Solution

NARAIN

Class 11 - Chemistry
1. N2h4 is reductant (or, reducing agent)
¯
¯¯
ClO3 is an oxidant (or, oxidising agent).
Explanation:
The following steps are taken into consideration in order to identify an Oxidant or reductant in an equation-
Step 1
Assign oxidation numbers for each atom in the given equation after writing out all redox couples in the reaction. So, in the given
equation

i. O: N2 -2H4+1 ⟶ N+2 O-2

ii. R: Cl+5 O3-2 ⟶ Cl- (ON = -1)

Step 2
Identify which reactant is being oxidised ( the oxidation number increases when it reacts ) and which reactant is being reduced (
the oxidation number goes down.)
Thus, we find that in the given reaction,
i. The oxidation number of N in N2H4 increases from (-2) to (+2), therefore, N2H4 is the reductant.

ii. The oxidation number of Cl in ClO3- decreases from (+5) to (-1), therefore, ClO3- is the oxidant.

Note that -
i. a reductant (or, reducing agent ) reduces the other substances and loses electrons; therefore its oxidation state /number
increases.
ii. an oxidant (or, oxidising agent ) oxidises other substances and gains electrons; therefore its oxidation state/number decreases.
2. A standard hydrogen electrode is also called as a " reversible electrode " because it can act both as anode as well as cathode in an
electrochemical cell.
Explanation :
Since the Standard reduction potential of Standard hydrogen electrode has been arbitrarily taken as zero , it is capable of
undergoing either oxidation or reduction reaction in a half cell of a galvanic cell, depending upon the nature or Standard reduction
potential of the other electrode being used in the cell

3.

Dividing the equation into two half reactions:

Oxidation half reaction: SO → SO 2−

3
2−

Reduction half reaction: MnO −

4
→ Mn2+
Balancing oxidation and reduction half reactions separately as:
Oxidation half reaction
2− 2−
SO → SO
3 4

SO
2−

3
→ SO + 2e-
2−

Since the reaction occurs in acidic medium,

SO
2−
3
→ SO
2−
4
0 + 2e- + 2H+
SO3
2−
+ H2O → SO 2−

4
+ 2H+ + 2e- ...(i)
Reduction half reaction
MnO
−

4
→ Mn2+
MnO
−

4
+ 5e- → Mn2+
MnO
−

4
+ 8H+ + 5e- → Mn2+ + 4H2O ...(ii)
To balance the electrons, multiply eq. (i) by 5 and eq. (ii) reduction half-reaction by 2.

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 Then add both the equation
2MnO
−

4
+ 5SO 2−

3
+ 6H+→ 2Mn2+ + 5SO 2−

4
+ 3H2O
4. In AgF2, the oxidation state of Ag is +2 which is most unstable. As the stable oxidation state of Ag is +1, it quickly accepts an
electron and thus achieves t a more stable +1 oxidation state as shown below,
Ag+2 + e- ⟶ Ag +1
Therefore, AgF2, if formed, will act as a strong oxidising agent.
5. It is based upon the relative positions of these metals in the activity series.
The correct order of the metals in which they displace each other from their salt solution is
Mg, Al, Zn, Fe,Cu
6. Let x be the oxidation state of Fe in Fe3O4.
3x + 4(-2) + 0
x = +8/3
Hence, the oxidation state of Fe in Fe3O4 is +8/3 but, oxidation number cannot be in the fraction.
Therefore Fe3O4 exists as a mixture of FeO and Fe2O3 in which Fe has an oxidation number of +2 and +3 respectively.
¯
¯¯
7. i. Cr(s) is oxidized and MnO4 is reduced.
ii. Cr(s) is a negative electrode,
MnO4 supported over inert Pt acts as a positive electrode.
¯
¯¯

8. The source of electrical energy in a galvanic cell is the spontaneous redox reaction taking place within the cell.
9. With the help of the standard reduction potential of element and its ions, we can identify the spontaneity of any reaction.
Since Eo of Cu2+ /Cu electrode (+ 0.34 V) is higher than that of H+/H2 electrode (0.0 V), H+ ions cannot oxidize Cu to Cu2+ ions
and hence, Cu does not dissolve in dil. HCI. In contrast, the electrode potential of NO ion, i.e., NO /NO electrode (+0.97 V) is
−

3
−

higher than that of a Cu electrode and hence, it can oxidize Cu to Cu2+ ions and hence Cu dissolves in dil. HNO3 and not in dil.
HCl.

10.

Dividing the equation into two half reactions:

Oxidation half reaction: I- → I2

Reduction half reaction: Cr O → Cr3+2

2−

Balancing oxidation and reduction half reactions separately as:

Oxidation half reaction
I- → I2

2I- → I2

2I- → I2 + 2e- ...(i)

Reduction half reaction
Cr2 O
2−

7
→ Cr3+
Cr2 O
2−

7
→ 2Cr3+
Cr2 O
2−

7
+ 6e- → 2Cr3+
Cr2 O7
2−
+ 14H- + 6e- → 2Cr3+ + 7H2O ...(ii)
Equalise the no. of electron by multiplying oxidation half-reaction by 3
[2I- → I2 + 2e- ] * 3 (oxidation half)

Cr2 O
2−

7
+ 14H- + 6e- → 2Cr3+ + 7H2O (reduction half)
Add two half-reaction to obtain the net reaction.
6I- + Cr 2 O7
2−
+ 14H- → 2Cr3+ + 3I2 + 7H2O

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11. I. Oxidation No. of C in CH3COOH
CH3COOH is an organic compound, wherein there are two C atoms having different oxidation states. The following rule is,
therefore, applicable to calculate the oxidation numbers of both the carbon atoms, separately -
The oxidation state of any chemically bonded carbon atom is assigned by,
i. adding -1 for each more electropositive atom,
ii. and +1 for each more electronegative atom ,
iii. zero for each carbon atom bonded directly with the C of interest.
Thus in CH3COOH, the oxidation numbers of different atoms are,

C-3H3+1C+3O-2O-2H+1
Thus, the oxidation numbers of the elements in CH3 COOH may be tabulated as below,

Element C H C O O H

Oxidation number -3 +1 +3 -2 -2 +1
So, The oxidation numbers of the two carbon atoms in CH3 COOH are 1 (for C linked with more electropositive atom H) & and 3
(for C bonded with more electronegative atom O), respectively.
II. The oxidation number of S in S2O82--

In S2 O8 2- there is one peroxy linkage, hence two out of eight oxygen atoms have (-1) oxidation state. Thus
2 x + 6 (-2 ) + 2 (-1)
= -2
(x represents the assumed oxidation number of S2O82-)
∴ 2x - 12 - 2 = 2
2x = 12
x = +6
So, the oxidation number of S in S2 O8-2 is +6
12. In a disproportionation reaction, the same species is simultaneously oxidised as well as reduced. Therefore, for such a redox
reaction to occur, the reacting species must contain either two or more than two positive or negative oxidation state including zero.
The element, in reacting species must present in intermediate states of higher and lower oxidation state, but in case of flourine,
fluorine does not show a positive oxidation state. That's why fluorine does not show a disproportionation reaction.
13. The electrode potential of Mg2+(aq)/Mg = -2.36 V is lower than the electrode potential of AI3+(aq)/AI electrode = -1.66 V.
Therefore, at Mg2+(ag)/Mg electrode oxidation process occur and it acts as the anode while at
Al3+(aq)/ Al reduction occurs and it acts as the cathode. In other words, Mg loses electrons and Al3+ ion accepts electrons. The
cell reaction is
3 Mg + 2 AI3+ ⟶ 3 Mg2+ + 2 AI
and E = E
0
cell
-E∘
3+ 2+
Mg |Mg
Al |Al

= -1.66 - (-2.36) = + 0.70 V

14. The arrangement of various elements in the order of increasing values of standard reduction potentials is called "Electrochemical
series"..
Charecteristics of electrochemical series :
The substances which are stronger reducing agents than hydrogen are placed above hydrogen in the series and have
negative values of standard electrode potentials.
All those substances which have positive values of reduction potentials and placed below hydrogen in the series are
weaker reducing agents.
The substances which are stronger oxidising agents than the H+ ion are placed below hydrogen in the series .
Activity of metals decreases from top to bottom.
The activity of nonmetals increases from top to bottom.
Metals at the top have the tendency to loose electrons readily . These are active metals .
15. Write the chemical formula for each reactant and product.

