NCERT CBSE Standard 11 Chemistry Chapter 8 Redox Reactions SKMClasses South Bangalore Subhashish Sir

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Solutions to Chapter 8 :

Redox Reactions

Must see https://zookeepersblog.wordpress.com/some-points-which-i-wish-all-my-new-prospective-students-know/

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Where there is oxidation, there is always reduction – Chemistry is essentially a study of redox systems.

Chemistry deals with varieties of matter and change of one kind of matter into the other. Transformation of matter from one kind into another occurs through the various types of reactions. One important category of such reactions is Redox Reactions. A number of phenomena, both physical as well as biological, are concerned with redox reactions. These reactions find extensive use in pharmaceutical, biological, industrial, metallurgical and agricultural areas. The importance of these reactions is apparent from the fact that burning of different types of fuels for obtaining energy for domestic, transport and other commercial purposes, electrochemical processes for extraction of highly reactive metals and non-metals, manufacturing of chemical compounds like caustic soda, operation of dry and wet batteries and corrosion of metals fall within the purview of redox processes. Of late, environmental issues like Hydrogen Economy (use of liquid hydrogen as fuel) and development of ‘Ozone Hole’ have started figuring under redox phenomenon.

8.1 CLASSICAL IDEA OF REDOX REACTIONS – OXIDATION AND REDUCTION REACTIONS Originally, the term oxidation was used to describe the addition of oxygen to an element or a compound. Because of the presence of dioxygen in the atmosphere (~20%), many elements combine with it and this is the principal reason why they commonly occur on the earth in the form of their oxides. The following reactions represent oxidation processes according to the limited definition of oxidation:

2Mg(s) + O₂(g) → 2MgO(s) (8.1)
S(s) + O₂ (g) → SO₂(g) (8.2)

In reactions (8.1) and (8.2), the elements magnesium and sulphur are oxidised on account of addition of oxygen to them. Similarly, methane is oxidised owing to the addition of oxygen to it.

CH₄ (g) + 2O₂ (g) → CO₂(g) + 2H₂O(l) (8.3)

A careful examination of reaction (8.3) in which hydrogen has been replaced by oxygen prompted chemists to reinterpret oxidation in terms of removal of hydrogen from it and, therefore, the scope of term oxidation was broadened to include the removal of hydrogen from a substance. The following illustration is another reaction where removal of hydrogen can also be cited as an oxidation reaction.

2H₂S(g) + O₂(g) → 2S(s) + 2H₂O(l) (8.4)

As knowledge of chemists grew, it was natural to extend the term oxidation for reactions similar to (8.1 to 8.4), which do not involve oxygen but other electronegative elements. The oxidation of magnesium with fluorine, chlorine and sulphur etc. occurs according to the following reactions :

Mg(s) + F₂(g) → MgF₂(s) (8.5) Mg(s) + Cl₂ (g) → MgCl₂ (s) (8.6) Mg(s) + S(s) → MgS(s) (8.7)

Incorporating the reactions (8.5 to 8.7) within the fold of oxidation reactions encouraged chemists to consider not only the removal of hydrogen as oxidation, but also the removal of electropositive elements as oxidation. Thus the reaction :

2K₄[Fe(CN)₆](aq) + H₂O₂(aq) →2K₃[Fe(CN)₆](aq) + 2KOH(aq)

is interpreted as oxidation due to the removal of electropositive element potassium from potassium ferrocyanide before it changes to potassium ferricyanide. To summarise, the term ‘oxidation’ is defined as the addition of oxygen/electronegative element to a substance or removal of hydrogen electropositive element from a substance.

In the beginning, reduction was considered as removal of oxygen from a compound. However, the term reduction has been broadened these days to include removal of oxygen/electronegative element from a substance or addition of hydrogen electropositive element to a substance.

According to the definition given above, the following are the examples of reduction processes:

(removal of electronegative element, chlorine from ferric chloride)

CH₂ = CH₂(g) + H₂(g) → H₃C-CH₃(g) (8.10)

(addition of hydrogen)
2HgCl₂(aq) + SnCl₂(aq) → Hg₂Cl₂(s)+SnCl₄(aq) (8.11)
(addition of mercury to mercuric chloride)

In reaction (8.11) simultaneous oxidation of stannous chloride to stannic chloride is also occurring because of the addition of electronegative element chlorine to it. It was soon realised that oxidation and reduction always occur simultaneously (as will be apparent by re-examining all the equations given above), hence, the word ‘redox’ was coined for this class of chemical reactions.

Problem 8.1

In the reactions given below, identify the species undergoing oxidation and reduction:
(i) H₂S(g) + Cl₂(g) → 2HCl(g) + S(s)
(ii)3Fe₃O₄(s) + 8Al(s) → 9Fe(s) + 4Al₂OM₃(s)
(iii) 2Na(s) + H₂(g) → 2NaH (s)

Solution
(i) H₂S is oxidised because a more electronegative element, chlorine is added to hydrogen (or a more electropositive element, hydrogen has been removed from S). Chlorine is reduced due to addition of hydrogen to it.
(ii) Aluminium is oxidised because oxygen is added to it. Ferrous ferric oxide (Fe₃O₄) is reduced because oxygen has been removed from it.
(iii) With the careful application of the concept of electronegativity only we may infer that sodium is oxidised and hydrogen is reduced.

Reaction (iii) chosen here prompts us to think in terms of another way to define redox reactions.

8.2 REDOX REACTIONS IN TERMS OF ELECTRON TRANSFER REACTIONS
We have already learnt that the reactions
2Na(s) + Cl₂(g) → 2NaCl(s) (8.12)
2Na(s) + O₂(g) → Na2O(s) (8.13)
2Na(s) + S(s) → Na₂S(s) (8.14)
are redox reactions because in each of these reactions sodium is oxidised due to the addition of either oxygen or more electronegative element to sodium. Simultaneously, chlorine, oxygen and sulphur are reduced because to each of these, the electropositive element sodium has been added. From our knowledge of chemical bonding we also know that sodium chloride, sodium oxide and sodium sulphide are ionic compounds and perhaps better written as Na⁺Cl^–(s), (Na⁺)₂O^2-(s), and (Na⁺)₂S_2-(s). Development of charges on the species produced suggests us to rewrite the reactions (8.12 to 8.14) in the following manner :

For convenience, each of the above processes can be considered as two separate steps, one involving the loss of electrons and the other the gain of electrons. As an illustration, we may further elaborate one of these, say, the formation of sodium chloride.
2Na(s) → 2Na⁺(g) + 2e^–
Cl₂(g) + 2e^– → 2 Cl^–(g)

Each of the above steps is called a half reaction, which explicitly shows involvement of electrons. Sum of the half reactions gives the overall reaction :
2Na(s) + Cl₂(g) → 2Na⁺Cl^–(s) or 2NaCl(s)

Reactions 8.12 to 8.14 suggest that half reactions that involve loss of electrons are called oxidation reactions. Similarly, the half reactions that involve gain of electrons are called reduction reactions. It may not be out of context to mention here that the new way of defining oxidation and reduction has been achieved only by establishing a correlation between the behaviour of species as per the classical idea and their interplay in electron-transfer change. In reactions (8.12 to 8.14) sodium, which is oxidised, acts as a reducing agent because it donates electron to each of the elements interacting with it and thus helps in reducing them. Chlorine, oxygen and sulphur are reduced and act as oxidising agents because these accept electrons from sodium. To summarise, we may mention that

Oxidation: Loss of electron(s) by any species.
Reduction: Gain of electron(s) by any species.
Oxidising agent : Acceptor of electron(s).
Reducing agent : Donor of electron(s).

Problem 8.2 Justify that the reaction : 2Na(s) + H₂(g) → 2NaH(s) is a redox change.
Solution Since in the above reaction the compound formed is an ionic compound, which may also be represented as Na⁺H^–(s), this suggests that one half reaction in this process is :
2Na(s) → 2Na⁺(g) + 2e^– and the other half reaction is: H₂(g) + 2e^– → 2H^–(g) This splitting of the reaction under examination into two half reactions automatically reveals that here sodium is oxidised and hydrogen is reduced, therefore, the complete reaction is a redox change.

8.2.1 Competitive Electron Transfer Reactions Place a strip of metallic zinc in an aqueous solution of copper nitrate as shown in Fig. 8.1, for about one hour. You may notice that the strip becomes coated with reddish metallic copper and the blue colour of the solution disappears. Formation of Zn²⁺ ions among the products can easily be judged when the blue colour of the solution due to Cu²⁺ has disappeared. If hydrogen sulphide gas is passed through the colourless solution containing Zn²⁺ ions, appearance of white zinc sulphide, ZnS can be seen on making the solution alkaline with ammonia.
The reaction between metallic zinc and the aqueous solution of copper nitrate is :
Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) (8.15)

In reaction (8.15), zinc has lost electrons to form Zn²⁺ and, therefore, zinc is oxidised. Evidently, now if zinc is oxidised, releasing electrons, something must be reduced, accepting the electrons lost by zinc. Copper ion is reduced by gaining electrons from the zinc. Reaction (8.15) may be rewritten as:

At this stage we may investigate the state of equilibrium for the reaction represented by equation (8.15). For this purpose, let us place a strip of metallic copper in a zinc sulphate solution. No visible reaction is noticed and attempt to detect the presence of Cu²⁺ ions by passing H₂S gas through the solution to produce the black colour of cupric sulphide, CuS, does not succeed. Cupric sulphide has such a low solubility that this is an extremely sensitive test; yet the amount of Cu²⁺ formed cannot be detected. We thus conclude that the state of equilibrium for the reaction (8.15) greatly favours the products over the reactants.
Let us extend electron transfer reaction now to copper metal and silver nitrate solution in water and arrange a set-up as shown in Fig. 8.2. The solution develops blue colour due to the formation of Cu²⁺ ions on account of the reaction:

Here, Cu(s) is oxidised to Cu²⁺(aq) and Ag⁺(aq) is reduced to Ag(s). Equilibrium greatly favours the products Cu²⁺ (aq) and Ag(s).
By way of contrast, let us also compare the reaction of metallic cobalt placed in nickel sulphate solution. The reaction that occurs here is :

At equilibrium, chemical tests reveal that both Ni²⁺(aq) and Co²⁺(aq) are present at moderate concentrations. In this case, neither the reactants [Co(s) and Ni²⁺(aq)] nor the products [Co²⁺(aq) and Ni (s)] are greatly favoured.

This competition for release of electrons incidently reminds us of the competition for release of protons among acids. The similarity suggests that we might develop a table in which metals and their ions are listed on the basis of their tendency to release electrons just as we do in the case of acids to indicate the strength of the acids. As a matter of fact we have already made certain comparisons. By comparison we have come to know that zinc releases electrons to copper and copper releases electrons to silver and, therefore, the electron releasing tendency of the metals is in the order: Zn>Cu>Ag. We would love to make our list more vast and design a metal activity series or electrochemical series. The competition for electrons between various metals helps us to design a class of cells, named as Galvanic cells in which the chemical reactions become the source of electrical energy. We would study more about these cells in Class XII.

8.3 OXIDATION NUMBER

A less obvious example of electron transfer is realised when hydrogen combines with oxygen

2H₂(g) + O₂(g) → 2H₂O(l) (8.18)

Though not simple in its approach, yet we can visualise the H atom as going from a neutral (zero) state in H₂ to a positive state in H₂O, the O atom goes from a zero state in O₂ to a dinegative state in H₂O. It is assumed that there is an electron transfer from H to O and consequently H₂ is oxidised and O₂ is reduced.

However, as we shall see later, the charge transfer is only partial and is perhaps better described as an electron shift rather than a complete loss of electron by H and gain by O. What has been said here with respect to equation (8.18) may be true for a good number of other reactions involving covalent compounds. Two such examples of this class of the reactions are:
H₂(s) + Cl₂(g) → 2HCl(g) (8.19) and,
CH₄(g) + 4Cl₂(g) → CCl₄(l) + 4HCl(g) (8.20)

In order to keep track of electron shifts in chemical reactions involving formation of covalent compounds, a more practical method of using oxidation number has been developed. In this method, it is always assumed that there is a complete transfer of electron from a less electronegative atom to a more electonegative atom. For example, we rewrite equations (8.18 to 8.20) to show charge on each of the atoms forming part of the reaction :

It may be emphasised that the assumption of electron transfer is made for book-keeping purpose only and it will become obvious at a later stage in this unit that it leads to the simple description of redox reactions.

Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules formulated on the basis that electron in a covalent bond belongs entirely to more electronegative element.

It is not always possible to remember or make out easily in a compound/ion, which element is more electronegative than the other. Therefore, a set of rules has been formulated to determine the oxidation number of an element in a compound/ion. If two or more than two atoms of an element are present in the molecule/ion such as Na₂S₂O₃Cr₂O₇^2-, the oxidation number of the atom of that element will then be the average of the oxidation number of all the atoms of that element. We may at this stage, state the rules for the calculation of oxidation number. These rules are:

1. In elements, in the free or the uncombined state, each atom bears an oxidation number of zero. Evidently each atom in H₂, O₂, Cl₂, O₃, P₄, S₈, Na, Mg, Al has the oxidation number zero.
2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. Thus Na+ ion has an oxidation number of +1, Mg²⁺ ion, +2, Fe³⁺ ion, +3, Cl^–ion, -1, O^2-ion, -2; and so on. In their compounds all alkali metals have oxidation number of +1, and all alkaline earth metals have an oxidation number of +2. Aluminium is regarded to have an oxidation number of +3 in all its compounds.
3. The oxidation number of oxygen in most compounds is -2. However, we come across two kinds of exceptions here. One arises in the case of peroxides and superoxides, the compounds of oxygen in which oxygen atoms are directly linked to each other. While in peroxides (e.g., H₂O₂, Na₂O₂), each oxygen atom is assigned an oxidation number of -1, in superoxides (e.g., KO₂, RbO₂) each oxygen atom is assigned an oxidation number of -(½). The second exception appears rarely, i.e. when oxygen is bonded to fluorine. In such compounds e.g., oxygen difluoride (OF₂) and dioxygen difluoride (O₂F₂), the oxygen is assigned an oxidation number of +2 and +1, respectively. The number assigned to oxygen will depend upon the bonding state of oxygen but this number would now be a positive figure only.
4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compounds containing two elements). For example, in LiH, NaH, and CaH₂, its oxidation number is -1.
5. In all its compounds, fluorine has an oxidation number of -1. Other halogens (Cl, Br, and I) also have an oxidation number of -1, when they occur as halide ions in their compounds. Chlorine, bromine and iodine when combined with oxygen, for example in oxoacids and oxoanions, have positive oxidation numbers.
6. The algebraic sum of the oxidation number of all the atoms in a compound must be zero. In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Thus, the sum of oxidation number of three oxygen atoms and one carbon atom in the carbonate ion, (CO₃)^2- must equal -2. By the application of above rules, we can find out the oxidation number of the desired element in a molecule or in an ion. It is clear that the metallic elements have positive oxidation number and nonmetallic elements have positive or negative oxidation number. The atoms of transition elements usually display several positive oxidation states. The highest oxidation number of a representative element is the group number for the first two groups and the group number minus 10 (following the long form of periodic table) for the other groups. Thus, it implies that the highest value of oxidation number exhibited by an atom of an element generally increases across the period in the periodic table. In the third period, the highest value of oxidationnumber changes from 1 to 7 as indicated below in the compounds of the elements.

A term that is often used interchangeably with the oxidation number is the oxidation state. Thus in CO₂, the oxidation state of carbon is +4, that is also its oxidation number and similarly the oxidation state as well as oxidation number of oxygen is -2. This implies that the oxidation number denotes the oxidation state of an element in a compound.

