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Revision Notes Class 11 Chemistry Chapter 8 - Redox Reactions

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Class 11 Chemistry Chapter 8 Redox Reactions focuses on comprehending the classical concept of Redox Reactions, which includes oxidation and reduction reactions as well as other subjects such as electrode processes, oxidation number, and electron transfer reactions. These CBSE revision notes for class 11 Chemistry Chapter 8 Redox Reactions are available for free download on the official website of Vedantu to assist students grasp the themes and prepare for the tests.


Students may use these notes to revise the whole chapter and become familiar with all of the major themes at the same time. Students may also utilise this to study more effectively, remain ahead in class, and do well on tests.


Students can ultimately examine the class 11 Chemistry Chapter 8 Redox Reactions CBSE revision notes PDF from Vedantu.


Topics Covered in Class 11 Chemistry Chapter 8 - Redox Reactions

The main topics covered in NCERT Solutions for Class 11 Chemistry Chapter 8 are given below.

  1. Classical Idea of Redox Reactions – Oxidation And Reduction Reactions

  2. Redox Reactions In Terms of Electron Transfer Reactions

  3. Competitive Electron Transfer Reactions

  4. Oxidation Number

  5. Types of Redox Reactions

  6. Balancing of Redox Reactions

  7. Redox Reactions as The Basis For Titrations

  8. Limitations of Concept of Oxidation Number

  9. Redox Reactions And Electrode Processes


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Redox Reactions Class 11 Notes Chemistry - Basic Subjective Questions

Section – A (1 Mark Questions)

1. Define oxidation reaction in terms of classical concept.

Ans. Addition of oxygen/electronegative element to a substance or removal of hydrogen/electropositive element from a substance is called oxidation reaction in terms of classical concept.


2. Define reduction reaction in terms of classical concept.

Ans. Removal of oxygen / electronegative element from a substance or addition of hydrogen/electropositive element to a substance is called reduction reaction in terms of classical concept.


3. What is oxidation in terms of electron transfer?

Ans. Oxidation is a process in which loss of electrons takes place from a substance.


4. What is meant by reduction in terms of electrons transfer ?

Ans. Reduction is a process in which gain of electrons take place in a substance.


5. Write formula for Mercury(II) chloride.

Ans. The formula for Mercury(II) chloride is HgCl2.


6. Define equivalent mass.

Ans. It is the number of parts by weight of the substance that combines or displaces, directly or indirectly, 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine. 


7. At what concentration of Zn2+ (aq) will its electrode potential become equal to its standard electrode potential?

Ans. At 1M concentration, electrode potential of Zn2+ (aq) will become equal to its standard electrode potential.


8. What is the Stock notation of chromium trioxide?

Ans. Stock notation of chromium trioxide is Cr2(III)O3.


9. What is the oxidation number of nitrogen in HN3?

Ans. HN3 is a neutral compound. So, sum of oxidation states of all elements is equal to zero. Therefore, 1 + 3x = 0

or $x=-\frac{1}{3}$

Hence, the oxidation number of N is $-\frac{1}{3}=-0.33$


10. $\text{CuSo}_4\left ( \text{aq} \right )+\text{Zn}(\text{s})\rightarrow \text{Cu}(\text{s})+\text{ZnSo}_4\left ( \text{aq} \right )$

This reaction is an example of ________.

Ans. It is a metal displacement reaction as Zn displaces Cu from CuSO4 solution.


Section – B (2 Marks Questions)

11. Name the different types of redox reaction.

Ans. The different types of redox reactions are

(i)  Combination reactions

(ii) Decomposition reactions

(iii) Displacement reactions

(iv) Disproportionation reaction


12. (a) Define disproportionation reaction.

(b) Identify the  type of given reaction.

$2\text{H}_2\text{O}_2\rightarrow 2\text{H}_2\text{O}+\text{O}_2$

Ans. (a) In a disproportionation reaction, the same element in one oxidation state is oxidized as well as reduced simultaneously.

(b) The decomposition of hydrogen peroxide is an example of disproportionation reaction where oxygen is oxidized and reduced simultaneously.

$2\overset{+1}{\text{H}}_2\overset{-1}{\text{O}}_2\rightarrow2\overset{+1}{\text{H}}_2\overset{-2}{\text{O}}+\overset{0}{\text{O}}_2$


13. Calculate oxidation number of underlined element : 

(i)K2CrO4 (ii) H2SO4

Ans. (i) K2CrO4

Let O.N of Cr is x

2(+1) + x + 4(−2) = 0

+ 2 + x − 8 = 0

x = + 6

O.N. of Cr = + 6

(ii) H2SO4

Let O.N of S is x

2 (+1) +x + 4 (−2) = 0

+2 + x −8 = 0

x = + 6


14. Check the feasibility of the following redox reaction with the help of electrochemical series.

$\text{Ni}(\text{s})+2\text{Ag}^+(\text{aq})\rightarrow\text{Ni}^{2+}(\text{aq})+2\text{Ag}(\text{s})$

Ans. The Eo value of Ni2+|Ni is −0.25 V while that of Ag+|Ag is +0.80 V. This means that nickel is placed below silver in the series and can easily reduce Ag+ ions to silver by releasing electrons. Thus, the redox reaction is feasible.


15. Justify that the following reactions are redox reactions :

(a)$\text{CuO}(\text{s})+\text{H}_2(\text{g})\rightarrow\text{Cu}(\text{s})+\text{H}_2\text{O}(\text{g})$

(b)$\text{Fe}_2\text{O}_3(\text{s})+\text{3CO}(\text{g})\rightarrow\text{2Fe}(\text{s})+\text{3CO}_2(\text{g})$

Ans. (a) $\overset{+2}{\text{C}}\text{u}\overset{-2}{\text{O}}(\text{s})+\overset{0}{\text{H}_2}(\text{g})\rightarrow\overset{0}{\text{C}}\text{u}(\text{s})+\overset{+1}{\text{H}}_2\overset{-2}{\text{O}}(\text{g})$

Here, O is removed from CuO, therefore, it is reduced to Cu while O is added to H2 to form H2O, therefore, it is oxidized. Further, O.N. of Cu decreases from +2 in CuO to 0 in Cu but that of H increases from 0 in H2 to +1 in H2O. Therefore, CuO is reduced to Cu but H2 is oxidized to H2O. Thus, this is redox reaction.

(b)$\overset{+3}{\text{Fe}_2}\overset{-2}{\text{O}_3}(\text{s})+3\overset{+2}{\text{C}}\text{O}(\text{g})\rightarrow\overset{0}{\text{Fe}}(\text{s})+3\overset{+4}{\text{C}}\text{O}_2(\text{g})$

Here O.N. of Fe decreases from +3 in Fe2O3 to 0 in Fe while that of C increases from +2 in CO to +4 in CO2. Further, oxygen is removed from Fe2O3 and added to CO, therefore, Fe2O3 is reduced while CO is oxidized. Thus, this is a redox reaction.


16. The following reaction represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidizing action.

$\text{CI}_2(\text{g})+2\text{OH}^{-}(\text{aq})\rightarrow\text{CIO}^{-}(\text{aq})+\text{CI}^{-}(\text{aq})+\text{H}_2\text{O}\text{(l)}$

Ans. $\overset{0}{\text{C}}\text{l}_2(\text{g})+2\overset{-2}{\text{O}}\overset{+1}{\text{H}^{-}}(\text{aq})\rightarrow\overset{+1}{\text{C}}\text{l}\overset{-2}{\text{O}^{-}}(\text{aq})+\overset{-1}{\text{Cl}^{-}}(\text{aq})+\overset{+1}{\text{H}_2}\overset{-2}{\text{O}}(\text{l})$

In this reaction, O.N. of Cl increases from 0 (in Cl2) to +1 (in ClO) and decreases to −1 (in Cl). Therefore, Cl2 is both oxidized to ClOand reduced to Cl.  Therefore, Cl2 bleaches the substances due to oxidizing action of hypochlorite, ClOions. In hypochlorite ion (ClO) Cl can decrease its oxidation number from +1 to 0 or −1.


17. PbO and PbO2 reacts with HCl according to following chemical equations :

(i) $\text{2PbO+4HCl}\rightarrow \text{2PbCl}_2+2\text{H}_2\text{O}$

(ii) $\text{2PbO+4HCl}\rightarrow \text{PbCl}_2+\text{Cl}_2+2\text{H}_2\text{O}$

Why do these compounds differ in their reactivity?

Ans. (i) $2\overset{+2}{\text{Pb}}\text{O+4HCl}\rightarrow2\overset{+2}{\text{Pb}}\text{Cl}_2+2\text{H}_2\text{O}$

(ii) $\overset{+4}{\text{Pb}}\text{O}_2+4\text{HCl}\rightarrow\overset{+2}{\text{Pb}}\text{Cl}_2+\overset{0}{\text{Cl}_2}+2\text{H}_2\text{O}$

In reaction (i), O.N. of none of the atoms undergo a change. Therefore, it is not a redox reaction. It is an acid-base reaction, because PbO is a basic oxide which reacts with HCl which is an acid.

The reaction (ii) is a redox reaction in which PbO2 gets reduced from +4 oxidation state to +2 oxidation state and acts as an oxidizing agent.


18. Define :

(i) Electrochemical series

(ii) Redox titrations

Ans. (i) Electrochemical series is a series of chemical elements/ions arranged in order of their standard electrode potentials.

(ii) A redox titration is a type of titration which relies on a redox reaction between the analyte and titrant. It may involve the use of an indicator.


