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Mole Concept-2Oxidation & Reduction
Let us do a comparative study of oxidation and reduction :
Oxidation Reduction
1. Addition of Oxygen 1. Removal of Oxygen
e.g. 2Mg + O2 2MgO e.g. CuO + C Cu + CO
2. Removal of Hydrogen 2. Addition of Hydrogen
e.g. H2S + Cl
2 2HCl + S e.g. S + H
2 H
2S
3. Increase in positive charge 3. Decrease in positive charge
e.g. Fe2+ Fe3+ + e– e.g. Fe3+ + e– Fe2+
4. Increase in oxidation number 4. Decrease in oxidation number
(+2) (+4) (+7) (+2)
e.g. SnCl2 SnCl
4e.g. MnO
4– Mn2+
5. Removal of electron 5. Addition of electron
e.g. Sn2+ Sn4+ + 2e– e.g. Fe3+ + e– Fe2+
Oxidation Number
It is an imaginary or apparent charge developed over atom of an element when it goes from its
elemental free state to combined state in molecules.
It is calculated on basis of an arbitrary set of rules.
It is a relative charge in a particular bonded state.
In order to keep track of electron-shifts in chemical reactions involving formation of compounds, a
more practical method of using oxidation number has been developed.
In this method, it is always assumed that there is a complete transfer of electron from a less
electronegative atom to a more electronegative atom.
Rules governing oxidation numberThe following rules are helpful in calculating oxidation number of the elements in their different
compounds. It is to be remembered that the basis of these rule is the electronegativity of the
element .
Fluorine atom :
Fluorine is most electronegative atom (known). It always has oxidation number equal to –1 in all its
compounds
Oxygen atom :
In general and as well as in its oxides , oxygen atom has oxidation number equal to –2.
In case of (i) peroxide (e.g. H2O
2,, Na
2O
2) is –1,
(ii) super oxide (e.g. KO2) is –1/2
(iii) ozonide (e.g. KO3) is –1/3
(iv) in OF2
is + 2 & in O2F
2is +1
Hydrogen atom :
In general, H atom has oxidation number equal to +1. But in metallic hydrides ( e.g. NaH, KH), it is –1.
Halogen atom :
In general, all halogen atoms (Cl, Br , I) have oxidation number equal to –1.
But if halogen atom is attached with a more electronegative atom than halogen atom, then it will
show positive oxidation numbers.
e.g.3
5
ClOK
,3
5
OHI
,4
7
OHCI
,5
3KBrO
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Metals :
(a) Alkali metal (Li , Na, K, Rb, .......) always have oxidation number +1
(b) Alkaline earth metal (Be , Mg , Ca .......) always have oxidation number +2.
(c) Aluminium always has +3 oxidation number
Note : Metal may have negative or zero oxidation number
Oxidation number of an element in free state or in allotropic forms is always zero
e.g.0
3
0
4
0
8
0
2 O,P,S,O
Sum of the oxidation numbers of atoms of all elements in a molecule is zero.
Sum of the oxidation numbers of atoms of all elements in an ion is equal to the charge on the ion .
If the group number of an element in modern periodic table is n, then its oxidation number may vary
from
(n – 10) to (n – 18) (but it is mainly applicable for p-block elements )
e.g. N- atom belongs to 15th group in the periodic table, therefore as per rule, its oxidation number
may vary from
–3 to +5 (2
3
3
NO,HN
, 32
3
ON
,2
4
ON
,52
5
ON
)
The maximum possible oxidation number of any element in a compound is never more than the
number of electrons in valence shell.(but it is mainly applicable for p-block elements )
Calculation of average oxidation number :
Example-1 Calculate oxidation number of underlined element :
(a) Na2S
2O
3(b) Na
2S
4O
6
Solution. (a) Let oxidation number of S-atom is x. Now work accordingly with the rules given before .
(+1) × 2 + (x) × 2 + (–2) ×3 =0
x = + 2
(b) Let oxidation number of S-atom is x
(+1) × 2 + (x) × 4 + (–2) × 6 = 0
x = + 2.5
It is important to note here that Na2S
2O
3have two S-atoms and there are four S-atom in Na
2S
4O
6.
However none of the sulphur atoms in both the compounds have + 2 or + 2.5 oxidation number, it is
the average of oxidation number, which reside on each sulphur atom. Therefore, we should work to
calculate the individual oxidation number of each sulphur atom in these compounds.
Calculation of individual oxidation numberIt is important to note that to calculate individual oxidation number of the element in its compound one should
know the structure of the compound and use the following guidelines.
Formula :
Oxidation Number = Number of electrons in the valence shell – Number of electrons taken up after bonding
Guidelines : It is based on electronegativity of elements.
1. If there is a bond between similar type of atom and each atom has same type of hybridisation, then
bonded pair electrons are equally shared by each element.
Example : Calculate oxidation number of each Cl-atom in Cl2molecule
Structure :
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: Number of electrons in the valence shell = 7.
Number of electrons taken up after bonding = 7.
oxidation number = 7 – 7 = 0.
: similarly, oxidation number = 7 – 7 = 0
2. If there is a bond between different type of atoms :
e.g. A – B (if B is more electronegative than A)
Then after bonding, bonded pair of electrons are counted with B - atom .
Example : Calculate oxidation number of each atom in HCl molecule
Structure :
Note : Electron of H-atom is now counted with Cl-atom, because Cl-atom is more electronegative
than H-atom
H : Number of electrons in the valence shell = 1
Number of electrons taken up after bonding = 0
Oxidation number of H = 1 – 0 = + 1
Cl : Number of electrons in the valence shell = 7
Number of electrons taken up after bonding = 8
Oxidation number of Cl = 7– 8 = – 1
Example-2 Calculate individual oxidation number of each S-atom in Na2S
2O
3(sodium thiosulphate) with the
help of its structure .
Solution. Structure :
Note : (central S-atom) is sp3 hybridised (25% s-character) and (terminal S-atom) is sp2
hydbridised (33% s-character). Therefore, terminal sulphur atom is more electronegative
than central sulphur atom. Now, the shared pair of electrons are counted with terminal S-
atom.
, S-atom : Number of electrons in the valence shell = 6
Number of electrons left after bonding = 0
Oxidation number of central S-atom = 6 – 0 = + 6
, S-atom : Number of electrons in the valence shell = 6
Number of electrons left after bonding = 8
Oxidation number of terminal S-atom = 6 – 8 = – 2
Now, you can also calculateAverage Oxidation number of S =2
)2(6 = + 2 (as we have calculated before)
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Miscellaneous Examples :
In order to determine the exact or individual oxidation number we need to take help from
the structures of the molecules. Some special cases are discussed as follows:
The structure of CrO5is
O O
O O
Cr||O
From the structure, it is evident that in CrO5
there are two peroxide linkages and one double bond. The
contribution of each peroxide linkage is –2. Let the oxidation number of Cr is x.
x + (–2)2 + (–2) = 0 or x = 6
Oxidation number of Cr = + 6 Ans
The structure of H2SO
5is H O O S
O
OOH
From the structure, it is evident that in H2SO
5, there is one peroxide linkage, two sulphur-oxygen double
bonds and one OH group. Let the oxidation number of S = x.