Group	1	2	13	14	15	16	17
Element	Na	Mg	Al	Si	P	S	Cl
Compound	NaCl	MgSO₄	AlF₃	SiCl₄	P₄O₁₀	SF₆	HClO₄
Higher oxidation number state of the group element	+1	+2	+3	+4	+5	+6	+7

The oxidation number state of a metal in a compound is sometimes presented according to the notation given by German chemist, Alfred Stock. It is popularly known as Stock notation. According to this, the oxidation number is expressed by putting a Roman numeral representing the oxidation number in parenthesis after the symbol of the metal in the molecular formula. Thus aurous chloride and auric chloride are written as Au(I)Cl and Au(III)Cl₃. Similarly, stannous chloride and stannic chloride are written as Sn(II)Cl₂ and Sn(IV)Cl₄. This change in oxidation number implies change in oxidation state, which in turn helps to identify whether the species is present in oxidised form or reduced form. Thus, Hg₂(I)Cl₂ is the reduced form of Hg(II) Cl₂.

Problem 8.3 Using Stock notation, represent the following compounds :HAuCl₄, Tl₂O, FeO, Fe₂O₃, CuI, CuO, MnO and MnO₂.

Solution By applying various rules of calculating the oxidation number of the desired element in a compound, the oxidation number of each metallic element in its compound is as follows:
HAuCl₄ → Au has 3
Tl₂O → Tl has 1
FeO → Fe has 2
Fe₂O₃ → Fe has 3
CuI → Cu has 1 CuO → Cu has 2
MnO → Mn has 2
MnO₂ → Mn has 4

Therefore, these compounds may be represented as:
HAu(III)Cl₄, Tl₂(I)O, Fe(II)O, Fe₂(III)O₃, Cu(I)I, Cu(II)O, Mn(II)O, Mn(IV)O₂.

The idea of oxidation number has been invariably applied to define oxidation, reduction, oxidising agent (oxidant), reducing agent (reductant) and the redox reaction. To summarise, we may say that:
Oxidation: An increase in the oxidation number of the element in the given substance.
Reduction: A decrease in the oxidation number of the element in the given substance.
Oxidising agent: A reagent which can increase the oxidation number of an element in a given substance. These reagents are called as oxidants also.
Reducing agent: A reagent which lowers the oxidation number of an element in a given substance. These reagents are also called as reductants.
Redox reactions: Reactions which involve change in oxidation number of the interacting species.

Problem 8.4 Justify that the reaction:
2Cu₂O(s) + Cu₂S(s) → 6Cu(s) + SO₂(g) is a redox reaction. Identify the species oxidised/reduced, which acts as an oxidant and which acts as a reductant.

Solution

Let us assign oxidation number to each of the species in the reaction under examination. This results into:

We therefore, conclude that in this reaction copper is reduced from +1 state to zero oxidation state and sulphur is oxidised from -2 state to +4 state. The above reaction is thus a redox reaction. Further, Cu₂O helps sulphur in Cu₂S to increase its oxidation number, therefore, Cu(I) is an oxidant; and sulphur of Cu₂S helps copper both in Cu₂S itself and Cu₂O to decrease its oxidation number; therefore, sulphur of Cu₂S is reductant.

8.3.1 Types of Redox Reactions

1. Combination reactions
A combination reaction may be denoted in the manner:
A + B → C
Either A and B or both A and B must be in the elemental form for such a reaction to be a redox reaction. All combustion reactions, which make use of elemental dioxygen, as well as other reactions involving elements other than dioxygen, are redox reactions. Some important examples of this category are:

2. Decomposition reactions
Decomposition reactions are the opposite of combination reactions. Precisely, a decomposition reaction leads to the breakdown of a compound into two or more components at least one of which must be in the elemental state. Examples of this class of reactions are:

It may carefully be noted that there is no change in the oxidation number of hydrogen in methane under combination reactions and that of potassium in potassium chlorate in reaction (8.28). This may also be noted here that all decomposition reactions are not redox reactions. For example, decomposition of calcium carbonate is not a redox reaction.

3. Displacement reactions In a displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be denoted as: X + YZ → XZ + Y Displacement reactions fit into two categories: metal displacement and non-metal displacement.

(a) Metal displacement:

A metal in a compound can be displaced by another metal in the uncombined state. We have already discussed about this class of the reactions under section 8.2.1. Metal displacement reactions find many applications in metallurgical processes in which pure metals are obtained from their compounds in ores. A few such examples are:

In each case, the reducing metal is a better reducing agent than the one that is being reduced which evidently shows more capability to lose electrons as compared to the one that is reduced.
(b) Non-metal displacement: The non-metal displacement redox reactions include hydrogen displacement and a rarely occurring reaction involving oxygen displacement. All alkali metals and some alkaline earth metals (Ca, Sr, and Ba) which are very good reductants, will displace hydrogen from cold water.

Less active metals such as magnesium and iron react with steam to produce dihydrogen gas:

Many metals, including those which do not react with cold water, are capable of displacing hydrogen from acids. Dihydrogen from acids may even be produced by such metals which do not react with steam. Cadmium and tin are the examples of such metals. A few examples for the displacement of hydrogen from acids are:

Reactions (8.37 to 8.39) are used to prepare dihydrogen gas in the laboratory. Here, the reactivity of metals is reflected in the rate of hydrogen gas evolution, which is the slowest for the least active metal Fe, and the fastest for the most reactive metal, Mg. Very less active metals, which may occur in the native state such as silver (Ag), and gold (Au) do not react even with hydrochloric acid.

In section (8.2.1) we have already discussed that the metals – zinc (Zn), copper (Cu) and silver (Ag) through tendency to lose electrons show their reducing activity in theorder Zn> Cu>Ag. Like metals, activity series also exists for the halogens. The power of these elements as oxidising agents decreases as we move down from fluorine to iodine in group 17 of the periodic table. This implies that fluorine is so reactive that it can replace chloride, bromide and iodide ions in solution. In fact, fluorine is so reactive that it attacks water and displaces the oxygen of water :

It is for this reason that the displacement reactions of chlorine, bromine and iodine using fluorine are not generally carried out in aqueous solution. On the other hand, chlorine can displace bromide and iodide ions in an aqueous solution as shown below:

As Br₂ and I₂ are coloured and dissolve in CCl₄, can easily be identified from the colour of the solution. The above reactions can be written in ionic form as:

Reactions (8.41) and (8.42) form the basis of identifying Br^– and I^– in the laboratory through the test popularly known as ‘Layer Test’. It may not be out of place to mention here that bromine likewise can displace iodide ion in solution:

The halogen displacement reactions have a direct industrial application. The recovery of halogens from their halides requires an oxidation process, which is represented by:

2X^– → X2 + 2e^– (8.44)

here X denotes a halogen element. Whereas chemical means are available to oxidise Cl^–, Br^– and I^–, as fluorine is the strongest oxidising agent; there is no way to convert F^– ions to F₂ by chemical means. The only way to achieve F2 from F^– is to oxidise electrolytically, the details of which you will study at a later stage.

4. Disproportionation reactions

Disproportionation reactions are a special type of redox reactions. In a disproportionation reaction an element in one oxidation state is simultaneously oxidised and reduced. One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state; and both higher and lower oxidation states of that element are formed in the reaction. The decomposition of hydrogen peroxide is a familiar example of the reaction, where oxygen experiences disproportionation. Here the oxygen of peroxide, which is present in -1 state, is converted to zero oxidation state in O₂ and decreases to -2 oxidation state in H₂O. Phosphorous, sulphur and chlorine undergo disproportionation in the alkaline medium as shown below :

The reaction (8.48) describes the formation of household bleaching agents. The hypochlorite ion (ClO^–) formed in the reaction oxidises the colour-bearing stains of the substances to colourless compounds.

It is of interest to mention here that whereas bromine and iodine follow the same trend as exhibited by chlorine in reaction (8.48), fluorine shows deviation from this behaviour when it reacts with alkali. The reaction that takes place in the case of fluorine is as follows:

2F₂(g) + 2OH^–<(aq) → 2 F^–(aq) + OF₂(g) + H₂O(l) (8.49)

(It is to be noted with care that fluorine in reaction (8.49) will undoubtedly attack water to produce some oxygen also). This departure shown by fluorine is not surprising for us as we know the limitation of fluorine that, being the most electronegative element, it cannot exhibit any positive oxidation state. This means that among halogens, fluorine does not show a disproportionation tendency.

Problem 8.5 Which of the following species, do not show disproportionation reaction and why ?
ClO^–, ClO^–₂, ClO^–₃, and ClO^–₄
Also write reaction for each of the species that disproportionates.

Solution

Among the oxoanions of chlorine listed above, ClO^–₄ does not disproportionate because in this oxoanion chlorine is present in its highest oxidation state that is, +7. The disproportionation reactions for the other three oxoanions of chlorine are as follows:

Problem 8.6 Suggest a scheme of classification of the following redox reactions
(a) N₂(g) + O₂(g) → 2NO(g)
(b) 2Pb(NO₃)₂(s) → 2PbO(s) + 2NO₂(g) +½O₂ (g)
(c) NaH(s) + H₂O(l) → NaOH(aq) + H₂ (g)
(d) 2NO₂(g) + 2OH^–(aq) → NO₂^–(aq) + NO^–₃(aq)+H₂O(l)

Solution

In reaction (a), the compound nitric oxide is formed by the combination of the elemental substances, nitrogen and oxygen; therefore, this is an example of combination redox reactions. The reaction (b) involves the breaking down of lead nitrate into three components; therefore, this is categorised under decomposition redox reaction. In reaction (c), hydrogen of water has been displaced by hydride ion into dihydrogen gas. Therefore, this may be called as displacement redox reaction. The reaction (d) involves disproportionation of NO₂ (+4 state) into NO₂ (+3 state) and NO₃(+5 state). Therefore reaction (d) is an example of disproportionation redox reaction.

The Paradox of Fractional Oxidation Number
Sometimes, we come across with certain compounds in which the oxidation number of a particular element in the compound is in fraction. Examples are: C₃O₂ [where oxidation number of carbon is (4/3)], Br₃O₈ [where oxidation number of bromine is (16/3)] and Na₂S₄O₆ (where oxidation number of sulphur is 2.5). We know that the idea of fractional oxidation number is unconvincing to us, because electrons are never shared/transferred in fraction. Actually this fractional oxidation state is the average oxidation state of the element under examination and the structural parameters reveal that the element for whom fractional oxidation state is realised is present in different oxidation states. Structure of the species C₃O₂, Br₃O₈ and S₄O₆^-2 reveal the following bonding situations: The element marked with asterisk in each species is exhibiting the different oxidation state (oxidation number) from rest of the atoms of the same element in each of the species. This reveals that in C₃O₂, two carbon atoms are present in +2 oxidation state each, whereas the third one is present in zero oxidation state and the average is 4/3. However, the realistic picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br₃O₈, 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. In the same fashion, in the species S₄O₆^2-, each of the two extreme sulphurs exhibits oxidation state of +5 and the two middle sulphurs as zero. The average of four oxidation numbers of sulphurs of the S₄O₆^2- is 2.5, whereas the reality being + 5,0,0 and +5 oxidation number respectively for each sulphur.We may thus, in general, conclude that the idea of fractional oxidation state should be taken with care and the reality is revealed by the structures only. Further, whenever we come across with fractional oxidation state of any particular element in any species, we must understand that this is the average oxidation number only. In reality (revealed by structures only), the element in that particular species is present in more than one whole number oxidation states. Fe₃O₄, Mn₃O₄, Pb₃O₄ are some of the other examples of the compounds, which are mixed oxides, where we come across with fractional oxidation states of the metal atom. However, the oxidation states may be in fraction as in O₂⁺ and O₂^– where it is +½ and -½ respectively.

The Paradox of Fractional Oxidation Number

Sometimes, we come across with certain compounds in which the oxidation number of a particular element in the compound is in fraction. Examples are: C₃O₂ [where oxidation number of carbon is (4/3)], Br₃O₈ [where oxidation number of bromine is (16/3)] and Na₂S₄O₆ (where oxidation number of sulphur is 2.5). We know that the idea of fractional oxidation number is unconvincing to us, because electrons are never shared/transferred in fraction. Actually this fractional oxidation state is the average oxidation state of the element under examination and the structural parameters reveal that the element for whom fractional oxidation state is realised is present in different oxidation states. Structure of the species C₃O₂, Br₃O₈ and S₄O₆^-2 reveal the following bonding situations:
The element marked with asterisk in each species is exhibiting the different oxidation state (oxidation number) from rest of the atoms of the same element in each of the species. This reveals that in C₃O₂, two carbon atoms are present in +2 oxidation state each, whereas the third one is present in zero oxidation state and the average is 4/3. However, the realistic picture is +2 for two terminal carbons and zero for the middle carbon. Likewise in Br₃O₈, 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. In the same fashion, in the species S₄O₆^2-, each of the two extreme sulphurs exhibits oxidation state of +5 and the two middle sulphurs as zero. The average of four oxidation numbers of sulphurs of the S₄O₆^2- is 2.5, whereas the reality being + 5,0,0 and +5 oxidation number respectively for each sulphur.We may thus, in general, conclude that the idea of fractional oxidation state should be taken with care and the reality is revealed by the structures only. Further, whenever we come across with fractional oxidation state of any particular element in any species, we must understand that this is the average oxidation number only. In reality (revealed by structures only), the element in that particular species is present in more than one whole number oxidation states. Fe₃O₄, Mn₃O₄, Pb₃O₄ are some of the other examples of the compounds, which are mixed oxides, where we come across with fractional oxidation states of the metal atom. However, the oxidation states may be in fraction as in O₂⁺ and O₂^– where it is +½ and -½ respectively.

Problem 8.7 Why do the following reactions proceed differently ?
Pb₃O₄ + 8HCl → 3PbCl₂ + Cl₂ + 4H₂O and Pb₃O₄ + 4HNO₃ → 2Pb(NO₃)₂ + PbO₂ + 2H₂O

Solution

Pb₃O₄ is actually a stoichiometric mixture of 2 mol of PbO and 1 mol of PbO₂. In PbO₂, lead is present in +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. PbO₂ thus can act as an oxidant (oxidising agent) and, therefore, can oxidise Cl^– ion of HCl into chlorine. We may also keep in mind that PbO is a basic oxide. Therefore, the reaction

Pb₃O₄ + 8HCl → 3PbCl₂ + Cl₂ + 4H₂O
can be splitted into two reactions namely:

Since HNO₃ itself is an oxidising agent therefore, it is unlikely that the reaction may occur between PbO₂ and HNO₃. However, the acid-base reaction occurs between PbO and HNO₃ as:

2PbO + 4HNO₃→ 2Pb(NO₃)₂ + 2H₂O

It is the passive nature of PbO₂ against HNO₃ that makes the reaction different from the one that follows with HCl.

8.3.2 Balancing of Redox Reactions

Two methods are used to balance chemical equations for redox processes. One of these methods is based on the change in the oxidation number of reducing agent and the oxidising agent and the other method is based on splitting the redox reaction into two half reactions – one involving oxidation and the other involving reduction. Both these methods are in use and the choice of their use rests with the individual using them.

(a) Oxidation Number Method: In writing equations for oxidation-reduction reactions, just as for other reactions, the compositions and formulas must be known for the substances that react and for the products that are formed. The oxidation number method is now best illustrated in the following steps:

Step 1: Write the correct formula for each reactant and product.
Step 2: Identify atoms which undergo change in oxidation number in the reaction by assigning the oxidation number to all elements in the reaction.
Step 3: Calculate the increase or decrease in the oxidation number per atom and for the entire molecule/ion in which it occurs. If these are not equal then multiply by suitable coefficients so that these become equal. (If you realise that two substances are reduced and nothing is oxidised or vice-versa, something is wrong. Either the formulas of reactants or products are wrong or the oxidation numbers have not been assigned properly).

Step 4: Ascertain the involvement of ions if the reaction is taking place in water, add H⁺ or OH^– ions to the expression on the appropriate side so that the total ionic charges of reactants and products are equal. If the reaction is carried out in acidic solution, use H⁺ ions in the equation; if in basic solution, use OH^– ions.

