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Chemical reactions involve transfer of electrons from one
chemical substance to another. These electron – transfer
reactions are termed as oxidation-reduction or redox-
reactions.
Molecular and Ionic equations
(1) Molecular equations : When the reactants and
products involved in a chemical change are written in molecular
forms in the chemical equation, it is termed as molecular
equation.
Example :
2 2 2 2
MnO 4 HCl MnCl 2 HO Cl
In above example the reactants and products have been
written in molecular forms, thus the equation is termed as
molecular equation****.
(2) Ionic equations : When the reactants and products
involved in a chemical change are ionic compounds, these will
be present in the form of ions in the solution. The chemical
change is written in ionic forms in chemical equation, it is
termed as ionic equation. Example ,
MnO 4 H 4 Cl
2
2 2
2
Mn 2 Cl 2 HO Cl
In above example the reactants and products have been
written in ionic forms, thus the equation is termed as ionic
equation****.
(3) Spectator ions : In ionic equations, the ions which
do not undergo any change and equal in number in both
reactants and products are termed as spectator ions and are not
included in the final balanced equations. Example,
Zn 2 H 2 Cl
Zn H 2 Cl
2
2
( Ionic
equation )
Zn 2 H
2
2
Zn H
( Final ionic
equation )
In above example, the
Cl ions are the spectator ions
and hence are not included in the final ionic balanced equation.
Oxidation-reduction and Redox reactions
(1) Oxidation : Oxidation is a process which involves;
addition of oxygen, removal of hydrogen, addition of non-metal,
removal of metal, Increase in + ve valency, loss of electrons and
increase in oxidation number.
(i) Addition of oxygen : 2 Mg + O 2
2 MgO
(ii) Removal of hydrogen : H 2 S+Cl 2 2 HCl + S
(iii) Addition of Non-metal : Fe + S FeS
(iv) Removal of metal : 2 KI+H 2 O 2 2 KOH+I 2
(v) Increase in +ve valency :
Fe Fe e
2 3
(vi) Loss of electrons (also known as de-electronation)
4
M
3
M
2
M
1
M
0
M
1
M
2
M
3
M
4
M
Loss of electrons
(a)
H H e
0
( Formation of proton )
(b)
MnO MnO e
4
2
4
( De-electronation of
2
4
MnO )
(c)
2 Fe 2 Fe 6 e
0 3
( De-electronation of iron )
(vii) Increase in oxidation number
(a)
0 2
Mg Mg ( From 0 to +2)
(b)
3
6
3
4
6
2
Fe ( CN ) Fe ( CN ) (From +2 to +3)
Redox Reactions
Chapter
(c)
0
2
2 Cl Cl
(From – 1 to 0)
(2) Reduction : Reduction is just reverse of oxidation.
Reduction is a process which involves; removal of oxygen,
addition of hydrogen, removal of non-metal, addition of metal,
decrease in + ve valency, gain of electrons and decrease in
oxidation number.
(i) Removal of oxygen : CuO C Cu CO
(ii) Addition of hydrogen : Cl H 2 HCl
2 2
(iii) Removal of non-metal
2 2 2 2 4
2 HgCl SnCl HgCl SnCl
(iv) Addition of metal :
2 2 2
HgCl Hg HgCl
(v) Decrease in +ve valency
(a)
3 2
Fe Fe (+ ve valency decreases )
(b)
4
6
3
6
[ Fe ( CN )] [ Fe ( CN )] (– ve valency increase s)
(vi) Gain of electrons (also known as electronation)
4
M
3
M
2
M
1
M
0
M
1
M
2
M
3
M
4
M
+e
+e
+e
+e
+e
+e
+e
+e
Gain of electrons
(a) ( ) 2 ()
2
Zn aq e ZnS
( Electronation of
2
Zn )
(b)
2 0
Pb 2 e Pb
( Electronation of
2
Pb )
(c)
4
6
3
6
[ Fe ( CN )] e [ Fe ( CN )]
( Electronation of
3
6
[ Fe ( CN )] )
(vii) Decrease in oxidation number
(a)
2 0
Mg Mg
( From +2 to 0)
(b)
4
6
3
6
Fe ( CN ) Fe ( CN ) ( From +3 to +2)
(c)
Cl 2 Cl
0
2
( From 0 to – 1)
(3) Redox-reactions
(i) An overall reaction in which oxidation and reduction
takes place simultaneously is called redox or oxidation-
reduction reaction****. These reactions involve transfer of
electrons from one atom to another. Thus every redox reaction
is made up of two half reactions ; One half reaction represents
the oxidation and the other half reaction represents the
reduction.