(+ 1) + (– 2) + x + (–2) 2+ (–2) + 1 = 0
or x + 2 – 8 = 0 or x – 6 = 0 or x = 6
Oxidation number of S in H2SO
5is + 6 Ans.
Paradox of fractional oxidation number
Fractional oxidation number is the average of oxidation state of all atoms of element under examination and
the structural parameters reveal that the atoms of element for whom fractional oxidation state is realised a
actually present in different oxidation states. Structure of the species C3O
2, Br
3O
8and S
4O
62– reveal the
following bonding situations :
The element marked with asterisk (*) in each species is exhibiting different oxidation number from
rest of the atoms of the same element in each of the species. This reveals that in C3O
2, 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
the middle carbon.
OC*CCO202
Structure of C3O
2
(Carbon suboxide)
Likewise in Br3O
8, eachof the two terminalbromine atoms are present in +6 oxidationstate and themiddle
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 S4O
62–, average oxidation number of S is + 2.5, whereas the
reality being +5,0,0 and +5 oxidation number respectively for respective sulphur atoms.
In general, the conclusion is that the idea of fractional oxidation state should be taken with care
and the reality is revealed by the structures only.
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Oxidising and reducing agentOxidising agent or Oxidant :
Oxidising agents are those compounds which can oxidise others and reduce itself during the chemical
reaction. Those reagents in which for an element, oxidation number decreases or which undergoes
gain of electrons in a redox reaction are termed as oxidants.
e.g. KMnO4, K
2Cr
2O
7, HNO
3, conc.H
2SO
4etc are powerful oxidising agents .
Reducing agent or Reductant :
Reducing agents are those compounds which can reduce other and oxidise itself during the chemical
reaction. Those reagents in which for an element, oxidation number increases or which undergoes
loss of electrons in a redox reaction are termed as reductants.
e.g. K , Na2S
2O
3etc are the powerful reducing agents.
Note : There are some compounds also which can work both as oxidising agent and reducing agent
e.g. H2O
2, NO
2–
HOW TO IDENTIFY WHETHER APARTICULAR SUBSTANCE IS AN OXIDISING OR AREDUCING AGENT
Redox reactionA reaction in which oxidation and reduction simultaneously take place is called a redox reaction
In all redox reactions, the total increase in oxidation number must be equal to the total decrease in oxidation
number.
e.g. 10 4
2
SOFe
+5
4KMnO2
+ 8H2SO
4 342
3
SOFe5
+ 4
2
SOMn2
+ K2SO
4+ 8H
2O
Disproportionation Reaction :A redox reaction in which same element present in a particular compound in a definite oxidation state is
oxidized as well as reduced simultaneously is a disproportionation reaction.
Disproportionation reactions are a special type of redox reactions. One of the reactants 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. For example :
)aq(OH21
22
)(OH22
2
+ )g(O2
0
)s(S0
8+ 12OH¯(aq) )aq(S4
22
+ )aq(OS2
2232
+ 6H
2O ()
)g(Cl0
2+ 2OH¯(aq) )aq(ClO
1 + )aq(Cl
1 + H
2O ()
Consider the following reactions :
(a) 2KClO3
2KCl + 3O2
KClO3
plays a role of oxidant and reductant both. Here, Cl present in KClO3
is reduced and O
present in KClO3
is oxidized. Since same element is not oxidized and reduced, so it is not a
disproportionation reaction, although it looks like one.
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(b) NH4NO
2 N
2+ 2H
2O
Nitrogen in this compound has -3 and +3 oxidation number, which is not a definite value. So it is not
a disproportionation reaction. It is an example of comproportionation reaction, which is a class of
redox reaction in which an element from two different oxidation state gets converted into a single
oxidation state.
(c)5
3KClO4
7
4KClO3
+1–
KCl
It is a case of disproportionation reaction and Cl atom is disproportionating.
List of some important disproportionation reactions
1. H2O
2 H
2O + O
2
2. X2+ OH–(dil.) X¯ + XO¯ (X = Cl, Br, I)
3. X2+ OH–(conc.) X¯ + XO
3¯
F2
does not undergo disproportionation as it is the most electronegative element.
F2
+ NaOH(dil.) F– + OF2
F2
+ NaOH(conc.) F– + O2
4. (CN)2+ OH– CN– + OCN–
5. P4
+ OH– PH3
+ H2PO
2¯
6. S8+ OH– S2– + S
2O
32–
7. MnO4
2– MnO4¯ + MnO
2
8. NH2OH N
2O + NH
3
NH2OH N
2+ NH
3
9. Oxyacids of Phosphorus ( +1, +3 oxidation number)
H3PO
2 PH
3+ H
3PO
3
H3PO
3 PH
3+ H
3PO
4
10. Oxyacids of Chlorine( Halogens)( +1, +3, +5 Oxidation number)
ClO– Cl– + ClO2
–
ClO2
– Cl– + ClO3
–
ClO3
– Cl– + ClO4
–
11. HNO2
NO + HNO3
Reverse of disproportionation is called Comproportionation. In some of the disproportionation
reactions, by changing the medium (from acidic to basic or reverse), the reaction goes in backward
direction and can be taken as an example of Comproportionation reaction.
¯ + O3¯ + H+
2+ H
2O
Balancing of redox reactionsAll balanced equations must satisfy two criteria.
1. Atom balance (mass balance ) :
There should be the same number of atoms of each kind on reactant and product side.
2. Charge balance :
The sum of actual charges on both sides of the equation must be equal.
There are two methods for balancing the redox equations :
1. Oxidation - number change method
2. Ion electron method or half cell method
Since First method is not very much fruitful for the balancing of redox reactions, students are advised
to use second method (Ion electron method ) to balance the redox reactions
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Ion electron method :
By this method redox equations are balanced in two different medium.
(a) Acidic medium (b) Basic medium
Balancing in acidic medium
Students are adviced to follow the following steps to balance the redox reactions by Ion electron
method in acidic medium
Example-3 Balance the following redox reaction :
FeSO4+ KMnO
4+ H
2SO
4 Fe
2(SO
4)
3+ MnSO
4+ H
2O + K
2SO
4
Solution. Step– Assign the oxidation number to each element present in the reaction.
4
262
OSFe
+ 4
2–71
MnOK
+ 4
26
2
1
OSH
34
26
2
3
)OS(Fe
+ 4
262
OSMn
+2
2
1
OH
Step :
Now convert the reaction in Ionic form byeliminating the elements or species, which are not undergoing
either oxidation or reduction.