Step 5 : Make the numbers of hydrogen atoms in the expression on the two sides equal by adding water (H₂O) molecules to the reactants or products. Now, also check the number of oxygen atoms. If there are the same number of oxygen atoms in the reactants and products, the equation then represents the balanced redox reaction.
Let us now explain the steps involved in the method with the help of a few problems given below:

Problem 8.8 Write the net ionic equation for the reaction of potassium dichromate(VI), K₂Cr₂O₇ with sodium sulphite, Na₂SO₃, in an acid solution to give chromium(III) ion and the sulphate ion.

Solution
Step 1: The skeletal ionic equation is:
Cr₂O₇^2-(aq)+ SO₃^2-(aq)→ Cr³⁺(aq) + SO₄^2-(aq) Step 2: Assign oxidation numbers for Cr and S

This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant.
Step 3: Calculate the increase and decrease of oxidation number, and make them equal:

Step 4: As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add 8H⁺ on the left to make ionic charges equal
Cr₂O₇^2-(aq) + 3SO₃^2-(aq)+ 8H⁺→ 2Cr³⁺(aq)+ 3SO₄^2-(aq)
Step 5: Finally, count the hydrogen atoms, and add appropriate number of water molecules (i.e., 4H₂O) on the right to achieve balanced redox change. Cr₂O₇^2-(aq) + 3SO₃^2-(aq)+ 8H⁺(aq) → 2Cr³⁺(aq) + 3SO₄^2-(aq) +4H₂O(l)

Problem 8.9
Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balanced ionic equation for the reaction.

Solution
Step 1 : The skeletal ionic equation is : MnO₄(aq) + Br^–(aq) → MnO₂(s) + BrO₃^–(aq)
Step 2 : Assign oxidation numbers for Mn and Br the oxident and bromide ion is the reductant.
Step 3: Calculate the increase and decrease of oxidation number, and make the increase equal to the decrease.

Step 4: As the reaction occurs in the basic medium, and the ionic charges are not equal on both sides, add 2 OH^– ions on the right to make ionic charges equal.
2MnO₄^–(aq) + Br^–(aq) → 2MnO₂(s) + BrO₃(aq) + 2OH_–(aq)
Step 5: Finally, count the hydrogen atoms and add appropriate number of water molecules (i.e. one H₂O molecule) on the left side to achieve balanced redox change.
2MnO₄^–(aq) + Br^–(aq) + H₂O(l) → 2MnO₂(s)+BrO₃(aq)+2OH(aq)
(b) Half Reaction Method:
In this method, the two half equations are balanced separately and then added together to give balanced equation. Suppose we are to balance the equation showing the oxidation of Fe²⁺ ions to Fe³⁺ ions by dichromate ions (Cr₂O₇)^2- in acidic medium, wherein, Cr₂O₇^2- ions are reduced to Cr³⁺ ions. The following steps are involved in this task.

Step 1: Produce unbalanced equation for the reaction in ionic form :
Fe²⁺(aq) + Cr₂O₇^2-(aq) → Fe₃₊(aq) + Cr₃₊(aq) (8.50)

Step 2: Separate the equation into half-reactions:

Step 3: Balance the atoms other than O and H in each half reaction individually. Here the oxidation half reaction is already balanced with respect to Fe atoms. For the reduction half reaction, we multiply the Cr³⁺ by 2 to balance Cr atoms.
Cr₂O₇^2-(aq) → 2 Cr³⁺(aq) (8.53)

Step 4: For reactions occurring in acidic medium, add H₂O to balance O atoms and H⁺ to balance H atoms. Thus, we get :

Cr₂O₇^2- (aq) + 14H⁺ (aq) → 2 Cr³⁺(aq) + 7H²O (l)(8.54)

Step 5: Add electrons to one side of the half reaction to balance the charges. If need be, make the number of electrons equal in the two half reactions by multiplying one or both half reactions by appropriate coefficients. The oxidation half reaction is thus rewritten to balance the charge:
Fe²⁺ (aq) → Fe³⁺ (aq) + e^– (8.55)

Now in the reduction half reaction there are net twelve positive charges on the left hand side and only six positive charges on the right hand side. Therefore, we add six electrons on the left side.
Cr₂O₇^2-(aq) + 14H⁺ (aq) + 6e^– → 2Cr₃₊(aq) + 7H₂O (l) (8.56)

To equalise the number of electrons in both the half reactions, we multiply the oxidation half reaction by 6 and write as :

6Fe²⁺ (aq) → 6Fe³⁺(aq) + 6e^– (8.57)

Step 6: We add the two half reactions to achieve the overall reaction and cancel the electrons on each side. This gives the net ionic equation as :
6Fe₂₊(aq) + Cr₂O₇^2-(aq) + 14H⁺(aq) → 6 Fe³⁺(aq) + 2Cr³⁺(aq) + 7H₂O(l) (8.58)
Step 7: Verify that the equation contains the same type and number of atoms and the same charges on both sides of the equation. This last check reveals that the equation is fully balanced with respect to number of atoms and the charges. For the reaction in a basic medium, first balance the atoms as is done in acidic medium. Then for each H⁺ion, add an equal number of OH^– ions to both sides of the equation. Where H⁺ and OH^– appear on the same side of the equation, combine these to give H₂O. Problem 8.10 Permanganate(VII) ion, MnO₄^– in basic solution oxidises iodide ion, I^– to produce molecular iodine (I₂) and manganese (IV) oxide (MnO₂). Write a balanced ionic equation to represent this redox reaction. Solution Step 1: First we write the skeletal ionic equation, which is
MnO₄^–(aq) + I^–(aq) → MnO₂(s) + I₂(s)

Step 2: The two half-reactions are:

Step 3: To balance the I atoms in the oxidation half reaction, we rewrite it as:
2I^–(aq) → I₂(s) Step 4:
To balance the O atoms in the reduction half reaction, we add two water molecules on the right:
MnO₄^–(aq) → MnO₂(s) + 2H₂O(l)
To balance the H atoms, we add four H⁺ ions on the left:
MnO₄^–(aq) + 4H⁺(aq) → MnO₂(s) + 2H₂O (l)

As the reaction takes place in a basic solution, therefore, for four H⁺ ions, we add four OH^– ions to both sides of the equation:

MnO₄^–(aq) + 4H⁺ (aq) + 4OH^–(aq) →MnO₂ (s) + 2H₂O(l) + 4OH^– (aq) Replacing the H⁺ and OH^– ions with water, the resultant equation is: MnO₄^–(aq) + 2H₂O (l) → MnO₂(s) + 4OH^– (aq)

Step 5 : In this step we balance the charges of the two half-reactions in the manner depicted as:
2I^–(aq) → I₂ (s) + 2e^– MnO₄^–(aq) + 2H₂O(l) + 3e^– → MnO₂(s)+ 4OH^–(aq)
Now to equalise the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2.
6I^–(aq) → 3I₂(s) + 6e^– 2 MnO₄^– (aq) + 4H₂O (l) +6e^– → 2MnO₂(s) + 8OH^– (aq)

Step 6: Add two half-reactions to obtain the net reactions after cancelling electrons on both sides. 6I^–(aq) + 2MnO₄^–(aq) + 4H₂O(l) → 3I₂(s) + 2MnO₂(s) +8OH^–(aq)

Step 7: A final verification shows that the equation is balanced in respect of the number of atoms and charges on both sides.

8.3.3 Redox Reactions as the Basis for Titrations In acid-base systems we come across with a titration method for finding out the strength of one solution against the other using a pH sensitive indicator. Similarly, in redox systems, the titration method can be adopted to determine the strength of a reductant/oxidant using a redox sensitive indicator. The usage of indicators in redox titration is illustrated below:

(i) In one situation, the reagent itself is intensely coloured, e.g., permanganate ion, MnO₄^–. Here MnO₄^– acts as the self indicator. The visible end point in this case is achieved after the last of the reductant (Fe²⁺ or C₂O₄^2-) is oxidised and the first lasting tinge of pink colour appears at MnO₄ ^– concentration as low as 10^-6 mol dm^-3 (10^-6 mol L^-1). This ensures a minimal ‘overshoot’ in colour beyond the equivalence point, the point where the reductant and the oxidant are equal in terms of their mole stoichiometry.

(ii) If there is no dramatic auto-colour change (as with MnO₄^– titration), there are indicators which are oxidised immediately after the last bit of the reactant is consumed, producing a dramatic colour change. The best example is afforded by Cr₂O₇^2-, which is not a self-indicator, but oxidises the indicator substance diphenylamine just after the equivalence point to produce an intense blue colour, thus signalling the end point.
(iii) There is yet another method which is interesting and quite common. Its use is restricted to those reagents which are able to oxidise I^– ions, say, for example, Cu(II): 2Cu²⁺(aq) + 4I^–(aq) → Cu₂I₂(s) + I₂(aq) (8.59)
This method relies on the facts that iodine itself gives an intense blue colour with starch and has a very specific reaction with thiosulphate ions (S₂O₃^2-), which too is a redox reaction:

I₂(aq) + 2S₂O₃^2-(aq)→2I^–(aq) + S₄O₆^2-(aq) (8.60)

I₂, though insoluble in water, remains in solution containing KI as KI₃. On addition of starch after the liberation of iodine from the reaction of Cu₂₊ ions on iodide ions, an intense blue colour appears. This colour disappears as soon as the iodine is consumed by the thiosulphate ions. Thus, the end-point can easily be tracked and the rest is the stoichiometric calculation only.

8.3.4 Limitations of Concept of Oxidation Number As you have observed in the above discussion, the concept of redox processes has been evolving with time. This process of evolution is continuing. In fact, in recent past the oxidation process is visualised as a decrease in electron density and reduction process as an increase in electron density around the atom(s) involved in the reaction.

8.4 REDOX REACTIONS AND ELECTRODE PROCESSES The experiment corresponding to reaction (8.15), can also be observed if zinc rod is dipped in copper sulphate solution. The redox reaction takes place and during the reaction, zinc is oxidised to zinc ions and copper ions are reduced to metallic copper due to direct transfer of electrons from zinc to copper ion. During this reaction heat is also evolved. Now we modify the experiment in such a manner that for the same redox reaction transfer of electrons takes place indirectly. This necessitates the separation of zinc metal from copper sulphate solution. We take copper sulphate solution in a beaker and put a copper strip or rod in it. We also take zinc sulphate solution in another beaker and put a zinc rod or strip in it. Now reaction takes place in either of the beakers and at the interface of the metal and its salt solution in each beaker both the reduced and oxidized forms of the same species are present. These represent the species in the reduction and oxidation half reactions. A redox couple is defined as having together the oxidized and reduced forms of a substance taking part in an oxidation or reduction half reaction.

This is represented by separating the oxidised form from the reduced form by a vertical line or a slash representing an interface (e.g. solid/solution). For example in this experiment the two redox couples are represented as Zn²⁺/Zn and Cu²⁺/Cu. In both cases, oxidised form is put before the reduced form. Now we put the beaker containing copper sulphate solution and the beaker containing zinc sulphate solution side by side (Fig. 8.3). We connect solutions in two beakers by a salt bridge (a U-tube containing a solution of potassium chloride or ammonium nitrate usually solidified by boiling with agar agar and later cooling to a jelly like substance). This provides an electric contact between the two solutions without allowing them to mix with each other. The zinc and copper rods are connected by a metallic wire with a provision for an ammeter and a switch. The set-up as shown in Fig.8.3 is known as Daniel cell. When the switch is in the off position, no reaction takes place in either of the beakers and no current flows through the metallic wire. As soon as the switch is in the on position, we make the following observations:

1. The transfer of electrons now does not take place directly from Zn to Cu²⁺ but through the metallic wire connecting the two rods as is apparent from the arrow which indicates the flow of current.

2. The electricity from solution in one beaker to solution in the other beaker flows by the migration of ions through the salt bridge. We know that the flow of current is possible only if there is a potential difference between the copper and zinc rods known as electrodes here.

The potential associated with each electrode is known as electrode potential. If the concentration of each species taking part in the electrode reaction is unity (if any gas appears in the electrode reaction, it is confined to 1 atmospheric pressure) and further the reaction is carried out at 298K, then the potential of each electrode is said to be the Standard Electrode Potential. By convention, the standard electrode potential (E^Θ) of hydrogen electrode is 0.00 volts. The electrode potential value for each electrode process is a measure of the relative tendency of the active species in the process to remain in the oxidised/reduced form. A negative E_Θ means that the redox couple is a stronger reducing agent than the H⁺/H₂ couple. A positive E_Θ means that the redox couple is a weaker reducing agent than the H⁺/H₂ couple. The standard electrode potentials are very important and we can get a lot of other useful information from them. The values of standard electrode potentials for some selected electrode processes (reduction reactions) are given in Table 8.1. You will learn more about electrode reactions and cells in Class XII.

SUMMARY

Redox reactions form an important class of reactions in which oxidation and reduction occur simultaneously. Three tier conceptualisation viz, classical, electronic and oxidation number, which is usually available in the texts, has been presented in detail. Oxidation, reduction, oxidising agent (oxidant) and reducing agent (reductant) have been viewed according to each conceptualisation. Oxidation numbers are assigned in accordance with a consistent set of rules. Oxidation number and ion-electron method both are useful means in writing equations for the redox reactions. Redox reactions are classified into four categories: combination, decomposition displacement and disproportionation reactions. The concept of redox couple and electrode processes is introduced here. The redox reactions find wide applications in the study of electrode processes and cells.

EXERCISES

8.1 Assign oxidation number to the underlined elements in each of the following species:

8.2 What are the oxidation number of the underlined elements in each of the following and how do you rationalise your results ?

8.3 Justify that the following reactions are redox reactions:
(a) CuO(s) + H₂(g) → Cu(s) + H₂O(g)
(b) Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
(c) 4BCl₃(g) + 3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃ (s)
(d) 2K(s) + F₂(g) → 2K⁺F^– (s)
(e) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)

8.4 Fluorine reacts with ice and results in the change:
H₂O(s) + F₂(g) → HF(g) + HOF(g) Justify that this reaction is a redox reaction.

8.5 Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇^2- and NO₃^–.

Suggest structure of these compounds. Count for the fallacy.

8.6 Write formulas for the following compounds:
(a) Mercury(II) chloride
(b) Nickel(II) sulphate
(c) Tin(IV) oxide
(d) Thallium(I) sulphate
(e) Iron(III) sulphate (f) Chromium(III) oxide

8.7 Suggest a list of the substances where carbon can exhibit oxidation states from -4 to +4 and nitrogen from -3 to +5.

8.8 While sulphur dioxide and hydrogen peroxide can act as oxidising as well as reducing agents in their reactions, ozone and nitric acid act only as oxidants. Why ?
8.9 Consider the reactions:
(a) 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(aq) + 6O₂(g)
(b) O₃(g) + H₂O₂(l) → H₂O(l) + 2O₂(g)
Why it is more appropriate to write these reactions as :
(a) 6CO₂(g) + 12H₂O(l) → C₆H₁₂O₆(aq) + 6H₂O(l) + 6O₂(g)
(b) O₃(g) + H₂O₂ (l) → H₂O(l) + O₂(g) + O₂(g)
Also suggest a technique to investigate the path of the above (a) and (b) redox reactions.

8.10 The compound AgF₂ is unstable compound. However, if formed, the compound acts as a very strong oxidising agent. Why ?
8.11 Whenever a reaction between an oxidising agent and a reducing agent is carried out, a compound of lower oxidation state is formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidising agent is in excess. Justify this statement giving three illustrations.
8.12 How do you count for the following observations ?
(a) Though alkaline potassium permanganate and acidic potassium permanganate both are used as oxidants, yet in the manufacture of benzoic acid from toluene we use alcoholic potassium permanganate as an oxidant. Why ? Write a balanced redox equation for the reaction.
(b) When concentrated sulphuric acid is added to an inorganic mixture containing chloride, we get colourless pungent smelling gas HCl, but if the mixture contains bromide then we get red vapour of bromine. Why ?

