Topic to be covered

`=>` Introduction
`=>` Classical idea of redox reaction–oxidation and reduction reactions
`=>` Redox reactions in terms of electron transfer reactions
`=>` Competitive Electron Transfer Reactions


`=>` 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.


`=>` 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:

`color{red}(2Mg(s) + O_2 (g) → 2MgO(s))` ..............(8.1)

`color{red}(S(s) +O_2(g) → SO_2 (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

`color{red}(CH_4 (g) + 2O_2 (g) → CO_2 (g) + 2H_2O (l))` ........... (8.3)
careful examination of reaction (8.3) in which hydrogen has been replaced by oxygen prompted chemists to reinterpret oxidation in terms of removal of hydrogen from it and, therefore, the scope of term oxidation was broadened to include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.

`color{red}(2 H_2S(g) + O_2 (g) → 2 S (s) + 2 H_2O (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.

`=>` `color{green}("The oxidation of magnesium with fluorine, chlorine and sulphur etc. occurs according to the following reactions :")`

`color{red}(Mg(s)+F_2(g) → MgF_2(s))` ..........(8.5)

`color{red}(Mg(s) +Cl_2 (g) → MgCl_2(s))` .................(8.6)

`color{red}(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 :

`color{red}(2K_4[Fe(CN)_6](aq) +H_2O_2(aq) → 2K_3 [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.

`color { maroon} ® color{maroon} ul (" REMEMBER")`
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:

`color{red}(2HgO (s) overset(Delta)→ 2Hg(l) +O_2(g))` ........................... (8.8)

(removal of oxygen from mercuric oxide )

`color{red}(2FeCl_3 (aq) +H_2 (g) → 2FeCl_2 (aq) +2HCl (aq))` .....................(8.9)

(removal of electronegative element, chlorine from ferric chloride)

`color{red}(undersettext{(addition of hydrogen)}(CH_2 = CH_2 (g) +H_2(g)) → H_3C - CH_3 (g))` ..............(8.10)

`color{red}(2HgCl_2 (aq) + SnCl_2 (aq) → Hg_2Cl_2 (s)+SnCl_4 (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.
Q 3141880723

In the reactions given below, identify the species undergoing oxidation and reduction :

(i) `H_2S (g) +Cl_2 (g) → 2HCl (g) +S (s)`

(ii) `3Fe_3O_4 (s) +8Al(s) → 9 Fe (s) +4Al_2O_3(s)`

(iii) `2Na(s) +H_2(g) → 2NaH (s)`


(i) `H_2S` 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 `(Fe_3O_4)` 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.


`=>` `color{green}("We have already learnt that the reactions:")`

`color{red}(2Na(s) + Cl_2(g) → 2NaCl (s))` ...................... (8.12)
`color{red}(4Na(s) + O_2(g) → 2Na2O(s))` ......................... (8.13)
`color{red}(2Na(s) + S(s) → Na_2S(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 `color{red}(Na^+Cl^– (s), (Na^+)_2O^(2–)(s))`, and `color{red}((Na^+)_2 S^(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

`color{red}(2Na(s) →2 Na^+ (g) +2e^(-))`

`color{red}(Cl_2 (g) +2e^(-) → 2Cl^(-) (g))`

•Each of the above steps is called a half reaction, which explicitly shows involvement of electrons.

• `color{green}("Sum of the half reactions gives the overall reaction :")`

`color{red}(2 Na(s) + Cl_2 (g) → 2 Na^+ Cl^– (s))` or `color{red}(2 NaCl (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

`color{green}("Oxidation:")` Loss of electron(s) by any species.

`color{green}("Reduction:")` Gain of electron(s) by any species.

`color{green}("Oxidising agent :")` Acceptor of electron(s).

`color{green}("Reducing agent :")` Donor of electron(s).
Q 3151880724

Justify that the reaction : `2 Na(s) + H_2(g) → 2 NaH (s)` is a redox change.


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 :
`2 Na (s) → 2 Na^+(g) + 2e^-`

and the other half reaction is:

`H_2(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.

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 `color{red}(Zn^(2+))` ions among the products can easily be judged when the blue colour of the solution due to `color{red}(Cu^(2+))` has disappeared.

`=>` If hydrogen sulphide gas is passed through the colourless solution containing `color{red}(Zn^(2+))` ions, appearance of white zinc sulphide, `color{red}(ZnS)` can be seen on making the solution alkaline with ammonia.

`=>` `color{green}("The reaction between metallic zinc and the aqueous solution of copper nitrate is :")`

`color{red}(Zn(s) + Cu^(2+) (aq) → Zn^(2+) (aq) + Cu(s))` ....................... (8.15)

`=>` In reaction (8.15), zinc has lost electrons to form `color{red}(Zn^(2+))` 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.

`=>` `color{green}("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 `color{red}(Cu^(2+))` ions by passing `color{red}(H_2S)` gas through the solution to produce the black colour of cupric sulphide, `color{red}(CuS)`, does not succeed.

`=>` Cupric sulphide has such a low solubility that this is an extremely sensitive test; yet the amount of `color{red}(Cu^(2+))` 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 `color{red}(Cu^(2+))` ions on account of the reaction:

`=>` Here, `color{red}(Cu(s))` is oxidised to `color{red}(Cu^(2+)(aq))` and `color{red}(Ag^+(aq))` is reduced to `color{red}(Ag(s))`. Equilibrium greatly favours the products `color{red}(Cu^(2+) (aq))` and `color{red}(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 `color{red}(Ni^(2+)(aq))` and `color{red}(Co^(2+)(aq))` are present at moderate concentrations. In this case, neither the reactants [`color{red}(Co(s))` and `color{red}(Ni^(2+)(aq))`] nor the products [`color{red}(Co^(2+)(aq))` and `color{red}(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: `color{red}(Zn > Cu > Ag)`.