Fe2+ +
4
7
OMn Fe3+ + Mn2+
Step :
Now identify the oxidation / reduction occuring in the reaction
Step V : Spilt the Ionic reaction in two half, one for oxidation and other for reduction.
Fe2+ oxidationFe3+
2ductionRe4 MnMnO
Step V :
Balance the atom other than oxygen and hydrogen atom in both half reactions
Fe2+ Fe3+ MnO4
– Mn2+
Fe & Mn atoms are balanced on both side.
Step V :
Now balance O & H atom by H2O & H+ respectively by the following way : For one excess oxygen
atom, add one H2O on the other side and two H+ on the same side.
Fe2+ Fe3+ (no oxygen atom ) .................(i)
8H+ + MnO4
– Mn2+ + 4H2O ................(ii)
Step V :
Equation (i) & (ii) are balanced atomwise. Now balance both equations chargewise. To balance the
charge, add electrons to the electrically positive side.
Fe2+ oxidation Fe3+ + e– ............(1)
5e– + 8H+ + MnO4
– ductionRe Mn2+ + 4H
2O ............(2)
Step V :
The number of electrons gained and lost in each half -reaction are equalised by multiplying both the
half reactions with a suitable factor and finally the half reactions are added to give the overall balanced
reaction.
Here, we multiply equation (1) by 5 and (2) by 1 and add them :
Fe2+ Fe3+ + e– ..........(1) × 5
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OH4MnFe5MnOH8Fe5
1)2.........(OH4MnMnOH8e5
223
42
22
4
(Here, at his stage, you will get balanced redox reaction in Ionic form)
Step X :
Now convert the Ionic reaction into molecular form by adding the elements or species, which are
removed in step (2).
Now, by some manipulation, you will get :
5 FeSO4+ KMnO
4+ 4H
2SO
4
2
5Fe
2(SO
4)
3+ MnSO
4+ 4H
2O +
2
1K
2SO
4or
10FeSO4
+ 2KMnO4
+ 8H2SO
4 5Fe
2(SO
4)
3+ 2MnSO
4+ 8H
2O + K
2SO
4.
Balancing in basic medium :
In this case, except step VI, all the steps are same. We can understand it by the following example:
Example-4 Balance the following redox reaction in basic medium :
ClO– + CrO2
– + OH– Cl– + CrO4
2– + H2O
Solution. By using upto step V, we will get :
24
6Oxidation
2
3–ductionRe
1
OCrOCrClOCl
Now, students are advised to follow step VI to balance ‘O’ and ‘H’ atom.
2H+ + ClO– Cl– + H2O | 2H
2O+ CrO
2– CrO
42– + 4H+
Now, since we are balancing in basic medium, therefore add as many as OH– on both side of
equation as there are H+ ions in the equation.
2OH– + 2H+ + ClO– Cl– + H2O +2OH– 4OH– + 2H
2O + CrO
2–
CrO4
2– + 4H+ + 4OH–
Finally you will get Finally you will get
H2O + ClO– Cl– + 2OH– ...........(i) 4OH– + CrO
2– CrO
42– + 2H
2O ........... (ii)
Now see equation (i) and (ii) in which O and H atoms are balanced by OH– and H2O
Now from step VIII
2e– + H2O + ClO– Cl– + 2OH– ............. (i) ×3
4OH– + CrO2
– CrO4
2– + 2H2O + 3e– ............. (ii) ×2
–––––––––––––––––––––––––––––––––––––––––––––––––––
Adding : 3ClO– + 2CrO2
– + 2OH– 3Cl– + 2CrO4
2– + H2O
Concept of equivalentsEquivalent mass of element
Number of parts by mass of an element which reacts or displaces from a compound 1.008 parts by
mass of hydrogen, 8 parts by mass of oxygen and 35.5 parts by mass of chlorine, is known as the
equivalent weight of that element.
e.g. 2Mg + O2
2MgO
48g 32g
12g 8g
32 g of O2
reacts with 48 g of Mg
8 g of O2
=32
848= 12 g
Equivalent weight of Mg = 12
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Similarly, Zn + H2SO
4 ZnSO
4+ H
2
65.5 g 32.75
Equivalent weight of Zn =2
5.65= 32.75 g
Al +2
3Cl
2 AlCl
3
27 g2
3× 71 g
111.5 g chlorine reacts with 27 g of Al.
35.5 chlorine reacts with5.111
5.3527 = 9.0 g of Al
Equivalent weight of aluminium =3
27= 9.0
As we can see from the above examples that equivalent weight is the ratio of atomic weight and a factor
(say n-factor or valency factor) which is in above three cases is their respective valencies.
Equivalent weight (E) :
In general, Eq. wt. (E) = x
M
factorn
.wt.Mol
)f.v(factorvalency
weightMolecularorweightAtomic
Number of Equivalents = speciesthatof.wt.eq
speciesofmass
For a solution, Number of equivalents = N1V
1, where N is the normality and V is the volume in litres
Equivalent mass is a pure number which, when expressed in gram, is called gram equivalent mass.
The equivalent mass of substance may have different values under different conditions.
There in no hard and fast rule that equivalent weight will be always less than the molecular
mass.
Valency factor calculation :
For Elements :
Valency factor = valency of the element.
For Acids :
Valency factor = number of replaceable H+ ions per acid molecule
Example-5 HCl , H2SO
4H
3PO
4H
3PO
3
{see there are only two replaceable H+ions}
Solution. Valency factor 1 2 3 2
(assume 100% dissicoiation)
Eq. wt. (E) 1
M
2
M
3
M
2
M
Replaceable hydrogen atoms are those hydrogen atoms which are attached with the atoms of group
VI and group VII i.e. O,S,Se,Te, & F, Cl ,Br ,I.
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For Bases :
Valency factor = number of replacable OH– ions per base molecule.
Example-6 NaOH, KOH
Solution. v .f. 1 1
Eq. wt. 1
M
1
M
Bases may be defined as the substances in which OH group is/are directly attached with group I
elements (Li,Na, K,Rb,Cs), group II elements (Be, Mg,Ca,Ba ) or group III elements (Al, Ga,In,Tl),
transition metals, non-metallic cations like PH4
+ , NH4
+ etc.
Acid - base reaction :
In case of acid base reaction, the valence factor is the actual number of H+ or OH– replaced in the
reaction. The acid or base may contain more number of replaceble H+ or OH– than actually replaced
in reaction.
v. f. for base is the number of H+ ion from the acid replaced by each molecule of the base
Example-7 2NaOH + H2
SO4
Na2
SO4
+ 2H2O
Base Acid
Solution. Valency factor of base = 1
Here, two molecule of NaOH replaced 2H+ ion from the H2SO
4. Therefore, each molecule of NaOH
replaced only one H+ ion of acid, so v.f. = 1.
v. f. for acid is the number of OH– replaced from the base by each molecule of acid
Example-8 NaOH + H2SO
4 NaHSO
4+ H
2O
Base Acid
Solution. Valency factor of acid = 1
Here, one of molecule of H2SO
4replaced one OH– from NaOH. Therefore, valency factor for H
2SO
4
is one
Eq. wt. of H2SO
4=
1
wt.Mol
Salts :
(a) In non-reacting condition
Valency factor = Total number of positive charge or negative charge present in the compound.
