In our daily life we come across processes like rusting of iron articles, fading of colour of clothes, burning of combustible substances such as cooking gas, wood, coal etc. All such processes fall in the category of specific type of chemical reactions called reduction-oxidation reactions or Redox reactions. A large number of industrial processes like electroplating, manufacture of caustic soda, extraction of metals like aluminium and sodium etc. are based on redox reactions. Redox reactions form the basis of electrochemical and electrolytic cells.
Oxidation and Reduction Reactions
Classical Concepts
Oxidation is a process of addition of oxygen or removal of hydrogen.
Reduction is the process of removal of oxygen or addition of hydrogen.
(i) Reaction of PbO on carbon.

Here oxygen is being removed from lead oxide(PbO) and is being added to carbon C. Therefore PbO is reduced while C is oxidised.
(ii) Reaction of H2S and Cl2.

Here hydrogen being removed from hydrogen sulphide (H2S) and is being added to chlorine (Cl2). Thus H2S is oxidised and Cl2 is reduced.
Electronic concept of Oxidation and Reduction Electrochemical reactions occur as a result of transference of electrons from one species to the other. For example, if magnesium is burnt in oxygen it gets oxidised to magnesium oxide (MgO). In the formation of magnesium oxide, two electrons from magnesium atom are transferred to oxygen atom.

The process of transference of electrons is described as redox process.
It is a process in which an atom or group of atoms taking part in chemical reaction loses one or more electrons. The loss of electrons result in the increase of positive or decrease in negative charge of the species.
For example :
Na * Na+ + e ( increase of positive charge)
Mg * Mg2+ + 2 e ( increase in positive charge)
Fe2+ * Fe3+ + e ( increase of positive charge)
MnO42 * MnO4 + e ( decrease in negative charge)
[Fe(CN)6]4 * [Fe(CN)6]3  + e (decrease in negative charge)

 The species which undergo the loss of electrons during the reactions are called reducing agents or reductants.
 MnO42, Fe2+, and Mg are reducing agents in the above examples.
It is a process in which an atom or group of atoms taking part in a chemical reaction gains one or more electrons. The gain of electrons results in the decrease of positive charge or increase in negative charge of the species.
For example,
Ag+ + e * Ag (decrease in positive charge)
Fe3+ + e * Fe2+ (decrease in positive charge)
[Fe(CN)6]3 + e * [Fe(CN)6] 4 (increase in negative charge)
MnO4 + e * MnO42 (increase in negative charge)
 The species which undergo gain of electrons during the reactions are called oxidising agents or oxidants.
 In the above example, Ag+, Fe3+, [Fe(CN)6]3 ions are oxidising agents.
Simultaneous occurrence of Oxidation and Reduction
In any process, oxidation can occur only if reduction is also taking place side by side and vice versa. Thus neither oxidation nor reduction can occur alone. That is why chemical reactions involving reduction-oxidation are called redox reactions. During the redox reaction there is transference of electrons from reducing agent to the oxidising agent as shown below:

For example, consider a reaction between zinc and copper ions.
Zn (s) + Cu2+(aq) * Zn2+(aq) + Cu(s)
In this reaction, zinc lose electrons and are oxidised to zinc ions(Zn2+) whereas cupric ions (Cu2+) gain electrons and are reduced to copper atoms. The cupric ions act as oxidising agent and zinc act as reducing agent. In fact, the oxidising agent gets reduced while reducing agents oxidised during redox reactions.
To sum up
Oxidation : Loss of electrons.
Reduction : Gain of electrons
Oxidising agent : Species which gains electrons
Reducing agent : Species which loses electrons.
01. In the reactions given below identify the species undergoing oxidation and reduction.
(i) H2S(g) + Cl2(g)  2 HCl(g) + S (s)
(ii) 3 Fe3O4(s) + 8 Al(s)  9 Fe(s) + 4 Al2O3(s)
(iii) 2 Na(s) + H2(g)  2 NaH(s)

Answer 01
(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) In the above reaction the compound formed is an ionic compound, which may be represented as Na+H(s). This suggests that one half of the reaction in this process is :
Na(s)  Na+(g) + e
And the other half of the reaction is :
H2(g) + 2 e  2 H
Thus sodium is oxidised and hydrogen is reduced.
Competitive Electron Transfer Reactions
Place a strip a strip of metallic zinc in an aqueous solution of copper nitrate as shown in Fig for about one hour.

Redox reaction between zinc and aqueous solution of copper nitrate occurring in a beaker

It can be noted that the strip becomes coated with reddish metallic copper and 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 solution of copper nitrate is :
Zn (s) + Cu2+(aq) * Zn2+(aq) + Cu(s) …. (1)
In the reaction , zinc has lost electrons to form Zn2+ and therefore , zinc is oxidised. Copper ion is reduced by gaining electrons from the zinc. The reaction (1) may be rewritten as :

At this stage, we may investigate the state of equilibrium for the reaction represented by equation(1). For this purpose, let us place a strip of metallic copper in a zinc sulphate solution. No visible reaction is noticed and the 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 extremely sensitive test ; yet the amount of Cu2+ formed cannot be detected. Thus we conclude that the state of equilibrium for the reaction (1) greatly favours the products over the reactants.
Let us extend electron transfer reaction to copper metal and silver nitrate solution in water and arrange a set-up as shown in Fig.

