REDOX REACTION
Calculation of oxidation number
1.
One of the following has both positive and negative oxidation states
(1) F
2.
1
3
(2) – 1
(3) + 1
(4) – 3
(2) + 4
(3) + 3
(4) + 6
The oxidation number of Phosphorus in Mg2P2O7 is :
(1) + 3
5.
(4) Na
The oxidation number of cobalt in K3[Co(NO2)6] is
(1) 0
4.
(3) He
Oxidation number of nitrogen in (NH4)2SO4 is
(1) –
3.
(2) Cl
(2) + 2
(3) + 5
(4) – 3
(3) – 2
(4) – 1
(3) + 4
(4) + 8
Which of the following is a redox reaction?
(1) NaCl + KNO3 ⎯⎯→ NaNO3 + KCl
(2) CaC2O4 + 2 HCl ⎯⎯→ CaCl2 + H2C2O4
(3) Mg (OH)2 + 2 NH4Cl ⎯⎯→ MgCl2 + 2 NH4OH
(4) Zn + 2 AgCN ⎯⎯→ 2 Ag + Zn (CN)2
6.
The oxidation number of Oxygen in Na2O2 is :
(1) + 1
7.
The oxidation state of osmium (Os) in OsO4 is
(1) + 7
8.
(2) + 6
In which of the following compounds, the oxidation number of iodine is fractional ?
(1) F7
9.
(2) + 2
(2)
–
3
(3) F5
(4) F3
Phosphorus has the oxidation state of +3 in
(1) Phosphorous acid
(2) Orthophosphoric acid
(3) Hypophosphorous acid
(4) Metaphosphoric acid
10.
In which of the following compounds, nitrogen has an oxidation state of –1 ?
(1) N2O
(2) NO2–
(3*) NH2OH
(4) N2H4
Balancing of redox reactions
11.
A reducing agent is a substance :
(1) in which an element undergoes increase in oxidation number.
(2) in which an element undergoes decrease in oxidation number.
(3) which gains electron(s)
12.
Which of the following is a redox reaction:
(1) 2 CrO42– + 2H+ → Cr2O72– + H2 O
(2) CuSO4 + 4 NH3 → [Cu (NH3)4] SO4
→ Na2S4O6 + 2Na
(4) Cr2O72– + 2OH– → 2 CrO42– + H2 O
(3) 2Na2 S2O3 +
13.
(4) which shares electron(s)
2
Which reaction does not represent auto redox or disproportionation reaction :
(1) Cl2 + OH– ⎯⎯→ Cl– + ClO3– + H2O
(3) 2Cu+ ⎯⎯→ Cu2+ + Cu
14.
(4) (NH4)2Cr2O7 ⎯⎯→ N2 + Cr2O3 + 4H2O
The compound that can work both as an oxidising as well as a reducing agent is :
(1) KMnO4
15.
(2) 2H2O2 ⎯⎯→ H2O + O2
(2) H2O2
(3) HNO3
(4) K2Cr2O7
When KMnO4 acts as an oxidising agent and ultimately forms MnO42–, MnO2, Mn2O3 and
Mn2+, then the number of electrons transferred in each case is :
(A) 4, 3, 1, 5
16.
(B) 1, 5, 3, 7
(C) 1, 3, 4, 5
Consider the following reaction:
3Br2 + 6CO32 – + 3H2O ⎯⎯→ 5Br – + BrO3– + 6 HCO3–
Which of the following statements is true regarding this reaction:
(1) Bromine is oxidized and the carbonate radical is reduced.
(2) Bromine is reduced and the carbonate radical is oxidized.
(3) Bromine is neither reduced nor oxidize
(4) Bromine is both reduced and oxidized.
(D) 3, 5, 7, 1
17.
Consider the reaction,
Zn + Cu2+ ⎯⎯→ Zn2+ + Cu
With reference to the above, which one of the following is the correct statement ?
(1) Zn is reduced to Zn2+
(2) Zn is oxidised to Zn2+
(3) Zn2+ is oxidised to Zn
18.
In the reaction X – + XO3– + H+ ⎯⎯→ X2 + H2O, the molar ratio in which X – and XO3– react is
(1) 1 : 5
19.
(4) Cu2+ is oxidised to Cu.