Example-9 Na2CO
3, Fe
2(SO
4)3
FeSO4.7H
2O
Solution. V.f. 2 2×3 = 6 2
Eq.wt.2
M
6
M
2
M
Note : In case of hydrated salt, positive/negative charge of water molecule is not counted.
(b) In reacting condition
Example-10 Na2
CO3
+ HCl NaHCO3
+ NaCl
Base Acid
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Solution. It is an acid base reaction, therefore valency factor for Na2CO
3is one while in non-reacting condition,
it will be two.
(c) Equivalent weight of oxidising / reducing agents in a redox reaction
In case of redox change , v.f. = Total change in oxidation number per molecule .
Example-11 KMnO4
+ H2O
2 Mn2+ + O
2
Solution. Mn in KMnO4is going from +7 to +2 , so change in oxidation number per molecule of KMnO
4is 5. So
the valency factor of KMnO4is 5 and equivalent weight is
5
M.
Normality :Normality of a solution is defined as the number of equivalents of solute present in one litre (1000 mL)
solution.
Let V mL of a solution is prepared by dissolving W g of solute of equivalent weight E in water.
Number of equivalents of solute =E
W
VmL of solution containE
Wequivalents of solute
1000 mL solution will contain VE
1000W
equivalents of solute.
Normality (N) = VE
1000W
Normality (N) = Molarity x Valency factor
N × V (in mL) = M × V (in mL) × n
or
milliequivalents = millimoles × n
Example-12 Calculate the normality of a solution containing 15.8 g of KMnO4 in 50 mL acidic solution.
Solution. Normality (N) =VE
1000W
Here W = 15.8 g , V = 50 mL E =factorValency
KMnOofmassmolar 4= 158/5 = 31.6
So, normality = 10 N
Example-13 Calculate the normality of a solution containing 50 mL of 5 M solution of K2Cr2O7 in acidic
medium.
Solution. Normality (N) = Molarity × valency factor
= 5 x 6 = 30 N
Law of EquivalenceThe law states that one equivalent of an element combine with one equivalent of the other. In a chemical
reaction, equivalents and milli equivalents of reactants react in equal amount to give same number of equivalents
or milli equivalents of products separately.
Accordingly
(i) aA + bB mM + nN
meq of A = meq of B = meq of M = m.eq. of N
(ii) In a compound MxN
y
meq of MxN
y= meq of M = meq of N
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Example-14 Find the number of moles of KMnO4needed to oxidise one mole Cu
2S in acidic medium.
The reaction is KMnO4
+ Cu2S Mn2+ + Cu2+ + SO
2
Solution. From law of equivalence,
equivalents of Cu2S = equivalents of KMnO
4
moles of Cu2S × v.f. = moles of kMnO
4× v.f.
1 × 8 = moles of KMnO4
× 5 moles of KMnO4
= 8/5
( v.f. of Cu2S = 2 (2 – 1) + 1 (4 – (–2))) = 8 and v.f. of KMnO
4= 1 (7 –2) = 5)
Example-15 The number of moles of oxalate ions oxidized by one mole of MnO4
– ion in acidic medium are :
(A)2
5(B)
5
2(C)
5
3(D)
3
5
Solution. Equivalents of C2O
42– = equivalents of MnO
4–
x(mole) × 2 = 1 × 5
( v.f. of C2O
42– = 2 (4 – 3) = 2 and v.f. of MnO
4– = 1 (7 – 2) = 5).
x =2
5mole of C
2O
42– ions.
Drawbacks of Equivalent concept
Since equivalent weight of a substance (for example oxidising or reducing agent) may be variable
hence it is better to use mole concept.
e.g. 5e– + 8H+ + MnO4
– Mn2+ +2H2O
Eq.wt of MnO4
– =5
MnOof.wt.Mol–
4
e.g. 3e– +2H2O + MnO
4– MnO
2+4OH–
Eq.wt of MnO4
– =3
MnOof.wt.Mol 4
Thus, the number of equivalents of MnO4
– will be different in the above two cases but
number of moles will be same.
Normality of any solution depends on reaction while molarity does not.
For example :
Consider 0.1mol KMnO4
dissolved in water to make 1L solution. Molarity of this solution is 0.1 M.
However, its normality is NOT fixed. It will depend upon the reaction in which KMnO4
participates.
e.g. if KMnO4forms Mn2+, normality = 0.1 x 5 = 0.5 N. This same sample of KMnO
4, if employed in
a reaction giving MnO2as product (Mn in +4 state), will have normality 0.1 × 3 = 0.3 N.
The concept of equivalents is handy, but it should be used with care. One must never equate
equivalents in a sequence which involves same element in more than two oxidation states. Consider
an example, KIO3reacts with KI to liberate iodine and liberated Iodine is titrated with standard hypo
solution. The reactions are :
(i) O3
– + ¯ 2
(ii) 2
+ S2O
32– S
4O
62–+ ¯
meq of hypo = meq of I2
= meq of IO3 = meq of I
meq of hypo = meq of IO3.
This is wrong. Note that I2
formed by equation (i) has v.f. = 5/3 & in equation (ii) has v.f. = 2.
v.f. of I2in both the equation are different, therefore we cannot equate milli equivalents in sequence.
In this type of case, students are advised to use mole concept.
"manishkumarphysics.in" 13
CHEMISTRY
Example-16 How many millilitres of 0.02 M KMnO4
solution would be required to exactly titrate 25 mL of 0.2 M
Fe(NO3)
2solution in acidic medium ?
Solution. Method -1 : Mole concept method
Starting with 25 mL of 0.2 M Fe2+, we can write :
Millimoles of Fe2+ = 25 x 0.2 ........(1)
and in volume V (in milliliters) of the KMnO4,
Millimoles of MnO4¯ = V (0.02) ........(2)
The balanced reaction is :
MnO4¯ + 5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H
2O
This requires that at the equivalent point,
1
MnOofmoles.m –4 =
5
Feofmoles.m 2
1
)02.0(V=
5
)2.0)(25((from (1) & (2))
V = 50 mL.
Method -2 : Equivalent Method :
At the equivalence point,
milliequivalents of MnO4¯ = milliequivalents of Fe2+
M1
× vf1
× V1
= M2
× vf2
× V2
0.02 × 5 × V1
= 0.2 × 1 × 25 ( MnO4
– Mn2+ ; v.f. = 5, Fe2+ Fe3+ ; v.f. = 1)
V1
= 50 mL.