8.13 Identify the substance oxidised reduced, oxidising agent and reducing agent for each of the following reactions:
(a) 2AgBr(s) + C₆H₆O₂(aq) → 2Ag(s) + 2HBr (aq) + C₆H₄O₂(aq)
(b) HCHO(l) + 2[Ag (NH₃)₂]+(aq) + 3OH^–(aq) → 2Ag(s) + HCOO^–(aq) + 4NH₃(aq) + 2H₂O(l)
(c) HCHO (l) + 2Cu²⁺(aq) + 5OH^–(aq) → Cu₂O(s) + HCOO^–(aq) + 3H₂O(l)
(d) N₂H⁴(l) + 2H₂O₂(l) → N₂(g) + 4H₂O(l)
(e) Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l)

8.14 Consider the reactions :
2S₂O₃^2-(aq) + I₂(s) → S₄O₆^2-(aq) + 2I^–(aq) S₂O₃^2-(aq) + 2Br₂(l) + 5H₂O(l) → 2SO₄^2-(aq) + 4Br^–(aq) + 10H⁺(aq)
Why does the same reductant, thiosulphate react differently with iodine and bromine ?

8.15 Justify giving reactions that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant.
8.16 Why does the following reaction occur ?
XeO₆^4- (aq) + 2F^– (aq) + 6H⁺(aq) → XeO₃(g)+ F₂(g) + 3H₂O(l)

What conclusion about the compound Na₄XeO₆ (of which XeO₆^4- is a part) can be drawn from the reaction.

8.17 Consider the reactions:
(a) H₃PO₂(aq) + 4AgNO₃(aq) + 2H₂O(l) → H₃PO₄(aq) + 4Ag(s) + 4HNO₃(aq)
(b) H₃PO₂(aq) + 2CuSO₄(aq) + 2H₂O(l) → H₃PO₄(aq) + 2Cu(s) + H₂SO₄(aq)
(c) C₆H₅CHO(l) + 2[Ag(NH₃)₂]+(aq) + 3OH^–(aq) → C₆H₅COO^–(aq) + 2Ag(s) + 4NH₃(aq) + 2H₂O(l)
(d) C₆H₅CHO(l) + 2Cu²⁺(aq) + 5OH^–(aq) → No change observed.

What inference do you draw about the behaviour of Ag⁺ and Cu²⁺ from these reactions ?

8.18 Balance the following redox reactions by ion – electron method :
(a) MnO₄^–(aq) + I^– (aq) → MnO₂(s) + I₂(s) (in basic medium)
(b) MnO₄^–(aq) + SO₂ (g) → Mn²⁺ (aq) + HSO₄^– (aq) (in acidic solution)
(c) H₂O₂ (aq) + Fe²⁺ (aq) → Fe³⁺ (aq) + H₂O (l) (in acidic solution)
(d) Cr₂O₇^2- + SO₂(g) → Cr³⁺ (aq) + SO₄^2- (aq) (in acidic solution)

8.19 Balance the following equations in basic medium by ion-electron method and oxidation number methods and identify the oxidising agent and the reducing agent.
(a) P₄(s) + OH^–(aq) → PH₃(g) + HPO₂^– (aq)
(b) N₂H₄(l) + ClO₃^–(aq) → NO(g) + Cl^–(g)
(c) Cl₂O⁷ (g) + H₂O₂(aq) → ClO₂^–(aq) + O²(g) + H⁺

8.20 What sorts of informations can you draw from the following reaction ?
(CN)₂(g) + 2OH^–(aq) → CN^–(aq) + CNO^–(aq) + H₂O(l)

8.21 The Mn³⁺ ion is unstable in solution and undergoes disproportionation to give Mn²⁺, MnO₂, and H⁺ ion. Write a balanced ionic equation for the reaction.
8.22 Consider the elements : Cs, Ne, I and F (a) Identify the element that exhibits only negative oxidation state. (b) Identify the element that exhibits only postive oxidation state. (c) Identify the element that exhibits both positive and negative oxidation states. (d) Identify the element which exhibits neither the negative nor does the positive oxidation state.
8.23 Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water.
8.24 Refer to the periodic table given in your book and now answer the following questions:
(a) Select the possible non metals that can show disproportionation reaction. (b) Select three metals that can show disproportionation reaction.

8.25 In Ostwald’s process for the manufacture of nitric acid, the first step involves the oxidation of ammonia gas by oxygen gas to give nitric oxide gas and steam. What is the maximum weight of nitric oxide that can be obtained starting only with 10.00 g. of ammonia and 20.00 g of oxygen ?

8.26 Using the standard electrode potentials given in the Table 8.1, predict if the reaction between the following is feasible: (a) Fe³⁺(aq) and I^–(aq) (b) Ag⁺(aq) and Cu(s) (c) Fe³⁺ (aq) and Cu(s) (d) Ag(s) and Fe³⁺(aq) (e) Br₂(aq) and Fe²⁺(aq).
8.27 Predict the products of electrolysis in each of the following: (i) An aqueous solution of AgNO₃ with silver electrodes (ii) An aqueous solution AgNO₃ with platinum electrodes (iii) A dilute solution of H₂SO₄ with platinum electrodes (iv) An aqueous solution of CuCl₂ with platinum electrodes.
8.28 Arrange the following metals in the order in which they displace each other from the solution of their salts. Al, Cu, Fe, Mg and Zn.
8.29 Given the standard electrode potentials, K⁺/K = -2.93V, Ag⁺/Ag = 0.80V, Hg²⁺/Hg = 0.79V Mg²⁺/Mg = -2.37V. Cr³⁺/Cr = -0.74V arrange these metals in their increasing order of reducing power.
8.30 Depict the galvanic cell in which the reaction Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) +2Ag(s) takes place, Further show:
(i) which of the electrode is negatively charged,
(ii) the carriers of the current in the cell, and
(iii) individual reaction at each electrode

Answer to Some Selected Problems

8.25 15 g

I. Multiple Choice Questions (Type-I)

1. Which of the following is not an example of redox reaction?

(i) CuO + H₂ -> Cu + H₂O
(ii) Fe₂O₃ + 3CO -> 2Fe + 3CO₂
(iii) 2K + F₂ -> 2KF
(iv) BaCl₂ + H₂SO₄ -> BaSO₄ + 2HCl

2. The more positive the value of E^Θ, the greater is the tendency of the species to get reduced. Using the standard electrode potential of redox couples given below find out which of the following is the strongest oxidising agent.

E^Θ values: Fe³⁺/Fe²⁺ = + 0.77; I₂(s)/I^– = + 0.54;
Cu²⁺/Cu = + 0.34; Ag⁺/Ag = + 0.80V

(i) Fe³⁺
(ii) I₂(s)
(iii) Cu²⁺
(iv) Ag⁺

3. E^Θ values of some redox couples are given below. On the basis of these values choose the correct option.

E^Θ values : Br₂/Br^– = + 1.90; Ag⁺ /Ag(s) = + 0.80
Cu²⁺/Cu(s) = + 0.34; I₂(s)/I^– = + 0.54

(i) Cu will reduce Br^–
(ii) Cu will reduce Ag
(iii) Cu will reduce I^–
(iv) Cu will reduce Br₂

4. Using the standard electrode potential, find out the pair between which redox reaction is not feasible.

E^Θ values : Fe³⁺/Fe²⁺ = + 0.77; I₂/I^– = + 0.54;
Cu²⁺/Cu = + 0.34; Ag⁺/Ag = + 0.80 V

(i) Fe³⁺ and I^–
(ii) Ag⁺ and Cu
(iii) Fe³⁺ and Cu
(iv) Ag and Fe³⁺

5. Thiosulphate reacts differently with iodine and bromine in the reactions given below: 2S₂O₃^2– + I₂ → S₄O₆^2– + 2I^–
S₂O₃^2– + 2Br₂ + 5H₂O → 2SO₄^2– + 2Br^– + 10H⁺
Which of the following statements justifies the above dual behaviour of thiosulphate?

(i) Bromine is a stronger oxidant than iodine.
(ii) Bromine is a weaker oxidant than iodine.
(iii) Thiosulphate undergoes oxidation by bromine and reduction by iodine in these reactions.
(iv) Bromine undergoes oxidation and iodine undergoes reduction in these reactions.

6. The oxidation number of an element in a compound is evaluated on the basis of certain rules. Which of the following rules is not correct in this respect?

(i) The oxidation number of hydrogen is always +1.
(ii) The algebraic sum of all the oxidation numbers in a compound is zero.
(iii) An element in the free or the uncombined state bears oxidation number zero.
(iv) In all its compounds, the oxidation number of fluorine is – 1.

7. In which of the following compounds, an element exhibits two different oxidation states.

(i) NH₂OH
(ii) NH₄NO₃
(iii) N₂H₄
(iv) N₃H

8. Which of the following arrangements represent increasing oxidation number of the central atom?

(i) CrO₂^– , ClO₃^– , CrO₄^2–, MnO₄^–
(ii) ClO₃^–, CrO₄^2– , MnO₄^– , CrO₂^–
(iii) CrO₂^– , ClO₃^– , MnO₄^– , CrO₄^2–
(iv) CrO₄^2–, MnO₄^– , CrO₂^– , ClO₃^–

9. The largest oxidation number exhibited by an element depends on its outer electronic configuration. With which of the following outer electronic configurations the element will exhibit largest oxidation number?

(i) 3d¹4s²
(ii) 3d³4s²
(iii) 3d⁵4s¹
(iv) 3d⁵4s²

10. Identify disproportionation reaction

(i) CH₄ + 2O₂ → CO₂ + 2H₂O
(ii) CH₄ + 4Cl₂ → CCl₄ + 4HCl
(iii) 2F₂ + 2OH^– → 2F^– + OF₂ + H₂O
(iv) 2NO₂ + 2OH^– → NO₂^– + NO₃^– + H₂O

11. Which of the following elements does not show disproportionation tendency?

(i) Cl
(ii) Br
(iii) F
(iv) I

II. Multiple Choice Questions (Type-II)

In the following questions two or more options may be correct.

12. Which of the following statement(s) is/are not true about the following decomposition reaction.
2KClO₃ → 2KCl + 3O₂

(i) Potassium is undergoing oxidation
(ii) Chlorine is undergoing oxidation
(iii) Oxygen is reduced
(iv) None of the species are undergoing oxidation or reduction

13. Identify the correct statement (s) in relation to the following reaction:
Zn + 2HCl → ZnCl₂ + H₂

(i) Zinc is acting as an oxidant
(ii) Chlorine is acting as a reductant
(iii) Hydrogen ion is acting as an oxidant
(iv) Zinc is acting as a reductant

14. The exhibition of various oxidation states by an element is also related to the outer orbital electronic configuration of its atom. Atom(s) having which of the following outermost electronic configurations will exhibit more than one oxidation state in its compounds.

(i) 3s¹
(ii) 3d¹4s²
(iii) 3d²4s²
(iv) 3s²3p³

15. Identify the correct statements with reference to the given reaction
P₄ + 3OH^– + 3H₂O → PH₃ + 3H₂PO₂^–

(i) Phosphorus is undergoing reduction only.
(ii) Phosphorus is undergoing oxidation only.
(iii) Phosphorus is undergoing oxidation as well as reduction.
(iv) Hydrogen is undergoing neither oxidation nor reduction.

16. Which of the following electrodes will act as anodes, when connected to Standard Hydrogen Electrode?

(i)	Al/Al³⁺	E^Θ = –1.66
(ii)	Fe/Fe²⁺	E^Θ = – 0.44
(iii)	Cu/Cu²⁺	E^Θ = + 0.34
(iv)	F₂(g)/2F^–(aq)	E^Θ = + 2.87

III. Short Answer Type

17. The reaction
Cl₂(g) + 2OH^–(aq) → ClO^–(aq) + Cl^–(aq) + H₂O (l)
represents the process of bleaching. Identify and name the species that bleaches the substances due to its oxidising action.

18. MnO₄^2– undergoes disproportionation reaction in acidic medium but MnO₄^– does not. Give reason.

19. PbO and PbO₂ react with HCl according to following chemical equations :
2PbO + 4HCl → 2PbCl₂ + 2H₂O
PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O

Why do these compounds differ in their reactivity?

20. Nitric acid is an oxidising agent and reacts with PbO but it does not react with PbO₂. Explain why?
21. Write balanced chemical equation for the following reactions:

(i) Permanganate ion (MnO₄^–) reacts with sulphur dioxide gas in acidic medium to produce Mn²⁺ and hydrogensulphate ion.(Balance by ion electron method)
(ii) Reaction of liquid hydrazine (N₂H₄) with chlorate ion (ClO₃^–) in basic medium produces nitric oxide gas and chloride ion in gaseous state.(Balance by oxidation number method)
(iii) Dichlorine heptaoxide (Cl₂O₇) in gaseous state combines with an aqueous solution of hydrogen peroxide in acidic medium to give chlorite ion (ClO₂^–) and oxygen gas.(Balance by ion electron method)

22. Calculate the oxidation number of phosphorus in the following species.
(a) HPO₃^2– and (b) PO₄^3–

23. Calculate the oxidation number of each sulphur atom in the following compounds:

(a) Na₂S₂O₃
(b) Na₂S₄O₆
(c) Na₂SO₃
(d) Na₂SO₄

24. Balance the following equations by the oxidation number method.

(i) Fe²⁺ + H⁺ + Cr₂O₇^2– → Cr³⁺ + Fe³⁺ + H₂O
(ii) I₂ + NO₃^– → NO₂ + IO₃^–
(iii) I₂ + S₂O₃^2– → I^– + S₄O₆^2–
(iv) MnO₂ + C₂O₄^2– → Mn²⁺ + CO₂

25. Identify the redox reactions out of the following reactions and identify the oxidising and reducing agents in them.

(i) 3HCl(aq) + HNO₃(aq) → Cl₂(g) + NOCl (g) + 2H₂O (l)
(ii) HgCl₂(aq) + 2KI (aq) → HgI₂(s) + 2KCl (aq)

(iv) PCl₃ (l) + 3H₂O (l) → 3HCl (aq) + H₃PO₃ (aq)
(v) 4NH₃ + 3O₂ (g) → 2N₂ (g) + 6H₂O (g)

26. Balance the following ionic equations

(i) Cr₂O₇^2– + H⁺ + I^– → Cr³⁺ + I₂ + H₂O
(ii) Cr₂O₇^2– + Fe²⁺ + H⁺ → Cr³⁺ + Fe³⁺ + H₂O
(iii) MnO₄^– + SO₃^2– + H⁺ → Mn²⁺ + SO₄^2– + H₂O (iv) MnO₄^– + H⁺ + Br^– → Mn²⁺ + Br₂ + H₂O

IV. Matching Type

27. Match Column I with Column II for the oxidation states of the central atoms.

Column I		Column II
(i)	Cr₂O₇^2-	(a)	+ 3
(ii)	MnO₄^–	(b)	+ 4
(iii)	VO₃^–	(c)	+ 5
(iv)	FeF₆^3–	(d)	+6
		(e)	+ 7

28. Match the items in Column I with relevant items in Column II.

Column I		Column II
(i)	Ions having positive charge	(a)	+7
(ii)	The sum of oxidation number of all atoms in a neutral molecule	(b)	–1
(iii)	Oxidation number of hydrogen ion (H⁺)	(c)	+1
(iv)	Oxidation number of fluorine in NaF	(d)	0
(v)	Ions having negative charge	(e)	Cation
		(f)	Anion

V. Assertion and Reason Type

In the following questions a statement of assertion (A) followed by a statement of reason (R) is given. Choose the correct option out of the choices given below each question.

29. Assertion (A) : Among halogens fluorine is the best oxidant.
Reason (R) : Fluorine is the most electronegative atom.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.