Redox reaction between copper and aqueous solution of silver nitrate occurring in a beaker.
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)
Let us also compare the reaction of metallic cobalt placed in nickel sulphate solution. The reaction that occurs is :
…… (3)
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 products [Co2+(aq) and Ni(s)] are greatly favoured.
By comparison we know that zinc releases electrons to copper and copper releases electrons to silver . The electron releasing tendency of the metals is in the order :
Zn > Cu > Ag
The competition for electrons between various metals helps us to design a class of cells , named as galvanic cells in which chemical reactions become the source of electrical energy.
Oxidation half and Reduction half reactions
Every reaction can be split up into half reactions, one representing loss of electrons i.e., oxidation half-reaction , while other representing gain of electrons, i.e., reduction half reaction.
Some examples are given below :
Zn + Cu2+ * Zn2+ + Cu (complete reaction )
Zn * 2 e * Zn2+ (oxidation half reaction)
Cu2+ + 2 e * Cu (reduction half reaction)
Sn2+ + 2 Hg2+ * Sn4+ + Hg22+ (complete reaction)
Sn2+ * 2 e * Sn4+ (oxidation half)
2 Hg2+ + 2 e * Hg22+ (reduction half reaction)
Oxidation number (ON) of an element is defined as the residual charge which its atom appears to have when all other atoms from the molecule are removed as ions.
During the removal of atoms, the electrons are counted according to the following rules :
 Electrons shared between two similar atoms are divided equally between the sharing atoms. For example, in chlorine molecule(Cl2) the electron pair is equally shared between two chlorine atoms. Therefore, one electron is counted with each chlorine atom as shown below:

Now there is no net charge on each atom of chlorine. In other words, oxidation number of chlorine in Cl2 is zero.
 Electrons shared between two dissimilar atoms are counted with more electronegative atom. For example, in hydrogen chloride molecule chlorine is more electronegative than hydrogen. Therefore , the shared pair is counted towards chlorine atom as shown below:

As a result of this, chlorine gets one extra electron and acquires a unit negative charge. Hence oxidation number of chlorine is *1. On the other hand hydrogen atom without electron has a unit positive charge. Hence, oxidation number of hydrogen in hydrogen chloride is +1.
Thus, atoms can have positive zero or negative value of oxidation numbers depending up on their state of combination. In fact, oxidation number is the charge assigned to the atom in a species according to some arbitrary rules as described below.
General rules for assigning Oxidation Number to an atom
The following rules are employed for determining oxidation number of the atoms.
 The oxidation number of the element in the free or elementary state is always zero.
For example,
Oxidation number of helium in He = 0
Oxidation number of chlorine in Cl2 = 0
Oxidation number of sulphur in S8 = 0
Oxidation number of phosphorus in P4 = 0
 The oxidation number of the element in the monoatomic ion is equal to the charge on the ion.
For example, in K+Cl , the oxidation number of K is +1
while that of Cl is **. In the similar way, oxidation
number of all alkali metals is +*, while those of
alkaline earth metals is +2 in their compounds.
 The oxidation number of fluorine is1 in all its compounds.
 Hydrogen is assigned oxidation number of ** in its compounds except in metal hydrides like NaH, MgH2, CaH2, LiH in which its oxidation number is **.
 Oxygen is assigned oxidation number of *2 in most of its compounds, however in peroxides like H2O2, BaO2, Na2O2 etc. its oxidation number is 1 . Similarly the exception also occurs in compounds of fluorine like OF2 and O2 F2 in which the oxidation number of oxygen is+2 and *1 respectively.
 The algebraic sum of the oxidation numbers of all the atoms in neutral molecule is zero. But in the case of compound ion , the sum of the oxidation numbers of all its atoms is equal to the charge on the ion.
The names of compounds are written by the use of O.N’s of the metal atoms. The O.N’s are written in the form of Roman numerals in brackets after the name of the metal. For example, O.N of copper in Cu2O is +1 and that in CuO is +2. Thus their name are copper(I) oxide and copper(II)oxide. This system of naming the compounds by the use of the oxidation numbers of representative metal atoms was proposed by Albert Stock and is referred as System of Stock Notations after his name. Some more examples are given below.
Cr2O3 : Chromium (III)Oxide.
SnCl2 : Tin(II) chloride
V2O5 : Vanadium (V) Oxide.
SnCl4 : Tin(IV) Chloride
Fe2 (SO4)3 : Iron (III)sulphate
FeSO4 : Iron(II) sulphate
Although the Stock Notations are generally used for metals, yet some compounds of non-metals have been named by this system. Some examples are given below:
SO2 : Sulphur(IV) Oxide SO3 : Sulphur(VI) Oxide
2. Calculate the oxidation number of the underlined elements in the following species :
CO2, Cr2O72, Pb3O4, CH2Cl2, PO43, S2O32, HNO3, NH4+
Answer 02
CO2 : (x) + 2(2) = 0 ; x = +4
Cr2O72 : 2(x) + 7(2) =  2 ; x = +6
Pb3O4 : 3(x) + 4(2) = 0 ; x = 8/3 = +2 (2/3)
CH2Cl2 : (x) + 2(+1) + 2 (1) = 0 ; x = 0
PO43 : (x) + 4(2) =  3 ; x = +5
S2O32 : 2(x) + 3(2) =  2 ; x = +2
HNO3 : 1 + x + 3 (2) = 0 ; x = +5
3. Assign ON of underlined elements in each of the following