TitrationsTitration is a procedure for determining the concentration of a solution by allowing a carefully measured
volume to react with a standard solution of another substance, whose concentration is known.
Standard solution - It is a solution whose concentration is known and is taken in burette. It is also called
Titrant.
There are two type of titrants :
Primary titrants/standard - These reagents can be accurately weighed and their solutions are not
to be standardised before use.
Ex : Oxalic acid, K2Cr
2O
7, AgNO
3, CuSO
4, ferrous ammonium sulphate, hypo etc.
Secondary titrants/standard : These reagents cannot be accurately weighed and their solutions
are to be standardised before use.
Ex : NaOH, KOH, HCl, H2SO
4,
2, KMnO
4etc.
Titrate : Solution consisting of substance to be estimated, generally taken in a beaker .
Equivalence point : It is the point when number of equivalents of titrant added becomes equal to number of
equivalents of titrate.
At equivalence point :
n1V
1M
1= n
2V
2M
2
Indicator : An auxiliary substance added for physical detection of completion of titration at equivalence
point. It generally show colour change on completion of titration.
Type of Titrations :
Acid-base titrations (to be studided in Ionic equilibrium)
Redox Titrations
"manishkumarphysics.in" 14
CHEMISTRY
Some Common Redox Titrations
Table of Redox Titrations : (Excluding Iodometric / Iodimetric titrations)––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
Estimation By titrating Reactions Relation*between
of with OA and RA––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
1. Fe2+ MnO4¯ Fe2+ Fe3+ + e– 5Fe2+ MnO
4¯
MnO4
– + 8H+ + 5e– Mn2+ + 4H2O Eq. wt. of Fe2+ = M/1
2. Fe2+ Cr2O
72– Fe2+ Fe3+ + e– 6Fe2+ Cr
2O
72–
Cr2O
72– + 14H+ + 6e– 2Cr3+ + 7H
2O Eq.wt. of Cr
2O
72– = M/6
3. C2O
42– MnO
4¯ C
2O
42– 2CO
2+ 2e– 5C
2O
42– 2MnO
4¯
MnO4
– + 8H+ + 5e– Mn2+ + 4H2O Eq. wt. of C
2O
42– = M/2
4. H2O
2MnO
4¯ H
2O
2 2H+ + O
2+ 2e– 5H
2O
2 2MnO
4¯
MnO4
– + 8H+ + 5e– Mn2+ + 4H2O Eq.wt. of H
2O
2= M/2
5. As2O
3MnO
4– As
2O
3+ 5H
2O 2AsO
43– + 10H+ + 4e– Eq. wt. of As
2O
3= M/4
MnO4
– + 8H+ + 5e– Mn2+ + 4H2O
6. AsO33– BrO
3– AsO
33– + H
2O AsO
43– + 2H+ + 2e– Eq. wt. of AsO
33– = M/2
BrO3
– + 6H+ + 6e– Br– + 3H2O Eq.wt. of BrO
3– = M/6
Permanganate Titrations :
KMnO4is generally used as oxidising agent in acidic medium, generally provided by dilute H
2SO
4.
KMnO4
works as self indicator persistent pink color is indication of end point.
Mainly used for estimation of Fe2+ , oxalic acid ,oxalates, H2O
2etc.
Example-17 Write the balanced reaction of titration of KMnO4
Vs oxalic acid in presence of H2SO
4.
Solution. Reaction : 2KMnO4
+ 3H2SO
4+ 5H
2C
2O
4 K
2SO
4+ 2MnSO
4+ 8H
2O + 10CO
2
Redox Changes : C2
3+ 2C4+ +2e
2
ME
422 OCH
5e + Mn7+ Mn2+
5
ME
4KMnO
Indicator : KMnO4
acts as self indicator.
Example-18 Write the balanced reaction of titration of KMnO4
Vs ferrous ammonium sulphate in presence of
H2SO
4.
Solution. Reaction : 2KMnO4+ 10[FeSO
4(NH
4)
2SO
4. 6H
2O] + 8H
2SO
4
5Fe2(SO
4)
3+ 10(NH
4)
2SO
4+ K
2SO
4+ 2MnSO
4+ 68H
2O
Redox Changes : Fe2+ Fe3+ + e
1
ME
4FeSO
Mn7+ + 5e Mn2+
5
ME
4KMnO
Indicator : KMnO4
acts as self indicator
Iodometric/Iodimetric Titrations :
Compound containing iodine are widely used in titrations.
(i) Iodide ions can be oxidised to 2
by suitable oxidising agent
2¯ (aq) 2(s) + 2e¯
(ii) Iodine (V) ions, O3¯ , will oxidise ¯ to
2
O¯ (aq) + 5¯ (aq) + 6H+ (aq) 3
2(s) + 3H
2O ()
"manishkumarphysics.in" 15
CHEMISTRY
(iii) Thiosulphate ions, S2O
32– , can reduce iodine to iodide ions.