(2) 5 : 1
(3) 2 : 3
(4) 3 : 2
which of the following behaves as both oxidising and reducing agents ?
(1) H2SO4
(2) SO2
(3) H2S
(4) HNO3
Classical Concept of Equivalent weight / Mass, Equivalent weight,
n-factor and Normality for Acid, Base and Precipitate
20.
When N2 is converted into NH3, the equivalent weight of nitrogen will be
(1) 1.67
21.
(2) 2.67
(3) 3.67
(4) 4.67
In the ionic equation 2K+BrO3– + 12H+ + 10e– ⎯⎯→ Br2 + 6H2O + 2K+,
the equivalent weight of KBrO3 will be :
(1) M/5
22.
(2) M/2
(3) M/6
(4) M/4
In the conversion NH2OH ⎯⎯→ N2O,
the equivalent weight of NH2OH will be :
(1) M/4
23.
(3) M/5
(4) M/1
The equivalent weight of MnSO4 is half its molecular weight when it is converted into
(1) Mn2O3
24.
(2) M/2
(2) MnO4–
(3) MnO2
(4) MnO42–
The equivalent weight of Mohr's salt, FeSO4.(NH4)2SO4.6H2O is equal to
(1) Its molecular weight
(3) half-its molecular weight
(2) Atomic weight
(4) one-third its molecular weight
25.
In the reaction : Na2S2O3 + 4Cl2 + 5H2O ⎯⎯→ Na2SO4 + H2SO4 + 8HCl,
the equivalent weight of Na2 S2 O3 will be : (M = molecular weight of Na2S2O3)
(1) M/4
26.
(2) 1 : 2
(3) 2 : 1
(4) 1 : 4
(2) M/2
(3) M/5
(4) M/4
The equivalent weight of phosphoric acid (H3PO4) in the reaction :
(1) 59
29.
(4) M/2
If molecular weight of KMnO4 is 'M', then its equivalent weight in acidic medium would b
(1) M
28.
(3) M/1
Equivalent weight of carbon in CO and CO2 are in the ratio of :
(1) 1 : 1
27.
(2) M/8
(2) 49
(3) 25
(4) 98
When HNO3 is converted into NH3, the equivalent weight of HNO3 will be :
(1) M/2
(2) M/1
(3) M/6
(4) M/8
(M = molecular weight of HNO3)
30.
31.
Which of the following relations is incorrect for solutions ?
(1) 3 N Al2(SO4)3 = 0.5 M Al2(SO4)3
(2) 3 M H2SO4 = 6 N H2 SO4
(3) 1 M H3 PO4 = 1/3 N H3PO4
(4) 1 M Al2 (SO4)3 = 6 N Al2(SO4)3
28 NO3– + 3As2S3 + 4H2O 6 ࢮAsO43– + 28NO + 9SO42– + 8H+.
What will be the equivalent mass of As2S3 in above reaction : (Molecular mass of As2S3 = M)
28 NO3– + 3As2S3 + 4H2O 6 ࢮAsO43– + 28NO + 9SO42– + H+.
(1)
M
2
(2)
M
4
(3)
M
24
(4)
M
28
Titration
32.
If 25 mL of a H2SO4 solution reacts completely with 1.06 g of pure Na2CO3, what is the
normality of this acid solution :
(1) 1 N
33.
(2) 0.5 N
(3) 1.8 N
(4) 0.8 N
How many millilitres of 0.1N H2SO4 solution will be required for complete reaction with a
solution containing 0.125 g of pure Na2CO3 :
(1) 23.6 mL
(2) 25.6 mL
(3) 26.3 mL
(4) 32.6 mL
34.
Equivalent mass of a bivalent metal is 32.7. Molecular mass of its chloride is :
(1) 68.2
35.
(2) 103.7
(3) 136.4
(4) 166.3
10 mL of 1 N HCl is mixed with 20 mL of 1 M H2SO4 and 30 mL of 1 M NaOH. The resultant
solution has :
(1) 20 meq of H+ ions
(2) 20 meq of OH–
(3) 0 meq of H+ or OH–
(4) 30 milli moles of H+
Hydrogen peroxide, Hardness of water
36.
The volume strength of 1.5 N H2O2 solution is :
(1) 4.8 V
37.