Answer 03
(a) +5 (b) +6 (c) +5 d) +6
(e)  1 (f) +3 (g) +6 (h) +6
4. Using stock notation, represent the following compounds :
(i) HAuCl4 (ii) Tℓ2O iii) FeO
(iv) Fe2O3 (v) CuI vi) MnO (vii) MnO2
Answer 04
i) HAu(III)Cl4 ii) Tℓ(I)2O iii) Fe(II)O
iv) Fe2(III)O3 v) Cu(I)I vi) Mn(II)O vii) Mn(IV)O2
5. Justify that the following reactions are redox reactions :
(a) CuO(s) +H2(g)  Cu(s) + H2O(g)
(b) Fe2O3(s) + 3 CO(g)  2 Fe(s) + 3 CO2(g)
(c) 4 BCl3(g) + 3 LiAlH4(s)  2 B2H6(g) + 3 LiCl(s)
+ 3 AlCl3(s)
(d) 4 NH3(g) + 5 O2(g)  4NO(g) + 6 H2O(g)
Answer 05

Oxidation and Reduction in terms of Oxidation Number
Oxidation is defined as a chemical process in which oxidation number of the element increases.
Reduction is defined as the chemical process in which oxidation number of the element decreases.
Consider the reaction between hydrogen sulphide and bromine to give hydrogen bromide and sulphur.

In the above example, the oxidation number of bromine decreases from 0 to *1, thus it is reduced. The oxidation number of S increases from *2 to 0. Hence H2S is oxidised.
Oxidising agent
Oxidising agent is a substance which undergoes the decrease in oxidation number of one or more of its elements.
Reducing agent
Reducing agent is a substance which undergoes the increase in the oxidation number of one or more of its elements.
In the above example H2S is the reducing agent while Br2 is the oxidising agent.
6. Justify that the reaction :
2 Cu2O(s) + Cu2S(s)  6 Cu + SO2 (g)
is a redox reaction . Identify the species oxidised / reduced , which acts as an oxidant and which acts as a reductant.
Answer 06
Let us assign the ON to each of the species in the reaction.
+1 2 +1 2 0 +4 2
2 Cu2 O (s) + Cu2 S(s)  6 Cu(s) + S O2
In the 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 the ON , 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.
1. Combination reactions
A combination reaction ay be denoted in the manner:
A + B  C
Either A and B or both A and B must be in 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. 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 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(above) . This may 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 displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be noted as :
X + YZ  X Z + 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. 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 , 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. A few examples for the displacement of hydrogen from acids are :

The above reactions 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.
The metals Zn, Cu and Ag through tendency to lose electrons show their reducing activity in the order 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 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 :

The reactions (above) form the basis of identifying Br and I in the laboratory through the test popularly known as ‘layer test ‘. Bromine 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 :
2 X  X2 + 2e
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.
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 experience 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 in water.
Phosphorous, sulphur and chlorine undergo disproportionation in the alkaline medium as shown below :

……. (1)
The above reaction describes the formation of household bleaching agents. The hypochlorite ion (ClO ) formed in the reaction oxidises the colour bearing stains of the substance to colourless compounds.
Bromine and iodine follow the same trend as exhibited by chlorine in reaction (1), fluorine shows deviation from this behaviour when it reacts with alkali. The reaction that takes place in the case of fluorine is as follows.

( Fluorine in the above reaction will attack water to produce some oxygen also). This depature shown by fluorine is due its most electronegative character. It cannot exhibit any positive oxidation state. This means that among halogens , fluorine does not show disproportionation tendency.
The Paradox of Fractional Oxidation Number
Some times , 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)
The idea of fractional oxidation number is unconvincing, because electrons are never shared/transferred in fraction. Actually this fractional oxidation state is the average oxidation state of the element under examination and its structural parameters reveal that the element for whom fractional oxidation state is realised is present in different oxidation states. Structures of the species C3O2 , Br3O8 and Na2S4O6 reveal the following bonding situations :

Structure of Br3O8 (tribromooctaoxide)

Structure of S4O6 2 (tetrathionate ion)
The element marked with asterisk in each species is exhibiting the different oxidation state (oxidation number) from rest of 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 middle carbon. Likewise in Br3O8 , each of the 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 S4O62 , 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 S4O62 is 2.5 , whereas the reality being +5 , 0, 0 and +5 oxidation number respectively for each sulphur.
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 the particular species is present in more than one whole number oxidation states. Fe3O4 , Mn3O4 , and 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.
Some other examples of oxidation numbers
The knowledge of bonding and individual molecular structure is also important aspect and has to be considered for calculating the oxidation number of the elements. Some examples are discussed as follows :
(i) Oxidation numbers of two N atoms in NH4NO3 are different. NH4NO3 is made of NH4+ and NO3.
In NH4+ ion O.N of N is 3 while in NO3 ion , the O.N. of N is +5.
(ii) Bleaching powder , CaOCl2 is made up of Ca2+ , OCl and Cl. Hence O.N of two Cl atoms in CaOCl2 are +1 (in OCl) and 1 (in Cl ) respectively. However, the average O.N of Cl in CaOCl2 is 0.
(iii) In H2S2O5 (caro’s acid) , the two of the O atoms are involved in peroxide linkage.