2S2O
(aq)
(s) S
4O
62– + 2–
colourless black colourless
Iodometric Titrations (Titration Solution is of Na2S
2O
3. 5H
2O)
––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
S.No. Estimation of Reaction Relation between O.A. and R.A.––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
1. 2
2
+ 2Na2S
2O
3 2Na + Na
2S
4O
6
2 2 2Na
2S
2O
3
or 2
+ 2S2O
32– 2¯ + S
4O
62– Eq.wt. of Na
2S
2O
3= M/1
2. CuSO4
2CuSO4
+ 4K 2Cu + 2K2SO
4+
22CuSO
4
2 2 = 2Na
2S
2O
3
or 2Cu2+ + 4¯ 2Cu + 2
Eq.wt.of CuSO4
= M/1
white ppt
3. CaOCl2
CaOCl2+ H
2O Ca(OH)
2+ Cl
2CaOCl
2 Cl
2
2 2
Cl2
+ 2K 2KCl + 2
2Na2S
2O
3
Cl2
+ 2¯ 2Cl¯ + 2
Eq.wt. of CaOCl2
= M/2
4. MnO2
MnO2+ 4HCl(conc.) MnCl
2+ Cl
2+ 2H
2O MnO
2 Cl
2
2 2Na
2S
2O
3
Cl2
+ 2K 2KCl + 2
Eq.wt. of MnO2
= M/2
or MnO2
+ 4H+ + 2Cl¯ Mn2+ + 2H2O + Cl
2
Cl2
+ 2¯ 2
+ 2Cl¯
5. O3¯ O
3¯ + 5¯ + 6H+ 3
2+ 3H
2O O
3¯ 3
2 6 6Na
2S
2O
3
Eq.wt. of O3¯ = M/6
6. H2O
2H
2O
2+ 2¯ + 2H+
2+ 2H
2O H
2O
2
2 2 2Na
2S
2O
3
Eq.wt. of H2O
2= M/2
7. Cl2
Cl2
+ 2¯ 2Cl¯ + 2
Cl2
2 2 2Na
2S
2O
3
Eq.wt. of Cl2
= M/2
8. O3
O3
+ 6¯ + 6H+ 32
+ 3H2O O
3 3
2 6 6Na
2S
2O
3
Eq.wt. of O3
= M/6
9. ClO¯ ClO¯ + 2¯ + 2H+ H2O + Cl¯ +
2ClO¯
2 2 2Na
2S
2O
3
Eq.wt. of OCl– = M/2
10. Cr2O
72– Cr
2O
72– + 14H+ + 6¯ 3
2+ 2Cr3+ + 7H
2O Cr
2O
72– 3
2 6
Eq.wt. of Cr2O
72– = M/6
11. MnO4– 2MnO
4– + 10¯+ 16H+ 2MnO
4– + 5
2+ 8H
2O 2MnO
4¯ 5
2 10
Eq.wt. of MnO4¯ = M/5
12. BrO3– BrO
3– + 6¯ + 6H+ Br– + 3
2+ 3H
2O BrO
3– 3
2 6
Eq.wt. of BrO3
– = M/6
13. As(V) H2AsO
4+ 2¯+ 3H+ H
3AsO
3+ H
2O +
2H
3AsO
4
2 2
Eq.wt. of H3AsO
4= M/2
14. HNO2
2HNO2
+ 2¯ 2
+ 2NO + H2O 2HNO
2
2 2
Eq.wt. of HNO2
= M/1
15. HClO HClO + 2¯ + H+ Cl¯ + 2
+ H2O HClOI
22Na
2S
2O
3
Eq.wt. of HClO = M/2
"manishkumarphysics.in" 16
CHEMISTRY
Iodimetric Titrations––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
S.No. Estimation of Reaction Relation between O.A. and R.A.––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
1. H2S H
2S +
2 S + 2¯ + 2H+ H
2S
2 2
(in acidic medium) Eq.wt. of H2S = M/2
2. SO32– SO
32– +
2+ H
2O SO
42– + 2¯ + 2H+ SO
32–
2 2
(in acidic medium) Eq.wt. of SO3
2– = M/2
3. Sn2+ Sn2+ + 2
Sn4+ + 2¯ Sn2+ 2 2
(in acidic medium) Eq.wt. of Sn2+ = M/2
4. As(III) (at pH 8) H2AsO
3¯ +
2+ H
2O HAsO
42– + 2¯ + 3H+ H
2AsO
3–
2 2
Eq.wt. of H2AsO
3¯ = M/2
5. N2H
4N
2H
4+ 2
2 N
2+ 4H+ + 4¯ N
2H
4= 2
2 4
Eq.wt. of N2H
4= M/4
Example-19 The sulphur content of a steel sample is determined by converting it to H2S gas, absorbing the H
2S
in 10 mL of 0.005 M I2and then back titrating the excess I
2with 0.002 M Na
2S
2O
3. If 10 mL Na
2S
2O
3
is required for the titration, how many milligrams of sulphur are contained in the sample?
Reactions :
H2S + I
2 S + 2I– + 2H+
I2+ 2S
2O
32– 2I– + S
4O
62–
Solution. Used millimoles of I2= (m.moles of I
2taken initially) –
2
usedhypoofmoles.m
= 0.005 × 10 – 0.002 ×2
10
= 0.04 = millimoles of H2S
weight of sulphur = 0.04 × 10–3 × 32 × 103 mg = 1.28 mg.
Hydrogen peroxide (H2O
2)
H2O
2can behave both like oxidising and reducing agent in both the mediums (acidic and basic).
Oxidising agent : (H2O
2H
2O)
(a) Acidic medium : 2e– + 2H+ + H2O
2 2H
2O
v.f. = 2
(b) Basic medium : 2e– + H2O
2 2OH–
v.f = 2
Reducing agent : (H2O
2 O
2)
(a) Acidic medium : H2O
2 O
2+ 2H+ + 2e–
v.f = 2
(b) Basic medium : 2OH– + H2O
2O
2+ 2H
2O + 2e–
v.f = 2
Note : Valency factor of H2O
2is always equal to 2.
Volume strength of H2O
2: Strength of H
2O
2is represented as 10V , 20 V , 30 V etc.
20V H2O
2means one litre of this sample of H
2O
2on decomposition gives 20L of O
2gas at STP.
"manishkumarphysics.in" 17
CHEMISTRY
Decomposition of H2O
2is given as :
H2O
2 H
2O +
2
1O
2
1 mole2
1× 22.4 L O
2at STP
= 34g = 11.2 L O2at STP
To obtain 11.2 litre O2
at STP, at least 34 g H2O
2must be decomposed.
For 20 L O2, we should decompose atleast
2.11
34×20 g H
2O
2
1 L solution of H2O
2contains
2.11
34×20 g H
2O
2
1 L solution of H2O
2contains
2.11
34×
17
20equivalents of H
2O
2(
2
34
2
ME
22OH = 17)
Normality of H2O
2=
2.11
34×
17
20=
6.5
20
Normality of H2O
2(N) =
5.6
OHofstrengthVolume 22
22OHM =.f.v
N22OH
=2
N22OH
Molarity of H2O
2(M) =
11.2
OHofstrengthVolume 22
Strength (in g/L) : Denoted by S
Strength = Molarity × Mol. wt = Molarity × 34
Strength = Normality × Eq. weight = Normality × 17
Example-20 20 mL of H2O2 after acidification with dilute H2SO4 required 30 mL of12
NKMnO4 for complete
oxidation. Final the strength of H2O2 solution. [Molar mass of H2O2 = 34]
Solution. meq. of KMnO4 = meq. of H2O2
30 ×12
1= 20 × N
N =2012
30
=
8
1N
strength = N × equivalent mass =8
1× 17 = 2.12 g/L.
Hardness of water (Hard water does not give lather with soap)Temporary hardness - due to bicarbonates of Ca & Mg
Permanent hardness - due to chlorides & sulphates of Ca & Mg. There are some method by which we can
soften the water sample.
(a) By boiling : 2HCO3
– H2O + CO
2+ CO
32– or
By Slaked lime : Ca(HCO3)
2+ Ca(OH)
2 CaCO
3+ 2H
2O
Ca2+ + CO3
2– CaCO3
(b) By Washing Soda : CaCl2+ Na
2CO
3 CaCO
3+ 2NaCl
(c) By ion exchange resins : Na2R + Ca2+ CaR + 2Na+
(d) By adding chelating agents like (PO3
–)3etc.