19. Calculate average oxidation state of S in Tetrathionate ion.

Ans.

state of S in Tetrathionate ion

Tetrathionate ion has four Sulphur atoms bonded to oxygen as in the structure. Out of the four sulphur atoms, the two terminal sulphur atoms are connected to three oxygen atom and one homo sulphur atoms. Each terminal sulphur atoms forms five bonds with oxygen atoms and so the oxidation state will be +5. The bridging sulphur atom being homonuclear have zero oxidation state. 

Total oxidation state of the entire four sulphur atoms is ten.

So, average oxidation state of Sulphur $=\frac{10}{4}=2.5$


20. Which of the two $\text{CIO}^{-}_2$ or $\text{CIO}^{-}_4$ show disproportio-nation reaction and why?

Ans. The oxidation state of Cl in $\text{CIO}^{-}_2$ is +3 and in $\text{CIO}^{-}_4$, it is +7. So, chlorine is present in highest oxidation state of +7 in $\text{CIO}^{-}_4$ and it cannot increase its oxidation state. Hence $\text{CIO}^{-}_4$ does not disproportionate.

The disproportionation reaction of $\text{CIO}^{-}_2$ is 

$3\overset{+3}{\text{Cl}}\text{O}^{-}_2\rightarrow\overset{-1}{\text{Cl}^{-}}+2\overset{+5}{\text{Cl}}\text{O}^{-}_3$


PDF Summary - Class 11 Chemistry Redox Reactions Notes (Chapter 5)

Oxidation: 

  1. Oxygen is added. $2{\text{Mg}} + {{\text{O}}_2} \to 2{\text{MgO}}$

  2. Hydrogen is being removed. ${{\text{H}}_2}\;{\text{S}} + {\text{C}}{{\text{l}}_2} \to 2{\text{HCl}} + {\text{S}}$

  3. Positive charge increases. ${\text{F}}{{\text{e}}^{2 + }} \to {\text{F}}{{\text{e}}^{3 + }} + {{\text{e}}^ - }$

  4. Removal of electron ${\text{S}}{{\text{n}}^{2 + }} \to {\text{S}}{{\text{n}}^{4 + }} + 2{{\text{e}}^ - }$


Reduction: 

  1. Oxygen is being removed. ${\text{CuO}} + {\text{C}} \to {\text{Cu}} + {\text{CO}}$

  2. Hydrogen is added. ${\text{S}} + {{\text{H}}_2} \to {{\text{H}}_2}\;{\text{S}}$

  3. Positive charge decreases. ${\text{F}}{{\text{e}}^{3 + }} + {{\text{e}}^ - } \to {\text{F}}{{\text{e}}^{2 + }}$

  4. Addition of electron. ${\text{F}}{{\text{e}}^{3 + }} + {{\text{e}}^ - } \to {\text{F}}{{\text{e}}^{2 + }}$


Oxidation Number

  • When an element transitions from its elemental free state to its combined form in molecules, it develops an imaginary or seeming charge over each atom.

  • It is determined using an arbitrary set of rules.

  • In a certain bonded condition, it is a relative charge.

  • A more practical way of employing oxidation number to keep track of electron-shifts in chemical reactions involving the synthesis of compounds has been created.

  • The complete transfer of electrons from a less electronegative atom to a more electronegative atom is always expected in this procedure.


Rules Governing Oxidation Number

The following rules can be used to calculate the oxidation number of elements in various compounds. It is important to remember that the electronegativity of the element is the foundation of these rules.

Atom of Fluorine:

Fluorine is the most electronegative of all the elements (known). In all of its compounds, it has an oxidation number of –1.

Atom of Oxygen:

Oxygen atoms have an oxidation number of –2 in general, as well as in their oxides.

  1. peroxide (e.g. $\left. {{{\text{H}}_2}{{\text{O}}_2},{\text{N}}{{\text{a}}_2}{{\text{O}}_2}} \right)$ is $ - 1$

  2. super oxide (e.g. ${\text{K}}{{\text{O}}_2}$ ) is $ - 1/2$

  3. ozonide (e.g. ${\text{K}}{{\text{O}}_3}$ ) is $ - 1/3$

  4. in ${\text{O}}{{\text{F}}_2}$ is $ + 2{\text{ in }}{{\text{O}}_2}\;{{\text{F}}_2}$ is $ + 1$

Hydrogen Atom:

The hydrogen atom has an oxidation number of +1 in general. However, it is –1 in metallic hydrides (e.g. NaH, KH).

Halogen Atom:

In general, all halogen atoms (Cl, Br, and I) have an oxidation number of –1.

If a halogen atom is connected to a more electronegative atom than the halogen atom, the oxidation numbers will be positive.

e.g. ${\text{KCl}}{{\text{O}}_3},{\text{HI}}{{\text{O}}_3},{\text{HCl}}{{\text{O}}_4},{\text{KBr}}{{\text{O}}_3}$.

Metals:

  1. The oxidation number of alkali metals (Li, Na, K, Rb, etc.) is always +1.

  2. The oxidation number of alkaline earth metals (Be, Mg, Ca, etc.) is always +2.

  3. The oxidation number of aluminium is always +3.

Note that a metal's oxidation number might be negative or zero.

  1. In the free state or in allotropic forms, an element's oxidation number is always zero.

e.g. ${\text{O}}_2^0,\;{\text{S}}_8^0,{\text{P}}_4^0$

  1. The sum of all the oxidation numbers of the atoms in a molecule is zero.

  2. The charge on an ion is equal to the sum of the oxidation numbers of all the atoms in the ion.

  3. If an element's group number is n in the contemporary periodic table, its oxidation number can range from (n – 10) to (n – 18). (but it is mainly applicable for p-block elements).


Calculation of Average Oxidation Number: Solved Examples

Example-1 : Calculate oxidation number of underlined element

  1. ${\text{N}}{{\text{a}}_2}{\underline {\text{S}} _2}$

  2. ${\text{N}}{{\text{a}}_2}{\underline {\text{S}} _4}$

  1. Let  oxidation number of S-atom is ${\text{x}}$. Now work accordingly with the rules given before.

$  ( + 1) \times 2 + (x) \times 2 + ( - 2) \times 3 = 0 $     

$ x =  + 2 $

  1. Let oxidation number of $S$-atom is $x$

$ \therefore ( + 1) \times 2 + (x) \times 4 + ( - 2) \times 6 = 0 $     

$ x =  + 2.5 $ 

It is vital to note that ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{2}}}{{\text{O}}_{\text{3}}}$ has two S-atoms, whereas ${\text{N}}{{\text{a}}_{\text{2}}}{{\text{S}}_{\text{4}}}{{\text{O}}_{\text{6}}}$ has four S-atoms. However, none of the sulphur atoms in either compound has an oxidation number of +2 or + 2.5; instead, the average of the oxidation numbers on each sulphur atom is used. As a result, we should strive to determine each sulphur atom's specific oxidation number in these compounds.


Individual Oxidation Number Calculation:

It's vital to remember that in order to calculate the individual oxidation number of an element in its compound, you'll need to know the structure of the compound and follow the steps below.


This is the formula:

Number of electrons in the valence shell minus number of electrons taken up after bonding = oxidation number


Recommendations: It is based on an element's electronegativity.


1. Bonded pair electrons are evenly shared by each element if there is a bond between similar types of atoms and each atom has the same sort of hybridisation.


Consider the Following Scenario:

Calculate each Cl-oxidation atom's number in the ${\text{C}}{{\text{l}}_{\text{2}}}$ molecule.


Hybridization of Cl in chlorine molecule

I : Number of electrons in the valence shell $ = 7$

Number of electrons taken up after bonding $ = 7$

$\therefore $ oxidation number $ = 7 - 7 = 0$

II : $\quad $ similarly, oxidation number $ = 7 - 7 = 0$

If a bond exists between different types of atoms, such as A – B (if B is more electronegative than A), the bound pair of electrons are tallied with the B-atom after bonding.


Consider the following scenario:

Calculate each atom's oxidation number in the HCl molecule.


Structure of HCl

Electron of H-atom is now counted with Cl-atom, because Cl-atom is more electronegative than H-atom

${\text{H}}:\quad $ Number of electrons in the valence shell $ = 1$

Number of electrons taken up after bonding $ = 0$

Oxidation number of ${\text{H}} = 1 - 0 =  + 1$

${\text{Cl}}:\quad $ Number of electrons in the valence shell $ = 7$

Number of electrons taken up after bonding $ = 8$

Oxidation number of ${\text{Cl}} = 7 - 8 =  - 1$


Solved Examples Example – 2

Calculate individual oxidation number of each S-atom in ${\text{N}}{{\text{a}}_2}\;{{\text{S}}_2}{{\text{O}}_3}$ ( sodium thiosulphate) with the help of its structure.


Hybridisation of each ‘S’ atom in the sodium thiosulphate

I (the middle S-atom) is sp3 hybridised (25 percent s-character), while II (the terminal S-atom) is sp2 (33 percent s-character). As a result, the terminal sulphur atom has a higher electronegative charge than the central sulphur atom. With the terminal S-atom, the shared pair of electrons is now counted.

I, S-atom: Number of electrons in the valence shell $ = 6$

Number of electrons left after bonding $ = 0$

Oxidation number of central $S$-atom $ = 6 - 0 =  + 6$

II, S-atom : Number of electrons in the valence shell $ = 6$

Number of electrons left after bonding $ = 8$

Oxidation number of terminal S-atom $ = 6 - 8 =  - 2$

Now, you can also calculate Average Oxidation number of ${\text{S}} = \dfrac{{6 + ( - 2)}}{2} =  + 2$ (as we have calculated before)


Miscellaneous Examples:

In order to determine the exact or individual oxidation number we need to take help from the structures of the molecules. Some special cases are discussed as follows :


Structure of CrO5

From the structure, it is evident that in ${\text{Cr}}{{\text{O}}_5}$ there are two peroxide linkages and one double bond. The contribution of each peroxide linkage is $ - 2$. Let the oxidation number of ${\text{Cr}}$ is ${\text{x}}$.