The O.N of such O atoms is taken as 1 as per rules. Hence ON of sulphur can be calculated as follows.
2(+1) + x + 3(2) + 2(1) = 0
x = +6
(iv) In chromium peroxide , CrO5 , the four out of the five O atoms are involved in peroxide linkage ( OO ). The molecular structure of the compound is :
Hence O.N of Cr can be calculated as :
x + 4(1) + 1 (2) = 0
x = +6
(v) In co-ordination complex, K4[Fe(CN)6] . The individual O.N of C and N are not known but for calculating O.N of Fe , the sum of O.N of C and N in CN unit is taken to be 1. Thus O.N of Fe is calculated as 4(+1) + x + 6(1) = 0 or x = +2.
(vi) In molecules where co-ordinate bonds are present between the atoms the contribution of bond pair of co-ordinate bond is neglected if the acceptor atom is less electronegative. On the other hand , if the co-ordinate bond is directed towards more electronegative atom, it contributes +2 charge to less electronegative atom while 2 charge to more electronegative atom. For example, in the contribution of co-ordinate bond pair is neglected while in OS=O , it is not neglected.

The idea of fractional oxidation state is only a myth and reality is revealed by the structures only. 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 the element in particular species is always present in whole number oxidation state in more than one oxidation states. Fe3O4 , Mn3O4 , Pb3O4 are some of other examples of the compounds where we come across with fractional oxidation of the metal atom in each of these oxides.
07. Which of the following species , do not show disproprotionation reaction and why ? ClO, ClO2 , CLO3  and ClO4. Also write reaction for each of the species that disproportionates.
Answer 08

09. Suggest a scheme of the following redox reactions.
(a) N2(g) + O2(g)  2 NO(g)
(b) 2 Pb(NO3)2 (s)  2 PbO(s) + 2 NO2(g) + ½ O2(g)
(c) NaH(s) + H2O(ℓ)  NaOH(aq) + H2(g)
(d) 2 NO2(g) + 2 OH (aq) NO2 (aq) + NO3 (aq)
+ H2O(ℓ)
Answer 08

Problem 09

Distinction between oxidation number and valence
Carbon atom is tetravalent in its compounds. But the carbon atom has different oxidation numbers in different compounds. This means that valence and oxidation number have different meanings. This important points of distinction are as follows :
Valency Oxidation number
1.Valency of an element is the number of hydrogen atoms or twice the number of oxygen atoms which combine with one atom of the element. 1. Oxidation number is the charge assigned to an atom of a olecule or ion according to some arbitrary rules.

2. Valency of an element is only a number without any sign. For example, valency of hydrogen is 1 and that of oxygen is 2. 2. The oxidation number of an element has either positive or negative sign. For example the oxidation number of hydrogen is +1 while that of oxygen is 2in its compounds.
3. Except for noble gases , valency of no other element is zero. 3. Oxidation number of an element can be zero. For example, in C6H12O6 , the oxidation number of carbon is zero.
4. Valency of an element is always a whole number. 4. The oxidation number of element can be even fractional . For example in Na2S4O6 , the O.N of sulphur is 2.5.
5. In most of the cases , an element has fixed valency in all its compounds. For example, carbon is tetravalent. 5. The oxidation number of an element is usually different compounds. For example, carbon atom has oxidation number  4 (CH4) , 3(C2H6) , +2 (CHCl3) and 0 (C6H12O6).
6. Some elements may show variable valency but it has normally not more than two value. For example, phosphorus has valency of 3 and 5 in PCl3 and PCl5. 6. The same element may show a large number of variable oxidation states in different compounds.

Two methods are used for balancing chemical equations for redox reactions. The first method is based on the change in the oxidation number of the reductant and the oxidant and the second method is based on splitting the redox reaction into two half reactions one involving oxidation and another involving reduction. Both methods work very well .


Problem 11

Problem 12

In general , in a chemical reaction described by the balanced chemical equation :
a A + b B + ….  c C + d D + …..
A, B, …. are reactants and C, D, ….. are the products. The coefficients a, b, c and d are known as stoichiometric coefficients. These indicate the number of moles of the reactants used and the number of moles of the products obtained in the reaction.
Volumetric titrations based on redox reactions are carried out in laboratories. If we know the molarities of the reductant and oxidant and the volumes used in titration and n1 and n2 are their stoichiometric coefficients, then it is possible to use the equation :

for volumetric estimation.
For example, in the titration of FeSO4 with KMnO4 , the balanced equation is :
10 FeSO4 + 2 KMnO4 + 8 H2SO4  5 Fe2(SO4)3 + K2SO4
+ 2 MnSO4 + 8 H2O
If FeSO4 is the reagent ‘1’ and KMnO4 is reagent ‘2’ then this equation becomes