30. Assertion (A): In the reaction between potassium permanganate and potassium iodide, permanganate ions act as oxidising agent.
Reason (R) : Oxidation state of manganese changes from +2 to +7 during the reaction.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.

31. Assertion (A) : The decomposition of hydrogen peroxide to form water and oxygen is an example of disproportionation reaction.
Reason (R) : The oxygen of peroxide is in –1 oxidation state and it is converted to zero oxidation state in O₂ and –2 oxidation state in H₂O.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.

32. Assertion (A) : Redox couple is the combination of oxidised and reduced form of a substance involved in an oxidation or reduction half cell.
Reason (R) : In the representation E^Θ_{Fe³⁺/ Fe²⁺} and E^Θ_{Cu²⁺ / Cu} , Fe³⁺/ Fe²⁺ and Cu²⁺ / Cu are redox couples.

(i) Both A and R are true and R is the correct explanation of A.
(ii) Both A and R are true but R is not the correct explanation of A.
(iii) A is true but R is false.
(iv) Both A and R are false.

VI. Long Answer Type

33. Explain redox reactions on the basis of electron transfer. Give suitable examples.

34. On the basis of standard electrode potential values, suggest which of the following reactions would take place? (Consult the book for E^Θ value).

(i) Cu + Zn²⁺ → Cu²⁺ + Zn
(ii) Mg + Fe²⁺ → Mg²⁺ + Fe
(iii) Br2 + 2Cl^– → Cl₂ + 2Br^–
(iv) Fe + Cd²⁺ → Cd + Fe²⁺

35. Why does fluorine not show disporportionation reaction?
36. Write redox couples involved in the reactions (i) to (iv) given in question 34.
37. Find out the oxidation number of chlorine in the following compounds and arrange them in increasing order of oxidation number of chlorine.

NaClO₄, NaClO₃, NaClO, KClO₂, Cl₂O₇, ClO₃, Cl₂O, NaCl, Cl₂, ClO₂.
Which oxidation state is not present in any of the above compounds?

38. Which method can be used to find out strength of reductant/oxidant in a solution? Explain with an example.

ANSWERS

I. Multiple Choice Questions (Type-I)

1. (iv) 2. (iv) 3. (iv) 4. (iv) 5. (i) 6. (i) 7. (ii) 8. (i) 9. (iv) 10. (iv) 11. (iii)

II. Multiple Choice Questions (Type-II)

12. (i), (iv) 13. (iii), (iv) 14. (iii), (iv) 15. (iii), (iv) 16. (i), (ii)

III. Short Answer Type

17. Hypochloriteion

18. In MnO₄^– , Mn is in the highest oxidation state i.e. +7. Therefore, it does not undergo disproportionation. MnO₄^2– undergoes disproportionation as follows : 3MnO₄^2– + 4H⁺ → 2MnO₄^– + MnO₂ + 2H₂O

19. 2PbO + 4HCl → 2PbCl₂ + 2H₂O (Acid base reaction) PbO₂ + 4HCl → PbCl₂ + Cl₂ + 2H₂O (Redox reaction) (Hint : Note the oxidation number of lead in the oxides)

20. PbO is a basic oxide and simple acid base reaction takes place between PbO and HNO₃. On the other hand in PbO₂ lead is in + 4 oxidation state and cannot be oxidised further. Therefore no reaction takes place. Thus, PbO₂ is passive, only PbO reacts with HNO₃. 2PbO + 4HNO₃ → 2Pb (NO₃)₂ + 2H₂O (Acid base reaction)

22. (a) +3, (b) +5 23. (a) +2 (b) +5, 0, 0, +5 (c) +4
(d) +6
Justification : Write Lewis structure of each ion then assign electron pair shared between atoms of different electronegativity to more electronegative atom and distribute the electron pair shared between atoms of same element equally. Now count the number of electrons possessed by each atom. Find out the difference in number of electrons possessed by neutral atom and that possessed by atom in the compound. This difference is the oxidation number. If atom present in the compound possesses more electrons than the neutral atom, the oxidation number is negative. If it possesses less electrons then oxidation number is positive.

(i) Lewis structure of S₂O₄^2– can be written as follows :

Electron pair shared between sulphur and oxygen is assigned to oxygen atoms because of more electronegativity of oxygen. Thus each sulphur atom is deficient of two electrons with respect to neutral sulphur atom hence, each sulphur atom is in +2 oxidation state. Each oxygen atom gets two excess electrons hence, it is in –2 oxidation state. Lewis structure of S₄O₆^2– can be written as follows :

To find out oxidation state of each atom we distribute electrons of electron pair shared between two sulphur atoms equally (i.e. one electron is assigned to each sulphur atom). Both the electrons of electron pair shared between sulphur and oxygen atom are assigned to oxygen as oxygen is more electronegative. Thus we find that each of the central sulphur atoms obtain six electrons. This number is same as that in the outer shell of neutral sulphur atom hence oxidation state of central sulphur atoms is zero. Each of the sulphur atoms attached to oxygen atoms obtain only one electron as its share. This number is less by five electrons in comparison to the neutral sulphur atom. So, outer sulphur atoms are in +5 oxidation state. Therefore average oxidation state of sulphur atoms is :

(5 + 0 + 0 + 5)/4 = 10/4 = 2.5

By using the formula we obtain average oxidation state of the particular type of atoms. Real oxidation state can be obtained only by writing the complete structural formula. Similarly we can see that each oxygen atom is in – 2 oxidation state.

In the same way one can find out the oxidation state of each atom in SO₃^2– and SO₄^2– ions. Oxidation state of metal atoms will be +1 as these will lose one electron in each case.

IV. Matching Type

27. (i) → (d) (ii) → (e) (iii) → (c) (iv) → (a)
28. (i)→ (e) (ii) → (d) (iii) → (c) (iv) → (b) (v) → (f)

V. Assertion and Reason Type

29. (ii) 30. (iii) 31. (i) 32. (ii)

Some Useful Links

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TITRIMETRIC ANALYSIS(REDOX REACTIONS)

THE oxidation and reduction reactions in aqueous solutions involve the transfer of electrons from one species to another. In the oxidation of a substance electron(s) is (are) transfered from the species and in reduction, electron(s) is (are) gained by the species. Oxidation and reduction reactions occur simultaneously. A reaction, which involves simultaneous oxidation and reduction, is called a redox reaction. The titrations involving redox reaction are called redox titrations. You know that in acid-base titrations, indicators which are sensitive to pH change are employed to note the end point. Similarly, in redox titrations there is a change in oxidation potential of the system. The indicators used in redox reactions are sensitive to change in oxidation potential. The ideal oxidation-reduction indicators have an oxidation potential intermediate between the values for the solution being titrated and the titrant and these show sharp readily detectable colour change.

EXPERIMENT 6.1
Aim

To determine the concentration/molarity of KMnO4 solution by titrating it against a 0.1 M standard solution of oxalic acid.

Theory

In the present experiment, potassium permanganate acts as a powerful oxidising agent. Although KMnO4 acts as an oxidising agent in alkaline medium also, for quantitative analysis mostly acidic medium is used. The oxidising action of KMnO₄ in the acidic medium can be represented by the following equation:

MnO₄^– + 8H⁺ +5e^– → Mn²⁺ + 4H₂O

The acid used in this titration is dilute sulphuric acid. Nitric acid is not used as it is itself an oxidising agent and hydrochloric acid is usually avoided because it reacts with KMnO4 according to the equation given below to produce chlorine and chlorine which is also an oxidising agent in the aqueous solution.

2KMnO₄ + 16 HCl → 2KCl + 2 MnCl₂ + 5Cl₂ + 8 H₂O

Since, oxalic acid acts as a reducing agent, it can be titrated against potassium permanganate in the acidic medium according to the following equation:

Reactions of oxalic acid

A. Chemical equations

?????????????

B. Ionic equation

Reduction half reaction : MnO₄^– + 5e^– + 8H+ → Mn²⁺ + 4H₂O] x 2
Oxidation half reaction : C₂O₄^– → 2CO₂ + 2e^– ] x 5
_____________________________________________________________________
2 MnO₄^– + 5C₂O₄^2– + 16H⁺ → 2Mn²⁺ + 10CO₂ + 8H₂O
_____________________________________________________________________

In these equations, MnO₄^– is reduced to Mn²⁺ and C₂O₄^2– is oxidised to CO₂. The oxidation number of carbon in C₂O₄^2– changes from +3 to +4.

In these titrations, potassium permanganate acts as a selfindicator. Initially colour of potassium permanganate is discharged due to its reduction by oxalic acid. After complete consumption of oxalate ions, the end point is indicated by the appearance of a light pink colour produced by the addition of a little excess of
unreacted potassium permanganate. Further, during the titration of oxalic acid against potassium permanganate, warming of oxalic acid solution (50°–60°C) along with dilute H₂SO₄ is required. This is essential because the reaction takes place at higher temperature. During the titration, first manganous sulphate is formed which acts as a catalyst for the reduction of KMnO₄ by oxalic acid. Therefore, in the beginning the reaction rate is slow and as the reaction proceeds, the rate of the reaction increases.

Material Required

Procedure

A. Preparation of 0.1 M standard solution of oxalic acid Prepare 0.1M oxalic acid solution as mentioned in experiment 2.1(Unit 2, Class XI, Laboratory Manual)

B. Titration of oxalic acid solution against potassium permanganate solution

(i) Rinse and fill a clean burette with potassium permanganate solution. Remove the air bubble, if any, from the nozzle of the burette by releasing some solution through it. The burette used in the permanganate titration must have a glass stop
cock as rubber is attacked by permanganate ions.

(ii) Take 10 mL of 0.1 M oxalic acid solution in a conical flask and add half of the test tube full ( 5 mL) of 1.0 M H₂SO₄ to it to prevent the formation of any precipitate of manganese dioxide during the course of the titration.

(iii) Heat the oxalic acid solution upto 50°– 60°C before titrating it with potassium permanganate solution taken in the burette. To increase the visibility of the colour change, place the conical flask containing the solution to be titrated over a white glazed tile kept below the nozzle of the vertically fitted burette.

(iv) Note the initial reading of the volume of permanganate solution in the burette and add it in small volumes to the hot oxalic acid solution while swirling the contents of the flask gently. The violet colour of permanganate solution is discharged on reaction with oxalic acid. The end point is indicated by the appearance of permanent light pink colour due to a slight excess of permanganate solution.

(v) Repeat the titration till three concordant readings are obtained. Since the solution of KMnO₄ is of dark colour, the upper meniscus should be considered for noting the burette readings.

(vi) Record the readings as shown in observation Table 6.1 and calculate the strength of potassium permanganate solution in mols/litre.

Table 6.1 : Titration of potassium permanganate solution against standard oxalic acid solution

????????????

Calculations

(i) The strength of the unknown solution in terms of molarity may be determined by the following equation.
a₁M₁V₁ = a₂M₂V₂ (6.1)

For oxalic acid vs potassium permanganate titration:

a₁ = 2, (the number of electrons lost per formula unit of oxalic acid in a balanced equation of half cell reaction)
a₂ = 5, (the number of electrons gained per formula unit of potassium permanganate in the balanced equation of half cell reaction)

M₁ and M₂ are the molarities of oxalic acid and potassium permanganate solutions used in the titration.

V₁ and V₂ are the volumes of oxalic acid and potassium permanganate solutions.

On putting the value of a1 and a2 in equation 6.1 we get
Oxalic acid KMnO₄
2M₁V₁ = 5M₂V₂
M₂ = 2M₁V₁/5V₂ ……..(6.2)

We can calculate the molarity of potassium permanganate solution by using equation 6.2. Strength of the solution is given by the following equation:

Strength = Molarity x Molar mass

Result
(i) Molarity of KMnO₄ solution is _______.
(ii) Strength of KMnO₄ solution is _______.

Precautions

(a) Always rinse the burette and the pipette with the solutions to be taken in them.
(b) Never rinse the conical flask with the experimental solutions.
(c) Remove the air gaps if any, from the burette.
(d) Never forget to remove the funnel from the burette before noting the initial reading of the burette.
(e) No drop of the liquid should hang at the tip of the burette at the end point and while noting reading.
(f) Always read the upper meniscus for recording the burette reading in the case of all coloured solutions.
(g) Never use pipette and burette with a broken nozzle.
(h) Lower end of the pipette should always remain dipped in the liquid while sucking the liquid.
(i) Do not blow out the last drop of the solution from the jet end of the pipette.
(j) The strength of the solution must be calculated up to the fourth decimal place.
(k) Do not forget to heat the mixture of oxalic acid and H₂SO₄ solutions between 50°–60° C while titrating it against potassium permanganate.

Discussion Questions

(i) What specific name is given to the permanganate titrations?
(ii) Which indicator is used in the permanganate titration?
(iii) Why is a burette with pinch-cock regulator not used for the permanganate titration ?
(iv) Why do we heat oxalic acid solution containing sulphuric acid up to 50–60°C in the permanganate titration?

EXPERIMENT 6.2
Aim
To determine the concentration/molarity of KMnO₄ solution by titrating it against standard solution of ferrous ammonium sulphate.

Theory
Like oxalic acid, ferrous ammonium sulphate also acts as a reducing agent in the titration against potassium permanganate. The reaction which takes place is given below :

(a) Chemical equation

Reduction half reaction : 2KMnO₄ + 3H₂SO₄ → K₂SO₄ + 2MnSO₄ + 3H₂O + 5[O]
Oxidation half reaction : 2FeSO₄(NH₄)2SO₄ . 6H₂O + H₂SO₄ + [O] → Fe₂(SO₄)₃ + 2(NH₄)₂SO₄ + 13H₂O] x 5
_______________________________________________________________________________________
2KMnO₄ + 8H₂SO₄ + 10FeSO₄(NH₄)₂SO₄ . 6H₂O → K₂SO₄ + 2MnSO₄ + 5Fe₂(SO₄)₃ + 10(NH₄)2SO₄ + 68H₂O
_______________________________________________________________________________________

(b) Ionic equation
Reduction half reaction : MnO₄– + 5e^– + 8H⁺ → Mn²⁺ + 4H₂O
Oxidation half reaction : Fe²⁺ → Fe³⁺ + e^–] x 5
_______________________________________________________________________________________
MnO₄^– + 5Fe²⁺ + 8H⁺ → Mn²⁺ + 5Fe³⁺ + 4H₂O
_______________________________________________________________________________________

The oxidation number of iron in Mohr’s salt is +2. Iron is oxidised during the reaction and its oxidation number changes from +2 to +3. In this titration heating of ferrous ammonium sulphate solution is not required because reaction rate is very high even at room temperature. Also, at high temperatures, ferrous ions may be oxidised to ferric ions by oxygen of air and error may be introduced in the experiment.

Material Required

Procedure

A. Preparation of 0.05 M, standard solution of ferrous ammonium sulphate

(Molar mass of FeSO₄(NH₄)2SO₄ . 6H₂O = 392 g mol^–1).
(i) Weigh 4.9000 g of ferrous ammonium sulphate and transfer it into a 250 mL measuring flask through a funnel.
(ii) Transfer the solid sticking to the funnel with the help of distilled water into the flask and add dilute H₂SO₄ into the flask drop wise to get the clear solution.
(iii) Shake the flask till the substance dissolves and make the solution upto the mark.

B. Titration of ferrous ammonium sulphate against potassium permanganate solution

(i) Rinse and fill the clean burette with potassium permanganate solution. Remove air bubbles if any, from the burette tip by releasing some solution through it.
(ii) Take 10 mL of 0.05 M ferrous ammonium sulphate solution in a conical flask and add half test tube (≈ 5 mL) full of (1.0 M) H₂SO₄ to it.
(iii) Titrate the above solution with potassium permanganate solution till the colour of the solution changes to permanent pink. Swirl the content of the flask during the titration.
(iv) Repeat the titration, until three concordant readings are obtained.
(v) Record the readings as shown in observation Table 6.2 and calculate the strength of potassium permanganate solution in mols/litre.