(2) 8.4 V
(3) 3 V
(4) 8 V
Temporary hardness is due to bicarbonates of Mg2+ and Ca2+. It is removed by addition of
CaO as follows :
Ca(HCO3)2 + CaO ⎯⎯→ 2CaCO3 + H2O
Mass of CaO required to precipitate 2 g CaCO3 is :
(1) 2 g
38.
(3) 0.28 g
(4) 1.12 g
(3) 0.6 N
(4) 0.8 N
The normality of "1.12 V H2O2" solution is
(1) 0.4 N
39.
(2) 0.56 g
(2) 0.2 N
The mass of oxalic acid crystals (H2C2O4 . 2H2O) required to prepare 50 mL of a 0.2 N solution
is :
(1) 4.5 g
(2) 6.3 g
(3) 0.63 g
(4) 0.45 g
Equivalent Concept for Redox reactions, KMnO4 / K2Cr2O7 v/s Reducing Agents & their
Redox Titration
40.
0.16 gm of dibasic acid required 25 ml of decinormal NaOH solution for complete
neutralisation. The molecular weight of the acid will be :
(1) 128
41.
(2) 56
(3) 28
(4) 240
The number of moles of oxalate ions oxidized by one mole of MnO4– ion in acidic medium is
(1)
5
2
(2)
2
5
(3)
3
5
(4)
5
3
42.
43.
Which of the following solutions will exactly oxidize 25 mL of an acid solution of 0.1 M iron
( ) oxalate:
(1) 25 mL of 0.1 M KMnO4
(2) 25 mL of 0.2 M KMnO4
(3) 25 mL of 0.6 M KMnO4
(4) 15 mL of 0.1 M KMnO4
If equal volumes of 0.1 M KMnO4 and 0.1 M K2Cr2O7 solutions are allowed to oxidise Fe2+ to
Fe3+ in acidic medium, then Fe2+ oxidised will be :
(1) more by KMnO4
(2) more by K2Cr2O7
(3) equal in both cases
(4) cannot be determined.
Solutions
Calculation of oxidation number
1. Sol.
Oxidation state of F may be zero and – 1.
Oxidation state of Cl may be – 1 to + 7
Oxidation state of He is zero
Oxidation state of Na may be 0 and + 1
2. Sol.
x + 4 (+1) = +1
x = –3
3. Sol.
[Co(NO2)6]3–
x + 6 (–1) = – 3
x = +3
4.
Sol.
2(+2) + 2x + 7 (–2) = 0
x = +5
5.
Sol.
The oxidation states show a change only in reaction
6.
Sol.
2(+1) + 2 x = 0
x=–1
7.
Sol.
x + 4 (–2) = 0
x = +8
8.
sol.
–
3
3(x) = –1
x=–
1
3
9.
Sol.
Phosphorous acid
oxidation state of P = + 3
Orthophosphoric acid
oxidation state of P = + 5
Hypophosphorous acid
oxidation state of P = + 1
Metaphosphoric acid
oxidation state of P = + 5
10.
Sol.
NH2OH
x + 3(+1) + 1(–2) = 0
x = –1
11.
Sol.
A reducing agent itself oxidises & reduces others. so an element shows increase in
oxidation number in it & loss of electron occurs.
12.
Sol.
S oxidises from +2 to + 2·5 and I reduces from 0 to – 1. Hence 3rd reaction is a redox
reaction.
13.
Sol.
In 4th reaction, N undergoes oxidation while Cr undergoes reduction.
14.
Sol.
In H2O2 oxidation state of oxygen is –1.
It can undergo both oxidation as well as reduction
O– – e– → O (Oxidation)
O– + e– → O2– (reduction)
Hence it can act both as oxidizing as well as reducing agent.
15.
Sol.
16.
Sol.
17.
Sol.
Increase in oxidation number is oxidation & decrease in oxidation number is
reduction.
Br2 undergoes disproportionation, i.e. it undergoes both oxidation & reduction.
Here,
Zn → Zn2+ (oxidation)
Cu2+ → Cu (reduction)
18.
Sol.
V.f. = 1
X–
+
XO3– + H+ ⎯⎯→ X2 + H2O
V.f. = 5
Molar ratio = 5 : 1
19.
Sol.
SO2
Here S is in +4 state
It can be oxidized to +6 state as also reduced to –2 state.
