Redox Reactions Practice Questions - DPP for JEE

 Page 1


1. (d)
The oxidation state shows a change only in (d)
2. (b) Magnesium provides cathodic protection and prevent rusting or
corrosion.
3. (c) . In this reaction oxidation
number of S is decreasing from + 6 to +4 hence undergoing
reduction and for HI oxidation Number of I is increasing from –1
to 0 hence underegoing oxidation therefore H
2
SO
4 
is
 
acting as
oxidising agent.
4. (a) (i) Mn
n+
 + ne
–
  M, for this reaction, high negative value of E
° indicates lower reduction potential, that means M will be a good
reducing agent.
(ii) Element F         C l        Br          I
Reduction potential +2.87    +1.36     +1.06     +0.54
(E° volt)
As reduction potential decreases from fluorine to iodine, oxidising
nature also decreases from fluorine to iodine.
(iii)  The size of halide ions increases from F
–
 to I
–
. The bigger ion can
loose electron easily. Hence the reducing nature increases from HF
to HI.
5. (d) Order of decreasing electrode potentials of  Mg, K, Ba and Ca is
Mg > Ca > Ba > K
It can be explained by their standard reduction potentials.
Page 2


1. (d)
The oxidation state shows a change only in (d)
2. (b) Magnesium provides cathodic protection and prevent rusting or
corrosion.
3. (c) . In this reaction oxidation
number of S is decreasing from + 6 to +4 hence undergoing
reduction and for HI oxidation Number of I is increasing from –1
to 0 hence underegoing oxidation therefore H
2
SO
4 
is
 
acting as
oxidising agent.
4. (a) (i) Mn
n+
 + ne
–
  M, for this reaction, high negative value of E
° indicates lower reduction potential, that means M will be a good
reducing agent.
(ii) Element F         C l        Br          I
Reduction potential +2.87    +1.36     +1.06     +0.54
(E° volt)
As reduction potential decreases from fluorine to iodine, oxidising
nature also decreases from fluorine to iodine.
(iii)  The size of halide ions increases from F
–
 to I
–
. The bigger ion can
loose electron easily. Hence the reducing nature increases from HF
to HI.
5. (d) Order of decreasing electrode potentials of  Mg, K, Ba and Ca is
Mg > Ca > Ba > K
It can be explained by their standard reduction potentials.
Highly negative value of  shows the least value of electrode
potential.
6. (c) On balancing the given reaction, we find
3Na
2
HAsO
3
 + NaBrO
3
 + 6HCl
6NaCl + 3H
3
AsO
4
 + NaBr
7. (c) In SO
3
? – –
x + 3(– 2) = – 2; x = + 4
In S
2
O
4
? – –
? 2x + 4(– 2) = – 2
2x – 8 = – 2
2x = 6; x = + 3
In 
2x + 6(– 2) = – 2
2x = 10; x = + 5
hence the correct order is
S
2
O
4
? – –
 < SO
3
? – –
 < S
2
O
6
? – –
8. (a) CrO
2
 Cl
2
Let O. No. of Cr = x
? x + 2 (–2) + 2 (–1) = 0
x – 4 – 2 = 0
? x = + 6
9. (a) If an electronegative element is in its lowest possible oxidation
state in a compound or in free state. It can function as a powerful
reducing agent.
e.g. I
–
10. (b) In iodometry, K
2
Cr
2
O
7
 liberates I
2
 from iodides (NaI or KI). Which
is titrated with Na
2
S
2
O
3
 solution.
Page 3


1. (d)
The oxidation state shows a change only in (d)
2. (b) Magnesium provides cathodic protection and prevent rusting or
corrosion.
3. (c) . In this reaction oxidation
number of S is decreasing from + 6 to +4 hence undergoing
reduction and for HI oxidation Number of I is increasing from –1
to 0 hence underegoing oxidation therefore H
2
SO
4 
is
 