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 2+ + 2− 3+ 3+
Fe +H +Cr2 O → Cr +Fe + H2 O
7

+2 −2 +3 +3
+6
2+ + 2− 3+ 3+
Fe +H + Cr2 O → Cr + Fe + H2 O
7

i. Balance the increase and decrease in the reaction

+2 −2 +3 +3
+6
2+ + 2− 3+ 3+
6 Fe +H + Cr2 O → 2 Cr +6Fe + H2 O
7

ii. Balancing H and O atoms by adding H+ and H2O molecules

2+ + 2− 3+ 3+
6Fe +14H +Cr2 O → 2Cr +6Fe +7H2 O
7

16. Standard hydrogen electrode ( SHE ) is a standard equipment for measuring the relative electrode potentials of different electrodes
, and is therefore , also termed reference electrode. As per IUPAC norms its Standard electrode potential ( ΔEo ) is taken as 0.000
volt.
17. Mn cannot exceed its oxidation state beyond 7. In MnO4-, the oxidation state of Mn is +7. Thus, here manganese cannot undergo

oxidation and it can undergo reduction only. That is why disproportionation is not possible whereas in MnO42- manganese is in +6
oxidation state which can be oxidized as well as reduced.
18. Cathode and anode in a galvanic cell can be identified by the following steps ,
1 . Writing the two half cell reactions separately ,
2. study or trace out the nature / type of redox reactions occuring in both the half cells &
3. Inferring about the two electrodes as below ,
(a) At cathode there is gain of electron causing reduction reaction .
(b) At anode there is loss of electrons.causing oxidation reaction.
This can also be made out by writing the cell composition and looking at its design as given below ,
In electro-chemical cell
(i)anode is written on L.H.S while ,
(ii)cathode is written on R.H.S.
19. Let us write the oxidation number of each atom involved in the given reaction above its symbol as:

Here, we have observed that the oxidation number of F increases from 0 in F2 to + 1 in HOF. Also, the oxidation
number decreases from 0 in F2 to -1 in HF. Thus, in the above reaction, F is both oxidized and reduced. Hence, the given reaction
is a redox reaction.
20. 2Na (s) + H2 (g) → 2NaH (s) is a redox change.

2Na (s) → 2Na+ (g) + 2e- and the other half reaction is:
H2 (g) + 2e- → 2H- (g)
This splitting of the reaction into two half-reactions automatically reveals here that sodium is oxidized and hydrogen is reduced.
The half-reaction that involves loss of electron are oxidation reaction and half-reaction involves a gain of an electron are reduction
reaction. hence is the case of sodium and hydrogen atoms respectively. Therefore. the complete reaction is a redox change.
21. Calculations:
i. In H2SO4
Let the oxidation number of S in H2SO4 be x.
Write the oxidation number of each atom above its symbol.
+ 1 x -2

H2 S O4

Calculate the sum of the oxidation numbers of all the atoms, and equate it with zero. Thus,
2(+1) + x + 4(-2) = 0
or, x - 6 = 0
∴ x = + 6

Thus the oxidation number of sulphur in H2SO4 is (+6)

ii. In Na2SO4
Write the oxidation number of each atom its symbol, assuming an oxidation number of S as x.
+ 1 x -2

Na2 S O4

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 Calculate the sum of the oxidation numbers of all the atoms, and equate it with zero,
2(+1) + x + 4(-2) = 0
2+x-8=0
or,( x - 6) = 0
∴ x = + 6

Thus. the oxidation number of S in Na2 SO4 is (+6)

22. Oxidation involves loss of one or more electrons by a species during a reaction.
23. To balance any chemical reaction follow the following step:
Step 1: Find out the oxidation state of each and every element participate in a redox reaction.
Step 2: Identify the oxidized and reduced species and balance them.
Step 3: Balance the O-atom by the help of a water molecule in case of acidic medium or by the help of OH- ion in case of the
basic medium.
Step 4: Balance the unequal H-atom by using H+ ion.
Step 5: Balance the unequal charge by using an electron.
The skeletal ionic equation is,
MnO
−

4
(aq) + Br- (aq) ⟶ MnO2(s) + BrO (aq) −

The balanced chemical reaction is,

− - - −
2MnO (aq) + Br (aq) + H2 O(I) → 2Mn O2 (s) + BrO3 (aq) + 2O H (aq)
4

24. To show disproportionation reaction, an element must be present in its intermediate oxidation state, so that it can oxidised and
reduced simultaneously during the reaction. In ClO4- ion, Chlorine atom exhibits +7 oxidation state which is its maximum
oxidation state therefore it will not get further oxidised, it will get reduced only to form either its minimum oxidation state(-1) or
any other state.
In other ions, Chlorine are in its intermediate state therefore they get oxidised and reduced simultaneously during the reaction to
form maximum(+7) or minimim(-1) oxidation state
25. From Standard reduction potential data we know higher the negative value of Eo , higher is the reducing tendency of an atom and
vice versa.
i. Since E° of Zn is more negative than that of Fe, therefore, Zn will be oxidized to Zn2+ ions while Fe2+ ions will be reduced to
Fe. In other words, Fe will not reduce Zn2+ ions.
ii. Since E° of Fe is more negative than that of Ni, therefore, Fe will be oxidized to Fe2+ ions while Ni2+ ions will be reduced to
Ni. Thus, Fe reduces Ni2+ ions.
26. The increasing order of oxidation numbers of iodine in the given compounds is,
HI < I2 < ICl < HIO2 < KIO3
The oxidation numbers of iodine in the given compounds are tabulated as below in their increasing order -
HI -1

I2 0

ICl +1

HIO2 +3

KIO3 +5

∴ the increasing order of O.N. of iodine is,

HI < I2 < ICl < HIO2 < KIO3
27. In a redox reaction the number of millimoles of any oxidising substance = number of millimoles of any reducing agent.
Number of millimoles of K2Cr2O7 present in 24 mL of 0.5 M solution = 24 × 0.5 = 12
The balanced chemical equation for the redox reaction is
K2Cr2O7 + 6 (NH4)2 SO4 FeSO4.6H2O + 7H2SO4 ⟶ K2SO4 + 6 (NH4 )2SO4 + 3 Fe2(SO4)3 + Cr2 (SO4)3 + 43 H2O
From the balanced equation, 6 moles Mohr’s salt are oxidised by = 1 mole K2Cr2O7
∴ 12 millimoles of Mohr’s salt will be oxidised by = 1

6
× 12 = 2 millimoles K2Cr2O7

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28. (i) It completes the internal circuit in a galvanic cell.
(ii) It maintains the electrical neutrality within the electrolytes of both the half cells of a galvanic cell.
29. Let the oxidation number of C in CH3CH2OH is x
x+3+x-2-2+1=0
2x = 0 ⇒ x = 0
Therefore, average oxidation number of C is zero.
Let see the structure of CH3CH2OH for understanding

Now, oxidation number of C1(i.e. CH​3​- carbon toword left) atom: x - 2 - 2 + 1 = 1 ⇒ x = +2

Oxidation number of C2 (i.e. -COOH carbon toword right) atom: x + 3 = + 1 ⇒ x = -2
That's why average oxidation number of C in CH₃COOH is zero.
30. EMF of a cell is the difference between the electrode potentials ( reduction potentials ) of the cathode and anode in a galvanic cell
when no current is drawn through the cell.
31. A species which loses electrons as a result of oxidation is a reducing agent.
The best reducing agent is Li2+ ion (Lithium ion ).
Explanation :
Li2+ ion is considered as best reducing agent because , it has greatest negative Standard electrode potential ( ΔEo ) as expressed
below,
Li2+ + e- --------> Li(s)

Eo / V
= -3.05
32. A redox couple consists of oxidised and reduced form of the same substance taking part in the oxidation or reduction half
reaction.
33. i. Reaction of carbon with oxygen in which C is a reducing agent while O2 is an oxidizing agent.

a. Reaction of Carbon with oxygen: If excess of carbon is burnt in a limited supply of O2, CO is formed in which the
oxidation state of C is +2.
2Co(s) + O2 ⟶ 2C+2O(g)
(excess )
b. If, however, excess of O2 is used, the initially formed CO gets oxidized to CO2 in which oxidation state of C is + 4.
+2 +4

2C (s) O2 (g) ⟶ 2C O(g); C (s) + O2 (g) + CO2 (g)

(Excess) (Excess)

ii. Reaction of phosphorus with chlorine:

In the reaction of Phophorus with chlorine, P4 is a reducing agent while Cl2 is an oxidizing agent.

a. when excess of P4 is used, PCl3 is formed in which the oxidation state of P is +3.

P04(s) + 6 Cl2 (g) ⟶ 4P3+ + Cl3

b. If, however, excess of Cl2 is used, the initially formed PCl3 reacts further to form PCl5 in which the oxidation state of P is
+5
P04 (s) + 6Cl2 (g) ⟶ 4P+3 Cl3

P04 (s) + 10Cl2 ⟶ 4 P+5 Cl5

iii. Reaction of sodium with oxygen:
Na is a reducing agent while O2 is an oxidizing agent.

a. When excess of Na is used, sodium oxide is formed in which the oxidation state of O is -2.
4Na(s) + O2(g) ⟶ 2 Na2O-2

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 b. If, however, excess of O2 is used, Na2O2 is formed in which the oxidation state of O is -1 which is greater than -2.
−2 −1

4N a(s) + O2 (g) ⟶ N a2 O(g); 2N a(s) + 2O2 (g) ⟶ N a2 O2 (s)

(Excess) (Excess)

34. i.

