Oxidation and Reduction

INTRODUCTION

Redox reactions are the chemical reactions in which reduction and oxidation occurs simultaneously. In straight words addition of oxygen is oxidation and removal of hydrogen is reduction.

ELECTRONIC CONCEPT OF OXIDATION AND REDUCTION

Oxidation is a process in which an atom or an ion loses one or more electrons

$Cu\to C{u}^{+}+e^{-};Ca\to C{a}^{+2}+2e^{-}$

Reduction is a process in which an atom or an ion gains one or more electrons

$Z{n}^{+2}+2e^{-}\to Zn;C{d}^{+4}+2e^{-}\to C{d}^{+2}$

${\left[{\mathsf{Fe}(\mathsf{CN})}_6^{}\right]}^{3-}+e^{-}\to {\left[Fe{\left(CN\right)}_6\right]}^{4-}$

REDOX REACTION

Redox reactions may be regarded as electron transfer reactions in which the electrons are transferred from one reactant to the other. As the result substance which losses electrons is called a reducing agent or reducctant while another which accepts the electrons is called an oxidizaing agent or oxidant.

REDOX REACTIONS IN AQUEOUS SOLUTION

In aqueous solutions, the spontaneous redox reactions can be carried out directly as well as indirectly.

Direct Redox Reactions: Redox reactions in which oxidation and reduction takes place in the same vessel are called direct redox reactions. In such reactions, the transfer of electrons from reductant to oxidant occurs over a very short distance (within molecular diameters). For example, if a zinc rod is placed in a solution of copper sulphate in a beaker, a spontaneous reaction occurs and following changes will be observed.

Observations for redox reaction occurring in a beaker

$\bullet$ Zinc rod starts dissolving and loses its mass gradually.

$\bullet$The blue color of the solution starts fading.

$\bullet$ Copper metal either starts settling at the bottom of the beaker or depositing on the zinc rod.

$\bullet$The reaction is exothermic i.e., it takes place with the evolution of heat.

$\bullet$ The solution remains electrically neutral throughout.

$\bullet$ The reaction will not continue indefinitely but stops after some time.

The overall reaction taking place in the beaker may be represented as:

Discarding the common SO₄^2– ions

$\text{Zn(s)+ C}{\text{u}}^{\text{2+}}\text{(aq) }\to \text{ Z}{\text{n}}^{\text{2+}}\text{(aq) + Cu(s) }................\text{(i)}$

Zinc loses electrons and changes to Zn²⁺ ions which go into the solution. As a result, the mass of zinc rod decreases. The electrons lost by zinc rod are gained by Cu²⁺_(aq) ions and they change into Cu_(s) atoms which settle down at the bottom of beaker in the form of precipitate. It may be noted that in the direct redox reaction, the chemical energy appears as heat.

Similarly, when we dip a copper strip in a silver nitrate solution, copper gets oxidised and go into the solution whereas Ag+ ions accept electrons and get reduced.

The reaction may be written as:

$\text{Cu(s) + 2A}{\text{g}}^{\text{+}}\text{(aq) }\to \text{ 2Ag(s) + C}{\text{u}}^{\text{2+}}\text{(aq) }...\text{(ii)}$

Thus, copper is oxidised to Cu²⁺ and Ag⁺ is reduced to Ag_(s). Therefore, copper acts as reducing agent or reductant while silver acts as oxidising agent or oxidant in equation (ii) and copper acts as an oxidising agent in equation (i). The reason being that electron-donating ability of zinc is more than that of Cu while electron donating ability of Cu is more than that of Ag.

