecules can get oxidized. But the oxidation of SO^2-₄ 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:

CuCl_2(aq)  →  Cu2⁺_(aq) + 2Cl^-_(aq)

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.

Cu2⁺_(aq) + 2e^-  →  Cu_(aq)  ;  E° = +0.34V

H₂O_(l)  + 2e^-  →  H_2(g) + 2OH^-  ;  E° = -0.83V

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^-.

2Cl^-_(aq) →  Cl_2(g)  + 2e^-  ;  E° = -1.36V

2H₂O_(l)  →  O_2(g) + 4H⁺_(aq) + 4e^-   ;  E° =  -1.23V

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.

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

The lesser 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.

The galvanic cell corresponding to the given redox reaction is:

(i) Zn electrode is negatively charged because at this electrode, Zn oxidizes to Zn²⁺ and 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:

Zn_(s)  →  Zn²⁺_(aq)  +  2e^-

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

Ag⁺_(aq)  +  e^-  →  Ag_(s)