$\therefore x + ( - 2)2 + ( - 2) = 0$ or $x = 6$

$\therefore $ Oxidation number of ${\text{Cr}} =  + 6$ Ans.


Structure of H2SO5

From the structure, it is evident that in ${{\text{H}}_2}{\text{S}}{{\text{O}}_5}$, there is one peroxide linkage, two sulphur-oxygen double bonds and one ${\text{OH}}$ group.

Let the oxidation number of $S = x$.

$\therefore ( + 1) + ( - 2) + x + ( - 2)2 + ( - 2) + 1 = 0$

or $x + 2 - 8$ or $x - 6 = 0\quad $ or $x = 6$

$\therefore $ Oxidation number of ${\text{S}}$ in ${{\text{H}}_2}{\text{S}}{{\text{O}}_5}$ is $ + 6$ Ans.


Paradox of Fractional Oxidation Number:

The structural parameters demonstrate that the atoms of the element for whom fractional oxidation state is realised are truly present in distinct oxidation states, and the fractional oxidation number is the average of oxidation state of all atoms of the element under investigation. The following bonding scenarios are shown by the species' ${{\text{C}}_3}{{\text{O}}_2},{\text{B}}{{\text{r}}_3}{{\text{O}}_8}$ structure:

In each species, the element highlighted with an asterisk (*) has a different oxidation number than the rest of the atoms of the same element. This demonstrates that two carbon atoms in ${{\text{C}}_{\text{3}}}{{\text{O}}_{\text{2}}}$ are each in the +2 oxidation state, while the third is in the zero oxidation state, giving a total of +4/3. The realistic picture, on the other hand, is +2 for two terabytes.


Oxidation of carbon in the compounds

Likewise in ${\text{B}}{{\text{r}}_3}{{\text{O}}_8}$, each of the two terminal bromine atoms are present in $ + 6$ oxidation state and the middle bromine* is present in $ + 4$ oxidation state. Once again the average, that is different from reality, is $ + 16/3$.


Structure of bromine

Oxidising and Reducing Agent: 

Oxidising Agent or Oxidant : 

Oxidising agents are substances that can oxidise others while reducing themselves during a chemical process. Oxidants are chemicals that cause an element's oxidation number to decrease or gain electrons in a redox process. e.g. ${\text{KMn}}{{\text{O}}_4},\;{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7},{\text{HN}}{{\text{O}}_3}$, conc. ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ etc are powerful oxidising agents.


Reducing Agent or Reductant : 

Reducing agents are substances that can both reduce and oxidise other molecules during a chemical reaction. Reductants are reagents that increase the oxidation number of an element or cause the element to lose electrons in a redox reaction. e.g. ${\text{KI}},{\text{N}}{{\text{a}}_2}\;{{\text{S}}_2}{{\text{O}}_3}$ etc are the powerful reducing agents.


Oxidizing and reducing agents

Redox Reaction:

A redox reaction is one in which both oxidation and reduction occur at the same time. The overall increase in oxidation number must match the total reduction in oxidation number in all redox processes.e.g. $10{\text{FeS}}{{\text{O}}_4} + 2{\text{KMn}}{{\text{O}}_4} + 8{{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to 5{\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3} + 2{\text{MnS}}{{\text{O}}_4} + {{\text{K}}_2}{\text{S}}{{\text{O}}_4} + 8{{\text{H}}_2}{\text{O}}$


Disproportionation Reaction :

A disproportionation reaction is a redox reaction in which the same element present in a specific molecule in a specific oxidation state is oxidised and reduced at the same time.


Disproportionation reactions are a sort of redox reaction that is unique. In a disproportionation reaction, one of the reactants must always contain an element that may exist in at least three oxidation states.


$2{{\text{H}}_2}{{\text{O}}_2}({\text{aq}}) \to 2{{\text{H}}_2}{\text{O}}(l) + {\mathop {\text{O}}\limits^0 _2}(\;{\text{g}})$


${{\text{S}}_{\text{8}}}({\text{s}}) + 12{\text{O}}{{\text{H}}^ - }({\text{aq}}) \to 4\;{{\text{S}}^{2 - }}({\text{aq}}) + 2\;{{\text{S}}_2}{\text{O}}_3^{2 - }({\text{aq}}) + 6{{\text{H}}_2}{\text{O}}(l)$

Consider the Following Reations:

  1. \[2{\text{KCl}}{{\text{O}}_3} \to 2{\text{KCl}} + 3{{\text{O}}_2}\]

${\text{KCl}}{{\text{O}}_3}$ plays a role of oxidant and reductant both. Here, ${\text{Cl}}$ present in ${\text{KCl}}{{\text{O}}_3}$ is reduced and ${\text{O}}$ present in ${\text{KCl}}{{\text{O}}_3}$ is oxidized. Since same element is not oxidized and reduced, so it is not a disproportionation reaction, although it looks like one.

  1. ${\text{N}}{{\text{H}}_4}{\text{N}}{{\text{O}}_2} \to {{\text{N}}_2} + 2{{\text{H}}_2}{\text{O}}$

Nirogen in this compound has $ - 3$ and $ + 3$ oxidation number, which is not a definite value. So it is not a disporportionation reaction. It is an example of comproportionation reaction, which is a class of redox reaction in which an element from two different oxidation state gets converted into a single oxidation state.

List of Some Important Disproportionation Reactions:

  1. ${{\text{H}}_2}{{\text{O}}_2} \to {{\text{H}}_2}{\text{O}} + {{\text{O}}_2}$

  2. ${{\text{X}}_2} + {\text{O}}{{\text{H}}^ - }$(dil.) $ \to {{\text{X}}^ - } + {\text{X}}{{\text{O}}^ - }({\text{X}} = {\text{Cl}},{\text{Br}},{\text{I}})$

F2 does not undergo disproportionation as it is the most electronegative element.

${{\text{F}}_2} + {\text{NaOH (dil}}{\text{.)}} \to {{\text{F}}^ - } + {\text{O}}{{\text{F}}_2}  $

{{\text{F}}_2} + {\text{NaOH (conc}}{\text{.)}} \to {{\text{F}}^ - } + {{\text{O}}_2} $

  1. ${({\text{CN}})_2} + {\text{O}}{{\text{H}}^ - } \to {\text{C}}{{\text{N}}^ - } + {\text{OC}}{{\text{N}}^ - }$

  2. ${{\text{P}}_4} + {\text{O}}{{\text{H}}^ - } \to {\text{P}}{{\text{H}}_3} + {{\text{H}}_2}{\text{PO}}_2^ - $

  3. ${{\text{S}}_8} + {\text{O}}{{\text{H}}^ - } \to {{\text{S}}^{2 - }} + {{\text{S}}_2}{\text{O}}_3^{2 - }$

  4. Oxyacids of Chlorine (Halogens) $( + 1, + 3, + 5$ Oxidation number)

$ {\text{Cl}}{{\text{O}}^ - } \to {\text{C}}{{\text{l}}^ - } + {\text{ClO}}_2^ -  $

$  {\text{ClO}}_2^ -  \to {\text{C}}{{\text{l}}^ - } + {\text{ClO}}_3^ -  $

$  {\text{ClO}}_3^ -  \to {\text{NO}} + {\text{HN}}{{\text{O}}_3} $

  1. ${\text{HN}}{{\text{O}}_2} \to {\text{NO}} + {\text{HN}}{{\text{O}}_3}$

Comproportionation is the inverse of disproportionation. By shifting the medium (from acidic to basic or reverse) in some disproportionation reactions, the reaction goes in the opposite direction and can be considered a Comproportionation reaction.

${{\text{I}}^ - } + {\text{IO}}_3^ -  + {{\text{H}}^ + } \to {{\text{I}}_2} + {{\text{H}}_2}{\text{O}}$


Balancing of Redox Reactions:

All balanced equations must meet two requirements.

  1. Atom balance (mass balance): On the reactant and product sides, there should be the same number of atoms of each species.

  2. Charge balance: On both sides of the equation, the sum of real charges must be equal.


The redox equations can be balanced in two ways:

  1. Number change technique of oxidation

  2. Half-cell method or ion electron method


Because the first approach is not particularly effective in balancing redox reactions, students are urged to use the second way (Ion electron method) to do so.


Method of Ion Electrons:

Redox equations in two different mediums are adjusted using this method.

  1. Acidic Medium and 

  2. Basic Medium


Balancing in Acidic Medium:

Students are advised to balance redox reactions using the ion electron technique in an acidic medium by following the procedures below.


Balance the following redox reaction :

${\text{FeS}}{{\text{O}}_4} + {\text{KMn}}{{\text{O}}_4} + {{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3} + {\text{MnS}}{{\text{O}}_4} + {{\text{H}}_2}{\text{O}} + {{\text{K}}_2}{\text{S}}{{\text{O}}_4}$

Step-I Assign the oxidation number to each element present in the reaction ${\text{FeSO}}_4^{ + 2} + {\text{KMn}}{{\text{O}}_4} + {{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to {\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3} + {\text{MnS}}{{\text{O}}_4} + {{\text{H}}_2}{\text{O}}$

Step II : Now convert the reaction in Ionic form by eliminating the elements or species, which are not undergoing either oxidation or reduction.

${\text{F}}{{\text{e}}^{2 + }} + {\text{MnO}}_4^{ - 7} \to {\text{F}}{{\text{e}}^{3 + }} + {\text{M}}{{\text{n}}^{2 + }}$

Step III : Now identify the oxidation/reduction occuring in the reaction.


Reduction and oxidation

Step IV : Spilt the Ionic reaction in two half, one for oxidation and other for reduction.