(ii) Types of redox reaction
(a) Direct redox reaction : The reactions in which
oxidation and reduction takes place in the same vessel are called
direct redox reactions.
(b) Indirect redox reaction : The reactions in which
oxidation and reduction takes place in different vessels are
called indirect redox reactions. Indirect redox reactions are the
basis of electro-chemical cells.
(c) Intermolecular redox reactions : In which one
substance is oxidised while the other is reduced.
For example, 2 Al FeO AlO 2 Fe
2 3 2 3
Here, Al is oxidised to
2 3
AlO while
2 3
FeO is reduced to
Fe.
(d) Intramolecular redox reactions : In which one
element of a compound is oxidised while the other is reduced.
For example,
3 2
2 KClO 2 KCl 3 O
Here,
5
Cl in
3
KClO
is reduced to
1
Cl in KCl while
2
O in
3
KClO
is oxidised to
0
2
O.
Oxidising and Reducing agents
(1) Definition : The substance (atom, ion or molecule)
that gains electrons and is thereby reduced to a low valency state
is called an oxidising agent , while the substance that loses
electrons and is thereby oxidised to a higher valency state is
called a reducing agent****.
Or
An oxidising agent is a substance, the oxidation
number of whose atom or atoms decreases while a reducing
agent is a substance the oxidation number of whose atom
increases.
(2) Important oxidising agents
(i) Molecules made up of electronegative elements.
Example : O 2 , O 3 and X 2 (halogens).
(ii) Compounds containing an element which is in the
highest oxidation state.
Example : , ,
4 2 2 7
KMnO KCrO
2 2 7
NaCrO, ,
3 , 2 4
CrO HSO
3 3 3 2 4 3 2 2 2
HNO , NaNO , FeCl , HgCl , KClO , SO , CO , HO etc.
(iii) Oxides of elements, , , , , ,
3 2 4 10
MgOCuOCrO CO PO etc.
(iv) Fluorine is the strongest oxidising agent.
(3) Important reducing agents
(i) All metals e.g. Na, Zn, Fe, Al, etc.
(ii) A few non-metals e.g. C, H 2
, S etc.
(iii) Hydracids : HCl, HBr, HI, H 2 S etc.
(iv) A few compounds containing anelement in the lower
oxidation state (ous).
Example : , ,
2 4
FeCl FeSO SnCl HgCl CuO
2 2 2 2
, , etc.
(v) Metallic hydrides e.g. NaH, LiH etc.
(vi) Organic compounds like HCOOH and ( COOH ) 2
and
their salts, aldehydes, alkanes etc.
(vii) Lithium is the strongest reducing agent in solution.
(viii) Cesium is the strongest reducing agent in absence of
water. Other reducing agents are
2 2 3
NaSO and KI.
(ix) Hypo prefix indicates that central atom of compound
has the minimum oxidation state so it will act as a reducing
agent.
Example :
3 2
HPO (hypophosphorous acid).
(4) Substances which act as oxidising as well as
reducing agents
Examples : H 2 O 2 , SO 2 , H 2 SO 3 , HNO 2 , NaNO 2 , Na 2 SO 3 , O 3
etc_._
(5) Tips for the identification of oxidising and
reducing agents
(i) If an element is in its highest possible oxidation state
in a compound, the compound can function as an oxidising
agent.
Example :
4 2 2 7 3 2 4 4
KMnO , KCrO , HNO , HSO , HClO etc.