N2 ⎯⎯→ NH3
20.
Sol.
0
–3
|________|
n=3 2=6
21.
Sol.
(V.f.) BrO3– = 5
Eq wt = M/5
Eq.wt = mol.wt./ n factor =
28
= 4.67
6
22.
Sol.
V.f. of NH2OH = 2
Eq wt = M/2
23.
Sol.
Equivalent wt. =
M
l.e., valency factor = 2
2
MnSO4 has valence factor 2 in the following reaction.
+2
+4
MnSO 4 ⎯⎯
→ MnO 2
v.f. = 2
24.
Sol.
Valency factor = 1
Equivalent weight =
25.
Sol.
M
1
+6
+2
Na2 S2 O3 ⎯⎯→ Na2 S O 4
the total change in oxidation number = 4 × 2 = 8
ENa2S2O3 =
26.
Sol.
mol. wt.
M
=
V. f
8
In CO
16 g O
12 g C
or 8 g O 6 g C
In CO2
32 g O
12 g C
8gO 3gC
Ratio 6 : 3 = 2 : 1.
27.
Sol.
In acidic medium KMnO4 shows following reaction –
+7
MnO 4– + 8H+ + e– → Mn2+ + 4H2O
V.f. = +5
Equivalent weight =
molecular wt. M
=
v.f.
5
28.
Sol.
Equivalent wt of acid =
Equivalent wt. =
29.
Sol.
molecular wt.
No. of H+ replaced per acid molecule
98
=1
1
+5
−3
HNO3 ⎯⎯
→ NH3
V.f. of HNO3 = 8
Eq. wt. = M/8.
30.
Sol.
Normality = Molarity × v.f.
1M H3PO4 = 3N H3 PO4
31.
Sol.
V.F. of As2S3 is 28.
Reason of this is change in oxidation state of both As and S.
V.F. = 2(5 – 3) + 3[6 – (–2) ] = 4 + 24 = 28.
Eq. wt = M/28
32.
Sol.
Eq. of Na2CO3 = Eq. of H2SO4
1.06
106
2=
25
N
1000
N = 0.8 N
33.
Sol.
m. eq. of H2SO4 = m. eq. of Na2CO3
0.1 ×
V
0.125
=
2
106
1000
V = 23.6 mL
34.
Sol.
EMCl2 = E
M+2
+ ECl−
= 32.7 + 35.5 = 68.2
Molecular mass = 2 × 68.2 = 136.4
= 2 × 68.2 = 136.4
35.
(1) Explanation : No. of meq of H+ = 10 × 1 + 20 × 2 = 50
Sol.
No. of meq of OH– = 30 × 1 = 30
No. of meq of H+ left unreacted = 50 – 30 = 20 meq
Hence , (1) is correct, (2), (3) and (4) are ruled out.
36.
Sol.
1.5 N H2O2
VS = ?
VS = Normality × 5.6 = 1.5 × 5.6 = 8.4 V
37.
Sol.
mole of CaCO3 = 2/100 = 0.02
so mole of CaO =
1
× 0.02 = 0.01
2
so mass of CaO = 0.01 × 56 = 0.56 g
sol(2)
39.
Sol.
H2C2O4. 2H2O = 2 + 24 + 64 + 36 = 126 and Equivalent wt. =
0.2 =
W
1000
126
50
2
40.
Sol.
Eq. of Acid = Eq. of Base
Mass
= MV
E
0.16
1
=
E
10
E = 64
25
1000
W = 0.63 g
126
2
[
H2SO4 , N = 2 M]
E=
M
VF
64 =
M
2
M = 218
41.
Sol.
Equivalents of C2O42– = equivalents of MnO4–
x(mole) × 2 = 1 × 5
x=
5
mole
2
42.
Sol.
Milli equivalents of FeC2O4 = 0.1 × 3 × 25 = 7.5
From choice (4), milli equivalents of KMnO4 = 0.1 × 5 × 15 = 7.5
m. eq. of FeC2O4 = m. eq. of KMnO4
43.
Sol.
m eq of KMnO4 = 0.1 × 5 × V = 0.5 V
& m eq K2Cr2O7 = 0.1 × 6 × V = 0.6 V
So, K2Cr2O7 will oxidise more Fe2+