${\text{F}}{{\text{e}}^{2 + }}\mathop  \to \limits^{{\text{ Oxidation }}} {\text{F}}{{\text{e}}^{3 + }}\mid {\text{MnO}}_4^ - \mathop  \to \limits^{{\text{ Reduction }}} {\text{M}}{{\text{n}}^{2 + }}$

Step V : Balance the atom other than oxygen and hydrogen atom in both half reactions.

${\text{F}}{{\text{e}}^{2 + }} \to {\text{F}}{{\text{e}}^{3 + }}\mid {\text{MnO}}_4^ -  \to {\text{M}}{{\text{n}}^{2 + }}$

Step VI :Now balance ${\text{O and H}}$ atom by ${{\text{H}}_2}{\text{O and }}{{\text{H}}^ + }$ respectively by the following way : 

For one excess oxygen atom, add one ${{\text{H}}_2}{\text{O}}$ on the other side and two ${{\text{H}}^ + }$on the same side.

${\text{F}}{{\text{e}}^{2 + }} \to {\text{F}}{{\text{e}}^{3 + }}\quad $ (no oxygen atom)

$8{{\text{H}}^ + } + {\text{MnO}}_4^ -  \to {\text{M}}{{\text{n}}^{2 + }} + 4{{\text{H}}_2}{\text{O}}$

Step VII : Equation (i) & (ii) are balanced atomwise. Now balance both equations chargewise. To balance the charge, add electrons to the electrically positive side.

${\text{F}}{{\text{e}}^{2 + }}\mathop  \to \limits^{{\text{ oxidation }}} {\text{F}}{{\text{e}}^{3 + }} + {{\text{e}}^ - }$

$5{{\text{e}}^ - } + 8{{\text{H}}^ + } + {\text{MnO}}_4^ - \mathop  \to \limits^{{\text{ Reduction }}} {\text{M}}{{\text{n}}^{2 + }} + 4{{\text{H}}_2}{\text{O}}$

Step VIII : The number of electrons gained and lost in each half-reaction are equalised by multiplying both the half reactions with a suitable factor and finally the half reactions are added to give the overall balanced reaction. Here, we multiply equation (1) by 5 and (2) by 1 and add them :

${\text{F}}{{\text{e}}^{2 + }} \to {\text{F}}{{\text{e}}^{3 + }} + {{\text{e}}^ - }\quad  \ldots  \ldots  \ldots .(1) \times 5$

$\dfrac{{5{{\text{e}}^ - } + 8{{\text{H}}^ + } + {\text{MnO}}_4^ -  \to {\text{M}}{{\text{n}}^{2 + }} + 4{{\text{H}}_2}{\text{O}}\quad  \ldots  \ldots  \ldots ..(2) \times 1}}{{5{\text{F}}{{\text{e}}^{2 + }} + 8{{\text{H}}^ + } + {\text{MnO}}_4^ -  \to 5{\text{F}}{{\text{e}}^{3 + }} + {\text{M}}{{\text{n}}^{2 + }} + 4{{\text{H}}_2}{\text{O}}}}$

Step IX : Now convert the ionic reaction into molecular form by adding the elements or species, which are removed in step (2). Now, by some manipulation, you will get :

$5{\text{FeS}}{{\text{O}}_4} + {\text{KMn}}{{\text{O}}_4} + 4{{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to \dfrac{5}{2}{\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3} + {\text{MnS}}{{\text{O}}_4} + 4{{\text{H}}_2}{\text{O}} + \dfrac{1}{2}\;{{\text{K}}_2}{\text{S}}{{\text{O}}_4}$

$10{\text{FeS}}{{\text{O}}_4} + 2{\text{KMn}}{{\text{O}}_4} + 8{{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to 5{\text{F}}{{\text{e}}_2}{\left( {{\text{S}}{{\text{O}}_4}} \right)_3} + 2{\text{MnS}}{{\text{O}}_4} + 8{{\text{H}}_2}{\text{O}} + {{\text{K}}_2}{\text{S}}{{\text{O}}_4}$

Balancing in Basic Medium :

In this case, except step VI, all the steps are same. We can understand it by the following example :

Balance the following redox reaction in basic medium :

${\text{Cl}}{{\text{O}}^ - } + {\text{CrO}}_2^ -  + {\text{O}}{{\text{H}}^ - } \to {\text{C}}{{\text{l}}^ - } + {\text{CrO}}_4^{2 - } + {{\text{H}}_2}{\text{O}}$

By using upto step ${\text{V}}$, we will get:

$ \mathop {{\text{Cl}}{O^ - }}\limits^{ + 1} \mathop  \to \limits^{{\text{ Reduction }}} {\text{C}}{{\text{l}}^{ + 3}} $

 $ {\text{CrO}}_2^ - \mathop  \to \limits^{{\text{ Oxidation }}} {\text{CrO}}_4^{2 - }     $


$2{{\text{H}}^ + } + {\text{Cl}}{{\text{O}}^ - } \to {\text{C}}{{\text{l}}^ - } + {{\text{H}}_2}{\text{O}}\mid 2{{\text{H}}_2}{\text{O}} + {\text{CrO}}_2^ -  + 4{{\text{H}}^ + }$

Now, since we are balancing in basic medium, therefore add as many as OH– on both side of equation as there are H+ ions in the equation.

$2{\text{O}}{{\text{H}}^ - } + 2{{\text{H}}^ + } + {\text{Cl}}{{\text{O}}^ - } \to {\text{C}}{{\text{l}}^ - } + {{\text{H}}_2}{\text{O}} + 2{\text{O}}{{\text{H}}^ - }$

Finally you will get

${{\text{H}}_2}{\text{O}} + {\text{Cl}}{{\text{O}}^ - } \to {\text{C}}{{\text{l}}^ - } + 2{\text{O}}{{\text{H}}^ - }$

$4{\text{O}}{{\text{H}}^ - } + 2{{\text{H}}_2}{\text{O}} + {\text{CrO}}_2^ -  \to {\text{CrO}}_4^{2 - } + 4{{\text{H}}^ + } + 4{\text{O}}{{\text{H}}^ - }$

Finally you will get

$4{\text{O}}{{\text{H}}^ - } + {\text{CrO}}_2^ -  \to {\text{CrO}}_4^{2 - } + 2{{\text{H}}_2}{\text{O}}$

Now see equation (i) and (ii) in which ${\text{O}}$ and ${\text{H}}$ atoms are balanced by ${\text{OH}}$ and ${{\text{H}}_2}{\text{O}}$ Now from step VIII

$ 2e^ {-} + H_{2}O + ClO^ {-} \to Cl^{-} + 2OH^{-} \cdots  \cdots  \times 3 $

$  4{\text{O}}{{\text{H}}^ - } + {\text{Cr}}{{\text{O}}_2} - {\text{CrO}}_4^{2 - } + 2{{\text{H}}_2}{\text{O}} + 3{{\text{e}}^ - } \cdots  \cdots  \times 2    $

$  {\text{Adding : }}3{\text{Cl}}{{\text{O}}^ - } + 2{\text{CrO}}_2^ -  + 2{\text{OH}} \to 3{\text{C}}{{\text{l}}^ - } + 2{\text{CrO}}_4^{2 - } + {{\text{H}}_2}{\text{O}}  $


Concept of Equivalents:

Equivalent Mass of Element

The equivalent weight of an element is the number of parts by mass of that element that reacts or displaces from a compound containing 1.008 parts by mass of hydrogen, 8 parts by mass of oxygen, and 35.5 parts by mass of chlorine.


${{\text{ e}}{\text{.g}}{\text{. }}}&{2{\text{Mg}} + {{\text{O}}_2} \to 2{\text{MgO}}} $

$  {}&{48\;{\text{g}}32\;{\text{g}}} $

$  {}&{12\;{\text{g}}8\;{\text{g}}} $

$  {\because \quad }&{32\;{\text{g of }}{{\text{O}}_2}{\text{ reacts with }}48\;{\text{g of Mg}}} $

$\therefore \quad 8\;{\text{g}}$ of ${{\text{O}}_2} = \dfrac{{48 \times 8}}{{32}} = 12\;{\text{g}}$

$\therefore \quad $ Equivalent weight of ${\text{Mg}} = 12$


Equivalent Weight (E) :

In general, Eq. wt. (E) $ = \dfrac{{{\text{ Atomic weight or molecular weight }}}}{{{\text{ valency factor (v}}{\text{.f) }}}} = \dfrac{{{\text{ Mol}}{\text{. wt}}{\text{. }}}}{{n - {\text{ factor }}}} = \dfrac{M}{x}$

  • Number of equivalents = N1V1 for a solution, where N is the normalcy and V is the volume in litres.

  • Equivalent mass is a pure number that is called gramme equivalent mass when given in grammes.

  • Under different situations, the equivalent mass of a substance may have different values.

  • There is no clear and fast rule that the molecular mass will always be less than the comparable weight.

  • Number of Equivalents $ = \dfrac{{{\text{ mass of species }}}}{{{\text{ eq}}{\text{. wt}}{\text{.of that species }}}}$


Valency Factor Calculation: 

For elements, the valency factor equals the element's valency.

For acids, the valency factor is the number of hydrogen ions that can be replaced per acid molecule. Example:

$ {\text{NaOH}},{\text{  KOH}}    $

$  {\text{v}}{\text{.f}}{\text{.}}\quad 1{\text{        1 }} $

$  {\text{Eq}}{\text{. wt}}{\text{. }}\dfrac{{\text{M}}}{1}\quad \dfrac{{\text{M}}}{1} $


Acid - Base Reation :

The valence factor in an acid-base reaction is the actual number of hydrogen or hydroxide ion exchanged in the reaction. There may be more hydrogen or hydroxide ion replaceable in the acid or base than is actually replaced in the reaction. The number of hydrogen ions from the acid that are replaced by each molecule of the base is denoted by v. f.