Table 6.2 : Titration of potassium permanganate solution against standard ferrous ammonium sulphate solution

????????????

Calculations

The strength of unknown solution in terms of molarity may be determined by the following equation :
a₁M₁V₁ = a₂M₂V₂

M₁ and M₂ are the molarities of ferrous ammonium sulphate and potassium permanganate solutions and V₁ and V₂ are volumes of ferrous ammonium sulphate and potassium permanganate solutions, respectively.

a₁ = 1, (the number of electrons lost per formula unit of ferrous ammonium sulphate in the half cell reaction)
a₂ = 5, (the number of electrons gained per formula unit of potassium permanganate in a half cell reaction)
Strength can be calculated by the formula given below :
Strength = Molarity x Molar mass

Result
The strength of the given potassium permanganate solution is _____ g/L.

Precautions
(a) Always use a fresh sample of ferrous ammonium sulphate to prepare its standard solution.
(b) Other precautions are same as that in Experiment 6.1.

Discussion Questions

(i) Why is ferrous ammonium sulphate solution not heated before titration?
(ii) Why is nitric acid or hydrochloric acid not used in permanganate titration? Explain.
(iii) Why is dilute sulphuric acid added while preparing a standard solution of ferrous ammonium sulphate ?
(iv) How will you prepare 100 mL of 0.1 M standard solution of ferrous ammonium sulphate?
(v) Why is KMnO₄ not regarded as a primary standard?
(vi) What type of titrations are given the name redox titrations? Name some other redox titrations?

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Simplified Knowledge Management Classes

Must see https://zookeepersblog.wordpress.com/some-points-which-i-wish-all-my-new-prospective-students-know/
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Do you want to make money working at home ?

see http://skmclasses.weebly.com/jobs.html
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you will get most videos. I say most because I do not upload all videos that I make. I have many more videos which are not in the net.

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The following Videos are available for you ( As of Now ). These explain tricky Physics and Mathematics Numericals.

Eventually I will try to give Videos for full course here for you.

These covers PU ( Pre University courses, school / college ) courses, IIT JEE, AIEEE ( All India Engineering Entrance Examination ) , CET ( Combined Engineering Test ), AIPMT ( All India Pre Medical Test ), ISc ( Intermediate Science / Indian School Certificate Exam ), CBSE ( Central Board Secondary Exam ), Roorkey Joint Entrance Test Questions ( Discontinued since 2002 ), APhO ( Asian Physics Olympiad ), IPhO ( International Physics Olympiad ), IMO ( International Mathematics Olympiad ) , NSEP ( National Standard Exam in Physics ), RMO ( Regional Math Olympiad , India ), INMO ( Indian National Maths Olympiad ), Irodov Solutions, Prof. H C Verma ( Concepts of Physics ) Solutions etc.

( You can see the history of Indian Participation in various Olympiads at ->
https://zookeepersblog.wordpress.com/indian-participation-in-ipho-icho-ibo-and-astronomy-olympiad/ )

[ In each of these videos there is at-least 1 or more errors. Please tell me about those ]

search for videos in http://www.skmclasses.kinja.com
You should get to see all the Uploaded videos. Though we have many more study videos.

Thanks and Regards
Zookeeper ;-D Subhashish Chattopadhyay

[ I suggest you see the videos starting with 1- first then starting with 2- ….. in that sequence. ]

[ Tell your friends about this link if you liked the videos ]

In case of doubts or suggestions, Please send me email at mokshya@gmail.com

search for videos in http://www.skmclasses.kinja.com

Answers to -> Frequently Asked Questions ( FAQ ) [ commonly asked intelligent Questions 🙂 ]

1 ) How do I prepare for IIT ?

Ans : – See the videos made by me ( search for videos in http://www.skmclasses.kinja.com
Though we have many more which have not been uploaded ). While watching the videos, take notes and try to solve the problems yourself by pausing the video. Tell me if any calculation is wrong. See the videos with 1- first then 2- and so on. Write to IAPT Kothrud, Pune office to buy ( 150 Rs approx ) the book with previous papers of NSEP ( National Standard Exam in Physics – The 1st level ), INPhO ( Indian National Physics Olympiad – 2nd level ). Prepare with these and see how much you are scoring. You can guess your ALL INDIA rank easily from NSEP, and INPhO rank. Since 1998 the IIT JEE toppers have been mostly representing India in IPhO.

2 ) Which codec and Player do I use to see the videos ?

Ans : – You can use GOM Player, or VLC Player. You have to have good speakers with filters or good earphones with filters. We have checked mostly it is OK with these. ( If you are depending only on your embedded speakers of computer /screen / keyboard then there may be extra distortions. As these speakers are often not of good Quality. Also install latest KL Codecs ) In any case reduce the volume see the board, imagine sitting in the last bench and solving the problems of your own. See if your solution differs anywhere with the scribbles on the board.

3 ) Why are you giving these ( high Quality ) lecture for free ?

Ans : Well there are lot of good things free in this world. Linux, My-SQL, Open-Office ….. Go to sourceforge and get thousands of high quality software free along with source code. Yes all officially free …. Why do you think Richard Stallman, Zimmerman, ….. etc are considered Guru philosophers ? In Punjab and Gurudwaras worldwide there are so many Langars where you get better food than Restaurants. ….. why ? Why do you have Dharmasalas and subsidized rest rooms near hospitals / Famous Temples / various places ? in Iftar party anyone can eat for free …. why ?

I am teaching since 1989 I have observed most students can do much better if they have the self motivation to solve and practice. Cheap books are available in second hand bookstalls, where you get thousands of Numericals to solve ….. but most students will like to blow their time going and coming for tuition, travel time …. TV for hours and hours watching cricket / Tennis games, playing computer games …. My free lectures are not going to make much difference in spending of unnecessary money for coaching ….. I know very well , how much people enjoy …. ! spending unnecessarily !!

–

Do you know that there are NO poor / needy students in Bangalore.
–

Sometime back I had tried to teach for IIT JEE FREE. Discussed with a few NGOs and social service guys. Arranged rooms but got only 1 student. We had informed many people in many ways to inform students …. We did not get students who are ready to learn for free. So I am sure these lectures are NOT FREE. If anyone learns from these, s/he changes and that’s the gain / benefit. This change ( due to learning ) is very costly …. Most do not want to learn ………..

search for videos in http://www.skmclasses.kinja.com
You will get most videos. I say most because I do not upload all videos that I make. I have many more videos which are not in the net.

🙂

4 ) How can I get all your lectures ?

Ans : – Apart from my lectures there are approx 700 GB of PCM ( Phy, Chem, Math ) lectures. It takes approx 3 years of continuous download from scattered sources. I have ( 20,000 )Thousands of these. You can take ALL of them from me in an external 1 TB hard disk, instead of spending so much money and time again for downloading. These cover ( by Various Professors ) everything of Chemistry, Physics, Maths… Lot of this is from outside India … as foreigners have much wider heart than Indians ( as most of GNU / open source software have been developed by Non-Indians ). I observed the gaps in these videos, and thus I am solving IIT, APhO, Roorkey, IPhO Numericals. Videos made by me along with these videos gives a complete preparation.

Send me a mail at mokshya@gmail.com to contact me.

search for videos in http://www.skmclasses.kinja.com
You will get most videos. I say most because I do not upload all videos that I make. I have many more videos which are not in the net.

🙂

5 ) How do you get benefited out of this ?

Ans :- If anyone learns we all will have better people in this world. I will have better “ YOU “.
🙂

6 ) Why do you call yourself a Zookeeper ?

Ans :- This is very nicely explained at https://zookeepersblog.wordpress.com/z00keeper-why-do-i-call-myself-a-zoookeeper/

🙂

7 ) Where do you stay ?

Ans :- Presently I am in Bangalore.

🙂

8 ) If I need videos in a few topics can you make them for me ?

Ans :- We actively answers doubts at doubtpoint.
see http://skmclasses.weebly.com/doubtpoint.html
In case you appreciate our time and efforts involved in answering complicated Questions, then get Quality answers at doubtpoint.

🙂

9 ) Why did you write an article saying there are No Poor students ?

Ans :- There are lots of NGOs and others working for rural / poor children education at lower classes. While very less effort is on for std 9 till 12. Also see the answer in question number ( 3 ) above. In more than 2 decades of teaching I never met a Poor child who was seriously interested in ( higher ) studies. As I have a mind / thinking of a ” Physicist “, I go by ” Experimental Observation “.

It is not about what is being said about poor in media / TV etc, or ” what it should be ” ( ? ) …. It is about what I see happening. Also to add ( confuse ? you more )…. You must be knowing that in several states over many years now girl students have better ( by marks as well as by pass percentage ) result in std 10 / Board Exams….. well but NEVER a girl student came FIRST in IIT JEE … why ? [ The best rank by a Girl student is mostly in 2 digits, very rarely in single digit ] ????? So ????

🙂

10 ) How much do I have to study to make it to IIT ?

Ans :- My experience of Teaching for IIT JEE since 1989, tells me, Total 200 hours per subject ( PCM ) is sufficient. If you see my Maths and Physics videos, each subject is more than 200 hours. So if someone sees all the videos diligently, takes notes and remembers, …… Done.

🙂

11 ) What is EAMCET ?

Ans :- Engineering Agriculture and Medicine Common Entrance Test is conducted by JNT University Hyderabad on behalf of APSCHE. This examination is the gateway for entry into various professional courses offered in Government/Private Colleges in Andhra Pradesh.

12 ) In your videos are you covering other Exams apart from IIT ?

Ans : – Yes. See many videos made by solving problems of MPPET, Rajasthan / J&K CET, UPSEAT ( UPES Engineering Aptitude Test ), MHCET, BCECE ( Bihar Combined Entrance Competitive Examination Board ), WB JEE etc

🙂

13 ) What is SCRA ?

Ans : – Special Class Railway Apprentice (SCRA) exam is conducted by Union Public Service Commission (UPSC) board, for about 10 seats.That translates into an astonishing ratio of 1 selection per 10,000 applicants. The SCRA scheme was started in 1927 by the British, to select a handful of most intelligent Indians to assist them in their Railway Operations, after training at their Railway’s largest workshop, i.e. Jamalpur Workshop, and for one year in United Kingdom. The selected candidates were required to appear in the Mechanical Engineering Degree Examination held by Engineering Council (London).

Thanks for your time. To become my friend in google+ ( search me as mokshya@gmail.com and send friend request )

Read http://edge.org/responses/what-scientific-concept-would-improve-everybodys-cognitive-toolkit
🙂
The following video is a must see for full CO2 cycle, plates of Earth, Geological activities, stability of weather

🙂
Article in Nature says CO2 increase is good for the trees
http://thegwpf.org/science-news/6086-co2-is-greening-the-planet-savannahs-soon-to-be-covered-by-forests.html
🙂
http://climaterealists.com/index.php?id=9752

BBC documentary Crescent and Cross shows the 1000 years of fight between Christians and Muslims. Millions have been killed in the name of Religion. To decided whose GOD is better, and which GOD to follow. The fight continues.
–

Summary of Women

🙂
The Virus of Faith

🙂
The God delusion

🙂
cassiopeia facts about evolution

–

Intermediate Fossil records shown and explained nicely Fossils, Genes, and Embryos http://www.youtube.com/watch?v=fdpMrE7BdHQ

The Rise Of Narcissism In Women

🙂
13 type of women whom you should never court
http://timesofindia.indiatimes.com/life-style/relationships/man-woman/13-Women-you-should-never-court/articleshow/14637014.cms

🙂
Media teaching Misandry in India

http://www.youtube.com/watch?v=-M2txSbOPIo

Summary of problems with women
http://problemwithwomentoday.blogspot.in/2009/12/problem-with-women-today-what-in-hell.html

🙂
Eyeopener men ? women only exists
http://www.youtube.com/watch?v=6ZAuqkqxk9A

🙂
Each of you is an Activist in some way or other. You are trying to propagate those thoughts, ideas that you feel concerned / excited about.
–

Did you analyze your effectiveness ?

Culturomics can help you

😀
see how biased women are. Experimental proof. Women are happy when they see another woman is beating a man ( see how women misbehave with men )

🙂
http://www.youtube.com/watch?v=LlFAd4YdQks

–

see detailed statistics at

An eye opener in Misandry

My sincere advice would be to be EXTREMELY careful ( and preferably away ) of girls. As girls age; statistically certain behavior in them has been observed. Most Male can NOT manage those behaviors… Domestic violence, divorce etc are rising very fast. Almost in all cases boys / males are HUGE loosers. Be extremely choosy ( and think from several angles ) before even talking to a girl.
🙂
https://zookeepersblog.wordpress.com/save-the-male/

🙂

How women manipulate men
http://www.angryharry.com/esWomenManipulateMen.htm

Gender Biased Laws in India
https://zookeepersblog.wordpress.com/biased-laws/

🙂

Violence against Men

🙂

Only men are victimised

Men are BETTER than women
http://www.menarebetterthanwomen.com/
🙂

see

🙂

Male Psychology

Women are more violent than men
http://www.independent.co.uk/news/uk/home-news/women-are-more-violent-says-study-622388.html

🙂

In the year 2010, 168 men ended their lives everyday ( on average ). More husbands committed suicide than wives.
🙂

http://www.rediff.com/news/report/ncrb-stats-show-more-married-men-committing-suicide/20111028.htm

It is EXTREMELY unfortunate that media projects men as fools, women as superiors, Husbands as servants, and replaceable morons. In ad after ad worldwide from so many companies, similar msg to disintegrate the world is being bombarded. It is highly unacceptable misandry

🙂

It is NOT at all funny that media shows violence against MEN. Some advertisers are trying to create a new ” Socially acceptable culture ” of slapping Men ( by modern city women ). We ( all men ) take objection to these advertisements.
We oppose this Misandry bad culture. Please share to increase awareness against Men bashing

🙂

Think what are you doing … why are you doing ?

