$>$ Oxidation
Oxidation is defined as the addition of oxygen/electronegative element to a substance or
removal of hydrogen/ electropositive element from a substance.
For example,
$$
begin{aligned}
2 mathrm{Mg}(s)+mathrm{O}_{2}(g) & longrightarrow & 2 mathrm{MgO}(s) \
2 mathrm{H}_{2} mathrm{~S}(g)+mathrm{O}_{2}(g) & longrightarrow & 2 mathrm{~S}(mathrm{~s})+2 mathrm{H}_{2} mathrm{O}(l) \
mathrm{Mg}(s)+mathrm{Cl}_{2}(g) & longrightarrow & mathrm{MgCl}_{2}(s)
end{aligned}
$$
$>$ Reduction
Reduction is defined as the removal of oxygen/electronegative element from a substance
or addition of hydrogen or electropositive element to a substance.
For example,
$$
2 mathrm{HgO}(s) stackrel{Delta}{longrightarrow} 2 mathrm{Hg}(l)+mathrm{O}_{2}(g)
$$
(Removal of oxygen from mercuric oxide)
$$
2 mathrm{FeCl}_{3}(a q)+mathrm{H}_{2}(g) longrightarrow 2 mathrm{FeCl}_{2}(a q)+2 mathrm{HCl}(a q)
$$
(removal of electronegative element, chlorine from ferricchloride)
$$
mathrm{CH}_{2}=mathrm{CH}_{2}(g)+mathrm{H}_{2}(g) longrightarrow mathrm{C}_{2} mathrm{H}_{6}(g)
$$
(addition of hydrogen)
$>$ Redox Reaction in Terms of Electron Transfer Reaction
A few examples of redox reaction on the basis of electronic concept are given below:
According to electronic concept every redox reaction consists of two steps known as half
reactions.
(i) Oxidation reaction: Half reactions that involve loss of electrons are called oxidation
reactions.
(ii) Reduction reaction: Half reactions that involve gain of electrons are called reduction
reactions.
$$
text { Oxidizing agent: Acceptor of electrons. }
$$
$$
text { Reducing agent: Donor of electrons. }
$$
$>$ Competitive Electron Transfer Reactions
To understand this concept let us do an experiment.
Place a strip of metallic zinc in an aqueous solution of copper nitrate as shown in Fig. After
one hour following changes will be noticed.
(i) Strips becomes coated with reddish metallic copper.
(ii) Blue color of the solution disappears.
(iii) If hydrogen sulphide gas is passed through the solution appearance of white ZnS can
be – seen on making the solution alkaline with ammonia.
$>$ Oxidation Number
It is the oxidation state of an element in a compound which is the charge assigned to an
atom of a compound is equal to the number of electrons in the valence shell of an atom
that are gained or lost completely or to a large extent by that atom while forming a bond
in a compound.
$>$ Rules for Assigning 0xidation Numbers
(i) The oxidation number of an element in its elementary form is zero. For example, $mathrm{H}_{2}, mathrm{O}_{2}, mathrm{~N}_{2}$ etc. have oxidation number equal to zero.
(ii) In a single monoatomic ion, the oxidation number is equal to the charge on the ion.
For example, Na+ ion has oxidation number of $+1$ and $mathrm{Mg}^{2+}$ ion has $+2$.
(iii) Oxygen has oxidation number – 2 in its compounds. However, there are some
exceptions.
Compounds such as peroxides. $mathrm{Na}_{2} mathrm{O}_{2}, mathrm{H}_{2} mathrm{O}_{2}$
oxidation number of oxygen $=-1$ In $mathrm{OF}_{2}$
0.N. of oxygen $=+2 mathrm{O}_{2} mathrm{~F}_{2}$
0.N. of oxygen $=+1$
(iv) In non-metallic compounds of hydrogen like $mathrm{HCl}, mathrm{
H}_{2} mathrm{~S}, mathrm{H}_{2} mathrm{O}$ oxidation number of
hydrogen $=+1$ but in metal hydrides oxidation number of hydrogen $=-1$
[LiH, NaH, CaH $_{2}$ etc.]
(v) In compounds of metals and non-metals metals have positive oxidation number while
non-metals have negative oxidation number. For example, In NaCl. Na has $+1$ oxidation
number while chlorine has $-1$.
(vi) If in a compound there are two non-metallic atoms the atoms with high
electronegativity is assigned negative oxidation number while other atoms have positive
oxidation number.
(vii) The algebraic sum of the oxidation number of all atoms in a compound is equal to
zero.
(viii) In poly atomic ion the sum of the oxidation no. of all the atoms in the ion is equal to
the net charge on the ion.
For example, in $left(mathrm{CO}_{3}right)^{-2}$ Sum of carbon atoms and three oxygen atoms is equal to $-2$.