"manishkumarphysics.in" 18
CHEMISTRY
Parts Per Million (ppm)When the solute is present in very less amount, then this concentration term is used. It is defined as the
number of parts of the solute present in every 1 million parts of the solution. ppm can both be in terms of
mass or in terms of moles. If nothing has been specified, we take ppm to be in terms of mass. Hence, a 100
ppm solution means that 100 g of solute is present in every 1000000 g of solution.
ppmA
= 610massTotal
Aofmass = mass fraction × 106
Measurement of Hardness :
Hardness is measured in terms of ppm (parts per million) of CaCO3or equivalent to it.
Hardness in ppm =610
solutionofmassTotal3
CaCOofmass
Example-21 0.00012% MgSO4and 0.000111% CaCl
2is present in water. What is the measured hardness of
water and millimoles of washing soda required to purify water 1000 L water ?
Solution. Basis of calculation = 100 g hard water
MgSO4
= 0.00012g =120
00012.0mole
CaCl2
= 0.000111g =111
000111.0mole
equivalent moles of CaCO3=
111
000111.0
120
00012.0mole
mass of CaCO3
=
111
000111.0
120
00012.0× 100 = 2 × 10–4 g
Hardness (in terms of ppm of CaCO3) =
64
10100
102
= 2 ppm
CaCl2+ Na
2CO
3 CaCO
3+ 2NaCl
NaSO4+ Na
2CO
3 MgCO
3+ Na
2SO
4
Required Na2CO
3for 100g of water =
111
000111.0
120
00012.0mole
= 2 × 10–6 mole
Required Na2CO
3for 1000 litre water =
100
210
100
102 66
mole ( d = 1g/mL)
=1000
20mole = 20 m mole
Strength of Oleum :
Oleum is SO3dissolved in 100% H
2SO
4. Sometimes, oleum is reported as more than 100% by weight, say
y% (where y > 100). This means that (y 100) grams of water, when added to 100 g of given oleum sample,
will combine with all the free SO3
in the oleum to give 100% sulphuric acid.
Hence, weight % of free SO3in oleum = 80(y 100)/18
"manishkumarphysics.in" 19
CHEMISTRY
Example-22 What volume of water is required (in mL) to prepare 1 L of 1 M solution of H2SO
4(density = 1.5g/mL)
by using 109% oleum and water only (Take density of pure water = 1 g/mL).
Solution. 1 mole H2SO
4in 1L solution = 98 g H
2SO
4in 1500 g solution = 98 g H
2SO
4in 1402 g water.
Also, in 109% oleum, 9 g H2O is required to form 109 g pure H
2SO
4& so, to prepare 98 g H
2SO
4,
water needed is 9/109 × 98 = 8.09 g.
Total water needed = 1402 + 8.09 = 1410.09 g = 1410.09 mL
Calculation of Available Chlorine from a sample of Bleaching Powder :
The weight of available Cl2
released from the given sample of bleaching powder on reaction with
dilute acids or CO2is called available chlorine.
224422 ClOHCaSOSOHCaOCl
2222 ClOHCaClHCl2CaOCl
222332 ClOHCOOCHCaCOOHCH2CaOCl
2322 ClCaCOCOCaOCl
Method of determination :
CaOCl2 + 2CH3COOH Ca(CH3COO)2 + H2O + Cl2(Sample of bleaching powder)
Cl2 + 2KI 2KCl + I2
I2 + 2Na2S2O3 indicatorasStarch Na2S4O6 + 2Nalv.f. = 2 v.f. = 1
End point is indicated by disappearance of blue colour.
Let M = Molarity of hypo (Na2S2O3) solution
millimoles of Cl2 produced = m.moles of I2 used by hypo
=2
VMwhere V = vol of hypo solution used in ml.
mass of Cl2 produced =2
10VM 3× 71
= 35.5 × M × V × 10–3
% of available chlorine =W
10VM5.35 3× 100
where W = amount of belaching powder taken in g.
or % of available Cl2 =W
VM3.55
Example-23 3.55 g sample of bleaching powder suspended in H2O was treated with enough acetic acid and K
solution. Iodine thus liberated required 80 mL of 0.2 M hypo for titration. Calculate the % of available
chlorine.
Solution. % of Cl2 =55.3
802.055.3 = 16%
"manishkumarphysics.in" 20
CHEMISTRY
MISCELLANEOUS SOLVED PROBLEMS (MSPS)
1. Calculate individual oxidation number of each S-atom in Na2S
4O
6(sodium tetrathionate) with the help of its
structure .
Sol.
2. Find the average and individual oxidation number of Fe & Pb in Fe3O
4& Pb
3O
4, which are mixed oxides.
Sol. (i) Fe3O
4is mixture of FeO & Fe
2O
3in 1 : 1 ratio
so, individual oxidation number of Fe = +2 & +3
& average oxidation number =3
)3(2)2(1 = 8/3
(ii) Pb3O
4is a mixture of PbO & PbO
2in 2 : 1 molar ratio
so, individual oxidation number of Pb are +2 & +4
& average oxidation number of Pb = 3/83
)4(1)2(2
3. Balance the following equations :
(a) H2O
2+ MnO
4– Mn+2 + O
2(acidic medium)
(b) Zn + HNO3(dil) Zn(NO
3)
2+ H
2O + NH
4NO
3
(c) Cr3
+ KOH + Cl2
K2CrO
4+ KO
4+ KCl + H
2O.
(d) P2H
4 PH
3+ P
4
(e) Ca3(PO
4)
2+ SiO
2+ C CaSiO
3+ P
4+ CO
Ans. (a) 6H+ + 5H2O
2+ 2MnO
4– 2Mn+2 + 5O
2+ 8H
2O
(b) 4Zn + 10HNO3(dil) 4Zn(NO
3)
2+ 3H
2O + NH
4NO
3
(c) 2Cr3
+ 64KOH + 27Cl2
2K2CrO
4+ 6KO
4+ 54KCl + 32H
2O.
(d) 6P2H
4 8PH
3+ P
4
(e) 2Ca3(PO
4)
2+ 6SiO
2+ 10C 6CaSiO
3+ P
4+ 10CO
4. Find the valency factor for following acids
(i) CH3COOH (ii) NaH
2PO
4(iii) H
3BO
3
Ans. (i) 1 (ii) 2 (iii) 1
5. Find the valency factor for following bases :
(i) Ca (OH)2
(ii) CsOH (iii)Al(OH)3
Ans. (i) 2 (ii) 1 (iii) 3
6. Find the valence factor for following salts :
(i) K2SO
4.Al
2(SO
4)3.24H
2O
(ii) CaCO3
Ans. (i) 8 (ii) 2
7. Find the valency factor for following redox reactions :
(i)