and can be used to calculate the unknown quantity , if the other quantities are known.
Redox Reactions as Basis for Titrations
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 C2O42 ) is oxidized 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 oxidant are equal in terms of their mol stoichiometry.
ii) If there is no dramatic auto-colour change ( as with MnO4 titration) , there are indicators which are oxidized immediately after the last bit of the reactant is consumed, producing a dramatic colour change. The best example is afforded by Cr2O72 , which is not a self indicator, but oxidizes the indicator substance diphenylamine just after the equivalence point to produce an intense blue colour, thus signaling the end point.
iii) There is another method which is used to those reagents which are able to oxidize I ions, for example , Cu(II) :
2 Cu2+(aq) + 4 I  Cu2I2 (aq) + I2(aq)
This method relies on the fact that iodine itself gives an intense blue colour with starch and has a very specific reaction with thiosulphate ions (S2O32 ), which too is a redox reaction.
I2(aq) + 2 S2O32 (aq)  2 I(aq) + S4O62 (aq)
I2 though insoluble in water , remains in solution containing KI as K I3.
On addition of starch after 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 be easily be tracked and the rest is the stiochiometric calculation only.
Limitations of Concept of Oxidation Number
According to the concept of oxidation number , oxidation means increase in oxidation number by the loss of electrons while reduction implies decrease in oxidation number by gain of electrons. However, actually during oxidation , there is a decrease in electron density around the atom (or atoms) undergoing oxidation. At the same time, there is an increase in electron density around atom(or atoms) undergoing reduction. This may be regarded as the limitations of the concept.
13. How many grams of potassium dichromate are required to oxidise 20.0 g of Fe2+ in FeSO4 to Fe3+ if the reaction is carried out in an acidic solution ?
Answer 13
When dissolved in water K2Cr2O7 dissociates into K+ and Cr2O72 ions and FeSO4 dissociates as Fe2+ and SO42 ions. The skeleton equation is :
Fe2+ + Cr2O72 + 14 H+  6F3+ + 2 Cr3+ + 7 H2O
From the balanced equation,
1 mol K2Cr2O7  6 mol FeSO4
Formula mass of K2Cr2O7 = 2 x 39 + 2 x 52 + 7 x 16 = 294
Formula mass of FeSO4 = 56 + 32 + 4 x 16 = 152 g
1 mol K2Cr2O7  6 mol FeSO4
294 g 6 x 152 g
6 x 152 g FeSO4 required K2Cr2O7 = 294 g
20 f FeSO4 required K2Cr2O7 = (294 x 20) / [6 x 152]
= 6.45 g
K2Cr2O7 required to oxidize 20 g FeSO4 = 6.45 g
14. How many millilitres of 0.125 M KMnO4 are required to react completely with 25.0 ml of 0.250 M FeSO4 solution in acidic medium ?

Answer 14
The balanced chemical equation:
MnO4 + 5 Fe2+ + 8 H+  Mn2+ + 5 Fe3+ + 4 H2O
From the balanced equation , it is clear that :
1 mol KMnO4  5 mol FeSO4
158 g 5 x 152 g
Moles of FeSO4 present in 25 mL of 0.25 M solution =
= [0.25 x 25 ] / 1000 = 0.00625 mol
Let us determine the number of moles of KMnO4 that must react
5 mol FeSO4 react with KMnO4 = 1 mol
0.00625 mol of FesO4 react with KmnO4 = (1 x 0.00625)/ 5
= 0.00125 mol
Now we to calculate the volume of 0.125 M KMnO4 solution which contain 0.00125 mol. According to the definition of molarity 0.125 M solution means that 0.125 mol of KMnO4 is present in 100 mL.
 0.00125 mol of KMnO4 is present in =
= [1000 x 0.00125] / 0.125
= 10 mL
15. How many millilitres of 0.025 M K2Cr2O7 are required to react completely with 25.0 ml of 0.20 M solution of FeSO4 ?
Answer 15
The balanced chemical equation is :
1 mol K2Cr2O7  6 mol FeSO4
Moles of FeSO4 present in 25 mL of 0.2 M solution =
= [0.2 x 25 ] / 1000 = 0.005 mol
Now 6 mol of FeSO4 requires K2Cr2O7 = 1 mol
 0.005 mol FeSO4 will require K2Cr2O7 =
= (1 x 0.005) / 6 = 0.00083 mol
Let us calculate the volume of K2Cr2O7 solution containing 0.000833 mol.
0.0025 mol of K2Cr=O7 are present in = 1000 mL
 0.000833 mol of K2Cr2O7 are present in =
= [1000 x 0.000833] / 0.025 = 33.3 mL
In aqueous solutions , the spontaneous redox reactions can be carried out directly as well as indirectly. The energy liberated during direct and indirect reactions appear in different forms.
Direct Redox Reactions
A redox reaction in which oxidation and reduction take place in the same vessel is called direct redox reaction. In such reactions, the transference of electrons from the reducing agent occurs over a very short distance(generally within the molecular diameters). For example, if a zinc rod is immersed in copper sulphate solution taken in a beaker, a spontaneous reaction takes place and the following observations are made :