acting as
oxidising agent.
4. (a) (i) Mn
n+
 + ne
–
  M, for this reaction, high negative value of E
° indicates lower reduction potential, that means M will be a good
reducing agent.
(ii) Element F         C l        Br          I
Reduction potential +2.87    +1.36     +1.06     +0.54
(E° volt)
As reduction potential decreases from fluorine to iodine, oxidising
nature also decreases from fluorine to iodine.
(iii)  The size of halide ions increases from F
–
 to I
–
. The bigger ion can
loose electron easily. Hence the reducing nature increases from HF
to HI.
5. (d) Order of decreasing electrode potentials of  Mg, K, Ba and Ca is
Mg > Ca > Ba > K
It can be explained by their standard reduction potentials.
Highly negative value of  shows the least value of electrode
potential.
6. (c) On balancing the given reaction, we find
3Na
2
HAsO
3
 + NaBrO
3
 + 6HCl
6NaCl + 3H
3
AsO
4
 + NaBr
7. (c) In SO
3
? – –
x + 3(– 2) = – 2; x = + 4
In S
2
O
4
? – –
? 2x + 4(– 2) = – 2
2x – 8 = – 2
2x = 6; x = + 3
In 
2x + 6(– 2) = – 2
2x = 10; x = + 5
hence the correct order is
S
2
O
4
? – –
 < SO
3
? – –
 < S
2
O
6
? – –
8. (a) CrO
2
 Cl
2
Let O. No. of Cr = x
? x + 2 (–2) + 2 (–1) = 0
x – 4 – 2 = 0
? x = + 6
9. (a) If an electronegative element is in its lowest possible oxidation
state in a compound or in free state. It can function as a powerful
reducing agent.
e.g. I
–
10. (b) In iodometry, K
2
Cr
2
O
7
 liberates I
2
 from iodides (NaI or KI). Which
is titrated with Na
2
S
2
O
3
 solution.
Here, one mole of K
2
Cr
2
O
7
 accepts 6 mole of electrons.
11. (c) 
Hence H
2
SO
3
 is the reducing agent because it undergoes oxidation.
12. (d) 
All undergo disproportionation
13. (d) H
2
S, the oxidation state of S is – 2. So it cannot accept more
electrons because on accepting 2 electrons S accquires a noble gas
configuration. So, it can acts only as a reducing agent by loosing
electron. On the other hand, the oxidation state of S in SO
2
 is + 4
which is an intermediate oxidation state of sulphur so it can 
reduce as well oxidise.
14. (b) The compound undergo oxidation itself and reduces others is
known as reducing agent. In this reaction O. N. of Ni changes from
0 to + 2 and hence Ni acts as a reducing agent.
15. (c) A reaction, in which a substance undergoes simultaneous oxidation
and reduction, is called disproportionation reaction. In these
reactions, the same substance simultaneously acts as an oxidising
agent and as a reducing agent. Here Cl undergoes simultaneous
oxidation and reduction.
16. (b) Na
2
S
4
O
6
 has the structure :
Page 4


1. (d)
The oxidation state shows a change only in (d)
2. (b) Magnesium provides cathodic protection and prevent rusting or
corrosion.
3. (c) . In this reaction oxidation
number of S is decreasing from + 6 to +4 hence undergoing
reduction and for HI oxidation Number of I is increasing from –1
to 0 hence underegoing oxidation therefore H
2
SO
4 
is
 