V 2 O 5 Vanadium (V)
oxide
CuO Copper (II) oxide
SnO 2 Tin (IV) oxide FeCl 3 Iron (III) chloride
(4) Rules for the determination of oxidation
number of an atom : The following rules are followed in
ascertaining the oxidation number of an atom,
(i) If there is a covalent bond between two same atoms
then oxidation numbers of these two atoms will be zero. Bonded
electrons are symmetrically distributed between two atoms.
Bonded atoms do not acquire any charge. So oxidation numbers
of these two atoms are zero.
A : A or A – A A
For example, Oxidation number of Cl in Cl 2
, O in O 2
and
N in N 2
is zero.
(ii) If covalent bond is between two different atoms then
electrons are counted towards more electronegative atom. Thus
oxidation number of more electronegative atom is negative and
oxidation number of less electronegative atom is positive. Total
number of charges on any element depends on number of
bonds.
A – B
A
:
A – B A
:
The oxidation number of less electronegative element (A)
is + 1 and + 2 respectively.
(iii) If there is a coordinate bond between two atoms then
oxidation number of donor atom will be + 2 and of acceptor
atom will be – 2.
A B
A
2+
:
(iv) The oxidation number of all the atoms of different
elements in their respective elementary states is taken to be
zero. For example, in N , Cl , H , P , S , O , Br , Na , Fe , Ag 2 2 2 4 8 2 2
etc. the oxidation number of each atom is zero.
(v) The oxidation number of a monoatomic ion is the
same as the charge on it. For example, oxidation numbers of
2
Na , Mg and
3
Al ions are + 1, + 2 and + 3 respectively while
those of
2
Cl , S and
3
N ions are – 1, – 2 and – 3 respectively.
(vi) The oxidation number of hydrogen is + 1 when
combined with non-metals and is – 1 when combined with active
metals called metal hydrides such as LiH, KH, MgH 2 , CaH 2 etc.
(vii) The oxidation number of oxygen is – 2 in most of its
compounds, except in peroxides like
2 2 2
HO , BaO etc. where it
is – 1. Another interesting exception is found in the compound
OF
2
( oxygen difluoride ) where the oxidation number of oxygen
is + 2. This is due to the fact that fluorine being the most
electronegative element known has always an oxidation number
of – 1.
(viii) In compounds formed by union of metals with non-
metals, the metal atoms will have positive oxidation numbers
and the non-metals will have negative oxidation numbers.
For example,
(a) The oxidation number of alkali metals ( Li, Na, K etc.)
is always +1 and those of alkaline earth metals ( Be, Mg, Ca etc) is +
(b) The oxidation number of halogens ( F, Cl, Br, I ) is
always – 1 in metal halides such as KF, AlCl 3 , MgBr 2 , CdI 2. etc.
(ix) In compounds formed by the union of different
elements, the more electronegative atom will have negative
oxidation number whereas the less electronegative atom will have
positive oxidation number.
For example,
(a) N is given an oxidation number of – 3 when it is
bonded to less electronegative atom as in NH 3
and NI 3
, but is
given an oxidation number of + 3 when it is bonded to more
electronegative atoms as in NCl 3.
(b) Since fluorine is the most electronegative element
known so its oxidation number is always – 1 in its compounds
i.e. oxides, interhalogen compounds etc.
(c) In interhalogen compounds of Cl, Br, and I; the more
electronegative of the two halogens gets the oxidation number
of – 1. For example, in BrCl 3 , the oxidation number of Cl is – 1
while that of Br is +3.
(x) For neutral molecule, the sum of the oxidation
numbers of all the atoms is equal to zero. For example, in NH 3
the sum of the oxidation numbers of nitrogen atom and 3
hydrogen atoms is equal to zero. For a complex ion, the sum of
the oxidation numbers of all the atoms is equal to charge on the
ion. For example, in
2
4
SO ion, the sum of the oxidation
numbers of sulphur atom and 4 oxygen atoms must be equal to
(xi) It may be noted that oxidation number is also
frequently called as oxidation state. For example, in H 2
O , the
oxidation state of hydrogen is +1 and the oxidation state of
oxygen is – 2. This means that oxidation number gives the
oxidation state of an element in a compound.
