Class 11 Redox Reaction Notes

  March 23, 2023    

 

REDOX REACTION

REDOX REACTIONS

Oxidation and reduction are complementary to each other and one cannot take place alone. So both oxidation and reduction will occur simultaneously it is obvious that if a substance takes electrons there must be another substance to give up these electrons. The reaction which involve oxidation and reduction are called redox reactions Redox reaction can be split into two half reactions namely oxidation half reaction (where oxidation takes place) and reduction half reaction (where reduction takes place)

 

eg.. Redox reaction

2Fe+3 + Sn2+ à 2Fe+2 +  Sn+4

Oxidation half reaction

Sn2+ à Sn+4 + 2e-

Reduction half reaction

Fe+3 +  e- à  Fe+2

 

OXIDATION NUMBER OR OXIDATION STATE

Oxidation number for an element is the arbitrary charge present on one atom when all other atoms bonded to it are removed. The removal of other atoms (dissociation of bonds between them) is done by assuming that the bonds are either ionic or purely covalent.

So when a bond between two identical atoms is broken, the bonded electrons are distributed equally between these two atoms. Similarly, when a bond between two different atoms is broken (one is more electronegative than other) bonded electrons are assumed to be retained by more electronegative atom.

For example, if we consider a molecule of HCI the CI atom is more electronegative than H-atom, therefore, the bonded electrons will go with more electronegative chlorine atom resulting in formation of H and Crions. So oxidation number of H and Cl in HCI are +1 and -1 respectively.

 The following points are important to determine the oxidation number of an element.

(1) The oxidation number of an atom in pure elemental form is considered to be zero eg H, O, Na, Mg

(2) Oxidation number of any element in simple monoatomic ion will be equal to the charge on that on for example, oxidation number of Na in Na is +1

 (3) Oxidation number of fluorine in its compound with other elements is always-1

(4) Oxidation number of oxygen is generally-2 but in case of peroxide (0,2) and superoxide (0,) oxygen has oxidation number-1 and respectively. In a compound OF, the oxidation number of oxygen is +2

 

(5) The oxidation number of alkali metals (Na,K) and alkaline earth metals (Ca, Mg) are +1 and +2 respectively in their compounds .

(6) The oxidation number of halogens is generally-1 when they are bonded to less electronegative elements.

(7) Oxidation number of hydrogen is generally +1 in most of its compounds but in case of metal hynde (NaH CaH,) the oxidation number of hydrogen is -1

(8) The algebraic sum of the oxidation numbers of all the atoms in a neutral compound is zero. In an in the algebraic sum of oxidation numbers is equal to the charge on that ion.

 

Special Example of Oxidation State Determination

1.      Oxidation number of sulphur in peroxo mono sulphuric acid (H₂SO)

Let us draw its structure.

 


 Here, we have to consider oxidation number of H = +1

    Oxidation number of oxygen in peroxy linkage = -1

    Oxidation number of oxygen in oxide form = -2

Here, two oxygen atoms are in form of peroxide and three oxygen atoms are in form of oxide

2(+1) + x + 3(-2) + 2(-1) = 0 X = +6

 

2.      Calculate the oxidation number of Cr in CrO5

Let us draw the structure.

 


 

 It is obvious from the above structure that CrO5 has butterfly structure in which four oxygen atoms are in form of peroxide and one oxygen in form of oxide.

      x+4(-1)+1(-2)=0

  Oxidation number of Cr in CrO5 is + 6.

 

Oxidising and Reducing Agent

A substance which undergoes oxidation acts as a reducing agent while a substance which undergoes reduction acts as an oxidising agent. For example, we take a redox reaction,

            Zn + Cu2+à  Zn+2 + Cu

In this reaction, Zn is oxidised to Zn2+, so Zn is reducing agent and Cu2+ is reduced to Cu so Cu² is an oxidising agent.

 

Important points for Identification of oxidising and Reducing agent :

1. An elements is in its highest possible oxidation state in a compound. It can behave as an oxidising agent. For example, KMnO4,, K2 Cr2O7, H₂SO4,  HNO3 , HCIO4

2. An element is in its lowest possible oxidation state in a compound. It can behave as a reducing agent. For example, H₂S

3. An element is in its intermediate oxidation state in a compound. It can behave both reducing as well as oxidising agent. For example, H2,O2 , HNO2 , H₂SO3 , so2

4. If highly electronegative element is in its higher oxidation state in a compound, that compound can behave as a powerful oxidising agent. For example, KCIO4 ,  KCIO3 ,  KIO3

5. If an electronegative element is in its lowest possible oxidation state in a compound or free state. It can behave as a powerful reducing agent. For example, I-, Br-, N3- , etc.