Here, oxidation state of S gets changed from -2 (H2S) to 0 (S), and Fe3+ changed to Fe2+. So oxidation of Sulphur atom takes
place while Fe get reduced.
The balanced reaction is:
2Fe3+ + H2S ⟶ 2Fe2+ + S + 2H+
ii.

Here, oxidation state of I- changes from -1 to 0 in I2 and IO is reduced to I2 . On solving, we get following balanced
−

equations
2IO
−

3
+ 12 I- + 12H+ ⟶ 7I2 + 6H2O
iii.

Here, Bi(s) is oxidised to Bi 3+

while NO
−
3
is reduced to NO2. The balanced reaction is:

Bi(s) + 3NO + 6H+ ⟶ 3NO2 + Bi

−
3
3+
+ 3H2O
35. It is believed that the photosynthesis reaction occurs in two steps. In the first step, H2O decomposes to give H2 and O2 in the
presence of chlorophyll and the H2 produced reduces CO2, to C6H12O6 in the second step. During the second step, some H2O
molecules are also produced and therefore, the reaction occurs as:
a. i. 12H2O (I) ⟶ 12H2(g) + 6O2 (g)
ii. 6CO2(g) + 12H2(g) ⟶ C6H12O6 (s) + 6H2O(I)
iii. 6CO2(g) + 12H2O(I) ⟶ C6H12O6 (s) + 6H2O(I) + 6O2(g)

Therefore, it is more appropriate to write the reaction for photosynthesis as (III) because it means that 12 molecules of H2O
are used per molecule of carbohydrate and 6H2O molecules are produced per molecule of carbohydrate during the process.
b. O2 is written two times in the product which suggests that 0, is being obtained from the two reactants as:
O3 (g) ⟶ O2(g) + O (g)
H2 O2 (l)+O(g)⟶H2 O(l)+ O2 (g)

O3 (g)+ H2 O2 (l)⟶H2 O(l)+ O2 (g)+ O2 (g)

The path of the reaction can be studied by using H2O18 in reaction (a) or by using H2O18 or O318 in reaction (b).
x +1 −2

36. C H O
2 6

2x + 6 (+1) + 1 (-2) = 0
2x = - 4 or x = -2
Therefore, the average oxidation number of C is -2.
Let us consider the structure of ethanol CH3CH2OH
H H

| |

2 1
H − C − C − OH
| |
H H

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 Oxidation number of C1 atom = 1 (+1) + 2 (+1) + x + 1 (-1) = 0
[C1 atom in CH3CH2OH is attached to one CH3 group (oxidation number = + 1), two H atoms and one -OH group (oxidation
number = -1)]
x=-2
Oxidation number of C2 atom = 3 (+1) + x + 1 (-1) = 0
x = -2
[C2 atom in CH3CH2OH is attached to three H-atoms and one -CH2OH group (oxidation number = - 1)].
+1x

37. K I ;
3

1
Oxidation number of K is +1. 1 (+1) + 3x = 0 or x = − 3
1
Therefore, the average oxidation number of I is − . 3

It is wrong because oxidation number can never be fractional. Let us consider the structure of KI3.
+1

K (I − I ← I )
−1
in this structure, coordinate bond is formed between I2 molecule and I-1 ion.
Hence, the oxidation number of three I atoms in KI3 are 0, (in I2) and - 1 respectively.
38. The redox system is used. In redox systems, the titration method can be adapted to determine the strength of a reductant/oxidant
using a redox-sensitive indicator. The usage of indicators in redox titration is illustrated below:
In one situation, the reagent itself is intensely coloured, e.g. permanganate ion MnO . Here, MnO acts as the self indicator.
−

4
−

The visible endpoint, in this case, is achieved after the last of the reductant (Fe2+ or C2 O
2−

4
) is oxidised and the first lasting tinge

of pink colour appears at MnO-4 concentration as low as 10-6 mol dm-3 and (10-6 mol L-1). This ensures a minimal 'overshoot' in
colour beyond the equivalence point, the point where the reductant and the oxidant are equal in terms of their mole stoichiometry.
39. i. The given redox reaction is ,
Zn (s) 2Ag+ (aq) ⟶ Zn2+ (aq) + 2 Ag (s)
Since Zn (s) gets oxidized, to Zn2+ (aq) ions, and Ag+ (aq) ions gets reduced to Ag (s) metal, therefore, oxidation occurs at the
zinc electrode (acting as anode) and reduction occurs at the silver electrode (as cathode). Thus, the galvanic cell corresponding
to the above redox reaction is depicted as:
Zn(s) | Zn2+ (aq) || Ag+ (aq) | Ag(s)
ii. a. Since oxidation occurs at the zinc electrode, therefore, electrons accumulate on the zinc electrode,/ anode. Hence, zinc
electrode is negatively charged.
b. Electrons move from Zn anode to Ag cathode in the external circuit. Since the direction of current in the external circuit is
opposite to that of the electrons so,
The carriers of current are silver cathode and Zinc anode through an external circuit in a direction from silver cathode to
zinc anode.
c. The reactions occurring at the two electrodes are
At anode:
Zn (s) ⟶ zn2+ (aq) + 2e-
At cathode
Ag+ (aq) + e- ⟶ Ag (s)
40. It is because Br2 is a stronger oxidising agent than I2. Accordingly,

considering the oxidation numbers of S in S2 O32-, S2O 2−

4
SO
2−

4
, we find that:
The average oxidation number of S in S O = + 2
2
2−
3

the oxidation number of S in S O = + 2.5

4
2−

The oxidation number of S in SO = + 6 2−

Hence, it is inferred that,

Since, (i) Br2 is a stronger oxidizing agent than l2, it oxidizes S of S 2 O3
2−
to a higher oxidation state of +6 and hence forms SO 2−

ion. (ii)
l2, however, being weaker oxidizing agent oxidizes S of S 2−
2 O3 ion to lower oxidation of +2.5 in s 4 O6
2−
ion.
It is because of this reason that thiosulphate reacts differently with Br2 and l2.
41. Please try yourself. The balanced equations are :

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 i. P (s) + 6OH (aq) ⟶ PH (g) + 3H PO (aq)
4
−
3 2
−

ii. 3N H (l) + 4ClO (aq) ⟶ 6NO(g) + 4Cl (aq) + 6H O(l)

2 4
−

3
−
2

iii. Cl O (g) + 4H O (aq) + 2OH (aq) ⟶ 2ClO − (aq) + 4O

2 7 2 2
−
2 2 (g) + 5H2 O(l)

+2 +2

42. a. 2PbO +4HCl → 2Pb Cl2 +2H2 O

In reaction (a), the oxidation number of none of the atoms undergoes a change. Therefore, it is not a redox reaction. It is an
acid-base reaction because PbO is a basic oxide that reacts with HCl acid.
+4 +2 0

b. Pb O 2 +4HCl → Pb Cl2 + Cl 2 +2H2 O

The reaction (b) is a redox reaction in which PbO2 gets reduced and acts as an oxidizing agent.
43. i. Halogens have a strong tendency to accept electrons. Therefore, they are strong oxidizing agents. Their relative oxidizing
power is however, measured in terms of their electrode potentials.
Since the electrode potentials of halogens decrease in the order :
F2 < (2.87V) > Cl2(+1.36V) > Br2(+1.09V) > I2(+0.54V),
Therefore, their oxidizing power decreases in the same order.
This is evident from the observation that

a. F2 oxidizes Cl- to Cl2, Br- to Br2, I- to I2

b. Cl2 oxidizes Br- to Br2 and I- to I2 but not F- to F2.

c. Br2, however oxidizes I- to I2. But Br2 fails to oxidise F- to F2 and Cl- to Cl2
The related above reactions are,
F2(g) + 2Cl-(aq) → 2F-(aq) + Cl2(g); F2(g) + 2Br-(aq) → 2F-(aq) + Br2(l)

F2(g) + 2I-(aq) → 2F-(aq) + I2(s); Cl2(g) + 2Br-(aq) → 2Cl-(aq) + Br2(l)

Cl2(g) + 2I-(aq) → 2Cl-(aq) + I2(s) and Br2(l) + 2I- → 2Br-(aq) + I2(s)

Thus, F2 is the best oxidant.

ii. Conversely, halide ions have a tendency to lose electrons and hence can act as reducing agents. Since the electrode potentials
of halide ions decrease in the following order,
I-(-0.54 V) > Br-(-1.09V) > CI-(-1.36V) > F- (-2.87 V),
Therefore, the reducing power of the halide ions or their corresponding hydrohalic acids decreases in the same order: HI >
HBr > HCl > HF.
Thus, hydroiodic acid is the best reductant. This is supported by the following reactions.
For example,
HI and HBr reduces H2SO4 to SO2 while HCI and HF do not.
2HBr + H2SO4 → Br2 + SO2 + 2H2O; 2HI + H2SO4 → I2 + SO2 + 2H2O

Further I- reduces Cu2+ to Cu+ but Br- does not.