Zn-CuSO4 Reaction in a beaker

i) Zinc rod starts dissolving and loses its mass gradually.
ii) The blue colour of CuSO4 solution slowly fades.
iii) A reddish brown precipitate of copper settles at the bottom of the beaker.
iv) The reaction is exothermic i.e., it takes place with the evolution of heat.
v) The solution remains electrically neutral.
vi) The reaction does not continue indefinitely but stops after some time.
The overall reaction taking place in the beaker may be represented as :
Zn(s)+Cu2+(aq) +SO42(aq) * Zn2+(aq) + SO42(aq)+ Cu(s)
Zinc loses electrons and changes into Zn2+ions. As a result the mass of the zinc rod decreases. The electrons lost by zinc rod are gained by Cu2+ ions and they change into Cu(s) atoms which settles down at the bottom of the beaker in the form of a precipitate. It may be noted that in the direct redox reaction, the chemical energy appears as heat.
If a copper rod is dipped in ZnSO4 solution, no reaction takes place. But if copper rod is dipped in the aqueous solution of silver nitrate, the silver ions oxidise copper to copper ions according to the following reaction.
Cu(s) + 2 Ag+(aq) * 2 Ag(s) + Cu2+(aq)
Indirect redox reactions and electrochemical cells
Redox reactions in which oxidation and reduction takes place in different vessels are called indirect redox reactions. In such reactions the transference of electrons between reducing agent and oxidising agent takes place indirectly through the conducting wires. The decrease of energy during indirect redox reactions appears as electrical energy. In other words, these reactions involve the conversion of chemical energy into electrical energy. The arrangement for carrying out the indirect redox reaction is referred to as electrochemical cell.
Thus, an electrochemical cell is a device in which chemical energy is converted into electrical energy. These are called Galvanic Cells.
In order to understand as to how the indirect redox reaction is used to produce electrical energy, consider one of the simplest electrochemical cells involving the reaction between zinc and copper sulphate. In its simplest form, a zinc rod is placed in ZnSO4 solution and copper rod is placed in CuSO4 solution in two separate beakers. The two metallic rods which act as electrodes are connected by conducting wire through a Galvanometer. The two solutions are joined by an inverted U-tube known as Salt bridge. The tube is filled with aqueous solution of some electrolyte such as KCl, KNO3 or NH4Cl to which gelatin or agar-agar has been added to convert it into semi-solid paste. A schematic diagram of this cell has been shown in Fig.

Electrochemical cell

The following observations are made when the key is inserted.
i) There is a deflection in the Galvanometer which indicates
the flow of electrons through the connecting wires.
ii) The direction of deflection indicates the flow of electrons
from zinc to copper.
iii) The zinc rod loses mass whereas the copper rod gains
iv) The concentration of zinc sulphate solution increases and
that of copper solution decreases.
v) No evolution of heat is observed during the reaction.
vi) The solutions in the two beakers remain electrically neutral throughout.
vii) The reaction as well as flow of electron stops after some time.

From these observations, it is clear that zinc atoms undergo oxidation by the loss of electrons.
Zn(s) * Zn2+(aq) + 2 e (oxidation)
The electrons liberated during oxidation are pushed through connecting wire to copper rod where these are picked by Cu2+ions which get reduced to copper atoms. The copper atoms so formed deposited on the copper rod. This is why copper rod gains weight.
Cu2+(aq) + 2 e * Cu(s) (Reduction)
The overall reaction taking place in the cell is :
The electrode at which oxidation takes place is called anode and the electrode at which reduction takes place is called cathode. In the above electrochemical cell oxidation occurs at the zinc rod, hence copper is the cathode. Anode electrode constitutes * ve terminal and the cathode constitutes +ve terminal of the cell.
In electrochemical cell a salt bridge serves two very important functions.
i) It allows the flow of current by completing the circuit.
ii) It maintains electrical neutrality.
The transference of electrons from anode to copper cathode leads to the development of a net positive charge around the anode due to the formation of Zn2+ ions and net negative charge around the cathode due to the deposition of Cu2+ ion as Cu(s) on the cathode. The positive charge accumulated around the anode will prevent electrons from it and the negative charge (due to excess of SO42) collected around the cathode will prevent the acceptance of electrons from the anode. As the transference of electron stops, no current will flow. The salt bridge restores the electrical neutrality of the solutions in the two compartments. It contains concentrated solution of an inert electrolyte, the ions of which are not involved in electrochemical reactions. The anions of the electrolyte(e.g. Cl in the case of salt bridge containing KCl) migrate to anode compartment and cations (i.e., K+ ions)migrate to the cathode compartment so that the solutions in these compartments become neutral and the flow of electron continues. Thus the salt bridge helps to prevent the accumulation of charges and maintains the flow of current.