 

TYPES OF REDOX REACTIONS

1.      Combination Reactions 

                 3Mg+N₂ à Mg3+2 N-32

 

2.      Decomposition Reaction

For a decomposition reaction to be a redox reaction, at least one of the components formed after the break-down of the compound must be in the elemental state Some common examples of redox decomposition reactions are

                                       2H2O à 2H2 +02

It may be noted that all decomposition reactions are not redox reactions. For example, decomposition of CaCO(s) is not a redox reaction because it does not involve change in oxidation number of elements

3. Displacement Reactions

(a) Metal displacement reaction

 

CuSO4 + Zn à  ZnSo4  + Cu

(b) Non-metal displacement reactions

(i) All alkali metals and some alkaline earth metals which are good reducing agents will displace hydrogen from cold water.

2Na + 2H2O à 2NaOH + H2

(ii) Less active metals such as Mg and Fe react with steam to produce dihydrogen gas.

Mg + 2H2O à  Mg(OH)2 + H2

(iii) Many metals including those which do not react with cold water displace H, from acids.

Zn + 2HCI à ZnCl2 +H₂

Metals like Cd and Sn which do not react with steam also react with acids to displace H, gas:

Cd+ 2HCl à  CdCl2 + H2

 

(c) Reactivity of non-metals

Their reactivity depends upon their oxidising power Fluorine is the strongest oxidising agent in aqueous solution, it can displace Cl, from Cr ions, Br, from Br ions and I, from I ions.

2X-  (aq) + F2 (aq) + X2 (g) + 2F- (aq)

Where X-  =  Cl- , Br- , I-

2H2O + 2F2 à 4HF + O2

Br2 + 2I- à 2Br- + I2

4. Disproportionation Reaction

A reaction in which the atoms of same element is simultaneously oxidised as well as reduced is called a disproportionation reaction

2H2O2 à 2H2O + O2

 

Redox Reactions can also be classified as

1. Intermolecular redox reactions: When reaction takes place between two different substances in such a way that one of them is oxidised and the other is reduced then it is intermolecular redox reaction.

2FeCI3  +  SnCl2 à  2FeCl2 +  SnCl4

2. Intramolecular redox reaction: One element of a compound is oxidised and other element of same compound is reduced in intramolecular redox reaction

 

(NH4)2 Cr2O7 à  N2 + Cr₂O3 + 4H2O

 

THE PARADOX OF FRACTIONAL OXIDATION STATES

(1) Carbon Suboxide (C3O2)

 


The average oxidation number of carbon atoms is 2+2+0/3 = 4/3

 

(ii) Tetrathionate ion (S4O6-2)

 


 

 The average oxidation number of sulphur atom in the ion is , 5+5+0+0/4 = 2.5

 

BALANCING OF REDOX REACTIONS

Oxidation Number Method

In this method, number of electrons lost in cxidation must be equal to number of electrons gained in reduction

Rules:

Step-1: Write the skeletal equation of all the reactants and products of the reaction.

 Step-2: Indicate the oxidation number of each element and identify the elements undergoing change in Oxidation number.

Step-3: Equalize the increase or decrease in oxidation number by multiplying both reactants and products undergoing change in oxidation number by a suitable integer

Step-4: Balance all atoms other than H and O, then balance O atom by adding water molecules to the side short of O-atoms

Step-5: In case of ionic reactions.

(a) For acidic medium

First balance O atoms by adding H2O molecules to the side deficient in O atoms and then

balance H-atoms by adding H+  ions to the side deficient in H atoms.

(b) For basic medium

First balance O atoms by adding H,O molecules to whatever side deficient in O atoms. The

H atoms are then balanced by adding HO molecules equal in number to the deficiency of H

atoms and an equal number of OH ions are added to the opposite side of the equations

Remove the duplication, if any

 

lon-Electron Method

Step 1 : Write the skeleton equation and indicate the oxidation number of all the elements which appear in the skeletal equation above their respective symbols

Step-2 Find out the species which are oxidised and which are reduced.

Step-3 Split the skeleton equation into two half reactions, te oxidation half reaction and reduction half reaction

 

Step-4  : Balance the two half reaction equations separately by the rules described below

 (i) in each half reaction, 1 balance the atoms of the elements which have undergone a change in oxidation number.

(ii) Add electrons to whatever side is necessary to make up the difference in oxidation number in each half reaction

(iii) Balance oxygen atoms by adding required number of HO molecules to the side deficient O atoms (iv) In the acidic medium. H atoms are balanced by adding H ions to the side deficient in H atoms. However in the basic medium, H atoms are balanced by adding H,O molecules equal in number to the deficiency of H atoms and an equal number OH ions are included in the opposite side of the equation. Remove the duplication if any .