2Cu2+ (aq) + 4I-(aq) → Cu2I2(s) + I2(aq);

Cu2+(aq) + 2Br- → No reaction

So, HI is a stronger reductant than HBr.
Further, among HCl and HF, HCl is a stronger reducing agent than HF because HCl reduces MnO2 to Mn2+ but HF does not.
MnO2(s) + 4HCl(aq) → MnCl2(aq) + Cl2(g) + 2H2O
MnO2(s) + 4HF(l) No reaction
Thus, the reducing character of hydrohalic acids decreases in the order: HI > HBr > HCl > HF.
44. i. CuO(s) + H (g) ⟶ Cu(s) + H O(g)
2 2

In this case, O.N. of Cu decreases from +2 (in CuO) to 0 (in Cu) and that of H increase from 0 (in H2) to +1 (in H2 O).
Therefore, CuO is reduced to Cu while H2 is oxidised to H2O. Thus, this is a redox reaction.
+ 3 −2 + 2 − 2 0 + 4 − 2

ii. Fe 2 O3 + 3C O → 2Fe(s) + 3C O
2

O.N. of Fe decrease from +3 (in Fe2O3) to 0 (in Fe) and that of C increases from +2 (in CO) to +4 (in CO2). Therefore, Fe2O
is reduced while CO is oxidised. Thus, this is a redox reaction.

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 +3 −1 +1 +3 +1 −3 +1 +1 −1 +3 −1

iii. 4 B Cl 3
(s) + 3Li Al H4 (s) → 2 B H6 (g) + 3Li Cl(s) + 3AI CI(s)
2 3

O.N. of B decrease from +3 (in BCI3) to -3 in B2H6 and that of H increases from -1 (in LiAIH4) to + 1 (in B2H6). Therefore,
BCI3 is reduced and LiAIH4 is oxidised. Thus, this is a redox reaction.
+1 −1
0 0

iv. 2 K(s) + F 2 (g) → 2K

+
+ F
−1
(s)

O.N. of K decrease from 0 (in K) to +1 (in K+) and that of F increases from 0 in (F2) to -1 (in F-). Hence, K has been oxidised
while F2 has been reduced. Therefore,it is a redox reaction.
−3 +1 0 +2 −2 +1 −2

v. 4 N H 3 (g) + 5O(g) → 4N O (g) + 6 H2 O (g)

2

In this case, O.N. of N decreases from -3 (in NH3) to +2 (in NO) and that of O decreases from 0 (in O2) to -2 (in H2O).
Therefore, NH3 has been oxidised while O2 has been reduced. Hence,it is a redox reaction.
45. i.

ii.

iii.

46. i. H2SO5,
By conventional method
+1 X −2

H S O
2 5

2(+1) + x + 5 (-2) = 0 or x = + 8
This is not possible because S cannot have O.N. more than 6. The fallacy is overcomed if we calculate its O.N. from its
structure as :

2 × (+1) + X + 2(−1) + 3 × (−2) = 0

for S
for H (for other O atom)

or x = + 6
ii. Cr2O 2 -
7

2 x + 7 (-2) = -2
2x = 12
This is correct and there is no fallacy.

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 iii. N . By conventional method.
−

x + 3(-2) = -1 or x = +5
From the structure.

x + 1 (−1) + 1 (−2) + 1 (−2) = 0 or x = +5

(for O) (for = O) (for→O)

Hence there is no fallacy about O.N. of N in N as +5 whether we calculate by conventional method or by chemical bonding.
−

47. i. HgCI2
ii. NiSO4
iii. SnO2
iv. TI2SO4
v. Fe2(SO4)3
vi. Cr2O3

48. In H2S4O6, let the oxidation number of S be x.

+1 x −2

H2 S 4 O 6

2 (+1) + 4x + 6 (-2) = 0
4x = + 10 or = + = + 2.5
10

Let us consider the structure of H2S4O6

O O

|| 0 0 ||
+5
+5
H − O − S − S − S − S − OH
|| ||

O O

In H2S4O6 the oxidation number of each of two S-atoms which are linked with each of the other by a single bond (in the centre) is
zero and each of the remaining two S-atoms both side is +5. Hence, the oxidation number of 4 S-atoms in H2S4O6 is +5, 0, 0, and
+5 respectively.
49. i. F.
Fluorine being the most electronegative element shows only a -ve oxidation state of -1.
ii. Cs.
Alkali metals because of the presence of a single electron in the valence shell, exhibit an oxidation state of + 1 only.
iii. I
Due to the presence of d -orbitals and seven electrons in the valence shell of iodine (I), it shows an oxidation state of
a. -1, in compounds of iodine with more electropositive elements with an oxidation state of +1 (such as H, Na, K Ca, etc.)
b. positive oxidation states of +2, +5 +7 in its compounds with more electronegative elements (such as i.e., O, F, etc.).
c. Ne
Explanation:
It is an inert gas with high ionization enthalpy and high positive electron gain enthalpy. Hence, it exhibits neither +ve nor -ve
oxidation states in its compounds.
50. To balance a chemical redox reaction. first of all, find the oxidized and reduced species by identifying their oxidation number and
then balance them according to their loss and gain of electron individually.
i. Skeleton of equation
+2 −3 0 0 −2

Cu O + N H3 → C u + N + H2 O
2

The oxidation number of Cu decreases from +2 to 0 and oxidation number (O.N.) of N-atom increases from –3 to 0.
In order to balance the increase of O.N. with a decrease of O.N., there should be three atoms of copper and two atoms of
nitrogen. Hence balancing hydrogen and oxygen atoms we have,
3 CuO + 2 NH3 → 3Cu + N2 + 3H2O
ii. Writing K2MnO4 twice Oxidation number of Mn, we have the skeleton of the equation
+6 +4 +7

K2 M n O4 + H2 O → M n O2 + K M n O4 + K OH

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 The oxidation number of Mn in 1 mole K2MnO4 decreases from +6 to +4 (MnO2) and in the other 1 mole increases from +6
to +7 (KMnO4) i.e. 1 mole gains two electrons while the other loses 1 electron.
In order to balance the Oxidation number of Mn, 1 mol. K2MnO4 and KMnO4 are multiplied by 2. Hence
K2MnO4 + 2K2MnO4 + H2O → MnO2 + 2KMnO4 + KOH
In order to balance the number of K and H atoms, KOH is multiplied by 4 and H2O by 2.
3K2MnO4 + 2H2O → MnO2 + 2KMnO4 + 4KOH
51. The given equations for different reactions are :
+2−2 0 0 +1−2

i. CuO(s) + H(g) → C u(s) + H2 O (g)

2

As per above equation, it is noted that.

a. an atom of oxygen (O) is removed from CuO,
it is reduced to Cu, while
∴

b. O is added to H2 to form H2O

∴ it is oxidized.
Further,
Oxidation number. of Cu decreases from +2 in CuO to 0 in Cu,
oxidation number of H increases from 0 in H2 to +1 in H2O.
∴ CuO is reduced to Cu but H2 is oxidized to H2O.
Thus, the reaction is a redox reaction.
+ 3 -2 + 2 0 + 4

ii. Fe2 O3 (s) + 3 CO (g) → 2 F e(s) + 3C O (g)

2

In the above equation for reaction, it is seen that

a. The oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe, and
oxidation number of C increases from +2 in CO to +4 in CO2.
Further,
oxygen is removed from Fe2O3, and
added to CO to form CO2
therefore, Fe2O3 is reduced while CO is oxidized.
Thus, the given reaction is a redox reaction.
iii. Similarly, in the given equation ,
4B+3Cl-13(g) + 3Li+1 Al+3 H-14 (s) ⟶ 2 B-32 H-16 (g) + 3LI+1 Cl-1 (s) + 3 Al+3 Cl-13
Oxidation number of B decreases from +3 in BCl3 to -3 in B2H6
while,
oxidation number of H increases from -1 in LiAlH4 to +1 in B2H6.
Therefore, BCl3 is reduced and
LiAlH4 is oxidized.
Further, it is noted that,
H is added to B forming B2 H6 from BCl3 but is removed from LiAlH4,
therefore,
BCl3 is reduced while LiAlH4 is oxidised.
Thus,
the redox nature of above reaction is justfied.
52. i. 1. Let the oxidation state of Mn atom be x, other atom like O-atom have -2 while K-atom have +1.

1 × ( +1) + (1 × x) + 4 × (- 2) = 0
1 + x + (-8) = 0
1+x-8=0

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 x-7=0
∴ x = +7

Hence, the oxidation number of Mn in MnO4 is +7.

ii. N in NO . −

Let the oxidation number of N in NO be x and other atom like O- atom is -2.
−

∴ Sum of the oxidation numbers of all the atoms in NO = x + 3(-2) = x - 6

−

But the sum of oxidation numbers of all the atoms in NO ion is equal to the charge present on it i.e., -1
−

∴ x - 6 = -1
or x = +5
Thus, the oxidation numbers of N in NO is +5.
−

53. i. F: being most electronegative; shows only a -ve oxidation state of -1.
ii. Cs: Alkali metals have only one electron in their valence shell and hence exhibit only +1 oxidation state.
iii. Ne: It is an inert gas and therefore, does not exhibit-ve or +ve O.S.
iv. The oxidation state of Ne is zero. It exhibits neither negative nor positive oxidation states. Hence, Ne is the element that
exhibits neither the negative nor does the positive oxidation state.
54. i.