42alkaline
2neutral
2acidic
4
MnOK
MnO
Mn
KMnO
"manishkumarphysics.in" 21
CHEMISTRY
(ii) K2Cr
2O
7
acidic Cr3+
(iii) C2O
42 CO
2
(iv) Fe2+ Fe3+
Ans. (i) 5, 3, 1; (ii) 6 ; (iii) 2 ; (iv) 1
8. Calculate the normality of a solution obtained by mixing 50 mL of 5 M solution of K2Cr2O7 and 50 mL of
2 M K2Cr2O7 in acidic medium.
Sol. v.f. of K2Cr2O7 = 6
so Nf =21
2211
VV
VNVN
=5050
50625065
= 21 N
9. Calculate the normality of a solution containing 13.4 g of Sodium oxalate in 100 mL Sol.
Sol. Normality = litreinsolutionofvol
wt.eq/gin.wt
Here, eq. wt. of Na2C2O4 = 134/2 = 67
so N =1000/100
67/4.13= 2N
10. The number of moles of ferrous oxalate oxidised by one mole of KMnO4in acidic medium is :
(A)2
5(B)
5
2(C)
5
3(D)
3
5
Sol. Eq. of FeC2O
4= Eq. of KMnO
4
moles of FeC2O
4× 3 = moles of KMnO
4× 5
so, moles of FeC2O
4= 5/3 Ans. (D)
11. How many moles of KMnO4
are needed to oxidise a mixture of 1 mole of each FeSO4
& FeC2O
4in acidic
medium ?
(A)5
4(B)
4
5(C)
4
3(D)
3
5
Sol. Eq. of KMnO4
= Eq. of FeSO4
+ Eq. of FeC2O
4
moles of KMnO4
× 5 = moles of FeSO4
× 1 + moles of FeC2O
4× 3
moles of KMnO4
= 4/5 Ans. (A)
12. Asample of hydrazine sulphate [N2H
6SO
4] was dissolved in 100 mL water. 10 mL of this solution was treated
with excess of FeCl3Sol. Ferrous ions formed were estimated and it required 20 mL of M/50 KMnO
4solution
in acidic medium.
Fe3+ + N2H
4 N
2+ Fe2+ + H+
MnO4¯ + Fe2+ + H+ Mn2+ + Fe3+ + H
2O
(a) Write the balanced redox reactions.
(b) Estimate the amount of hydrazine sulphate in one litre of Sol.
Sol. (a) Given 4Fe3+ + N2H
4 N
2+ 4Fe2+ + 4H+
MnO4¯ +5Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H
2O
(b) In 10 mL solution, eq. of N2H
6SO
4= Eq. of Fe2+ = Eq. of KMnO
4
= 20 ×50
1× 5 × 10–3 = 2 × 10–3
v.f. of N2H
6SO
4= 4
so, weight of N2H
6SO
4in 1 L solution =
104
1000102 3
× 130 = 6.5 g.
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CHEMISTRY
13. Write the balanced redox reaction and calculate the equivalent weight of oxidising agent and reducing agent
for titration of K2Cr
2O
7Vs Ferrous ammonium sulphate.
Ans. The reaction : 6[FeSO4(NH
4)
2SO
4. 6H
2O] + K
2Cr
2O
7+ 7H
2SO
4
3Fe2(SO
4)
3+ Cr
2(SO
4)
3+ K
2SO
4+ 6(NH
4)
2SO
4+ 43H
2O
Redox changes :
1
ME
4FeSO ;
6
ME
722 OCrK
14. One litre of acidified KMnO4 solution containing 15.8 g KMnO4 is decolorized by passing sufficient
SO2. If SO2 is produced by FeS2, what is the amount of FeS2 required to give desired SO2 ?
Ans. 15 g.
Sol. v.f. of KMnO4
= 5 & v.f. of SO2
= 2
Now, Eq. of KMnO4
= Eq. of SO2
5/158
8.15= moles of SO
2× 2
so, moles of SO2
= 1/4
Now, applying POAC on S, we get :
2 × mole of FeS2
= 1 × moles of SO2
so, moles of FeS2
=8
1
2
1
4
1
so, weight of FeS2
=8
1× 120 = 15 g.
15. An aqueous solution containing 0.1 g KIO3
(formula weight = 214) and an excess of K was acidified with
HCl. The liberated 2consumed 45 mL of thiosulphate. The molarity of sodium thiosulphate solution is :
The reaction involved is :
O3
– + – + H+ 2
+ H2O
(A) 0.0623 M (B) 0.0313 M (C) 0.126 M (D) 0.252 M
Sol. O3
– + 5– + H+ 32
+ H2O
2Na2S
2O
3+
2 2Na + Na
2S
4O
6
Now, Moles of KO3
=214
1.0
So, Moles of 2
= 3 ×214
1.0
Now, Moles of Na2S
2O
3= 2 × 3 ×
214
1.0
M × VL
= 2 × 3 ×214
1.0 M ×
1000
45= 2 × 3 ×
214
1.0
Now, Molarity of hypo solution = 2 × 3 ×45
1000
214
1.0 = 0.0623 M Ans. (A)
16. A fresh H2O
2solution is labelled 11.2 V. This solution has the same concentration as a solution which is :
(A) 3.4% (w / w) (B) 3.4% (v / v) (C) 3.4% (w / v) (D) None of these
Sol. Molarity of H2O
2=
2.11
2.11
2.11
strength.vol = 1
Now, %(w/v) = mLinsolutionof.wt
ginsoluteof.wt× 100
= Molarity × Mol. wt. of solute ×10
1
= 1 × 34 ×10
1= 3.4% Ans. (C)
"manishkumarphysics.in" 23
CHEMISTRY
17. 100 mL each of 1 N H2O
2and 11.2 V H
2O
2solution are mixed, then the final solution is equivalent to :
(A) 3 M H2O
2solution (B) 0.5 N H
2O
2solution
(C) 25.5 g/L H2O
2solution (D) 2.55 g/L H
2O
2Sol.
Sol. Nfinal
=100100
1006.5
2.111001
VV
VNVN
21
2211
= 3/2 = 1.5 N
So, Molarity =2
5.1
.f.v
Normality = 0.75 M
Strength of solution in g/L = Molarity × Mol. wt. = 0.75 × 34 = 25.5 g/L Ans. (C)
18. Calculate the percentage of available chlorine in a sample of 3.55 g of bleaching powder which was
dissolved in 100 mL of water. 25 mL of this solution, on treatment with KI and dilute acid, required
20 mL of 0.125 N sodium thiosulphate Sol.
Ans. 10 %
Sol. CaOCl2 + H2O Ca(OH)2 + Cl2
Cl2 + 2K 2KCl + 2
2 + 2Na2S2O3 Na2S2O6 + 2Na
In 25 mL solution,
moles of Na2S2O3 =1000
20×
1
125.0= 25 × 10–4
So, moles of 2 =2
1× moles of Na2S2O3
=2
1× 25 × 10–4 = 12.5 × 10–4
So, in 100 mL solution, moles of Cl2 = 4 × 12.5 × 10–4 = 50 × 10–4
So, weight of Cl2 = 50 × 10–4 × 71 g
% of available Cl2 =55.3
711050 4
× 100 = 10%