Electrode potential

When a strip of a metal M is brought in contact with the solution of its own ions Mn+, it has either of the two tendencies.
(a) Tendency of oxidation
Metal atoms may lose electrons to form Mn+ in solution. In doing so, the metal strip develops a negative charge due to accumulation of electrons released. The solution at the same time develops a positive charge due to the formation of positive ions. The oxidation process does not continue indefinitely and ultimately, a state of equilibrium is established quickly.
M ⇌ Mn+ + n e
(b) Tendency of Reduction
The metal ions in solution may gain electrons from the strip, get reduced and are deposited on the strip as atoms. In doing so, the strip acquires positive charge and the solution develops negative charge due to the decrease in concentration of positive ion. Ultimately, a state of equilibrium is established as shown below :
Mn+ + n e ⇌ M
In either case, the equilibrium leads to the separation of charges and results into a potential being developed between the metal strip and its solution. The net charge separation and hence, the potential difference at equilibrium depends upon :
i) The nature of metal and its ion.
ii) Concentration of ions in the solution.
iii) Temperature.
The electrical potential difference set up between the metal and its solution is known as half cell electrode potential. It is a measure of tendency of the electrode in the half cell to lose or gain electrons. According to the present convention, the half reactions are always written as reduction half reactions and their potentials are represented by reduction potentials. It may be noted that :
i) Reduction potential(tendency to gain electrons) and oxidation potential (tendency to lose electrons) are numerically equal but have opposite signs.
ii) Reduction potential of an electrode generally increases with increase in concentration of its ions and decrease with the decrease in concentration of the ions in the electrolyte.
iii) The reduction potential of the electrode when concentration of the ions in the solution is 1 mol L 1 and temperature is 298 K is called Standard Reduction Potential and is represented by E0RED.
Representation of an Electrochemical Cell
The electrochemical cells can be represented according to certain universally accepted conventions as described below:
1. The half cell is usually represented as M*Mn+ or Mn+*M where M represents the symbol of the element and Mn+ represents its cations in electrolyte. The line (*) represents the interface between the two phases (may be solid * liquid or liquid * liquid or liquid * gas ). The concentration of the electrolyte may also be mentioned in brackets. For example,
let us consider copper rod in contact with cupric ions with concentration equal to 1.0 M.
(a) It can be represented as Cu2+(1M)  Cu if it constitutes reduction half cell.
(b) It can be represented as Cu * Cu2+ (1.0 M), if it constitutes oxidation half cell.
2. Anode half-cell (oxidation half-cell) is written on the left hand side whereas cathodic half cell (reduction half) is written on the right hand side.
3. The two half cells are separated by two vertical lines which indicate salt bridge.
For example, Zn-CuSO4 cell may be represented as :
Zn * Zn2+ (1.0 M) ** Cu2+ (1.0 M) * Cu
In cases where porous pot is used in separating the electrodes, a single vertical line is drawn between the two electrodes as shown below:
Zn * Zn2+ (1.0 M) * Cu2+ (1.0 M) * Cu
Sometimes negative and positive signs are also put on the electrodes to show the release or gain of electrons taking place on them. Anode is a negative terminal, while cathode is a positive terminal. The flow of electrons through the external circuit takes place from anode to cathode. For example:
(anode) (cathode)
*ve terminal +ve terminal
The electrochemical cell consists of two half cells. The electrodes in these half cells have different electrode potentials. When the circuit is completed, the loss of electrons occurs at the electrode having lower reduction potential, whereas the gain of electrons occur at electrode with higher reduction potential. The difference in the electrode potential of two electrodes of the cell is termed as Electromotive Force (EMF) or Cell Voltage. Mathematically, it can be represented as :
EMF = Ecathode * Eanode
= Eright * Eleft
= ER * EL
EMF of the cell may be defined as the potential difference between the two terminals of the cell when no current is drawn from it. It is measured with the help of potentiometer or vaccum tube voltmeter.
Absolute value of electrode potential of an electrode cannot be determined because a half cell by itself cannot cause movement of charges (flow of electrons) . It is due to the fact that once equilibrium is reached between the electrode and the solution in a half cell , no further displacement of charges can occur unless and until it is connected to another half-cell with different electrode potential. This difficulty is overcome by finding the electrode potentials of various electrodes relative to a reference electrode whose electrode potential is arbitrarily fixed. The common reference electrode used for this purpose is Standard Hydrogen Electrode (SHE) or Normal Hydrogen Electrode(NHE) whose potential is arbitrarily taken to be zero.

Normal hydrogen electrode consists of a platinum wire sealed into a glass tube and carrying a platinum foil at one end. The platinum foil is coated with finely divided platinum. The electrode is placed in a beaker containing an aqueous solution of some acid having one molar concentration of H+ ions. Hydrogen gas at one atmospheric pressure is continuously bubbled through the solution at a temperature of 298 K. The oxidation or reduction in NHE takes place at the platinum foil.
If NHE acts as anode then oxidation will take place at it as :
H2 (g) * 2 H+ (aq) + 2 e
In such a case it is represented as :
Pt, (1/2 )H2 * H+ (1 M)
If NHE acts as cathode, then reduction will take place at it as
2 H+ (aq) + 2 e * H2 (g)
In this case it is represented as :
H+ (1 M) * (1 / 2 ) H2, Pt
Pt is an inert electrode.
The electrode potentials of other electrodes are determined by coupling them with NHE. The electrode potential of the electrode determined as relative to standard hydrogen electrode under standard conditions is called Standard Electrode Potential. It is represented as E0. The standard conditions are 1 M concentration of ions in solution , at 298 K and 1 atmospheric pressure.
Measurement of Standard Electrode potentials
In order to determine the standard electrode potential of an electrode in standard conditions, it is connected to normal hydrogen electrode to constitute a cell.