Thus, here, H2S undergoes oxidation because oxidation number of Sulphur atom get increased from -2 to 0
while Cl2 undergoes reduction because oxidation number of Cl atom get decreased from 0 to -1.
ii.

Thus, here Fe3O4 undergoes reduction because oxidation number of Fe atom get decreased while AI atom undergoes
oxidation because Oxidation number of Al atom get increased from 0 to +3.
iii. 2 Na(s) + H2(g) → 2 NaH(s)
Since Na is more electro-positive than H so, here Na atom get oxidised by losing electron while H-atom get reduced by
accepting electron.

55. i. An aqueous solution, AgNO3 ionises to give Ag+ (aq) and NO3 (aq) ions.
¯
¯¯

AgNO3(aq) ⟶ Ag+(aq) + NO3 (aq)

¯
¯¯

Thus, when an electric current is passed through AgNO3 solution,

Ag+(aq) ions move towards the cathode

and
NO3 (aq) ions move towards the anode.
¯
¯¯

∴ at the cathode, either Ag+(aq) ions or H2O molecules may get reduced.
Which of these will actually get discharged would depend upon their electrode potentials given as below:
Ag+(aq) + e- ⟶ Ag(s); Eo = + 0.80 V
2H2O(l) + 2e- ⟶ H2(g) + 2OH-(aq); Eo = -0.83 V

Since the reduction electrode potential of Ag+ (aq) ions is higher than that of H2O molecules, therefore, at the cathode, Ag+

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 (aq) ions (rather than H2O molecules) are reduced/discharged.
Similarly, at the anode, either Ag metal of the anode or H2O molecules may be oxidised.
Their reduction electrode potentials are:
Ag(s) ⟶ Ag+ (aq) + e-; Eo = - 0.80 V
2H2O(l) ⟶ O2(g) + 4H+ (aq) + 4e-; Eo = -1.23 V
Since the electrode potential of Ag is much higher than that of H2O,
∴ the anode metal (ie. Ag) gets oxidized and (not the H2O molecule).
¯
¯¯
It may, however, be mentioned here that the oxidation potential of NO3 ions is even lower than that of H2O since more bonds
¯
¯¯
are to broken during the reduction of NO3 ions than those in H2O.

Thus, when an aqueous solution of AgNO3 is electrolysed, Ag from Ag anode dissolves while Ag+ (aq) ions present in the
solution get reduced and get deposited on the cathode,
ii. In aqueous solution, CuCl2 ionises as follows:

CuCl2(aq) ⟶ Cu2+(aq) + 2Cl-(aq)

On passing electric current, Cu2+(aq) ions move towards the cathode and Cl-(aq) ions move towards anode.
Thus, at the cathode, either Cu2+(aq) or H2O molecules can get reduced.
Their electrode potential are:
Cu2+(aq) + 2e- ⟶ Cu(s); Eo = +0.34 V
H2O(l) + 2e- ⟶ H2(g) + 2OH-; Eo = -0.83 V

Since the electrode potential of Cu2+(aq) ions is much higher than that of H2O,at the cathode, Cu2+(aq) ions are reduced (and
not H2O molecules.)

Similarly, at the anode either Cl-(aq) ions or H2O molecules are oxidized.
Their oxidation potentials are:
2Cl-(aq) ⟶ Cl2(g) + 2E-; ΔEo = -1.36 V

2H2O(l) ⟶ O2(g) + 4H+(aq) + 4e-; ΔEo = -1.23 v

56. Step 1: The skeletal ionic equation is: Cr O (aq) + SO (aq) → Cr (aq) + SO (aq)
2
2−

7
2−
3
3+ 2−

Step 2: Assign oxidation numbers for Cr and S Cr O (aq) + SO (aq) → Cr(aq) + SO (aq)
2
2−

7
2−

3
2−

This indicates that the dichromate ion is the oxidant and the sulphite ion is the reductant
Step 3: Calculate the increase and decrease of oxidation number, and make them equal from step-2
The oxidation state of chromium changes from +6 to +3. There is a decrease of +3 in the oxidation state of chromium on the right-
hand side of the equation while the left-hand side has +6 oxidation state. The oxidation state of sulphur changes from +4 to +6.
There is an increase of +2 in the oxidation state of sulphur on the right-hand side while left-hand side has +4 oxidation state.
Thus we get Cr O (aq) + 3SO (aq) → 2Cr (aq) + 3SO (aq)
2
2−

7
2−

3
3+ 2−

Step 4: As the reaction occurs in the acidic medium, and further the ionic charges are not equal on both the sides, add 8H+ on the
left to make ionic charges equal Cr O (aq) + 3SO (aq) + 8H → 2Cr (aq) + 3SO (aq)
2
2−

7
2−

3
+ 3+

n
2
4

Step 5: Finally, count the hydrogen atoms, and add an appropriate number of water molecules (i.e., 4H2O) on the right to achieve
balanced redox change. Cr 2−
2 O7 (aq) + 3SO
2−

3
(aq) + 8H
+
(aq) → 2Cr
3∗
(aq) + 3SO
2−

4
(aq) + 4H2 O(l)

57. a. Writing the oxidation number of on each atom,

+1−1 +1+5 −2 0 +3 −2 −1 +1 −2

3HCl(aq) + HN O3 → Cl 2 (g)+ N O Cl(g)+ 2H2 O (l)

Here, O.N. of Cl increases from -1 (in HCl) to 0 (in Cl2).

Therefore, Cl- is oxidized, and hence HCl acts as a reducing agent.

The O.N. of N decreases from +5 (in HNO3) to +3 (in NOCl) and therefore, HNO3 acts as an oxidizing agent.
Thus, the reaction is a redox reaction.
+2−1 +1−1 +2−1 +1−1

b. HgCl (aq)+ 2kI (aq) → HgI (aq)+ KCl (aq)

2 2

Here O.N. of none of the atoms undergo a change (oxidation/ reduction) and therefore, this is not a redox reaction.

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 +3−2 +2−2 Δ 0 +4−2

c. Fe O + 3CO(g) −
2 3 → 2F e(s) + 3CO (g) 2

Here, the oxidation number of Fe decreases from (in Fe2O3) to 0 (in Fe), and therefore, Fe2O3 acts as an oxidizing agent.
O.N. of C increases from +2 (in CO) to +4 (in CO2) and therefore, CO acts as a reducing agent. Thus, this is a redox reaction.
+3 −1 +1 −2 +1 −1

d. P Cl 3 (l) + 3 H 2 O (l) → 3H Cl (aq) + H3 O2 (aq)

Here oxidation number of none of the atoms undergo a change (oxidation/ reduction) and therefore, it is not a redox reaction.
-3 + 1 0 0 +1 −2

e. 4N H (g) + 3O (g) → 2N (g) + 6 H O (g)

3 2 2 2

Here, the oxidation number of N increases from -3 (in NH3) to 0 in (N2), and therefore, NH3 acts as a reducing agent. O.N. of
O decreases from 0 (in O2) to -2 (in H, O), and therefore, O2 acts as an oxidizing agent.
Thus, this reaction is a redox reaction.
58. The given equations for different reactions are as follows:
+2−2 0 0 +1−2

i. CuO(s) + H (g) → C u(s) + H O (g)

2 2

As per the above equation, it is noted that.

i. an atom of oxygen (O) is removed from CuO,
∴ it is reduced to Cu, while

ii. O is added to H2 to form H2O

∴ it is oxidized.
Further,
Oxidation number. of Cu decreases from +2 in CuO to 0 in Cu,
the oxidation number of H increases from 0 in H2 to +1 in H2O.
∴ CuO is reduced to Cu but H2 is oxidized to H2O.
Thus, the reaction is a redox reaction.
+ 3 -2 + 2 0 + 4

ii. Fe 2 O3 (s) + 3 CO (g) → 2 F e(s) + 3C O 2 (g)

In the above equation for the reaction, it is seen that

1. The oxidation number of Fe decreases from +3 in Fe2O3 to 0 in Fe, or the addition of electron involved so Fe2O3 is
reduced to Fe.
the oxidation number of C increases from +2 in CO to +4 in CO2.
therefore, Fe2O3 is reduced while CO is oxidized.
Thus, the given reaction is a redox reaction.
iii. Similarly, in the given equation,
4B+3Cl-13(g) + 3Li+1 Al+3 H-14 (s) ⟶ 2 B-32 H-16 (g) + 3LI+1 Cl-1 (s) + 3 Al+3 Cl-13
The oxidation number of B decreases from +3 in BCl3 to -3 in B2H6.
while the oxidation number of H increases from -1 in LiAlH4 to +1 in B2H6.
Therefore, BCl3 is reduced and LiAlH4 is oxidized to B2H6.
The redox nature of the above reaction is justified.
iv. Let us write the oxidation number of each element involved in the given reaction as:
K = 0 & +1
F = 0 & -1
The oxidation number of K increases from 0 in K to + 1 in KF i.e. K is oxidised to KF. on the other hand, the O.N. of F
decreases from 0 in F2 to -1 in KF i.e. F2 is being reduced to KF.
Hence, the given reaction is a redox reaction.
v. Let us write the oxidation number of each element involved in the given reaction as:
In NH3, N = -3 & H = + 1
In NO, N = + 2 & O = -2
In H2O, H = + 1 & O = -2
In O2, O = 0
Here, the oxidation number of N increases from -3 in NH3 to + 2 in NO. on the other hand, the O.N. of O2 decreases from 0 to

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 -2 i.e. O2 is reduced.
Hence, the given reaction is a redox reaction.
59. i. Electrolysis of aqueous solution of AgNO3 using silver electrodes:
+ −
AgNO 3 (s) + nH2 O ⟶ Ag (aq) + NO (aq)
3

+ −
H2 O ⇌ H + OH

At cathode: Ag+ ions have lower discharge potential than H+ ions.