Example:

$ 2{\text{NaOH}} + {{\text{H}}_2}{\text{S}}{{\text{O}}_4} \to ${\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4} + 2{{\text{H}}_2}{\text{O}}  $

$  {\text{Base          acid}} $

Valency factor of base $ = 1$

Here, two molecule of ${\text{NaOH}}$ replaced $2{{\text{H}}^ + }$ion from the ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$. Therefore, each molecule of ${\text{NaOH}}$ replaced only one ${{\text{H}}^ + }$ion of acid, so v.f $ = 1$

  • v. f. for acid is the number of ${\text{O}}{{\text{H}}^ - }$replaced from the base by each molecule of acid.


Salts :

  1. ln Non-Reacting Condition

The total amount of positive or negative charges present in the chemical is referred to as the valency factor. Example:

$  {{\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3},}&{{\text{F}}{{\text{e}}_2}{{\left( {{\text{S}}{{\text{O}}_4}} \right)}_3}} $

$  {{\text{ v}}{\text{.f}}{\text{. }}}&2&{2 \times 3 = 6} $

$  {{\text{ Eq}}{\text{. wt}}{\text{. }}\quad }&{\dfrac{{\text{M}}}{2}}&{\dfrac{{\text{M}}}{6}} $

  1. In Reacting Condition: Example

$ {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} + {\text{HCl}} \to {\text{NaHC}}{{\text{O}}_3} + {\text{NaCl}}$

$  {\text{base           acid}} $

Because it is an acid-base reaction, the valency factor for sodium carbonate is one, while it is two in the non-reacting state.

  1. Equivalent Weight of Oxidising / Reducing Agents in a Redox Reaction

v.f. = Total change in oxidation number per molecule in the case of a redox change.

Example: ${\text{KMn}}{{\text{O}}_4} + {{\text{H}}_2}{{\text{O}}_2} \to {\text{M}}{{\text{n}}^{2 + }} + {{\text{O}}_2}$

Mn in ${\text{KMn}}{{\text{O}}_4}$ is going from $ + 7$ to $ + 2$, so change in oxidation number per molecule of ${\text{KMn}}{{\text{O}}_4}$ is 5 . So the valency factor of ${\text{KMn}}{{\text{O}}_4}$ is 5 and equivalent weight is $\dfrac{{\text{M}}}{5}$.


Normality: 

The number of equivalents of solute present in one litre (1000 mL) solution is known as the solution's normality. Let V mL of solution be made by dissolving W g of equivalent weight E solute in water.

Number of equivalents of solute $ = \dfrac{W}{E}$

${\text{VmL}}$ of solution contain $\dfrac{{\text{W}}}{{\text{E}}}$ equivalents of solute

$1000\;{\text{mL}}$ solution will contain $\dfrac{{{\text{W}} \times 1000}}{{{\text{E}} \times {\text{V}}}}$ equivalents of solute

Normality $({\text{N}}) = \dfrac{{{\text{W}} \times 1000}}{{{\text{E}} \times {\text{V}}}}$

Normality $({\mathbf{N}}) = $ Molarity $ \times $ Valency factor ${\text{N}} \times {\text{V}}({\text{inmL}}) = {\text{M}} \times {\text{V}}($ in ${\text{mL}}) \times {\text{n}}$

or

milliequivalents = millimoles $ \times n$

Example:

Calculate the normality of a solution containing $15.8\;{\text{g}}$ of ${\text{KMn}}{{\text{O}}_4}$ in $50\;{\text{mL}}$ acidic solution.

Normality $({\mathbf{N}}) = \dfrac{{W \times 1000}}{{E \times V}}$

Here 

$ {\text{W}} = 15.8\;{\text{g}},\quad \;{\text{V}} = 50\;{\text{mL}}\quad   $

$  {\text{E}} = \dfrac{{{\text{ molar mass of KMn}}{{\text{O}}_4}}}{{{\text{ Valency factor }}}} = 158/5 = 31.6 $

So, normality $ = 10\;{\text{N}}$


Law of Equivalence: 

One equivalent of one element combines with one equivalent of the other, according to the law. Equivalents and milliequivalents of reactants react in the same proportion in a chemical reaction to produce the same number of equivalents or milli equivalents of products separately.

  1. $\quad {\text{aA}} + {\text{bB}} \to {\text{mM}} + {\text{nN}}$

meq of ${\text{A}} = $ meq of ${\text{B}} = $ meq of ${\text{M}} = {\text{m}}$.eq. of ${\text{N}}$

  1. In a compound ${{\text{M}}_{\text{x}}}{{\text{N}}_{\text{y}}}$

meq of ${{\text{M}}_{\text{x}}}{{\text{N}}_{\text{y}}} = $ meq of ${\text{M}} = $ meq of ${\text{N}}$


Example:

Find the number of moles of ${\text{KMn}}{{\text{O}}_4}$ needed to oxidise one mole ${\text{C}}{{\text{u}}_2}\;{\text{S}}$ in acidic medium. The reaction is ${\text{KMn}}{{\text{O}}_4} + {\text{C}}{{\text{u}}_2}\;{\text{S}} \to {\text{M}}{{\text{n}}^{2 + }} + {\text{C}}{{\text{u}}^{2 + }} + {\text{S}}{{\text{O}}_2}$

From law of equivalence,

equivalents of ${\text{C}}{{\text{u}}_2}\;{\text{S}} = $ equivalents of ${\text{KMn}}{{\text{O}}_4}$

moles of ${\text{C}}{{\text{u}}_2}\;{\text{S}} \times {\text{v}} \cdot {\text{f}} = $ moles of ${\text{kMn}}{{\text{O}}_4} \times {\text{v}} \cdot {\text{f}}$

$1 \times 8 = $ moles of ${\text{KMn}}{{\text{O}}_4} \times 5\quad  \Rightarrow $ moles of ${\text{KMn}}{{\text{O}}_4} = 8/5$

$\left( \therefore  \right.$ v.f. of ${\text{C}}{{\text{u}}_2}\;{\text{S}} = 2(2 - 1) + 1(4 - ( - 2)) = 8$ and ${\text{v}}.{\text{f}}$. of $\left. {{\text{KMn}}{{\text{O}}_4} = 1(7 - 2) = 5} \right)$


Example: The number of moles of oxalate ions oxidized by one mole of ${\text{MnO}}_4^ - $ion in acidic medium are :

  1. $\dfrac{5}{2}$

  2. $\dfrac{2}{5}$

  3. $\dfrac{3}{5}$

  4. $\dfrac{5}{3}$

Equivalents of ${{\text{C}}_2}{\text{O}}_4^{2 - } = $ equivalents of ${\text{MnO}}_4^ - $

${\text{x}}($ mole $) \times 2 = 1 \times 5$

$\left( \therefore  \right.$ v.f. of ${{\text{C}}_2}{\text{O}}_4^{2 - } = 2(4 - 3) = 2$ and v.f. of ${\text{MnO}}_4^ -  = 1(7 - 2) = 5$.

${\text{x}} = \dfrac{5}{2}$ mole of ${{\text{C}}_2}{\text{O}}_4^{2 - }$ ions.


Example: How many millilitres of $0.02{\text{MKMn}}{{\text{O}}_4}$ solution would be required to exactly titrate $25\;{\text{mL}}$ of $0.2{\text{MFe}}{\left( {{\text{N}}{{\text{O}}_3}} \right)_2}$ solution in acidic medium ?

At the equivalence point,

milliequivalents of ${\text{MnO}}_4^ -  = $milliequivalents of ${\text{F}}{{\text{e}}^{2 + }}$

$ {{\text{M}}_1} \times {\text{v}}{{\text{f}}_1} \times {{\text{V}}_1} = {{\text{M}}_2} \times {\text{v}}{{\text{f}}_2} \times {{\text{V}}_2}    $

$  0.02 \times 5 \times {{\text{V}}_1} = 0.2 \times 1 \times 25\quad    $

$  \because {\text{MnO}}_4^ -  \to {\text{M}}{{\text{n}}^{2 + }};{\text{v}} \cdot {{\text{f}}_2} = 5,{\text{F}}{{\text{e}}^{2 + }} \to {\text{F}}{{\text{e}}^{3 + }};{\text{v}}.{\text{f}}, = 1 $

$  \therefore \quad {{\text{V}}_1} = 50\;{\text{mL}}  $


Titrations: 

Titration is a method of estimating a solution's concentration by allowing a precisely measured volume to react with a known-concentration standard solution of another substance.


A standard solution is a known concentration solution that is taken in a burette. Titrant is another name for it.


There are Two Types of Titrants

Primary Titrants/Standards - These reagents can be precisely weighed, and their solutions do not require standardisation before use. Example: oxalic acid, ferrous ammonium sulfate.

Secondary Titrants/Standards: Because these reagents cannot be precisely weighed, their solutions must be standardised prior to use. Example: ${\text{NaOH,KOH}}$


Titrate: A solution containing the chemical to be measured, usually in a beaker.


The point at which the number of titrant equivalents added equals the number of titrate equivalents is known as the equivalence point.

At equivalence point:

${{\text{n}}_1}\;{{\text{V}}_1}{{\text{M}}_1} = {{\text{n}}_2}\;{{\text{V}}_2}{{\text{M}}_2}$


Indicators: 

Auxiliary material added to allow physical detection of titration completion at the equivalence point. When the titration is finished, the colour usually changes.

Titrations come in a variety of shapes and sizes.