Oxidation and reduction half reactions

Every redox reaction can be split up into two half – reactions one representing loss of electron i.e. oxidation half reaction while the other representing gain of electrons i.e. reduction half reaction like

$\bullet$ $\mathsf{Zn}+C{u}^{2+}\to Z{n}^{2+}+Cu$

$\mathsf{Zn}\to Z{n}^{2+}+2e^{-}(oxidationhalfreaction)$

$\mathsf{C}{\mathsf{u}}^{+2}+2e^{-}\to Cu(\mathrm{Re}ductionhalfreaction)$

$\bullet$ $S{n}^{2+}+2H{g}^{2+}\to S{n}_{}^{4+}+H{g}_2^{2+}$can spilt up into half reactions, as

$S{n}^{2+}\to S{n}^{4+}+2e^{-}(oxidationhalfreaction)$

${2Hg}_{}^{2+}+2e^{-}\to H{g}_2^{2+}+(\mathrm{Re}ductionhalfreaction)$

OXIDATION NUMBER OR OXIDATION STATE

Oxidation no. of an element may be defined as the charge which an atom of the element has in its ion or appears to have when present in the combined states with other atoms

Rules for assigning oxidation number

$\bullet$ The O.N of the element in the free or elementary state is always zero irrespective of its allotropic form

Oxidation number of helium is He = 0

Oxidation number of chlorine is $C{l}_2=0$

Oxidation number of sulphur in $S_8$= 0

Oxidation number of phosphorus in $P_4=0$

$\bullet$ The O. N. of the element in mono atomic ion is equal to the charge on the ion. For example, in $K^{+}C{l}^{-}$, the O.N. of K is +1 while that Cl is –1

$\bullet$ O.N. of all alkali metals is +1 while those of alkaline earth metals is +2 in all their compounds

$\bullet$ The O.N. of fluorine is always –1 in all its compounds. (Note that fluorine is most electronegative element, hence cannot attain positive oxidation state)

$\bullet$ Hydrogen is assigned +1 O.N. when it is attached to more electronegative element but it has –1 O.N. in metal hydrides like$NaH,Mg{H}_2,Ca{H}_2,LiHetc.$

$\bullet$ Oxygen is assigned O.N –2 in most of its compounds.

Exceptions: F₂O, hence O is in +2 oxidation state.

In super oxides like$K{O}_{2}ithas-\frac{1}{2}O.N.$

In peroxides like $H_2{O}_2,Ba{O}_2,N{a}_2{O}_{2}etc.$ O.N is –1.

In ozonide the oxidation number of oxygen is assigned -1/3.

$\bullet$ In accordance with principle of conservation of charge, the algebric sum of the oxidation number of all the atoms in molecule is zero. But in case of polyatomic ion the sum of O.N. of all its atoms is equal to charge on the ion.

$\bullet$ Maximum O.N. of an element can be equal to its group number. (Except O and F)

$\bullet$ Minimum O.N. of an element can be equal to (8-n) n is the group no. of the element (Except metals)

$\bullet$ The evaluation of O.N cannot be made directly in some cases e.g. HCN by using the above rule since we have no rule for oxidation No. of both N and C.

In all such cases evaluation of O.N should be made be taking contribution of covalent bonds and coordinate bonds

(a) When two atoms are attached with the help of single covalent bond then its contribution for less electronegative atom is +1 and for more electronegative atom

is –1.

(b) When covalent bond is present between the two same atoms then its contribution will be zero for both the atoms

(c) In case of Coordinate bond, it gives +2 value of oxidation number to less electronegative atom and -2 values to more electronegative atom when coordinate bond is directed from less electronegative atom to more electronegative atom.

(d) If coordinate bond is directed from more electronegative to less electronegative atom then its contribution will be zero for both the atoms.

Illustration 1: Determine the oxidation state of sulphur in Na₂S₄O₆

Solution: Oxidation number of Na= +1

Oxidation number of O = -2

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

$a=+\frac{5}{2}$

The value is the average oxidation number of S. The structure of Na₂S₄O₆ is

Thus oxidation number of each S atom forming coordinate bond is +5 whereas oxidation number of each S atom involved in pure covalent bond is zero.

Averege oxidation number = $\frac{+5+5+0+0}{4}=+\frac{5}{2}$