Hence Ag+ ions will be deposited as silver (in preference to H+ ions).
At anode: Since the silver electrode is attacked by NO ions, Ag anode will dissolve to form Ag+ ions in the solution.
−

3
+ −
Ag ⟶ Ag + e

ii. Electrolysis of aqueous solution of AgNO3 using platinum electrodes:

At cathode: same as above.
At anode: Since silver is not attackable, out of OH- and NO ions, OH- ions have lower discharge potential and hence OH-
−
3

ions will be discharged in preference to NO The OH- will decompose to give 02.
−

− −
OH (aq) ⟶ OH + e
−
4OH (aq) ⟶ 2H2 O(l) + O2 (g)

iii. Electrolysis of H2SO2 with Pt electrodes:

+ 2−
H2 SO4 (aq) ⟶ 2H (aq) + SO (aq)
4
+ −
H2 O ⇌ H + OH

At cathode: H +
+ e
−
⟶ H

H + H ⟶ H2 (g)

At anode: OH −
⟶ OH + e
−

4OH ⟶ 2H2 O + O2 (g)

iv. Electrolysis of aqueous solution of CuCl2 with platinum electrodes:

2+ −
CuCl2 (s) + (aq) ⟶ Cu (aq) + 2Cl (aq)
+ −
H2 O ⇌ H + OH

At cathode: Cu2+ will be reduced in

2+ −
Cu + 2e ⟶ Cu

At anode: Cl- ions will be oxidised in preference to OH-ions

− −
Cl ⟶ Cl + e

Cl + Cl ⟶ Cl2

Thus copper will be deposited on the cathode and Cl, will be liberated at anode.
60. a. In manufacture of benzoic acid from toluene, alcoholic potassium permanganate is used because of the following reasons ,
i. the cost of adding acid or a base is avoided , because in neutral medium OH- are produced in the reaction itsef. Thus ,the
process becomes cost effective .
ii. KMnO4 and alcohol , both being polar , are homogeneous to each other. Toluene and alcohol are also miscible with each
other forming a homogeneous solution .
Thus the reaction mixture becomes homgeneous .
iii. Further, kinetically reactions proceed at a faster rate in homogeneous medium . Hence in presence of alcohol , KMnO4 and
toluene can react at a faster rate .making the process still more economical.
The balanced redox reaction in neutral medium is given below:
− − −
MnO (aq) + 2H2 O(l) + 3e ⟶ MnO2 (s) + 4OH (aq)1 × 2
4

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 b.
When Conc. H SO is added to an inorganic mixture containing chloride, a pungent smelling gas HCl is produced
2 4

because a stronger acid displaces a weaker acid from its salt .

Since HCl is a very weak reducing agent, it can not reduce H SO to SO and hence HCl is not oxidized to Cl .
2 4 2 2

However, when the mixture contains bromide ion, the initially produced HBr being a strong reducing agent than HCl , it
reduces H SO to SO and is itself oxidized to produce red vapour of Br .
2 4 2 2

The reaction takes place in following two steps:

2NaBr + 2H2 SO4 ⟶ 2NaHSO 4 + 2HBr

2HBr + H2 SO4 ⟶ Br2 + SO2 + 2H2 O

61. Step 1: First we write the skeletal ionic equation, which is

MnO4
−
(aq) + I– (aq) → MnO2(s) + I2(s)
Step 2: The two half-reactions are:
Oxidation half-reaction : (-1)I–(aq) → (0)I2 (s)
Reduction half-reaction: MnO (aq) → MnO2(s) −

Step 3: To balance the I atoms in the oxidation half-reaction, we rewrite it as 2I– (aq) → I2 (s)
Step 4: To balance the O atoms in the reduction half-reaction, we add two water molecules on the right and to balance H atom we
add 4H+ ions on left :
MnO
−

4
(aq) + 4 H+(aq) → MnO2(s) + 2H2O (l)
As the reaction takes place in a basic solution, therefore, for four H+ ions, we add four OH– ions to both sides of the equation:
MnO
−

4
(aq) + 4H+ (aq) + 4OH–(aq) → MnO2 (s) + 2 H2O(l) + 4OH– (aq)
Replacing the H+ and OH– ions with water.
the resultant equation is:
MnO
−

4
(aq) + 2H2O (l) → MnO2(s) + 4 OH– (aq)
Step 5 :In this step we balance the charges of the two half-reactions in the manner depicted as:
2I– (aq) → I2 (s) + 2e–

MnO
−

4
(aq) + 2H2O(l) + 3e– → MnO2(s) + 4OH–(aq)
Now to equalise the number of electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2.
6I–(aq) → 3I2 (s) + 6e-

2 MnO (aq) + 4H2O (l) +6e– → 2MnO2(s) + 8OH– (aq)

−

Step 6: Add two half-reactions to obtain the net reactions after canceling electrons on both sides
6I–(aq) + 2MnO (aq) + 4H2O(l) → 3I2(s) + 2MnO2(s) + 8OH–(aq)
−

Step 7: Verify that the equation contains the same type and the same number of an atom and the same charges on both sides of the
equation. This last check reveals that the equation is fully balanced with respect to the number of atoms and the charges.
62. Step -1: Separate the equation into two half-reactions.
The oxidation number of various atoms are shown below:
+6 −2 +2 +1 +3 +3 −1 −2
2− 2 + + 3 + 3 +
C r2 O + F e + H → C r + F e + H 2 O
7

In this case, chromium undergoes reduction, oxidation number decreases from +6 (in Cr 2−
2 O7 ) to +3 (in Cr3+)
Fe2+ (O.N. = +2) changes to Fe3+ (O.N. = +3). The species undergoing oxidation are reduction are:
Oxidation half: Fe2+ → Fe3+
Reduction half; Cr O → Cr3+ 2
2−

Step - 2: Balance each half-reaction separately as:

Fe2+ → Fe3+
i. Balance all atoms other than H and 0. This step is not needed, because, it is already balanced.

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 ii. The oxidation number on left is +2 and on right is +3. To account for the difference, the electron is added to the right as:
Fe2+ → Fe3+ + e-
iii. Changes in already balanced.
iv. No need to add H or O.
The balanced half equation is:
Fe2+ → Fe3+ + e- ...(i)
Consider the second half equation
Cr2 O
2−

7
→ Cr3+
a. Balance the atoms other than H and O.
Cr2 O 2Cr3+ 2−

7
→

b. The oxidation number of chromium on the left is +6 and on the right is +3. Each chromium atom must gain three
electrons. Since there are two Cr atoms, add 6e- on the left.
Cr2 O
2−

7
+ 6e- → 2Cr3+
c. Since the reaction takes place in acidic medium add 14H+ on the left to equate the net charge on both sides.
+ 6e- + 14H+ → 2Cr3+
Cr2 O
2−

d. To balance Fl atoms, add 7H2O molecules on the right.

Cr2 O
2−

7
+ 6e- + 14H+ → 2Cr3++7H2O ...(b)
This is the balanced half equation.
Step - 3: To equalise the number of the electron in both reactions multiply oxidation half by 6
Fe2+ → [Fe33+ + e-] × 6
Step 4: add the two half-reaction to achieve the overall reaction and cancel the electron on each side.
2− − + 3+
Cr2 O +6e +14H ⟶2Cr +7H2 O
7

2+ 2− + 3+ 3+
6Fe + Cr2 O +14H ⟶6Fe +2Cr +7H2 O
7

The balanced equation is:

6Fe2+ + Cr 2 O7
2−
+ 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
63. Let the oxidation number of an underlined atom be x.
+1 +1 x −2

i. Na H 2 P O
4

1(+1) + 2(+1) + x + 4 (-2) = 0 ∴ x = +5

Oxidation number of P in NaH2PO4 is + 5
+1 +1 x −2

ii. Na H S O
4

1(+1) + 1(+1) + x + 4 (-2) = 0

x-6=0∴x=+6
+1 x −2

iii. N 4 P2 O
7

4(+1) + 2(x) + 7(-2) = 0

2 x - 10 = 0 ∴ x = + 5
+1 x −2

iv. K 4 Mn O
4

2(+1) + x + 4(-2) = 0
x-6=0∴x=+6
+2 x

v. Ca O
4

+ 2 + 2(x) = 0 ∴ x = + 1
vi. In NaBH4, H is present as hydride ion. Therefore, its oxidation number is -1.
Thus,
+1 x −1

Na B H4

∴ 1(+1) + x + 4(−1) = 0 or x = +3
Thus, the oxidation number of B in NaBH4 = + 3.