Types of Indicators:

  • Titrations of acid and base (to be studied in Ionic equilibrium)

  • Permanganate titration: 


In acidic media, ${\text{KMn}}{{\text{O}}_{\text{4}}}$ is commonly used as an oxidising agent, which is usually provided by dilute ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}.$ The persistent pink colour of ${\text{KMn}}{{\text{O}}_{\text{4}}}$ serves as a self-indicator of the termination point.

oxalic acid, oxalates, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ and other elements are commonly measured.


Example: Write the balanced reaction of titration of ${\text{KMn}}{{\text{O}}_4}$ Vs oxalic acid in presence of ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$.

Reaction:

$2{\text{KMn}}{{\text{O}}_4} + 3{{\text{H}}_2}{\text{S}}{{\text{O}}_4} + 5{{\text{H}}_2}{{\text{C}}_2}{{\text{O}}_4} \to {{\text{K}}_2}{\text{S}}{{\text{O}}_4} + 2{\text{MnS}}{{\text{O}}_4} + 8{{\text{H}}_2}{\text{O}} + 10{\text{C}}{{\text{O}}_2}$

Redox Changes ${\text{C}}_2^{3 + } \to 2{{\text{C}}^{4 + }} + 2{\text{e}}$ $\left( {{{\text{E}}_{{{\text{H}}_2}{{\text{c}}_2}{{\text{O}}_4}}} = \dfrac{{\text{M}}}{2}} \right)$

$5{{\text{e}}^ - } + {\text{M}}{{\text{n}}^{7 + }} \to {\text{M}}{{\text{n}}^{2 + }}\quad \left( {{{\text{E}}_{{\text{kMn}}{{\text{O}}_4}}} = \dfrac{{\text{M}}}{5}} \right)$

Indicator : ${\text{KMn}}{{\text{O}}_{\text{4}}}$ acts as self indicator.


Hydrogen Peroxide: 

In both mediums, ${{\text{H}}_{\text{2}}}{{\text{O}}_{\text{2}}}$ can act as an oxidising and reducing agent (acidic and basic).


Structure of H2O2

Oxidising Agent: $\left( {{{\text{H}}_2}{{\text{O}}_2} \to {{\text{H}}_2}{\text{O}}} \right)$

  1. Acidic Medium: $2{{\text{e}}^ - } + 2{{\text{H}}^ + } + {{\text{H}}_2}{{\text{O}}_2} \to 2{{\text{H}}_2}{\text{O}}$

${\text{v}} \cdot {\text{f}} = 2$

  1. Basic Medium: $2{{\text{e}}^ - } + {{\text{H}}_2}{{\text{O}}_2} \to 2{\text{O}}{{\text{H}}^ - }$

${\text{v}} \cdot {\text{f}} = 2$


Reducing Agent: $\left( {{{\text{H}}_2}{{\text{O}}_2} \to {{\text{O}}_2}} \right)$

  1. Acidic Medium: \[{{\text{H}}_2}{{\text{O}}_2} \to {{\text{O}}_2} + 2{{\text{H}}^ + } + 2{{\text{e}}^ - }\]

${{\text{v}}_{\text{f}}}{\text{f}} = 2$

  1. Basic Medium: $2{\text{O}}{{\text{H}}^ - } + {{\text{H}}_2}{{\text{O}}_2} \to {{\text{O}}_2} + 2{{\text{H}}_2}{\text{O}} + 2{{\text{e}}^ - }$

${\text{v}} \cdot {\text{f}} = 2$

Note: Valency factor of hydrogen peroxide is always 2.


Volume strength of ${{\text{H}}_2}{{\text{O}}_2}:$ Strength of ${{\text{H}}_2}{{\text{O}}_2}$ is represented as $10\;{\text{V}},20\;{\text{V}},30\;{\text{V}}$ etc. $20{\text{V}}{{\text{H}}_2}{{\text{O}}_2}$ means one litre of this sample of ${{\text{H}}_2}{{\text{O}}_2}$ on decomposition gives 20L of ${{\text{O}}_2}$ gas of STP. Decomposition of ${{\text{H}}_2}{{\text{O}}_2}$ is given as :

${{\text{H}}_2}{{\text{O}}_2} \to {{\text{H}}_2}{\text{O}} + \dfrac{1}{2}{{\text{O}}_2}$

1 mole $\quad \dfrac{1}{2} \times 22.4\;{\text{L}}{{\text{O}}_2}$ at STP

$ = 34\;{\text{g}}\quad  = 11.2\;{\text{L}}{{\text{O}}_2}$ at STP

To obtain 11.2 litre ${{\text{O}}_2}$ at STP, at least $34\;{\text{g}}{{\text{H}}_2}{{\text{O}}_2}$ must be decomposed.

For $20\;{\text{L}}{{\text{O}}_2}$, we should decompose atleast $\dfrac{{34}}{{11.2}} \times 20\;{\text{g}}{{\text{H}}_2}{{\text{O}}_2}$

$\therefore \quad 1\;{\text{L}}$ solution of ${{\text{H}}_2}{{\text{O}}_2}$ contains $\dfrac{{34}}{{11.2}} \times 20\;{\text{g}}{{\text{H}}_2}{{\text{O}}_2}$

$\therefore \quad 1\;{\text{L}}$ solution of ${{\text{H}}_2}{{\text{O}}_2}$ contains $\dfrac{{34}}{{11.2}} \times \dfrac{{20}}{{17}}$ equivalents of ${{\text{H}}_2}{{\text{O}}_2}\quad \left( {{{\text{E}}_{{{\text{H}}_2}{{\text{O}}_2}}} = \dfrac{{\text{M}}}{2} = \dfrac{{34}}{2} = 17} \right)$

Nomality of ${{\text{H}}_2}{{\text{O}}_2} = \dfrac{{34}}{{11.2}} \times \dfrac{{20}}{{17}} = \dfrac{{20}}{{5.6}}$

Normality of ${{\text{H}}_2}{{\text{O}}_2}(\;{\text{N}}) = \dfrac{{{\text{ Volume strength of }}{{\text{H}}_2}{{\text{O}}_2}}}{{5.6}}$

$\because \quad {{\text{M}}_{{{\text{H}}_2}{{\text{O}}_2}}} = \dfrac{{{{\text{N}}_{{{\text{H}}_2}{{\text{O}}_2}}}}}{{{\text{v}}.{{\text{f}}_.}}} = \dfrac{{{{\text{N}}_{{{\text{H}}_2}{{\text{O}}_2}}}}}{2}$

Molarity of ${{\text{H}}_2}{{\text{O}}_2}({\text{M}}) = \dfrac{{{\text{ Volume strength of }}{{\text{H}}_2}{{\text{O}}_2}}}{{11.2}}$ Strength (in ${\text{g}}/{\text{L}}):$ Denoted by ${\text{S}}$ Strength = Molarity $ \times $ Mol. ${\text{wt}} = $ Molarity $ \times 34$

Strength = Normality $ \times {\text{Eq}}.$ weight = Normality $ \times 17$


Example: $20\;{\text{mL}}$ of ${{\text{H}}_2}{{\text{O}}_2}$ after acidification with dilute ${{\text{H}}_2}{\text{S}}{{\text{O}}_4}$ required $30\;{\text{mL}}$ of $\dfrac{{\text{N}}}{{12}}{\text{KMn}}{{\text{O}}_4}$ for complete oxidation. Final the strength. of ${{\text{H}}_2}{{\text{O}}_2}$ solution. (Molar mass of $\left. {{{\text{H}}_2}{{\text{O}}_2} = 34} \right)$

meq. of ${\text{KMn}}{{\text{O}}_4} = $ meq. of ${{\text{H}}_2}{{\text{O}}_2}$

$ 30 \times \dfrac{1}{{12}} = 20 \times {{\text{N}}^\prime } $

$  {{\text{N}}^\prime } = \dfrac{{30}}{{12 \times 20}} $

$   = \dfrac{1}{8}\;{\text{N}}  $

strength $ = {{\text{N}}^\prime } \times $ equivalent mass $ = \dfrac{1}{8} \times 17 = 2.12\;{\text{g}}/{\text{L}}$


Hardness of water (Hard water does not give lather with soap):

${\text{Ca}}$ and ${\text{Mg}}$ bicarbonates cause temporary hardness.

${\text{Ca}}$ and ${\text{Mg}}$ chlorides and sulphates cause permanent hardness. We can soften the water sample using a few different methods.

  1. By boiling

$2{\text{HCO}}_3^ -  \to {{\text{H}}_2}{\text{O}} + {\text{C}}{{\text{O}}_2} + {\text{CO}}_3^{2 - }$

  1. By washing soda

${\text{CaC}}{{\text{l}}_2} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} \to {\text{CaC}}{{\text{O}}_3} + 2{\text{NaCl}}$


Parts Per Million(ppm): This concentration phrase is used when the solute is present in a very small amount. It's the number of parts of the solute in every million parts of the solution. ppm can be expressed in units of mass or moles. If nothing else is mentioned, we assume ppm refers to mass. As a result, a 100 ppm solution has 100 g of solute per 1000000g of solution.


${\text{pp}}{{\text{m}}_{\text{A}}} = \dfrac{{{\text{ mass of A}}}}{{{\text{ Total mass }}}} \times {10^6} = $ mass fraction $ \times {10^6}$


Measurement of Hardness: 

Hardness is measured in terms of ppm (parts per million) of ${\text{CaC}}{{\text{O}}_3}$ or equivalent to it.

${\text{ Hardness in ppm }} = \dfrac{{{\text{ mass of CaC}}{{\text{O}}_3}}}{{{\text{ Total mass of solution }}}} \times {10^6}$


Example: $0.00012\% {\text{MgS}}{{\text{O}}_4}$ and $0.000111\% {\text{CaC}}{{\text{l}}_2}$ is present in water. What is the measured hardness of water and millimoles of washing soda required to purify water $1000\;{\text{L}}$ water ?