Step-5: The two half reactions are then multiplied by suitable integers so that the total number of electrons gained in one half of the reaction is equal to the number of electrons lost in the other half reaction. The two half reactions are then added up. These rules are illustrated by the following examples

Step-6 (Verification): To verify whether the equation thus obtained is balanced or not, the total charge on either side of the equation must be equal.

 

VOLUMETRIC ANALYSIS

It is a method of determination of strength of one solution with the help of another solution of known strength. 

A quantitative analysis is one in which the amount or concentration of a particular substance in a sample is determined accurately and precisely

Volumetric analysis is a quantitative analysis.

The reaction used in volumetric analysis must fulfill the following conditions:

1. The chemistry of reaction must be known and it should be represented by a definite chemical equation

2. The reaction should be instantaneous and proceed to completion over a fairly wide range of concentration

 

LAW OF EQUIVALENCE

Law of equivalence is applicable to all the reactions which are used in volumetric analysis.

At the end point.

 Equivalents of one reactant  =  Equivalents of other reactant

 ( where. No of equivalents or weight / eq mass = NV or N-factor x MVor n factor x No of moles )

 

Back Titration: Let us assume that we are given a definite weight of an impure substance Z and percentage purity of Z in the sample is to be calculated. We are provided with two solutions X and Y, and concentration of both solutions are known to us. This type of titration will work, if the following conditions are satisfied:

(a) X and Y should react with each other.

(b) X and pure Z can react with each other.

(c) Product of X and Z should not react with Y. Since, we have to calculate percentage purity of Z So, it should not be left unreacted i.e in reaction of Z with X, Z should be limiting. Excess amount of X is determined by titrating the resulting mixture with Y

Double Titration: This is a titration of specific compound using different indicators. When the solution containing NaOH and Na,CO, is titrated using phenolphthalein indicator, the following reaction takes place at the phenolphthalein end point.

NaOH + HCl → NaCl + H₂O

Na2CO3 + HCI → NaHCO3 + NaCI

Here, Equivalents of NaOH+ equivalents of Na,CO, (n-fac 1) Equivalents of HCI - 0)

When methyl orange is used, Na,CO, is converted into NaCl + CO2 + H₂O

Hence, Equivalents of NaOH + Equivalents of Na2CO3 (n-fac = 2) = Equivalents of HCI (1)

Above titration can be carried out using phenolphthalein and methyl orange in continuation as well as separately. Accordingly, we apply law of equivalents to calculate percentage composition of the mixture with the help of equation (i) and (ii) if the HCI consumption in two different steps are given.

REDOX TITRATION

Oxidation-Reduction Titration.

The titrations based on oxidation-reduction reaction are called redox titrations

Permanganate Titration : These are the titrations in which potassium permanganate is used as an oxidising agent in acidic medium.

Indicator :  KMnO, as self-indicator

MnO4 + 8H+   5e-   à Mn2+   +    4H2O

Before the end point, the solution remains colourless but after equivalence point, addition of one drop of KMHO, imparts colour to the solution

 

Dichromate Titration

Here K2C2O7 is used as oxidising agent in acidic medium.

Cr2O7 + 14H+    +    5e-     à    2Cr3+     +     7H2O

Indicator K,Fe(CN)) a external indicator or diphenyl amine as intemal indicator

 

lodimetric Titration

These are the Titrations in which free iodine is used. As it is difficult to prepare the solution of iodine (volatile and less soluble in water), it is dissolved in Kl solution

Ki +I2 à Kl3

This solution is first standardised before use. This solution is used for the estimation

 

Indicator:  Starch

 

lodometric Titration

I2 + Na2S2O3    à    2NaL   +   Na2S4O6

2CuSO4     +   4KI      à     CU2L2    +     2K2SO4     +    I2    

 

Indicator : Starch

 

VOLUME STRENGTH OF H2O2

Volume strength of R_{2}*O_{2} solution is defined as volume 0 2 evolved at STP in mi that is obtained per ml of H2O2  solution te. 1 litre solution of H2O2 gives 10 Litre  of oxygen at STP volume strength of H2O2  is 10 volume

Let us say a H2O2 sample is labeled V volume

Then,

2H2O2   à   2H2O  + O2

22400 ml O2 is obtained by 68g H2O2

V ml O2 is obtained by 68/22400 X V g H2O2

 Then , 1000 ml H2o2 = 68/22.4 V  g /litre