Zn (s)  Zn2+ + 2 e 2 H+ + 2 e  H2
Zn electrode acts as the anode when it is
connected to NHE
If the electrode forms the negative side of the cell, it is allotted negative electrode potential and if it forms the positive side of the cell, it is allotted a positive value of electrode potential. The potential difference between the electrodes is determined with the help of voltmeter(more correctly by potentiometer). At the same time, direction of flow of current is also observed which gives the idea of direction of flow of electrons. Electron-flow occurs from anode to cathode.
Ecell = Ecathode * Eanode
Knowing the E0cell and electrode potential of one of the electrode(NHE), the electrode potential of the other electrode can be calculated.
For example, in order to find out standard electrode potential of zinc electrode containing 1 M concentration of Zn2+ ions it is connected to NHE as shown in Fig.
The voltmeter reading shows the potential difference (E0cell) of 0.76 V and indicates that electrons flow from zinc electrode to NHE. From this, it follows that zinc electrode act as anode and NHE acts as cathode.
E0cell = E0cathode * E0anode
= E0 2H+/ H2 * E0 Zn2+ / Zn
0.76 V = 0 * E0 Zn2+ / Zn
E0 Zn2+ / Zn = * 0.76 V
Similarly, when copper electrode is coupled with NHE, , the voltmeter reading shows a potential difference(E0cell) of 0.34 V and indicates that electrons flow from NHE towards copper electrode. Therefore, NHE in the cell acts as anode and copper electrode acts as cathode.

H2  2 H + 2 e Cu2+ + 2 e Cu(s)
Copper electrode acts as cathode
when it is connected to NHE

E0cell = E0cathode * E0anode
= E0 Cu2+ / Cu * E0 2H+/ H2
0.34 V = E0 Cu2+ / Cu * 0
E0 Cu2+ / Cu = +0.34 V
In a similar manner, the standard reduction potentials of other electrodes including metals as well as non-metals can be determined
For example, the standard reduction potentials of chlorine can be determined by using the electrode consisting of Cl2 gas at one atmospheric pressure and 298 K in equilibrium with 1 M solution of Cl ions.
Different metal / metal ion combinations have different values of electrode potentials. The various elements can be arranged in the order of increasing or decreasing values of their reduction potentials. The arrangements of various elements in the order of increasing values of standard reduction potentials is called Electrochemical Series or Activity Series. The electrochemical series consisting of some electrodes along with their respective reduction potentials with their respective reduction half reactions are given in the following TABLE.

Applications of Electrochemical series
Some important applications of electrochemical series are discussed below :
1. Calculation of the standard EMF of the Cell
From the electrochemical series, the standard reduction potentials of the electrode are found out. The electrode with higher reduction potential is taken as cathode and other as anode. The EMF of the cell is calculated as:
E0cell = E0ccathode * E0anode
However, if the reaction taking place in the cell is also to be determined the following steps are followed :
(a) Write reduction equation for both electrodes along with their reduction potential separately.
(b) Balance each reaction with respect to the number of atoms of each kind and electrical charges.
(c) Multiply the reactions by a suitable number so that the number of electrons involved in both the half reactions are equal. Reduction potential (E0) is not multiplied by an integer because it simply gives the relative tendency of reduction.
(d) Subtract the equation for reaction having lower reduction potential from the other reaction having higher reduction potential. The difference gives EMF of the cell.
2. Comparison of the reactivity of metals
The metals which occupy higher position in electrochemical series have lower reduction potentials. This indicates that ions of such metals will not be reduced to metals easily. On the contrary, such metals would be easily oxidised to their ions by losing electrons. Therefore, from the position of a metal in the electrochemical series, it is possible to predict the relative reactivities of metals. A metal having smaller reduction potential (i.e., greater negative reduction potential) can displace metals having greater reduction potential (i.e., smaller negative reduction potential).from the solutions of their salts. In other words, a metal occupying higher position in the series can displace the metals lying below it from the solutions of their salts.
Thus, we conclude that the metals occupying higher positions in electrochemical series are more reactive in displacing the other metals from the solutions of their salts. For example, zinc lies above copper in the series and hence, it can displace copper from copper sulphate solution. Copper cannot displace zinc from zinc sulphate solution because it lies below zinc in the series and hence it is less reactive.
3. Predicting the feasibility of a redox reaction
With the help of electrochemical series we can predict whether a given redox reaction will take place or not. From the given equation, the substances undergoing oxidation and reduction are identified. The substance undergoing oxidation will act as anode and the substance undergoing reduction will act as cathode. The EMF of this hypothetical cell is calculated as under :
E0cell = E0cathode * E0anode
If the EMF comes out to be positive , the given redox reaction will take place and if EMF comes out to be negative, the given redox reaction will not take place.
3. To Predict the Reaction of Metal with dilute acids to liberate hydrogen gas
Some metals like Zn, Fe, react with dil. HCl or H2SO4 to liberate H2 gas while some metals like Cu, Ag do not liberate H2 gas with dil HCl or dil H2SO4. A prediction about capacity of a given metal to produce H2 gas by its reaction with dilute acids can be easily made from the knowledge of electromotive series. Metals which lie above hydrogen in the electromotive series can reduce H+ ions to hydrogen and hence liberate hydrogen gas on reaction with dil. acids. In other words, metals having negative reduction potentials can displace hydrogen from acids. For example, zinc (E0 Zn2+ / Zn = * 0.76 V) lies above hydrogen in the series and hence it can displace hydrogen from dilute acids, whereas copper (E0 Cu2+ / Cu = +0.34 V) which is lying below hydrogen in the series cannot displace hydrogen from acids.


Atoms and Molecules

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