Basis of calculation $ = 100\;{\text{g}}$ hard water

${\text{MgS}}{{\text{O}}_4} = 0.00012\;{\text{g}} = \dfrac{{0.00012}}{{120}}\;{\text{mole}}$

${\text{CaC}}{{\text{l}}_2} = 0.000111\;{\text{g}} = \dfrac{{0.000111}}{{111}}\;{\text{mole}}$

equivalent moles of ${\text{CaC}}{{\text{O}}_3} = \left( {\dfrac{{0.00012}}{{120}} + \dfrac{{0.000111}}{{111}}} \right)$ mole

mass of ${\text{CaC}}{{\text{O}}_3} = \left( {\dfrac{{0.00012}}{{120}} + \dfrac{{0.000111}}{{111}}} \right) \times 100 = 2 \times {10^{ - 4}}\;{\text{g}}$

Hardness (in terms of ppm of $\left. {{\text{CaC}}{{\text{O}}_3}} \right) = \dfrac{{2 \times {{10}^{ - 4}}}}{{100}} \times {10^6} = 2{\text{ppm}}$

${\text{CaC}}{{\text{l}}_2} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} \to {\text{CaC}}{{\text{O}}_3} + 2{\text{NaCl}}$

$${\text{MgS}}{{\text{O}}_4} + {\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3} \to {\text{MgC}}{{\text{O}}_3} + {\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}$$

Required ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$ for $100\;{\text{g}}$ of water $ = \left( {\dfrac{{0.00012}}{{120}} + \dfrac{{0.000111}}{{111}}} \right)$ mole

$ = 2 \times {10^{ - 6}}{\text{ mole }}$

Required ${\text{N}}{{\text{a}}_2}{\text{C}}{{\text{O}}_3}$ for 1000 litre water $ = \dfrac{{2 \times {{10}^{ - 6}}}}{{100}} \times {10^6} = \dfrac{2}{{100}}$ mole $\quad (\because {\text{d}} = {\text{lg}}/{\text{mL}})$

$ = \dfrac{{20}}{{1000}}{\text{ mole }} = 20\;{\text{m mole }}$


Solved Examples: 

1. Calculate individual oxidation number of each S-atom in ${\text{N}}{{\text{a}}_2}\;{{\text{S}}_4}{{\text{O}}_6}$ (sodium tetrathionate) with the help of its structure.


Structure of sodium tetrathionate

2. Find the average and individual oxidation number of ${\text{Fe and\;Pb}}$ in ${\text{F}}{{\text{e}}_3}{{\text{O}}_4}{\text{ and\;P}}{{\text{b}}_3}{{\text{O}}_4}$, which are mixed oxides.

(i) ${\text{F}}{{\text{e}}_3}{{\text{O}}_4}$ is mixture of ${\text{FeO, F}}{{\text{e}}_2}{{\text{O}}_3}$ in 1: 1 ratio so, individual oxidation number of ${\text{Fe}} =  + 2, + 3$

average oxidation number $ = \dfrac{{1( + 2) + 2( + 3)}}{3} = 8/3$

(ii) ${\text{P}}{{\text{b}}_3}{{\text{O}}_4}$ is a mixture of ${\text{PbO, Pb}}{{\text{O}}_2}$ in 2: 1 molar ratio so, individual oxidation number of ${\text{Pb}}$ are $ + 2, + 4.$

average oxidation number of ${\text{Pb}} = \dfrac{{2( + 2) + 1( + 4)}}{3} = 8/3$


3. Calculate the normality of a solution obtained by mixing $50\;{\text{mL}}$ of $5{\mathbf{M}}$ solution of ${{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}$ and $50\;{\text{mL}}$ of $2{\text{M}}{{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7}$ in acidic medium.

v.f. of ${{\text{K}}_2}{\text{C}}{{\text{r}}_2}{{\text{O}}_7} = 6$

so ${{\text{N}}_{\text{f}}} = \dfrac{{{{\text{N}}_1}\;{{\text{V}}_1} + {{\text{N}}_2}\;{{\text{V}}_2}}}{{\;{{\text{V}}_1} + {{\text{V}}_2}}}$

$ = \dfrac{{5 \times 6 \times 50 + 2 \times 6 \times 50}}{{50 + 50}} = 21\;{\text{N}}$


4. Calculate the normality of a solution containing $13.4\;{\text{g}}$ of sodium oxalate in $100\;{\text{mL}}$ Sol.

Normality $ = \dfrac{{{\text{ wt}}{\text{. in }}g/eq{\text{. wt }}}}{{{\text{ vol of solution in litre }}}}$

Here, eq. wt. of ${\text{N}}{{\text{a}}_2}{{\text{C}}_2}{{\text{O}}_4} = 134/2 = 67$

so $\quad {\text{N}} = \dfrac{{13.4/67}}{{100/1000}} = 2\;{\text{N}}$


5. The number of moles of ferrous oxalate oxidised by one mole of ${\text{KMn}}{{\text{O}}_4}$ in acidic medium is :

  1. $\dfrac{5}{2}$

  2. $\dfrac{2}{5}$

  3. $\dfrac{3}{5}$

  4. $\dfrac{5}{3}$

Eq. of ${\text{Fe}}{{\text{C}}_2}{{\text{O}}_4} = {\text{Eq}}$. of ${\text{KMn}}{{\text{O}}_4}$

moles of ${\text{Fe}}{{\text{C}}_2}{{\text{O}}_4} \times 3 = $ moles of ${\text{KMn}}{{\text{O}}_4} \times 5$

so, moles of ${\text{Fe}}{{\text{C}}_2}{{\text{O}}_4} = 5/3$

Ans. (d)


6. How many moles of ${\text{KMn}}{{\text{O}}_4}$ are needed to oxidise a mixture of 1 mole of each ${\text{FeS}}{{\text{O}}_4}{\text{ and Fe}}{{\text{C}}_2}{{\text{O}}_4}$ in acidic medium?

  1. $\dfrac{4}{5}$

  2. $\dfrac{5}{4}$

  3. $\dfrac{3}{4}$

  4. $\dfrac{5}{3}$

Eq. of ${\text{KMn}}{{\text{O}}_4} = $ Eq. of ${\text{FeS}}{{\text{O}}_4} + {\text{Eq}}$. of ${\text{Fe}}{{\text{C}}_2}{{\text{O}}_4}$

moles of ${\text{KMn}}{{\text{O}}_4} \times 5 = $ moles of ${\text{FeS}}{{\text{O}}_4} \times 1 + $ moles of

${\text{Fe}}{{\text{C}}_2}{{\text{O}}_4} \times 3$

$\therefore $ moles of ${\text{KMn}}{{\text{O}}_4} = 4/5\quad $

Ans. (a)


7. A fresh ${{\text{H}}_2}{{\text{O}}_2}$ solution is labelled $11.2\;{\text{V}}$. This solution has the same concentration as a solution which is:

  1. $3.4\% (w/w)$

  2. $3.4\% ({\text{v}}/{\text{v}})$

  3. $3.4\% (w/v)$

  4. None of these

Molarity of ${{\text{H}}_2}{{\text{O}}_2} = \dfrac{{{\text{ vol}}{\text{. strength }}}}{{11.2}} = \dfrac{{11.2}}{{11.2}} = 1$

Now, $\% ({\text{w}}/{\text{v}}) = \dfrac{{{\text{ wt}}{\text{. of solute in g}}}}{{{\text{ wt}}{\text{. of solution in mL}}}} \times 100$

= Molarity $ \times $ Mol. wt. of solute $ \times \dfrac{1}{{10}}$

$ = 1 \times 34 \times \dfrac{1}{{10}} = 3.4\% $

Ans. (c)


8. $100\;{\text{mL}}$ each of $1\;{\text{N}}{{\text{H}}_2}{{\text{O}}_2}$ and $11.2\;{\text{V}}{{\text{H}}_2}{{\text{O}}_2}$ solution are mixed, then the final solution is equivalent to :

  1. $3{\text{M}}\;{{\mathbf{H}}_2}{{\mathbf{O}}_2}$ solution

  2. $0.5\;{\text{N}}{{\text{H}}_2}{{\text{O}}_2}$ solution

  3. $25.5\;{\text{g}}/{\text{L}}{{\text{H}}_2}{{\text{O}}_2}$ solution

  4. $2.55\;{\text{g}}/{\text{L}}{{\text{H}}_2}{{\text{O}}_2}$ 

Ans:

${{\text{N}}_{{\text{fral }}}} = \dfrac{{{{\text{N}}_1}\;{{\text{V}}_1} + {{\text{N}}_2}\;{{\text{V}}_2}}}{{\;{{\text{V}}_{\text{t}}} + {{\text{V}}_2}}} = \dfrac{{1 \times 100 + \left( {\dfrac{{11.2}}{{5.6}}} \right) \times 100}}{{100 + 100}} = 3/2 = 1.5\;{\text{N}}$

So, $\quad $ Molarity $ = \dfrac{{{\text{ Normality }}}}{{{\text{ v}}{\text{.f}}{\text{. }}}} = \dfrac{{1.5}}{2} = 0.75{\text{M}}$

Strength of solution in ${\text{g}}/{\text{L}} = $ Molarity $ \times $ Mol. wt. $ = 0.75 \times $ $34 = 25.5\;{\text{g}}/{\text{L}}\quad $


Class 11 Redox Reactions Revision Notes

Firstly, we can say that the chapter explains about the redox reaction. Also, the oxidation and reduction reaction occurs at the same time. Furthermore, in texts, three-tier conceptualization called as oxidation number, classical, and electronics are presented. Moreover, as per each conceptualization, we can see oxidation, reduction, an oxidizing agent (also called oxidant), and reducing agent (also called as reductant) redox. Also, a consistent set of rules were handed over according to their consistent set of rules. Besides, the oxidation numbers and ion-electron both are very useful in representing equations for the redox numbers.