Every Man must know this …

🙂
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distribution distribution energies molecules particular temperature IITJEE skmclasses.weebly.com graph bond enthalpy enthalpy change IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com breaking homolytic fission 1 mol bond molecules gaseous species carbanion organic ion IITJEE SKMClasses.weebly.com carbon atom hIITJEE skmclasses.weebly.com negative charge carbocation organic ion IITJEE SKMClasses.weebly.com carbon atom hIITJEE skmclasses.weebly.com positive charge catalyst substance IITJEE SKMClasses.weebly.com increases rate chemical reaction process cation positively charged ion cis trans isomerism special type isomerism IITJEE SKMClasses.weebly.com non hydrogen group skmclasses.weebly.com hydrogen atom C C=C double bond cis isomer ( Z isomer) IITJEE skmclasses.weebly.com H atoms on IITJEE SKMClasses.weebly.com carbon same side trans isomer E isomer H atoms carbon different bond compound substance formed IITJEE SKMClasses.weebly.com two IITJEE SKMClasses.weebly.com chemically bonded elements fixed ratio, usually chemical formula concentration amount solute mol IITJEE SKMClasses.weebly.com 1 dm 3 1000 cm solution coordinate bond shared pair electrons provided one bonding atoms called dative covalent bond covalent bond bond formed shared pair electrons cracking breaking long chained saturated hydrocarbonsIITJEE SKMClasses.weebly.com mixture shorter chained alkanesalkenes curly arrow symbol IITJEE SKMClasses.weebly.com reaction mechanisms IITJEE SKMClasses.weebly.com show movement electron Coaching ICWA Coaching CFA Coaching CFP Coaching CMAT Coaching School Tuitions CBSE School Tuitions Home Tuitions 9th STD Tuitions PUC Coaching 10th Std Tuitions College Tuitions Maths Tuitions Engineering Tuitions Accounts & Finance Tuitions MBA & BBA Coaching Microbiology & Biotech Tuitions Study Abroad GRE & SAT Coaching GMAT Coaching IELTS/TOEFL Coaching PTE Coaching proteins protonation pyridines pyrroles quinones quinolines radical reaction radicals rearrangement receptors reduction regioselectivity retro reaction rhodium ring closure ring contraction ring expansion ring opening ruthenium samarium scandium Schiff bases selenium self-assembly silicon sodium solid-phase synthesis solvent effects spectroscopy sphingolipids spiro compounds stereoselective synthesis stereoselectivity steric hindrance steroids Stille reaction substituent effects sulfates sulfonamides sulfones sulfoxides sulfur supported catalysis supramolecular tandem reaction tautomerism terpenoids thioacetals thiols tin titanium total synthesis transesterification transition metals transition states tungsten Umpolung vinylidene complexes vitamins Wacker reaction Wittig reaction ylides zeolites zinc BRST Quantization Effective field theories Field Theories Higher Dimensions Field Theories Lower Dimensions Large Extra Dimensions Lattice Quantum Field Theory Nonperturbative Effects Renormalization Group Renormalization Regularization skmclasses.weebly.com Renormalons Sigma Models Solitons Monopoles skmclasses.weebly.com Instantons Supersymmetric gauge theory Topological Field Theories 1/N Expansion Anyons Chern-Simons Theories Confinement Duality Gauge Field Theories Lattice Gauge Field Theories Scattering Amplitudes Spontaneous Symmetry Breaking Strong Coupling Expansion Topological States Matter Wilson ‘t Hooft skmclasses.weebly.comPolyakov loops Anomalies Field skmclasses.weebly.comString Theories BRST Symmetry Conformal skmclasses.weebly.com W Symmetry Discrete skmclasses.weebly.comFinite Symmetries Gauge Symmetry Global Symmetries Higher Spin Symmetry Space-Time Symmetries AdS-CFT Correspondence Black Holes String Theory Bosonic Strings Brane Dynamics Gauge Theories Conformal Field Models String Theory D-branes dS vacua string theory F-Theory Flux compactifications Gauge-gravity correspondence Holography skmclasses.weebly.comcondensed matter physics (AdS CMT) Holography skmclasses.weebly.comquark-gluon plasmas Intersecting branes models Long strings M(atrix) Theories M-Theory p-branes Penrose limit skmclasses.weebly.compp-wave background String Duality String Field Theory String theory skmclasses.weebly.comcosmic string Superstring Vacua Superstrings skmclasses.weebly.comHeterotic Strings Tachyon Condensation Topological Strings 2D Gravity Black Holes Classical Theories Gravity Higher Spin Gravity Lattice Models Gravity Models Quantum Gravity Spacetime Singularities Extended Supersymmetry Supergravity Models Superspaces Supersymmetric Effective Theories Supersymmetry skmclasses.weebly.com Duality Supersymmetry Breaking Differential skmclasses.weebly.comAlgebraic Geometry Integrable Hierarchies Non-Commutative Geometry Quantum Groups Statistical Methods Stochastic Processes Cosmology Theories beyond SM Solar skmclasses.weebly.comAtmospheric Neutrinos Thermal Field Theory Be Ansatz Boundary Quantum Field Theory Exact S-Matrix Quantum Dissipative Systems Random Systems B-Physics Beyond Standard Model Compactification skmclasses.weebly.com String Models CP violation Electromagnetic Processes skmclasses.weebly.com Properties GUT Heavy Quark Higgs Kaon LEP HERA skmclasses.weebly.com SLC Neutrino Physics Quark Masses skmclasses.weebly.comSM Parameters Rare Decays Standard Model Supersymmetric Standard Model Technicolor skmclasses.weebly.com Composite Models Chiral Lagrangians Deep Inelastic Scattering Higher Twist Effects Lattice QCD Parton Model Phase Diagram QCD Phenomenological Models QCD Quark-Gluon Plasma Resummation Sum Rules Aim Global Education Koramangala Computer Networking Training Cloud Computing Training JBOSS Training Juniper Certification Training L2 & L3 Protocol Training MCTS Training Engineering design Training CAD & CAM Training MATLAB Training PLC Training SCADA Training VLSI Design Multimedia & Design Training 2D Animation Training 3D Animation Training 4D Animation Training CorelDRAW Training VFX Training Web Technologies Training ASP.Net Training JQuery pair breaking formation covalent bond dative covalent shared pair electrons IITJEE SKMClasses.weebly.com hIITJEE skmclasses.weebly.com been provided one bonding atoms only IITJEE SKMClasses.weebly.com called coordinate bond dehydration elimination reaction IITJEE SKMClasses.weebly.com water removed saturated molecule IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com unsaturated molecule delocalised Electrons IITJEE SKMClasses.weebly.com shared IITJEE SKMClasses.weebly.com two atoms displacement reaction reaction IITJEE SKMClasses.weebly.com reactive element displaces less reactive element IITJEE SKMClasses.weebly.com aqueous solution latter’s ions displayed formula showing relative positioning atoms molecule skmclasses.weebly.com bonds IITJEE SKMClasses.weebly.com disproportionation oxidation skmclasses.weebly.com reduction element redox reaction dynamic equilibrium equilibrium IITJEE SKMClasses.weebly.com exists closed system IITJEE SKMClasses.weebly.com rate forward reaction equal IITJEE SKMClasses.weebly.com rate reverse reaction E/Z isomerism type stereoisomerism IITJEE SKMClasses.weebly.com different groups attached IITJEE SKMClasses.weebly.com carbon C=C double bond arranged differently space restricted rotation C=C bond electron configuration arrangement electrons IITJEE SKMClasses.weebly.com atom electronegativity measure attraction bonded atom skmclasses.weebly.com pair electrons covalent bond electron shielding repulsion IITJEE SKMClasses.weebly.com electrons different inner shells Shielding reduces net attractive force IITJEE SKMClasses.weebly.com positive nucleus outer shell electrons electrophile atom group atoms IITJEE SKMClasses.weebly.com attracted IITJEE SKMClasses.weebly.com electron rich centre atom IITJEE SKMClasses.weebly.com accepts pair electrons covalent bond electrophilic addition type addition reaction IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com electrophile attracted electron rich centre atom accepts pair electrons IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com new covalent bond elimination reaction removal molecule IITJEE SKMClasses.weebly.com saturated molecule IITJEE SKMClasses.weebly.com unsaturated molecule empirical formula simplest whole number ratio atoms IITJEE SKMClasses.weebly.com element present compound endothermic reaction reaction IITJEE SKMClasses.weebly.com enthalpy products greater enthalpy reactants resulting heat being taken surroundings enthalpy heat content IITJEE SKMClasses.weebly.com stored chemical system standard enthalpy change combustion enthalpy change IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com one mole substance reacts completely IITJEE SKMClasses.weebly.com oxygen under standard conditions reactants skmclasses.weebly.com products being IITJEE SKMClasses.weebly.com standard states (standard) enthalpy change formation enthalpy change IITJEE SKMClasses.weebly.com one mole compound formed IITJEE SKMClasses.weebly.com constituent elements IITJEE SKMClasses.weebly.com standard states under standard conditions (standard) enthalpy change reaction enthalpy change IITJEE SKMClasses.weebly.com accompanies reaction molar quantities expressed chemical equation under standard conditions reactants skmclasses.weebly.com products being IITJEE SKMClasses.weebly.com standard states enthalpy cycle diagram showing alternative routes IITJEE SKMClasses.weebly.com reactants products IITJEE SKMClasses.weebly.com allows indirect determination IITJEE SKMClasses.weebly.com enthalpy change IITJEE SKMClasses.weebly.com other known enthalpy changes using Hess’ law enthalpy profile diagram skmclasses.weebly.com reaction IITJEE SKMClasses.weebly.com compare enthalpy reactants IITJEE SKMClasses.weebly.com enthalpy products esterification reaction IITJEE SKMClasses.weebly.com alcohol IITJEE SKMClasses.weebly.com carboxylic acid IITJEE SKMClasses.weebly.com produce ester skmclasses.weebly.com water exothermic reaction IITJEE SKMClasses.weebly.com enthalpy products smaller enthalpy reactants, resulting heat loss IITJEE SKMClasses.weebly.com surroundings fractional distillation separation components liquid mixture skmclassesfractions IITJEE SKMClasses.weebly.com differ boiling point skmclasses.weebly.com hence chemical composition IITJEE SKMClasses.weebly.com distillation typically using fractionating column fragmentation process mass spectrometry IITJEE SKMClasses.weebly.com causes positive ion split skmclasses pieces one positive fragment ion functional group part organic molecule responsible skmclasses.weebly.com chemical reactions general formula simplest algebraic formula member homologous series. skmclasses.weebly.com example general formula alkanes giant covalent lattice dimensional structure atoms, bonded together strong covalent bonds giant ionic lattice three dimensional structure oppositely charged ions, bonded together strong ionic bonds giant metallic lattice three dimensional structure positive ions skmclasses.weebly.com delocalised electrons, bonded together strong metallic bonds greenhouse effect process IITJEE SKMClasses.weebly.com absorption subsequent emission infrared radiation atmospheric gases warms lower atmosphere planet’s surface group vertical column Periodic Table Elements group IITJEE SKMClasses.weebly.com similar chemical properties skmclasses.weebly.com atoms skmclasses.weebly.comnumber outer shell electrons Hess law reaction IITJEE SKMClasses.weebly.com one route skmclasses.weebly.com initial final conditions IITJEE SKMClasses.weebly.com skmclasses.weebly.com total enthalpy change skmclasses.weebly.com skmclasses.weebly.com route heterogeneous catalysis reaction IITJEE SKMClasses.weebly.com catalyst IITJEE skmclasses.weebly.com different physical state reactants; frequently, reactants IITJEE SKMClasses.weebly.com gases whilst catalyst solid heterolytic fission breaking covalent bond IITJEE SKMClasses.weebly.com both bonded electrons going IITJEE SKMClasses.weebly.com one atoms, forming cation (+ ion) skmclasses.weebly.com IITJEE SKMClasses.weebly.com anion ion homogeneous catalysis reaction catalyst skmclasses.weebly.com reactants physical state, IITJEE SKMClasses.weebly.com frequently aqueous gaseous state homologous series series organic compounds IITJEE SKMClasses.weebly.com skmclasses.weebly.com functional group, IITJEE SKMClasses.weebly.com successive member differing homolytic fission breaking covalent bond IITJEE SKMClasses.weebly.com one bonded electrons going IITJEE SKMClasses.weebly.com atom, forming two radicals hydrated Crystalline skmclasses.weebly.com containing water molecules hydrocarbon compound hydrogen skmclasses.weebly.com carbon hydrogen bond strong dipole attraction IITJEE SKMClasses.weebly.com electron deficient hydrogen atom (O H on different molecule hydrolysis reaction IITJEE SKMClasses.weebly.com water aqueous hydroxide ions IITJEE SKMClasses.weebly.com breaks chemical compound skmclasses two compounds initiation first step radical substitution IITJEE SKMClasses.weebly.com free radicals generated ultraviolet radiation intermolecular force attractive force IITJEE SKMClasses.weebly.com neighbouring molecules Intermolecular forces van der Waals’ forces induced dipole ces permanent dipole forces hydrogen bonds ion positively negatively charge atom covalently bonded group atoms molecular ion ionic bonding electrostatic attraction IITJEE SKMClasses.weebly.com oppositely charged ions first) ionisation energy IITJEE SKMClasses.weebly.com remove one electron IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com ion one mole gaseous 1+ ions IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com one mole gaseous 2+ ions second) ionisation energy IITJEE SKMClasses.weebly.com remove one electron IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com ion one mole gaseous 1+ ions IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com one mole gaseous 2+ ions successive ionisation measure energy IITJEE SKMClasses.weebly.com remove IITJEE SKMClasses.weebly.com electron Chemistry energy second ionisation energy energy IITJEE SKMClasses.weebly.com one electron IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com ion one mole gaseous 1+ ions IITJEE SKMClasses.weebly.com one mole gaseous 2+ ions isotopes Atoms skmclasses.weebly.com element IITJEE SKMClasses.weebly.com different numbers neutrons different masses le Chatelier’s principle system dynamic equilibrium subjected change position equilibrium will shift minimise change limiting reagent substance chemical reaction IITJEE SKMClasses.weebly.com runs out first lone pair outer shell pair electrons IITJEE SKMClasses.weebly.com involved chemical bonding mass nucleon number particles protons aneutrons) nucleus mechanism sequence steps showing path taken electrons reaction metallic bond electrostatic attraction IITJEE SKMClasses.weebly.com positive metal ions adelocalised electrons molar mass substance units molar mass IITJEE SKMClasses.weebly.com molar volume IITJEE SKMClasses.weebly.com mole gas. units molar volume IITJEE SKMClasses.weebly.com dm room temperature skmclasses.weebly.com pressure molar volume approximately 24.0 substance containing IITJEE skmclasses.weebly.com many particles thereIITJEE SKMClasses.weebly.com carbon atoms exactly 12 g carbon isotope molecular formula number atoms IITJEE SKMClasses.weebly.com element molecule molecular ion M positive ion formed mass spectrometry IITJEE SKMClasses.weebly.com molecule loses electron molecule small group atoms held together covalent bonds monomer small molecule IITJEE SKMClasses.weebly.com combines IITJEE SKMClasses.weebly.com monomers polymer nomenclature system naming compounds nucleophile atom group atoms attracted electron deficient centre atom donates pair electrons covalent bond nucleophilic substitution type substitution reaction IITJEE SKMClasses.weebly.com nucleophile attracted electron deficient centre atom, IITJEE SKMClasses.weebly.com donates pair electrons IITJEE SKMClasses.weebly.com new covalent bond oxidation Loss electrons IITJEE SKMClasses.weebly.com increase oxidation number oxidation number measure number electrons IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com atom uses bond IITJEE SKMClasses.weebly.com atoms another element. Oxidation numbers IITJEE SKMClasses.weebly.com derive d rules oxidising agent reagent IITJEE SKMClasses.weebly.com oxidises (takes electrons from) another species percentage yield period horizontal row elements Periodic Table Elements show trends properties across period periodicity regular periodic variation properties elements IITJEE SKMClasses.weebly.com atomic number position Periodic Table permanent dipole small charge difference across bond resulting IITJEE SKMClasses.weebly.com difference electronegativities bonded atoms permanent dipole dipole force attractive force IITJEE SKMClasses.weebly.com permanent dipoles neighbouring polar molecules pi