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 +1 x −2
vii.
H2 S2 O7

∴ 2(+1) + 2(x) + 7(−2) = 0 or x = +6

Thus, the oxidation number of S in H2S2O7 = + 6
+1 + 3 x − 2 +1 − 2
viii.
KAl(SO4 ) 12 (H2 O)
2

or
+1 + 3 + 2x + 8(−2) + 12(2 × 1 − 2) or x = +6
Thus, the oxidation number of s in KAl(SO4)2, 12H2O = +6
64. Step 1:
The two half-reactions involved in the given reaction are:
-1

Oxidation half-reaction: I (aq) → I

2
0
(s)

+7

Reduction half-reaction: Mn O −
(aq) → Mn O
+4

2
(aq)
4

Step 2:
Balancing I in the oxidation half-reaction, we have:
2l-(aq) ⟶ l2(s)

Now, to balance the charge, we add 2 e- to the RHS of the reaction.

2l-(aq) ⟶ l2(s) + 2e-
Step 3:
In the reduction half-reaction, the oxidation state of Mn has reduced from +7 to +4. Thus, 3 electrons are added to the LHS of the
reaction.
MnO-4(aq) + 3e-⟶ MnO2(aq)

Now, to balance the charge, we add 4 OH- ions to the RHS of the reaction as the reaction is taking place in a basic medium.
MnO-4(aq) + 3e- ⟶ MnO2(aq) + 4OH-
Step 4:
In this equation, there are 6 O atoms on the RHS and 4 O atoms on the LHS. Therefore, two water molecules are added to the
LHS.
MnO-4(aq) + 2H2O + 3e- ⟶ MnO2(aq) + 4OH-
Step 5:
Equalising the number of electrons by multiplying the oxidation half reaction by 3 and the reduction half reaction by 2, we have:
6l-(aq) ⟶ 3l2(s) + 2e-

2MnO-4(aq) + 4H2O + 6e- ⟶ 2MnO2(s) + 8OH-(aq)

Step 6:
Adding the two half reactions, we have the net balanced redox reaction as:
6l-(aq) + 2MnO-4(aq) + 4H2O(l) ⟶ 3l2(s) + 2MnO2(s) + 8OH-(aq)
65. i. VO : x + 2 (-2) = + 1 or x = +5
+

2
––
ii. –
UO
–
: x + 2 (-2) = + 2 or x = +6
2+
2

iii. Ba–Xe6 : +2(2) +x + 6(-2) or x = +8

–
6

iv. K – P O : 1 × 4 + 2x + 7 (-2) or x = +5
–
4 2 7

v. K S :1 × 2 + x or x = -2
2
–
–

66. i. Carbon has a high affinity towards oxygen so it is generally used as a reducing agent. However, in the case of highly reactive
metals, it cannot be used as a reducing agent.
ii. Ozone is a source of oxygen generally behaves as an oxidizing agent.
iii. As we know the addition of hydrogen is called reduction so nascent hydrogen behaves as a reducing agent.
iv. Nitric acid is a good oxidant i.e., oxidizing agents as it readily provides oxygen.
v. Chlorine can behave as oxidant as well as reductant depending upon the nature of the other reactant.
vi. SO2 can also behave as an oxidant as well as a reductant.

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67. Whenever a reaction between an oxidizing agent and a reducing agent is carried out, a compound of lower oxidation state is
formed if the reducing agent is in excess and a compound of higher oxidation state is formed if the oxidizing agent is in excess.
This can be illustrated as follows:
i. (Carbon) C is a reducing agent, while (oxygen) O2 acts as an oxidizing agent.
If an excess of C is burnt in the presence of an insufficient amount of O2, then CO will be produced, wherein the Oxidation
number of C is + 2.
C(excess) + O2→ CO
Let the oxidation number of C be x in CO.
x + 1(-2) = 0 {oxidation number of O = -2}
x=2
On the other hand, if C is burnt in an excess of O2, then CO2 will be produced wherein the oxidation number of C is + 4.
C + O2(excess) → CO2
Let the oxidation number of C be x in CO2.
X + 2(-2) = 0 {oxidation number of O = -2}
X=4
ii. (Potassium) P4 and F2 are reducing and oxidizing agents respectively.
If an excess of P4 is treated with F2, then PF3 will be produced, wherein the oxidation number is + 3.
P4(excess) + F2→ PF3
Let the oxidation number of P be x in PF3.
x + 3(-1) = 0 {O.N of F = -1}
x=3
However, P is treated with an excess of F2, then PF5 will be produced wherein the oxidation number of P is + 5.
P4 + F2(excess) → PF5
Let the oxidation number of P be x in PF5.
x + 5(-1) = 0 {.O.N. of F = -1}
x=5
iii. (potassium) K acts as a reducing agent, whereas (oxygen) O2 is an oxidizing agent.
If an excess of K reacts with O2 then K2O will be formed wherein the oxidation number of O is -2.
4K(excess) + O2→ 2K2O
Let the oxidation number of O be x in K2O.
2(+ 1) + x = 0 {.O.N. of K = + 1}
x = -2
However, if K reacts with an excess of O2, then K2O2 will be formed wherein the Oxidation number of O is -1.
2K + O2(excess) → K2O2
let the oxidation number of O be x in K2O2.
2(+ 1) + 2x = 0 {.O.N. of K = + 1}
x = -1
68. We need to write a skeletal equation for the reaction of chlorine with sulphur dioxide in water and balance the skeletal equation by
ion-electron method
The skeletal equation is:
Cl2 (aq) + SO2 (aq) + H2 O(l) ⟶ Cl-(aq) + SO2(aq)
In order to balance the skeletal equation by ion-electron method, the steps followed are, F
Write reduction half cell equation and oxidation half cell equation separately.
Reduction half equation is:
Cl2(aq) ⟶ Cl-(aq)
Balance Cl atoms,
0 −1
−
C l2 (aq) → 2 C l (aq)

Balance oxidation number of chlorine by adding electrons:

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 Cl2(aq) + 2e- ⟶ 2Cl-(aq) ....(i)
Oxidation half-equation:
+6
+4
2− −
SO (aq) → SO (aq) + 2e
4
2

Balance oxidation number of S by adding electrons:

2− −
SO 2 (aq) → SO (aq) + 2e
4

Balance charge by adding 4H+ ions

2− + −
SO 2 (aq) → SO (aq) + 4H (aq) + 2e
4

Balance O atoms by adding 2H2O at a side opposite to that of H+

SO 2 (aq) + 2H2 O(l) → SO
2−

4
(aq) + 4H
+ −
(aq) + 2e .... (ii)
Now adding eq. (i) and eq. (ii), we have,
Cl2(aq) + SO2(aq) + SO2(aq) + 2H2O(l) ⟶ 2Cl-(aq) + SO42-(aq) + 4H+(aq)
The above equation represents a balanced redox equation for the reaction of chlorine iwth sulphur dioxide in water.
69. a. Na2S2O3
Structure of Na2S2O3 is
1
S

2
Na − O− S −O − Na
||
O

The oxidation state of S1 is -2.

Let the oxidation state of S2 be x.
2 × (+1) + 3(-2) + x + 1 × (-2) = 0
(For Na) (For O) (For coordinate S)
+2 - 6 + x - 2 = 0
Thus, the oxidation states of two S atoms in Na2S2O3 are -2 and +6
b. Na2S4O6
O O

|| ||

1 2 3 4
Na − O− S − S − S − S −O − Na
|| ||

O O

For the left, oxidation number of S1 = (-2) + (-2) + (-1) = +5

S2 = 0
S3 = 0
S4 = +5
c. Na2SO3
Oxidation number of S is 2 × (+1) + x + 3 × (-2) = 0
x = +4
d. Na2SO4
Oxidation number of S is +2 + x + (-8) = 0
x = +6
70. a. MnO + 5S02 + 2H2O + H+ → 5HSO + Mn2+
−
4
−
4

Balancing by ion-electron method:

+7 −2 +4−2 +2 +1+6−2

MN O
−

4
+ SO2 → M n
2+
+ HSO
−

4
(Skeletal ionic equation)
The two half-reaction are
Oxidation half: SO2 → HSO −

Reduction half: MnO −

4
→ Mn2+
+ 2e-2ClO
−

Oxidation half: SO2 → HSO −

4
2

SO2 + 2H2O → HSO −

4
+ 3H- + 2e- ...(i)
(Add 2H2O molecules to balance O atoms in reduction half)

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 Reduction half:
MnO
−

4
+ 5e- → Mn2+
MnO
−

4
+ 8H+ + 5e- → Mn2+ + 4H2O ...(ii)
(Add 4H2O molecules to balance O atoms and H atoms)
Add oxidation and reduction half-reaction.
− + − 2+
[ MnO + 8H + 5e → Mn + 4H2 O] × 2
4
− + − 2+
[MnO +8H +5e → Mn +4H2 O]×2
4

− + − 2+
2MnO +5SO 2 +2H2 O+ H →5HSO +2Mn
4 4

a. 3N2H4 + 4ClO −

3
→ 6NO + 4Cl- + 6H2O
Balancing by oxidation number method:

6N2H4 + 8ClO −
3
→ 12NO + 8Cl- + 12H2O

b. Cl2O7(g) + 4H2O2(aq) → 2ClO (aq) + 3H2O(l) + 4O2(g) + 2H+(aq) −

Balancing by ion-electron method:

Oxidation half reaction:
H2O2 → O2 + 2e-

Adding 2H+ on the right side to balance H atoms and change.