Fluorine $left(mathrm{F}_{2}right)$ is so highly reactive non-metal that it displaces oxygen from water.
$$
begin{array}{cc}
+1-2 & 0 & +1-1 & 0 \
2 mathrm{H}_{2} mathrm{O}(a q) & +2 mathrm{~F}_{2}(g) longrightarrow & 4 mathrm{HF}(g)+mathrm{O}_{2}(g)
end{array}
$$
Disproportionation Reaction. In a disproportionation reaction an element in one
oxidation state is simultaneously oxidizes and reduced.
For example,
$$
begin{array}{l}
+1-1 \
2 mathrm{H}_{2} mathrm{O}_{2}(a q) longrightarrow 2 mathrm{H}_{2} mathrm{O}(l)+mathrm{O}_{2}(g)
end{array}
$$
Hence, the oxygen of peroxide, which is present in $-1$ oxidation state is connected to zero
oxidation state and in $mathrm{O}_{2}$ and in $mathrm{H}_{2} mathrm{O}$ decreases to $-2$ oxidation state.
$>$ Fractional Oxidation Numbers
Elements as such do not have any fractional oxidation numbers. When the same element
are involved in different bonding in a species, their actual oxidation states are whole
numbers but an average of these is fractional.
For example, In $mathrm{C}_{3} mathrm{O}_{2}$
<smiles>O=C=C=C=O</smiles> The average O.N. of carbon atoms $=frac{(2+2+0)}{3}=frac{4}{3}$.
Fractional 0.N. of a particular element can be calculated only if we know about the
structure of the compound or in which it is present.
$>$ Balancing of Redox Reactions
(i) Oxidation Number Method. Following steps are involved:
(ii) Write the correct formula for each reactant and product.
(b) By assigning the oxidation change in oxidation number can be identified.
(c) Calculate the increase and decrease in oxidation number per atom with respect to the
$=$ reactants. If more than one atom is present then multiply by suitable coefficient.
(d) Balance the equation with respect to all atoms. Balance hydrogen and oxygen atoms
also.
(e) If the reaction is carried out in acidic medium, use $mathrm{H}^{+}$ ions in the equation. If it is in
basic medium use $mathrm{OH}^{-}$ ions.
(f) Hydrogen atoms in the expression can be balanced by adding $left(mathrm{H}_{2} mathrm{O}right)$ molecules to the
reactants or products.
If there are the same number of oxygen atoms on the both side of equation then it
represents the balanced redox reaction.
(ii) Half Reaction Method. In this method two half equation are balanced separately and
then added together to give balanced equation.
Let us consider the oxidation of $mathrm{Fe}^{2+}$ ions to $mathrm{Fe}^{3+}$ ions by dichromate ions $left(mathrm{Cr}_{2} mathrm{O}_{7}^{2-}right)$ ions in acidic medium $left(mathrm{Cr}_{2} mathrm{O}_{7}right)^{2-}$ ions are reduced to $mathrm{Cr}^{3+}$ ions. Following steps are involved:
1. Write the unbalanced equation for the reaction in ionic form.
$$
mathrm{
Fe}^{2+}(a q)+mathrm{Cr}_{2} mathrm{O}_{7}^{2-}(a q) longrightarrow mathrm{Fe}^{3+}(a q)+mathrm{Cr}^{3+}(a q)
$$
2. Separate the equation in two half reactions:
Oxidation half: –
$$
begin{array}{c}
+2 \
mathrm{Fe}^{2+}(a q) & +3^{3+}(a q)
end{array}
$$
Reduction half:
$$
left.mathrm{Cr}_{2} mathrm{O}_{7}^{2-} longrightarrow mathrm{Cr}^{3+} a qright)
$$
3.
$$
mathrm{Cr}_{2} mathrm{O}_{7}^{2-}(a q) longrightarrow 2 mathrm{Cr}^{3+}(a q)
$$
4. Add $mathrm{H}_{2} mathrm{O}$ to balance $mathrm{O}$ atoms and $mathrm{H}^{+}$ to balance $mathrm{H}$ atoms.
$$
mathrm{Cr}_{2} mathrm{O}_{7}^{2-}(a q)+14 mathrm{H}^{+}(a q) longrightarrow 2 mathrm{Cr}^{3+}(a q)+7 mathrm{H}_{2} mathrm{O}(l)
$$
5. Add electrons to one side to balance the charges.
$$
begin{array}{c}
mathrm{Fe}^{2+}(a q) & longrightarrow mathrm{Fe}^{3+}(a q)+e^{-} \
mathrm{Cr}_{2} mathrm{O}_{7}^{2-}(a q)+14 mathrm{H}^{+}(a q)+6 e^{-} & longrightarrow 2 mathrm{Cr}^{3+}(a q)+7 mathrm{H}_{2} mathrm{O}(l)
end{array}
$$
To equalise the number of electrons in both half reactions, we multiply the oxidation half reaction by 6 .