(xii) In the case of representative elements, the highest
oxidation number of an element is the same as its group number
while highest negative oxidation number is equal to (8 – Group
number) with negative sign with a few exceptions. The most
common oxidation states of the representative elements are
shown in the following table,
Group Outer shell
configuration
Common oxidation numbers
(states)
except zero in free state
I A ns
1
II A ns
2
III A ns
2
np
1
+3, +
IV A ns
2
np
2
+4,+3,+2,+1, – 1, – 2, – 3, – 4
V A ns
2
np
3
+5,+3,+1, – 1, – 3
VI A ns
2
np
4
+6,+4,+2,– 2
VII A ns
2
np
5
+7,+5,+3, +1, – 1
(xiii) Transition metals exhibit a large number of
oxidation states due to involvement of ( n – 1 ) d electron
besides ns electron.
(xiv) Oxidation number of a metal in carbonyl complex is
always zero.
Example : Ni has zero oxidation state in
4
NiCO.
(xv) Those compounds which have only C , H and O the
oxidation number of carbon can be calculated by following
formula,
C
O H
n
n n
C
Oxidation numberof' '
Where,
O
n is the number of oxygen atom,
H
n is the number
of hydrogen atom,
C
n is the number of carbon atom.
For example, (a)
3
CHOH ; 4 , 1 , 1
H C O
n n n
Oxidation number of ‘ C ’ = 2
1
( 1 2 4 )
(b)
HCOOH ; 2 ,
H
n 2 , 1
O c
n n
Oxidation number of carbon = 2
1
( 2 2 2 )
(5) Procedure for calculation of oxidation number
: By applying the above rules, we can calculate the oxidation
numbers of elements in the molecules/ions by the following
steps.
(i) Write down the formula of the given molecule/ion
leaving some space between the atoms.
(ii) Write oxidation number on the top of each atom. In
case of the atom whose oxidation number has to be calculated
write x.
(iii) Beneath the formula, write down the total oxidation
numbers of each element. For this purpose, multiply the
oxidation numbers of each atom with the number of atoms of
that kind in the molecule/ion. Write the product in a bracket.
(iv) Equate the sum of the oxidation numbers to zero for
neutral molecule and equal to charge on the ion.
(v) Solve for the value of x.
Table : 13.2 Oxidation number of some elements in compounds, ions or chemical species
Element Oxidation
Number
Compounds, ions or chemical species
Sulphur (S) – 2 H 2 S , Zn S , NaH S , ( Sn S 3 )
2 –
, Ba S , C S 2
0 S , S 4 , S 8 , S CN
-
+ 4 S O
2
, H
2
S O
3
, ( S O
3
2 –
, S OCl 2
, NaH S O 3
, Ca [ H S O 3
]
2
, [ H S O
3
]
,
S F
4
+ 6 H
2
S O
4
, ( S O
4
2 -
, [ H S O 4
]
,
Ba S O 4,
KH S O
4,
S O
3
, S F
6
, H
2
S
2
O
7
, ( S
2
O
7
2 –
Nitrogen ( N ) – 3 N H 3 , ( N H 4 )
, Al N, Mg 3 N 2 , ( N )
3 –
, Ca 3 N 2 , C N
-
– 2 N 2 H 4 , ( N 2 H 5 )
– 1 N H 2 OH
, N
3
H
0 N
2
+ 1 N 2 O
+ 2 N O
+ 3 H N O
2
, ( N O
2
, NaN O 2
, N
2
O
3
, N F
3
+ 4 N O
2
+ 5 H N O 3 , ( N O 3 )
, K N O 3 , N 2 O 5
Chlorine ( Cl ) – 1 H Cl , Na Cl , Ca Cl 2 , Al Cl 3 , I Cl , I Cl 5 , SO Cl 2 , CrO 2 Cl 2 , K Cl , K 2 Pt Cl 6 , HAu Cl 4 , C Cl 4
0 Cl, Cl 2
- 1 HO Cl , NaO Cl , ( O Cl )
, Cl 2 O
, ( Cl O 2
, H Cl O 2
, K Cl O 3
, Na Cl O 3
, H Cl O 3
- 7 H Cl O 4 , Cl 2 O 7 , K Cl O 4 , ( Cl O 4 )
Xenon ( Xe ) + 6 Xe O 3
, Xe F 6
(6) Exceptional cases of evaluation of oxidation
numbers : The rules described earlier are usually helpful in
determination of the oxidation number of a specific atom in
simple molecules but these rules fail in following cases. In these
cases, the oxidation numbers are evaluated using the concepts of
chemical bonding involved.
