Redox Reactions

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.

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.

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.

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.

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.

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.

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.

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.

In KI_3, 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.

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.

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.

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.

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.

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.

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

In Fe₂O_3, 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 CO_2. Hence, the given reaction is a redox reaction.

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

In BCl₃, B = + 3 and Cl = -1

In LiAlH_4, Li = + 1, Al = + 3 and H = -1

In B₂H_6,B = -3 and H = + 1

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

In AlCl_3, 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 B₂H_6. 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.

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.

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

In NH_3, N = -3 & H = + 1

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

In H₂O, H = + 1 &O = -2

In O_2, 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.

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.

(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) Cr₂O₇^2-

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 Cr₂O₇^2-

The structure of Cr₂O₇^2- 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.

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.

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.

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.

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.

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.

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.

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’

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

(a) The process of photosynthesis involves two steps

Step 1:

H₂O decomposes to give H₂ and O₂.

2 H₂O_(l)→ 2H_2(g) + O_2(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 O_2.

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

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

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

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.

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 SO_2.

2NaCl + 2H₂SO₄→ 2NaHSO₄ + 2HCl

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.

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.

Oxidised substance:- HCHO

Reduced Substance:- Cu^{2 +}

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.

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.

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.

The average O.N. of S in S₂O₃^2- is + 2 while in S₄O₆^2- it is + 2.5.The O.N. of S in S₄O₆^2- is + 6. Since Br₂ is a stronger oxidizing agent than I₂, it oxidizes S of S₂O₃^2- to a higher oxidation state of + 6 and hence forms SO₄^2- ion. I₂, however, being weaker oxidizing agent oxidizes S of S₂O₃^2- ion to a lower oxidation state of + 2.5 in S₄O₆^2- ion. It is because of this reason that thiosulphate reacts differently with Br₂ and I₂.

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

F_2(aq) + 2Cl^-_(s)→ 2F^-_(aq) + Cl_2(g)

F_2(aq) + 2Br^-_(aq)→ 2F^-_(aq) + Br_2(l)

F_2(aq) + 2I^-_(aq)→ 2F^-_(aq) + I_2(s)

On the other hand, Cl_2, 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) → Cu₂I_2(s) + I_2(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 XeO₆^4- oxidises fluorine from -1 oxidation state to 0 oxidation state and reduces XeO₆^4- from +8 oxidation state to +6 oxidation state as shown in the equation below :

⁺⁸XeO^4- (aq) + 2F^-1(aq) + 6H⁺¹(aq) → ⁺⁶XeO₃(g) + F₂ (g) + 3H₂O (l)

Hence, we can easily conclude that XeO₆^4- is a stronger oxidising agent than F^-.

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⁺.

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

The oxidation half reaction is as below :

I^-_(aq)→ I_2(s)

The reduction half reaction is as below :

MnO^-_4(aq) → MnO_2(aq)

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

2I^-_(aq)→ I_2(aq)

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

2I^-_(aq)→ I_2(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^-→ MnO_2(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^-→ MnO_2(aq) + 4OH^-

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

MnO^-_4(aq) + 3e^- + 2H₂O → MnO_2(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)→ 3I_2(aq) + 6 e^-

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

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

6I^-_(aq) + 2MnO^-_4(aq) + 4H₂O → 3I_2(aq) + 2MnO_2(aq) + 8OH^-

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

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) + 5SO_2(g) + 2H₂O + H⁺_(aq)→ 2Mn²⁺_(aq) + 5HSO^-_4(aq)

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:-

H₂O_2(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:-

H₂O_2(aq) + 2Fe²⁺_(aq) + 2H⁺_(aq)→ 2 Fe³⁺_(aq) + 2H₂O

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

SO_2(g) + 2H₂O → SO₄^2-_(aq) + 4H⁺_(aq) + 2e^-

And the reduction half reaction as:-

Cr₂O₇^2-_(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:-

Cr₂O₇^2-_(aq) + 3 SO_2(g) + 2H⁺_(aq)→ 2Cr³⁺_(aq) + 3SO₄^2-_(aq) + H₂O

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: -

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:-

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:-

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:-

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 :-

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.

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: -

-

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.

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: -

-

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.

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.

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.

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.

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.

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.

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.

(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 SO₄^2- 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.

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

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.

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:-