bond (p bond reactive part double bond formed above skmclasses.weebly.com below plane bonded atoms sideways overlap p orbitalspolar covalent bond bond IITJEE SKMClasses.weebly.com permanent dipole polar molecule molecule IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com overall dipole skmclasses account dipoles across bonds polymer long molecular chain built monomer units precipitation reaction formation solid solution during chemical reaction Precipitates IITJEE SKMClasses.weebly.com formed IITJEE SKMClasses.weebly.com two aqueous solutions IITJEE SKMClasses.weebly.com mixed together principal quantum number n number representing relative overall energy orbital IITJEE SKMClasses.weebly.com increases distance nucleus sets orbitals IITJEE SKMClasses.weebly.com value IITJEE skmclasses.weebly.com electron shells energy levels propagation two repeated radical substitution IITJEE SKMClasses.weebly.com build up products chain reaction radical species unpaired electron rate reaction change concentration reactant product redox reaction reaction IITJEE SKMClasses.weebly.com reduction skmclasses.weebly.com oxidation take IITJEE SKMClasses.weebly.com reducing agent reagent IITJEE SKMClasses.weebly.com reduces (adds electron to) species reduction Gain electrons decrease oxidation number yield actual amount mol product theoretical amount mol product Chemistry reflux continual boiling skmclasses.weebly.com condensing reaction mixture ensure IITJEE SKMClasses.weebly.com reaction IITJEE SKMClasses.weebly.com without contents flask boiling dry relative atomic mass weighted mean mass atom element compared one twelfth mass IITJEE SKMClasses.weebly.com atom carbon relative formula mass weighted mean mass formula unit compared IITJEE SKMClasses.weebly.com one twelfth mass atom carbon relative isotopic mass mass atom isotope compared IITJEE SKMClasses.weebly.com one twelfth mass atom carbon relative molecular mass weighted mean mass molecule compared twelfth mass atom carbon 12 repeat unit specific arrangement atom s IITJEE SKMClasses.weebly.com occurs structure over over again. Repeat units IITJEE SKMClasses.weebly.com included brackets outside IITJEE SKMClasses.weebly.com symbol n Salt chemical compound formed IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com acid IITJEE SKMClasses.weebly.com H+ ion acid IITJEE skmclasses.weebly.com been replaced metal ion another positive ion such IITJEE skmclasses.weebly.com ammonium ion, NH saturated hydrocarbon IITJEE SKMClasses.weebly.com single bonds only shell group atomic orbitals IITJEE SKMClasses.weebly.com skmclasses.weebly.com principal quantum number known main energy level simple molecular lattice three dimensional structure molecules, bonded together weak intermolecular forces skeletal formula simplified organic formula, IITJEE SKMClasses.weebly.com hydrogen atoms removed alkyl chains, leaving carbon skeleton skmclasses.weebly.com associated functional groups species particle IITJEE SKMClasses.weebly.com part chemical reaction specific heat capacity, c energy IITJEE SKMClasses.weebly.com raise temperature 1 g substance 1 C spectator ions Ions present part chemical reaction standard conditions pressure 100 kPa 1 atmosphere stated temperature usually 298 K (25 °C), skmclasses.weebly.com concentration 1 mol dm reactions aqueous solutions standard enthalpies enthalpystandard solution solution known concentration Standard solutions normally IITJEE SKMClasses.weebly.com titrations IITJEE SKMClasses.weebly.com determine unknown information another substance Chemistry standard state physical state substance under standard conditions 100 kPa 1 atmosphere) skmclasses.weebly.com 298 K 25 C stereoisomers Compounds skmclasses.weebly.com structural formula IITJEE SKMClasses.weebly.com different arrangement atoms space stoichiometry molar relationship IITJEE SKMClasses.weebly.com relative quantities substances part reaction stratosphere second layer Earth’s atmosphere, containing ‘ozone layer’, about 10 km IITJEE SKMClasses.weebly.com 50 km above Earth’s surface structural formula formula showing minimal detail skmclasses.weebly.com arrangement atoms molecule structural isomers Molecules IITJEE SKMClasses.weebly.com skmclasses.weebly.com molecular formula different structural arrangements atoms subshell group skmclasses.weebly.com type atomic orbitals s, p, d f within shell substitution reaction reaction IITJEE SKMClasses.weebly.com atom group atoms replaced different atom group atoms termination step end radical substitution IITJEE SKMClasses.weebly.com two radicals combine IITJEE SKMClasses.weebly.com molecule thermal decomposition breaking chemical substance IITJEE SKMClasses.weebly.com heat skmclasses least two chemical substances troposphere lowest layer Earth’s atmosphere extending Earth’s surface about 7 km (above poles) about 20 km above tropics unsaturated hydrocarbon hydrocarbon containing carbon carbon multiple bonds van der Waals’ forces Very weak attractive forces IITJEE SKMClasses.weebly.com induced dipoles neighbouring molecules volatility ease IITJEE SKMClasses.weebly.com liquid turns skmclasses gas Volatility increases boiling point decreases water crystallisation Water molecules IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com essential part crystalline structure absolute zero – theoretical condition concerning system at zero Kelvin IITJEE SKMClasses.weebly.com system does IITJEE SKMClasses.weebly.com emit absorb energy (all atoms rest accuracy – how close value IITJEE SKMClasses.weebly.com actual true value IITJEE SKMClasses.weebly.com see precision acid compound that, IITJEE SKMClasses.weebly.com dissolved water pH less 7.0 compound IITJEE SKMClasses.weebly.com donates hydrogen ion acid anhydride compound IITJEE SKMClasses.weebly.com two acyl groups boundIITJEE SKMClasses.weebly.com single oxygen atom acid dissociation constant – IITJEE SKMClasses.weebly.com equilibrium constant skmclasses.weebly.com dissociation weak acid actinides – fifteen chemical elements IITJEE SKMClasses.weebly.com actinium (89) skmclasses.weebly.com lawrencium (103 activated complex – structure IITJEE SKMClasses.weebly.com forms because collisionIITJEE SKMClasses.weebly.com molecules new bondsvIITJEE SKMClasses.weebly.com formed activation energy – minimum energy IITJEE SKMClasses.weebly.com must be inputIITJEE SKMClasses.weebly.com chemical system activity series actual yield addition reaction – within organic chemistry, IITJEE SKMClasses.weebly.com two IITJEE SKMClasses.weebly.com molecules combineIITJEE SKMClasses.weebly.com IITJEE SKMClasses.weebly.com larger aeration mixing air skmclasses liquid solid alkali metals metals Group 1 on periodic table alkaline earth metals – metals Group 2 on periodic table allomer substance IITJEE SKMClasses.weebly.com hIITJEE skmclasses.weebly.comdifferent composition another skmclasses.weebly.comcrystalline structure allotropy elements IITJEE SKMClasses.weebly.com different structures skmclasses.weebly.com therefore different forms IITJEE skmclasses.weebly.com Carbon diamonds, graphite, skmclasses.weebly.com fullerene anion negatively charge ions anode – positive side dry cell battery cell aromaticity – chemical property conjugated rings IITJEE SKMClasses.weebly.com results unusual stability. See IITJEE SKMClasses.weebly.com benzene atom – chemical element IITJEE SKMClasses.weebly.com smallest form, skmclasses.weebly.com made up neutrons skmclasses.weebly.comprotons within nucleus skmclasses.weebly.comelectrons circling nucleus atomic mass unit atomic number number representing IITJEE SKMClasses.weebly.com element IITJEE SKMClasses.weebly.com corresponds IITJEE SKMClasses.weebly.com number protons within nucleus atomic orbital region IITJEE SKMClasses.weebly.com electron atom may be found atomic radius average atomic mass Avogadro’s law Avogadro’s number number particles mole substance ( 6.02×10^23 ) barometer deviceIITJEE SKMClasses.weebly.comIITJEE SKMClasses.weebly.com measure pressure atmosphere base substance IITJEE SKMClasses.weebly.com accepts proton skmclasses.weebly.com high pH; common example sodium hydroxide (NaOH biochemistry chemistry organisms boiling phase transition liquid vaporizing boiling point temperature IITJEE SKMClasses.weebly.com substance startsIITJEE SKMClasses.weebly.com boil boiling-point elevation process IITJEE SKMClasses.weebly.com boiling point elevated adding substance bond – attraction skmclasses.weebly.com repulsion IITJEE SKMClasses.weebly.com atoms skmclasses.weebly.com molecules IITJEE SKMClasses.weebly.com cornerstone Boyle’s law Brønsted-Lowrey acid chemical species IITJEE SKMClasses.weebly.com donates proton Brønsted–Lowry acid–base reaction Brønsted-Lowrey base – chemical species IITJEE SKMClasses.weebly.com accepts proton buffered solution – IITJEE SKMClasses.weebly.com aqueous solution consisting weak acid skmclasses.weebly.comits conjugate base weak base skmclasses.weebly.comits conjugate acid IITJEE SKMClasses.weebly.com resists changes pH IITJEE SKMClasses.weebly.com strong acids basesIITJEE SKMClasses.weebly.com added burette (IITJEE SKMClasses.weebly.com buret glasswareIITJEE SKMClasses.weebly.com dispense specific amounts liquid IITJEE SKMClasses.weebly.com precision necessary titration skmclasses.weebly.com resource dependent reactions example combustion catalyst chemical compoundIITJEE SKMClasses.weebly.comIITJEE SKMClasses.weebly.com change rate IITJEE SKMClasses.weebly.com speed up slow down reaction,IITJEE SKMClasses.weebly.com regenerated at end reaction cation – positively charged ion centrifuge equipmentIITJEE SKMClasses.weebly.comIITJEE SKMClasses.weebly.com separate substances based on density rotating tubes around centred axis cell potential force galvanic cell IITJEE SKMClasses.weebly.com pulls electron through reducing agentIITJEE SKMClasses.weebly.com oxidizing agent chemical Law certain rules IITJEE SKMClasses.weebly.com pertain IITJEE SKMClasses.weebly.com laws nature skmclasses.weebly.comchemistry – examples chemical reaction – change one IITJEE SKMClasses.weebly.com substances skmclassesanother multiple substances colloid mixture evenly dispersed substances such IITJEE skmclasses.weebly.comm milks combustion IITJEE SKMClasses.weebly.com exothermic reaction IITJEE SKMClasses.weebly.com oxidant skmclasses.weebly.comfuel IITJEE SKMClasses.weebly.com heat skmclasses.weebly.comoften light compound – substance IITJEE SKMClasses.weebly.com made up two IITJEE SKMClasses.weebly.com chemically bonded elements condensation phase changeIITJEE SKMClasses.weebly.com gasIITJEE SKMClasses.weebly.com liquid conductor material IITJEE SKMClasses.weebly.com allows electric flow IITJEE SKMClasses.weebly.com freely covalent bond – chemical bond IITJEE SKMClasses.weebly.com involves sharing electrons crystal solid IITJEE SKMClasses.weebly.com packed IITJEE SKMClasses.weebly.com ions, molecules atoms IITJEE SKMClasses.weebly.com orderly fashion cuvette glasswareIITJEE SKMClasses.weebly.com spectroscopic experiments. usually made plastic glass quartz skmclasses.weebly.comshould be IITJEE possible deionization removal ions, skmclasses.weebly.com water’s case mineral ions such IITJEE skmclasses.weebly.comsodium, iron skmclasses.weebly.comcalcium deliquescence substances IITJEE SKMClasses.weebly.com absorb water IITJEE SKMClasses.weebly.com atmosphereIITJEE SKMClasses.weebly.com liquid solutions deposition – settling particles within solution mixture dipole electric magnetic separation charge dipole moment – polarity polar covalent bond dissolution solvation – spread ions monosacharide double bond sharing two pairs electradodes Microcentrifuge Eppendorf tube IITJEE SKMClasses.weebly.com Coomassie Blue solution earth metal – see alkaline earth metal electrolyte solution IITJEE SKMClasses.weebly.com conducts certain amount current skmclasses.weebly.com split categorically IITJEE skmclasses.weebly.com weak skmclasses.weebly.comstrong electrolytes electrochemical cell using chemical reaction’s current electromotive force made electromagnetic radiation type wave IITJEE SKMClasses.weebly.com through vacuums IITJEE skmclasses.weebly.comwell IITJEE skmclasses.weebly.commaterial skmclasses.weebly.comclassified IITJEE skmclasses.weebly.com self-propagating wave electromagnetism fields IITJEE SKMClasses.weebly.com electric charge skmclasses.weebly.comelectric properties IITJEE SKMClasses.weebly.com change way IITJEE SKMClasses.weebly.com particles move skmclasses.weebly.com interact electromotive force device IITJEE SKMClasses.weebly.com gains energy IITJEE skmclasses.weebly.comelectric charges pass through electron – subatomic particle IITJEE SKMClasses.weebly.com net charge IITJEE SKMClasses.weebly.com negative electron shells – IITJEE SKMClasses.weebly.com orbital around atom’s nucleus fixed number electrons usually two eight electric charge measured property (coulombs) IITJEE SKMClasses.weebly.com determine electromagnetic interaction element IITJEE SKMClasses.weebly.com atom IITJEE SKMClasses.weebly.com defined IITJEE SKMClasses.weebly.com atomic number energy – system’s abilityIITJEE SKMClasses.weebly.com do work enthalpy – measure total energy thermodynamic system (usually symbolized IITJEE skmclasses.weebly.comH entropy – amount energy IITJEE SKMClasses.weebly.com available skmclasses.weebly.com work closed thermodynamic system usually symbolized IITJEE skmclasses.weebly.com S enzyme – protein IITJEE SKMClasses.weebly.com speeds up catalyses reaction Empirical Formula – IITJEE SKMClasses.weebly.com called simplest formula gives simplest whole -number ratio atoms IITJEE SKMClasses.weebly.com element present compound eppendorf tube – generalized skmclasses.weebly.comtrademarked term skmclasses.weebly.com type tube; see microcentrifuge freezing – phase transitionIITJEE SKMClasses.weebly.com liquidIITJEE SKMClasses.weebly.com solid Faraday constant unit electrical charge widelyIITJEE SKMClasses.weebly.com electrochemistry skmclasses.weebly.comequalIITJEE SKMClasses.weebly.com ~ 96,500 coulombs represents 1 mol electrons, Avogadro number electrons: 6.022 × 1023 electrons. 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33, Sector IV HSR Layout, Bangalore hydrolysis hydrosilylation hydrostannation hyperconjugation imides imines indium indoles induction inhibitors insertion iodine ionic liquids iridium iron isomerization The Brigade International School , Brigade Millenium JP Nagar Brigade Millenium, JP Nagar Bangalore ketones kinetic resolution lactams lactones lanthanides Lewis acids ligands lipids lithiation lithium macrocycles magnesium manganese Mannich bases medicinal chemistry metalation metallacycles metallocenes metathesis Michael addition Mitsunobu reaction molecular recognition molybdenum multicomponent reaction nanostructures natural products neighboring-group effects nickel nitriles nitrogen nucleobases nucleophiles nucleophilic addition nucleophilic National Centre For Excellence 154/1, “Victorian Enclave”, 5th Main, Malleshpalya, Bangalore aromatic substitution nucleosides nucleotides olefination oligomerization oligonucleotides oligosaccharides organometallic reagents osmium oxidation oxygen oxygenations ozonolysis palladacycles palladium peptides pericyclic reaction peroxides phase-transfer catalysis phenols pheromones phosphates phosphorus phosphorylation Adugodi Aga Abbas Ali Road Agaram Agrahara Dasara Halli Agrahara Dasarahalli Airport Exit Road Airport Main Road Airport Road Akkipet Ali Askar Road Alur Venkatarao Road Amarjyothi Layout Amruth Nagar Amrutha Halli Ananda Nagar Anandrao Circle Anche Palya Ane Palya Anekal Anjana Nagar Anubhava Nagar APMC Yard Arabic College Arakere Arcot Sreenivasachar Street Ashok Nagar Ashwath Nagar Attibele Attiguppe Austin Town Avala Halli Avenue Road B. Narayanapura Babusahib Palya Bagalagunte Bagalur Balaji Nagar Balepet Banashankari Banashankari 1st Stage Banashankari 2nd Stage Banashankari 3rd Stage Banaswadi Banaswadi Ring Road Bangalore G.P.O Bannerghatta Bannerghatta Road Bapuji Nagar Basappa Circle Basava Nagar Basavanagudi Basaveshwara Nagar Basaveshwara Nagar 2nd Stage Basaveshwara Nagar 3rd Block Basaveshwara Nagar 3rd Stage Basaveshwara Road Bazaar Street Begur BEL Road Bellandur Bellandur Outer Ring Road Bellary Road BEML Layout Benagana Halli Bendre Nagar Benson Town Bharati Nagar Bhattara Halli Bhoopasandra Bhuvaneshwari Nagar Bidadi Bileka Halli Bilekahalli Binny Mill Road Bismillah Nagar Bommana Halli Bommanahalli Kendriya Vidyalaya Malleswaram 18th Cross Malleswaram Bangalore Bommasandra Bommasandra Industrial Area Brigade Road Brindavan Nagar Brookefield Brunton Road BTM 1st Stage BTM 2nd Stage Bull Temple Road Palace Orchards/Sadashivnagar area located north city centre IITJEE SKMClasses.weebly.com property prices higher brackets possibly IITJEE SKMClasses.weebly.com up-market residential area in Bangalore M.G. 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