$$
6 mathrm{Fe}^{2+}(a q) longrightarrow 6 mathrm{Fe}^{3+}(a q)+6 e^{-}
$$
6. The net ionic reaction can be written as $6 mathrm{Fe}^{2+}(a q)+mathrm{Cr}_{2} mathrm{O}_{7}^{2-}(a q)+14 mathrm{H}^{+}(a q) longrightarrow 6 mathrm{Fe}^{3+}(a q)+2 mathrm{Cr}^{3+}(a q)+7 mathrm{H}_{2} mathrm{O}(l)$
7. Verify the equation so that same type of number of atoms and the same charges on both sides of the equation.
$>$ Redox Reactions as the Basis for Titration
Potassium Permanganate Titration: In these titrations potassium permanganate (pink in
color) acts as an oxidizing agent in the acidic medium while oxalic acid or some ferrous
salts acts as a reducing agents.
The ionic equation can be written as:
$2 mathrm{MnO}_{4}^{-}+16 mathrm{H}^{+}+5 mathrm{C}_{2} mathrm{O}_{4}^{2-} longrightarrow 2 mathrm{Mn}^{2+}+8 mathrm{H}_{2} mathrm{O}+10 mathrm{CO}_{2}$
$2 mathrm{MnO}_{4}^{-}+16 mathrm{H}^{+}+10 mathrm{Fe}^{2+} longrightarrow 2 mathrm{Mn}^{2+}+2 mathrm{H}_{2} mathrm{O}+10 mathrm{Fe}^{3+}$
These are the examples of redox titration.
On both these titrations, potassium permanganate itself acts as indicator. It is commonly
known as self-indicator. The appearance of pink color in the solution represents the end
points.
Potassium Dichromate Titration: In place of potassium permanganate, potassium
dichromate can also be used in the presence of dil. $mathrm{H}_{2} mathrm{SO}_{4 .}$ The ionic equation for the
redox reaction with FeSO $_{4}left(mathrm{Fe}^{2+}right.$ ions $)$ is given.
$$
mathrm{Cr}_{2} mathrm{O}_{7}^{2-}(a q)+14 mathrm{H}^{+}(a q)+6 mathrm{Fe}^{2+}(a q) longrightarrow 2 mathrm{Cr}^{3+}(a q)+6 mathrm{Fe}^{3+}(a q)+7 mathrm{H}_{2} mathrm{O}(l)
$$
The reaction is used in the estimation of ferrous ions in volumetric analysis. Sodium Thiosulphate Titration: The redox reaction between sodium thiosulphate $left(mathrm{S}_{2} mathrm{O}_{3}^{2-}right.$ ions) and $mathrm{I}_{2}$ are also an example of redox titration.
$$
mathrm{I}_{2}(a q)+2 mathrm{~S}_{2} mathrm{O}_{3}^{2-}(a q) longrightarrow 2 mathrm{I}^{-}(a q)+mathrm{S}_{4} mathrm{O}_{6}^{2-}(a q)
$$
This method based on the fact that iodine itself gives an intense blue colour with starch and has a very specific reaction with thiosulphate $left(mathrm{S}_{2} mathrm{O}_{3}{ }^{2-}right)$ ions.
$>$ Limitation of Concept of Oxidation Number
According to the concept of oxidation number, oxidation means increase in oxidation
number – by loss of electrons and reduction means decrease in oxidation number by the
gain of electrons. However, during oxid
ation there is decrease in electron density while
increase in electron density around the atom undergoing reduction.
$>$ Redox Reactions and Electrode Processes-Electrochemical Cells
A device in which the redox reaction is carried indirectly and the decrease in energy
appears as the electrical energy are called electrochemical cell.
Electrolytic Cell. The cell in which electrical energy is converted into chemical energy.
Example, when lead storage battery is recharged, it acts as electrolytic cell.
Redox Reactions and Electrode Processes. When zinc rod is dipped in copper sulphate
solution redox reaction begins hence, zinc is oxidized to $mathrm{Zn}^{2+}$ ions and $mathrm{Cu}^{2+}$ ions are
reduced to metal.
> Redox reaction: Reactions in which oxidation and reduction occur simultaneously are
called redox reactions.
> Oxidation: Involves loss of one or more electrons.
$>$ Reduction: Involves gain of one or more electrons.
$>$ Oxidizing agent: Accepting electrons.
$>$ Reducing agent: Losing electrons.
$>$ Electrochemical cell: It is a device in which redox reaction is carried indirectly and
decrease in energy gives electrical energy.
$>$ Electrode potential: It is the potential difference between the electrode and its ions in
solution.
$>$ Standard electrode potential: It is the potential of an electrode with respect to standard
hydrogen electrode.
$>$ Electrochemical series: It is activity series. It has been formed by arranging the metals in
order of increasing standard reduction potential value.