Type I****. In molecules containing peroxide linkage in
addition to element-oxygen bonds. For example,
(i) Oxidation number of S in H 2
SO
5
(Permonosulphuric acid or Caro's acid)
By usual method;
2 5
HSO
2 1 x 5 ( 2 ) 0 or x 8
But this cannot be true as maximum oxidation number
for S cannot exceed + 6. Since S has only 6 electrons in its
valence shell. This exceptional value is due to the fact that two
oxygen atoms in
2 5
HSO
shows peroxide linkage as shown
below,
O
H O S O O H
O
Therefore the evaluation of o.n. of sulphur here should be
made as follows,
2 × (+1) + x + 3 × (–2) + 2 × (–1)
(for H) (for S) (for O) (for O–O)
or 2 + x – 6 – 2 = 0 or x = + 6.
(ii) Oxidation number of S in H 2 S 2 O 8
(Peroxidisulphuric acid or Marshall's acid)
By usual method ;
2 2 8
HSO
1 × 2 + 2x + 8 (–2) = 0
2x = + 16 – 2 = 14 or x = + 7
Similarly Caro's acid, Marshall's acid also has a peroxide
linkage so that in which S shows +6 oxidation state.
O O
H O S O O S O H
O O
Therefore the evaluation of oxidation state of sulphur
should be made as follow,
2 × (+1) + 2 × (x) + 6 × (–2) + 2 × (–1) = 0
(for H) (for S) (for O) (for O–O)
or 2 + 2x – 12 – 2 = 0 or x = + 6.
(iii) Oxidation number of Cr in CrO 5
(Blue perchromate)
By usual method
5
CrO ; x – 10 = 0 or x = + 10
This cannot be true as maximum O. N. of Cr cannot be
more than + 6. Since Cr has only five electrons in 3d orbitals
and one electron in 4s orbital. This exceptional value is due to
the fact that four oxygen atoms in CrO 5 are in peroxide linkage.
The chemical structure of CrO 5
is
O
O O
Cr
O O
Therefore, the evaluation of o.n. of Cr should be made as
follows
x + 1 × (– 2) + 4 (–1) = 0
(for Cr ) (for O ) (for O–O )
or x – 2 – 4 = 0 or x = + 6.
Type II****. In molecules containing covalent and
coordinate bonds, following rules are used for evaluating the
oxidation numbers of atoms.
(i) For each covalent bond between dissimilar atoms the
less electronegative element is assigned the oxidation number of
- 1 while the atom of the more electronegative element is
assigned the oxidation number of – 1.
(ii) In case of a coordinate-covalent bond between similar
or dissimilar atoms but the donor atom is less electronegative
than the acceptor atom, an oxidation number of +2 is assigned
to the donor atom and an oxidation number of – 2 is assigned to
the acceptor atom.
Conversely, if the donor atom is more electronegative
than the acceptor atom, the contribution of the coordinate bond
is neglected. Examples,
(a) Oxidation number of C in HC N and HN
C
The evaluation of oxidation number of C cannot be made
directly by usual rules since no standard rule exists for oxidation
numbers of N and C.
In such cases, evaluation of oxidation number should be
made using indirect concept or by the original concepts of
chemical bonding.
(b) Oxidation number of carbon in H – N
C
The contribution of coordinate bond is neglected since the
bond is directed from a more electronegative N atom (donor) to
a less electronegative carbon atom (acceptor).
Therefore the oxidation number of N in HN C
remains
- 3 as it has three covalent bonds.
1 × (+ 1) + 1 × (– 3) + x = 0
(for H ) (for N ) (for C )
or 1 + x – 3 = 0 or x = + 2.