Most notably, we can categorize the oxidation reaction into four categories. These four categories are given as decomposition, combination, disproportionation, and displacement reactions. Also, this concept introduces the electrode and redox couple processes. Furthermore, the redox reactions find wide applications in the study of electrode and cell processes. Moreover, this chapter explains various rules of redox reactions. Also, it explains how to balance a redox reaction. Also, including on what role the titration plays in the redox reaction.


Sub-Topics Covered Under Redox Reactions

Let us look at some of the subtopics that fall under the Redox Reactions.

  • Balance Redox Reactions - This topic gives the overview of redox reaction and various steps to balance it

  • Classical Idea of Redox Reactions - This one defines the classical oxidation and redox reaction 

  • Oxidation Number - This topic highlights the oxidation, its importance, and its rules

  • Redox Reactions & Electrode Potential - This topic discusses the reactions and different types of cells

  • Types of Redox Reactions - This topic teaches us about various types of redox reactions

  • Redox Reactions as the Basis of Titrations - It describes the redox titration, including its types

  • Redox Reactions - Electron Transfer Reactions - This topic speaks about different electron transfer reactions


Some Important Points to Determine the Oxidation Number

  • The algebraic sum of oxidation numbers of the atoms in an uncharged or neutral compound is zero. The algebraic sum in an ion is equal to the charge on the ion

  • All the elements present in the elementary state have an oxidation number as zero. For example, He, Cl2, S8, P4

  • Because fluorine is the most electronegative element, it always contains an oxidation number of -1 in all of its compounds

  • The oxidation number for alkali metals is given as +1, and for the alkaline earth, metals are given as + 2.

  • The oxidation number of metal in amalgams is given as zero.


Important Questions from Redox Reactions (Short, Long & Practice)

Short Answer Type Questions

1. Define Oxidation number and Electrode potential.

2. Define Oxidation and Reduction in terms of electrons.

3. Why are articles made of Iron coated with Zinc to check their rusting?

4. What is the Oxidation number of alkali metals in its compounds?

5. Which reaction occurs at the cathode in a Galvanic cell?

6. What are the maximum and minimum Oxidation numbers of N?


Long Answer Type Questions

1. Starting with the correctly balanced half-reaction, write the overall net ionic equation for the following change:

Chloride ion is oxidised to Cl2 by MnO4- (in acid solution)

2. Give the rules on the basis of which oxidation numbers are assigned to various elements.

3. Write the method used for balancing redox reaction by oxidation number method.


Practice Questions

1. The half cell reactions with their oxidation potentials are:

Pb(s) → Pb2+ + 2e-, E°oxi = + 0.13 v.

Ag(s) → Ag+(aq) + e-, E°oxi = – 0.80 v.

Write the cell reaction and calculate its EMF.

2. How many millimoles of potassium dichromate is required to oxidise 24 cm3 of 0.5 M mohr’s salt solution in an acidic medium.

3. Calculate the cone of hypo (N2a2SO3 5H2O) solution in g dm-3 if 10.0 of this solution decolorized 15 ml of M/40 iodine solution.

4. Determine the volume M/8 KMnO4 solution required to react completely with 25.0 cm3 of M/4 FeSO4 solution in an acidic medium.

5. 16.6 gm of pure KI was dissolved in water and the solution was made up to one litre V cm3 of this solution was acidified with 20 cm3 of 2 MHCl the resulting solution required 10 cm3 of decinormal KIO3 for complete oxidation of I- ions to ICl. Find out the value of V.


Class 11 Redox Reactions Revision Notes

Firstly, we can say that the chapter explains about the redox reaction. Also, the oxidation and reduction reaction occurs at the same time. Furthermore, in texts, three-tier conceptualization called oxidation number, classical, and electronics are presented. Moreover, as per each conceptualization, we can see oxidation, reduction, an oxidising agent (also called oxidant), and reducing agent (also called reductant) redox. Also, a consistent set of rules were handed over according to their consistent set of rules. Besides, the oxidation numbers and ion-electrons both are very useful in representing equations for the redox numbers.


Most notably, we can categorise the oxidation reaction into four categories. These four categories are given as decomposition, combination, disproportionation, and displacement reactions. Also, this concept introduces the electrode and redox couple processes. Furthermore, the redox reactions find wide applications in the study of electrode and cell processes. Moreover, this chapter explains various rules of redox reactions. Also, it explains how to balance a redox reaction. Also, including on what role the titration plays in the redox reaction.


Importance of Redox Reactions Revision Notes

Vedantu, an online teaching and learning platform, offers free downloadable PDF chapter-by-chapter Chemistry Revision Notes, including brief keynotes for the CBSE board examination on the Redox Reactions topic. The review notes on Redox Reactions are extremely significant since they allow students to practise for their studies and have a better understanding of their board examinations. The Class 11 CBSE Revision notes are made for a few important subjects such as English, Maths, Hindi, Chemistry, Biology, and Physics. These core subjects might be very difficult for the students, and correspondingly preparing revision notes on every chapter will allow them to have a skillful studying pattern to achieve much better results and enjoy studying the subject.


Tips to Use Notes of Chapter 8 Chemistry Class 11 

The class 11 chemistry ch 8 notes are designed to ensure success in all examinations. However, you can increase your chances of getting higher marks by using the following tips:

  • Read the NCERT chemistry book to get a basic idea of the chapter and highlight the sections you have trouble understanding.

  • Read the class 11 chemistry chapter 8 notes to understand the sections you previously had trouble understanding.

  • Solve the questions given at the end of the chapter in the textbook.

  • Without copying from the textbook answer all the questions to the best of your understanding.

  • Compare your answers to the solutions available.


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Key Features of Revision Notes for Class 11 Chemistry Chapter 8 - Redox Reactions

  • All solutions of these Revision  Notes are curated by experts at Vedantu. 

  • Explanation to each solution is accurate and to the point.

  • Every solution is very clear and easy to understand as it is written in simple language.

  • Solutions are useful for students from an examination point of view. 

  • This PDF can be utilised while preparing for both board and competitive exams.

  • Reactions are explained in detail to make students understand concepts in a better way.

  • Revision  Notes for CBSE Class 11 Chemistry Chapter 8 are absolutely free and available in PDF format for download.


Conclusion

Revision Notes for Redox Reactions are now accessible on Vedantu in PDF format. Because these notes are created by professional teachers and students may rely on them. In addition, we have included more exam-related questions on this website. These notes and critical questions can be used by students to prepare for and revise these topics.

FAQs on Revision Notes Class 11 Chemistry Chapter 8 - Redox Reactions

1. Explain some top concepts of Redox Reactions?

Some of the top concepts of Redox Reactions can be listed as follows.

  • Redox Reactions

  • Classical View of Redox Reactions

  • Redox Reactions in terms of Electron Transfer

  • Rules to assign oxidation number to an atom

  • Types of Redox Reactions

2. Explain the types of Redox Reactions?

  • Decomposition Reactions - Chemical reactions, where a compound breaks into either two or more simple substances

  • Combination Reactions - The chemical reactions, where either two or more substances (compounds or elements) combine to form a single substance

  • Displacement Reactions - The reactions, where one ion or atom in a compound is replaced by another ion or atom of another element

3. Explain the key points of Revision Notes?

Let us look at some of the key points that are related to the Revision Notes.

  • The revision notes is provided for various subjects like Physics, Maths, Chemistry, Biology, and more

  • It is very helpful to revise the whole syllabus at the exam days

  • It covers all the important points and formulas

  • Even if you wish to take an overview of a chapter, this quick revision notes is here to fulfill it for you

4. Where can I download the Class 11 Chemistry Redox Reactions Revision Notes?

Students can get their Revision Notes on Class 11 Chemistry Redox Reactions from the state boards' respective official websites. Else, they can get all the board’s papers from the official website of Vedantu (vedantu.com).

5. How can Vedantu help students in preparing better for Chapter 8 of Class 11 Chemistry?

Chemistry can seem like an unnerving subject for some, but when you have Vedantu by your side, you can prepare for Chapter 8 of Class 11 Chemistry easily. Here is how Vedantu helps you:

  1. NCERT textbooks are your holy grail for scoring perfect marks in Chemistry. Vedantu provides perfect NCERT Solutions for you.

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All the study material is also available on the Vedantu app at free of cost.

6. What is the Class 11 Chapter 8 “Redox Reaction” about?

Chapter 8 "Redox Reactions" focuses on various redox reactions. The chapter aims to teach the students mechanism of these reactions via the electron transfer process. It also contains important definitions like oxidation, reduction, oxidant, reductant, etc. It teaches the students to identify these reactions and also classify them. It further talks about the real-time applications of these reactions. Students can find beneficial notes for this chapter on the page  Class 11 Chemistry Revision Notes for Chapter 8 for easy reference and exam preparation.

7. What is a redox reaction?

Redox reaction is a type of reaction in which oxidation and reduction reactions occur simultaneously. These are reactions that involve a change in the oxidation number of the involved species. Combination reactions, decomposition reactions, displacement reactions, and disproportionation reactions are various kinds of redox reactions. This chapter contains various important reactions. Students can refer to Vedantu's Class 11 Chapter 8 Revision Notes to study all the important reactions in one place.

9. What are the applications of redox reactions?

There are several real-time applications of redox reactions. These applications are:

  • Redox reactions are greatly used in the study of electrode processes and cells.

  • They are also utilised in pharmaceutical, biological, industrial, metallurgical and agricultural areas.

  • They are used in studying various phenomena like the "Hydrogen economy" and the ozone hole.

  • Used for manufacturing of chemical compounds like caustic soda, operation of dry and wet batteries, and corrosion of metals.