H2O2 → O2 + 2H+ + 2e-
Reduction half reaction:
Cl2O7 + 8e- → 2ClO −

Adding H2O and H+ to balance H and O atoms

Cl2O7 + 8e- + 6H+ → 2ClO + 3H2O −

Adding oxidation half and reduction half-reaction.

− +
[ H2 O2 → O2 + 2e + 2H ] × 4
− + −
Cl2 O7 +8e +6H →2ClO +3H2 O
2

− +
Cl2 O7 +4H2 O2 →2ClO +3H2 O+4O2 +2H
2

71. The balanced equation for the reaction along with its stoichiometry is:
1100k

4N H2 (g) + 5O2 (g) −−−→ 4N O(g) + 6H2 O(g)

4×17= 68g 5×32=160g Pt 4×30=120g

∴ stoichiometrically ,
68 g of NH3 will react will O2
= 160 g
∴ 10 g of NH3 will react with O2 (g)

= 160×10

68
g
= 23.53 g
But the amount of O2 which is actually available
= 20.0 g
So , this amount (ie. 20 g ) of available O2 is lesser than the amount which is needed (ie.23.6 g ).
Therefore, O2 is the limiting reagent for this reaction .
∴ calculations must be based upon the amount of O2 taken and not on the amount of NH3 taken.
Again , as per stoichiometry of the reaction / equation
160 g of O2 produces NO
= 120 g

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 ∴ 20 g of O2 will produce
NO = 120

60
× 20
= 15 g
Thus the maximum weight of nitric oxide that can be obtained is 15 g.
72. The oxidant and reductant in the following reactions are as follows:
i. Oxidant: H+
Reductant: Zn
ii. Oxidant: H2O2

Reductant: [Fe(CN)6]4-

iii. Oxidant: [Fe(CN)6]3-

Reductant: H2O2
iv. Oxidant: F2
Reductant: BrO −

v. Oxidant: NaClO3
Reductant: 2I2

73. For spontaneity of the reaction, Ecell value must be positive because the relation between standard Gibbs free energy and Ecell
value is given as ΔG = -n F Ecell
From the formula it is clear that higher the positive value of Ecell, more negative will be the value of ΔG and higher will be the
spontaneity.
To calculate Ecell, the following formula is used:

Ecell = Eocathode - Eoanode

Here both Eo values are of standard reduction potential.

It may be noted that whenever any half-reaction equation is multiplied by an integer, its E° is not multiplied by that integer.

i. Fe3+(aq) and I-(aq)

The possible reaction between Fe3+(aq) and I-(aq) is as follows:
2Fe3+ (aq) + 2I- (aq) ⟶ 2Fe2+ (aq) + I2 (s)
Oxidation half-reaction:
2I- (aq) ⟶ I2 (s) + 2e- ; E° = -0.54 V
Reduction half-reaction:
2Fe3+ (aq) + 2e- ⟶ 2Fe2+ (aq) ; E° = + 0.77 V
Overall reaction
2Fe3+ (aq) +2I- (aq) ⟶ 2Fe2+ (aq) + I2 (s); E° = + 0.23 V
Positive emf indicates that the reaction is feasible because ΔG will be negative.
ii. The possible reaction between Ag+ (aq) and Cu(s) is as follows:
2Ag+ (aq) + Cu (s) ⟶ Cu2+ (aq) + 2 Ag (aq)
Separate the equation into two half reactions and write electrode potential for each half-reaction.
Oxidation half-reaction:
Cu(s) ⟶ Cu2+(aq) + 2e- ; E° = - 0.34 V
Reduction half reaction:
2Ag+ (aq) + 2e- ⟶ 2Ag(s); E° = + 0.80 V
Overall reaction:
Cu(s) + 2Ag+(aq) ⟶ Cu2+(aq) + 2Ag; E° = + 0.46 V
Positive emf indicates that the reaction is feasible because ΔG will be negative.
iii. The possible reaction between Fe3+(aq) and Cu(s) occurs according to the following equation,
2Fe3+(aq) + Cu(s) ⟶ 2Fe2+ + Cu2+
Oxidation half-reaction:

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 Cu(s) ⟶ Cu2+ + 2e-; E° = -0.34 V
Reduction half-reaction:
2Fe3+(aq) + 2e- ⟶ 2Fe2+(aq) ; E° = + 0.77 V
Overall reaction:
Cu(s) + 2Fe3+(aq) ⟶ Cu2+(aq) + 2Fe2+ (aq); E° = + 0.43 V
Positive emf indicates that the reaction is feasible because ΔG will be negative.
iv. The possible reaction between Fe3+(aq) and Ag(s) occurs according to the following equation
Ag(s) + Fe3+ (aq) ⟶ Ag+ (aq) + Fe2+ (aq)
oxidation half-reaction
Ag(s) ⟶ Ag+ (aq) + e- ; Eo = -0.80 V
Reduction half-reaction
Fe3+(aq) + e- ⟶ Fe2+(aq) ; E° = + 0.77 V
Overall reaction
Ag(s) + Fe3+(aq) ⟶ Ag+(aq) + Fe2+(aq); E° = - 0.03 V
Negative emf indicates that the reaction is not feasible because ΔG will be positive.
v. The possible reaction between Br2(aq) and Fe2+ (aq) occurs according to the following equation.

Br2(aq) + 2Fe2+(aq) ⟶ 2Br- + 2Fe3+(aq)

Oxidation half reaction
2Fe2+(aq) ⟶ 2Fe3+(aq) + 2e- ; E° = - 0.77 V
Reduction half reaction
Br2(aq) + 2e- ⟶ 2Br-(aq); E° = + 1.09 V
Overall reaction
2Fe2+(aq) + Br2(aq) ⟶ 2Fe3+(aq) + 2Br- (aq); E° = + 0.32 V
Positive emf indicates that the reaction is feasible because ΔG will be negative.
74. i. In PbO2, lead is present in the +4 oxidation state, whereas the stable oxidation state of lead in PbO is +2. PbO2 thus can act as

an oxidant (oxidizing agent) and, therefore, can oxidise Cl– ion of (hydrochloric) HCl into chlorine. We know that PbO is a
basic oxide. Therefore, the reaction Pb3O4 + 8HCl → 3PbCl2 + Cl2 + 4H2O can be split into two reactions namely:
2PbO + 4HCl → 2PbCl2 + 2H2O (acid-base reaction)
PbO2 (+4) + 4HCl(–1) → PbCl2 (+2)+ Cl2 (0) +2H2O (redox reaction)
ii. Since ( nitric acid) HNO3 itself is an oxidizing agent, therefore, it is unlikely that the reaction may occur between PbO2 and
HNO3. However, the acid-base reaction occurs between PbO and HNO3 as 2PbO + 4HNO3 → 2Pb(NO3)2 + 2H2O
It is the passive nature of PbO2 against HNO3 that makes the reaction different from the one that follows with HCl.
75. The following steps are processed to write the desired balanced ionic equation,
The skeletal equation is,
Mn3+ (aq) ⟶ Mn2+ (aq) + MnO2 + H+ (aq)
Identify the oxidation and reduction of half equations from the skeletal equation
Oxidation half-equation:
3 + + 4
3 +
M n (aq) → Mn O (s)
2

Balancing half equations in the following steps,

i. balance O.N. by adding electrons at the suitable side ( ie. RHS )
Mn
3+
(aq) → MnO2 (s) + e-

ii. Balance charge by adding 4H+ ions on RHS,

3 +

Mn (aq) ⟶ MnO2(s) + 4H+(aq) + e-

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iii. Balance O atoms by adding 2H2O on a side opposite to that of H+ ion :
3 +

Mn (aq) + 2H2O(l) ⟶ MnO2(s) + 4H+(aq) + e- (i)

Reduction half equation:
3 + + 2
3 + 2 +
M n (aq) → M n

Balance O.N. by adding electrons on the suitable side (ie.LHS)

Mn3+ (aq) + e- ⟶ Mn2+ (aq) (ii)
Adding Eq. (i) and Eq. (ii), the balanced equation for the ionic reaction is obtained as under ,
2Mn3+ (aq) + 2H2O(l) ⟶ MnO2(s) + Mn2+ (aq) + 4H+(aq)

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