(c) Oxidation number of carbon in HC N
In HC N , N is more electronegative than carbon,
each bond gives an oxidation number of – 1 to N. There are
three covalent bonds, the oxidation number of N in
HC N is taken as – 3
Now HC N
+1 + x – 3 = 0 x = + 2
Peroxide linkage
Peroxide linkage
Peroxide linkage
Peroxide linkage
Type III. In a molecule containing two or more atoms
of same or different elements in different oxidation states.
(i) Oxidation number of S in Na 2 S 2 O 3
By usual method
2 2 3
NaSO
2 × (+1) + 2 × x + 3 (–2) = 0 or 2 + 2 x – 6 = 0
or x = 2.
But this is unacceptable as the two sulphur atoms in
Na 2
S
2
O
3
cannot have the same oxidation number because on
treatment with dil. H 2 SO 4 , one sulphur atom is precipitated
while the other is oxidised to SO 2.
Na SO HSO NaSO SO S HO
2 2 3 2 4 2 4 2 2
In this case, the oxidation number of sulphur is evaluated
from concepts of chemical bonding. The chemical structure of
Na 2 S 2 O 3 is
O
Na O S O Na
S
Due to the presence of a co-ordinate bond between two
sulphur atoms, the acceptor sulphur atom has oxidation number
of – 2 whereas the other S atom gets oxidation number of + 2.
2 × (+1) + 3 × (–2) + x × 1 + 1 × (– 2) = 0
(for Na ) (for O ) (for S ) (for coordinated S )
or + 2 – 6 + x – 2 = 0 or x = + 6
Thus two sulphur atoms in Na 2
S
2
O
3
have oxidation
number of – 2 and +6.
(ii) Oxidation number of chlorine in CaOCl 2
(bleaching powder)
In bleaching powder, Ca ( OCl ) Cl , the two Cl atoms
are in different oxidation states i.e. , one Cl
-
having
oxidation number of – 1 and the other as OCl
having
oxidation number of +1.
(iii) Oxidation number of N in NH 4 NO 3
By usual method N 2 H 4 O 3 ; 2 x + 4 × (+1) + 3 × (–1) = 0
2 x + 4 – 3 = 0 or 2 x = + 1 (wrong)
No doubt NH 4
NO
3
has two nitrogen atoms but one N has
negative oxidation number (attached to H ) and the other has
positive oxidation number (attached to O ). Hence the evaluation
should be made separately for
4
NH and
3
NO
4
NH x + 4 × (+1) = +1 or x = – 3
3
NO x + 3 (– 2) = – 1 or x = + 5.
(iv) Oxidation number of Fe in Fe 3
O
4
In Fe 3 O 4 , Fe atoms are in two different oxidation states.
Fe 3
O
4
can be considered as an equimolar mixture of FeO [iron
(II) oxide] and Fe 2 O 3 [iron (III) oxide]. Thus in one molecule of
Fe 3
O
4
, two Fe atoms are in + 3 oxidation state and one Fe atom
is in + 2 oxidation state.
(v) Oxidation number of S in sodium
tetrathionate ( Na 2
S
4
O
6
Its structure can be represented as follows,
O O
Na O S S S S ONa
O O
The two S - atoms which are linked to each other have
oxidation number zero. The oxidation number of other S - atoms
can be calculated as follows
Let oxidation number of S = x.
2 × x + 2 × 0 + 6 × ( – 2) = – 2
(for S ) (for S–S ) (for O )
x = + 5.
Balancing of oxidation-reduction reactions
Though there are a number of methods for balancing
oxidation – reduction reactions, two methods are very
important. These are,
(1) Oxidation number method
(2) Ion – electron method
(1) Oxidation number method : The method for
balancing redox reactions by oxidation number change method
was developed by Johnson****. In a balanced redox reaction, total
increase in oxidation number must be equal to the total decrease
in oxidation number. This equivalence provides the basis for
balancing redox reactions. This method is applicable to both
molecular and ionic equations. The general procedure involves
the following steps,
(i) Write the skeleton equation (if not given, frame it)
representing the chemical change.
(ii) Assign oxidation numbers to the atoms in the
equation and find out which atoms are undergoing oxidation
and reduction. Write separate equations for the atoms
undergoing oxidation and reduction.
(iii) Find the change in oxidation number in each
equation. Make the change equal in both the equations by
multiplying with suitable integers. Add both the equations.
(iv) Complete the balancing by inspection. First balance
those substances which have undergone change in oxidation
number and then other atoms except hydrogen and oxygen.
Finally balance hydrogen and oxygen by putting H 2 O molecules
wherever needed.
The final balanced equation should be checked to ensure
that there are as many atoms of each element on the right as
there are on the left.
(v) In ionic equations the net charges on both sides of the
equation must be exactly the same. Use H
ion/ions in acidic
reactions and OH
ion/ions in basic reactions to balance the
charge and number of hydrogen and oxygen atoms.
The following example illustrate the above rules,
Step : I Cu HNO CuNO NO HO
3 32 2 2
( Skeleton equation )
to lower state of oxidation. Such a reaction, in which a substance
undergoes simultaneous oxidation and reduction is called
disproportionation and the substance is said to
disproportionate****.
Following are the some examples of disproportionation ,
2
0
2
2
1
2 2 2
H O HO HO O
1
4
7
3
5
KClO KClO KCl
3
3
2
1
2 2
0
P NaOH HO NaH PO PH
(4) Cl NaOH NaCl NaClO HO
hot
conc
2
5
3
1
( .)
0
2
If an element is in its highest possible oxidation state in a
compound, it can act as an oxidising agent. for example,
KMnO 4 , K 2 Cr 2 O 7 , HNO 3 , H 2 SO 4 , HClO 4 etc.
If an element is in its lowest oxidation state in a compound,
it can act as a reducing agent. For example, H 2 S , H 2 C 2 O 4 ,
FeSO 4
, Na 2
S
2
O
3
, SO
2
, SnCl 2
, many metals etc.
The strength of oxyacids of chlorine decrease in the
order. HClO 4 > HClO 3 > HClO 2 > HClO
If highly electronegative element is in its highest
oxidation state in a compound that compound can act as
powerful oxidant. For example, KClO 4 , KClO 3 , KBrO 3 ,
KIO
3
etc.
If an element is in intermediate oxidation state in a
compound, it can act as both oxidising & reducing agent.
For example, H 2 O 2 , H 2 SO 3 , HNO 3 , SO 2 etc.
Oxidation, Reduction
2 2
HO reduces
4
MnO ion to [KCET (Med.) 2000]
(a)
Mn (b)
2
Mn
(c)
3
Mn (d)
Mn
2. When a sulphur atom becomes a sulphide ion
[AMU 1999]
(a) There is no change in the composition of atom
(b) It gains two electrons
(c) The mass number changes
(d) None of these
3. The ultimate products of oxidation of most of hydrogen
and carbon in food stuffs are [DCE 2001]
(a) HO
2
alone (b)
2
CO alone
(c) HO
2
and
2
CO (d) None of these
4. When P reacts with caustic soda, the products are
3
PH
and.
2 2
NaHPO This reaction is an example of
[IIT 1980; Kurukshetra CEE 1993; CPMT 1997]
(a) Oxidation
(b) Reduction
(c) Oxidation and reduction (Redox)
(d) Neutralization
5. Which one of the following does not get oxidised by
bromine water [MP PET/PMT 1988]
(a)
2
Fe to
3
Fe (b)
Cu to
2
Cu
(c)
2
Mn to
4
MnO (d)
2
Sn to
4
Sn
6. In the reaction.
2 2 2
H S NO HO NO S HS
2
is
(a) Oxidised (b) Reduced
(c) Precipitated (d) None of these
7. The conversion of
2
PbO to
3 2
Pb ( NO )is
(a) Oxidation
(b) Reduction
(c) Neither oxidation nor reduction
(d) Both oxidation and reduction
8. In the course of a chemical reaction an oxidant
[MP PMT 1986]
(a) Loses electrons
(b) Gains electrons
(c) Both loses and gains electron
(d) Electron change takes place
2
2 CuI Cu CuI , the reaction is [RPMT 1997]
(a) Redox (b) Neutralisation
(c) Oxidation (d) Reduction
decrease
increase
decrease
increase
decrease
increase
decrease
increase