acting as
oxidising agent.
4. (a) (i) Mn
n+
 + ne
–
  M, for this reaction, high negative value of E
° indicates lower reduction potential, that means M will be a good
reducing agent.
(ii) Element F         C l        Br          I
Reduction potential +2.87    +1.36     +1.06     +0.54
(E° volt)
As reduction potential decreases from fluorine to iodine, oxidising
nature also decreases from fluorine to iodine.
(iii)  The size of halide ions increases from F
–
 to I
–
. The bigger ion can
loose electron easily. Hence the reducing nature increases from HF
to HI.
5. (d) Order of decreasing electrode potentials of  Mg, K, Ba and Ca is
Mg > Ca > Ba > K
It can be explained by their standard reduction potentials.
Highly negative value of  shows the least value of electrode
potential.
6. (c) On balancing the given reaction, we find
3Na
2
HAsO
3
 + NaBrO
3
 + 6HCl
6NaCl + 3H
3
AsO
4
 + NaBr
7. (c) In SO
3
? – –
x + 3(– 2) = – 2; x = + 4
In S
2
O
4
? – –
? 2x + 4(– 2) = – 2
2x – 8 = – 2
2x = 6; x = + 3
In 
2x + 6(– 2) = – 2
2x = 10; x = + 5
hence the correct order is
S
2
O
4
? – –
 < SO
3
? – –
 < S
2
O
6
? – –
8. (a) CrO
2
 Cl
2
Let O. No. of Cr = x
? x + 2 (–2) + 2 (–1) = 0
x – 4 – 2 = 0
? x = + 6
9. (a) If an electronegative element is in its lowest possible oxidation
state in a compound or in free state. It can function as a powerful
reducing agent.
e.g. I
–
10. (b) In iodometry, K
2
Cr
2
O
7
 liberates I
2
 from iodides (NaI or KI). Which
is titrated with Na
2
S
2
O
3
 solution.
Here, one mole of K
2
Cr
2
O
7
 accepts 6 mole of electrons.
11. (c) 
Hence H
2
SO
3
 is the reducing agent because it undergoes oxidation.
12. (d) 
All undergo disproportionation
13. (d) H
2
S, the oxidation state of S is – 2. So it cannot accept more
electrons because on accepting 2 electrons S accquires a noble gas
configuration. So, it can acts only as a reducing agent by loosing
electron. On the other hand, the oxidation state of S in SO
2
 is + 4
which is an intermediate oxidation state of sulphur so it can 
reduce as well oxidise.
14. (b) The compound undergo oxidation itself and reduces others is
known as reducing agent. In this reaction O. N. of Ni changes from
0 to + 2 and hence Ni acts as a reducing agent.
15. (c) A reaction, in which a substance undergoes simultaneous oxidation
and reduction, is called disproportionation reaction. In these
reactions, the same substance simultaneously acts as an oxidising
agent and as a reducing agent. Here Cl undergoes simultaneous
oxidation and reduction.
16. (b) Na
2
S
4
O
6
 has the structure :
O.N. of two S* atoms are +5 each and that of other two S atoms is zero
each.
17. (a)
Hence, bromine is a stronger oxidising agent than I
2
, as it oxidises S of 
 whereas I
2
 oxidises it only into 
18. (a) Higher the value of reduction potential higher will be the oxidising
power whereas the lower the value of reduction potential higher
will be the reducing power.
19. (c) 
i.e. maximum change in oxidation number is observed in Cl (+5 to –1).
20. (d) Zinc gives H
2
 gas with dil H
2
SO
4
/HCl but not with HNO
3
 because
in HNO
3
, NO
3
? –
 ion is reduced and give NH
4
NO
3
, N
2
O, NO and NO
2
(based upon the concentration of HNO
3
)
Zn is on the top position of hydrogen in electrochemical series. So Zn
displaces H
2
 from dilute H
2
SO
4
 and HCl with liberation of H
2
.
Zn + H
2
SO
4
  ZnSO
4
 + H
2
Page 5


1. (d)
The oxidation state shows a change only in (d)
2. (b) Magnesium provides cathodic protection and prevent rusting or
corrosion.
3. (c) . In this reaction oxidation
number of S is decreasing from + 6 to +4 hence undergoing
reduction and for HI oxidation Number of I is increasing from –1
to 0 hence underegoing oxidation therefore H
2
SO
4 
is
 
acting as
oxidising agent.
4. (a) (i) Mn
n+
 + ne
–
  M, for this reaction, high negative value of E
° indicates lower reduction potential, that means M will be a good
reducing agent.
(ii) Element F         C l        Br          I
Reduction potential +2.87    +1.36     +1.06     +0.54
(E° volt)
As reduction potential decreases from fluorine to iodine, oxidising
nature also decreases from fluorine to iodine.
(iii)  The size of halide ions increases from F
–
 to I
–
. The bigger ion can
loose electron easily. Hence the reducing nature increases from HF
to HI.
5. (d) Order of decreasing electrode potentials of  Mg, K, Ba and Ca is
Mg > Ca > Ba > K
It can be explained by their standard reduction potentials.
Highly negative value of  shows the least value of electrode
potential.
6. (c) On balancing the given reaction, we find
3Na
2
HAsO
3
 + NaBrO
3
 + 6HCl
6NaCl + 3H
3
AsO
4
 + NaBr
7. (c) In SO
3
? – –
x + 3(– 2) = – 2; x = + 4
In S
2
O
4
? – –
? 2x + 4(– 2) = – 2
2x – 8 = – 2
2x = 6; x = + 3
In 
2x + 6(– 2) = – 2
2x = 10; x = + 5
hence the correct order is
S
2
O
4
? – –
 < SO
3
? – –
 < S
2
O
6
? – –
8. (a) CrO
2
 Cl
2
Let O. No. of Cr = x
? x + 2 (–2) + 2 (–1) = 0
x – 4 – 2 = 0
? x = + 6
9. (a) If an electronegative element is in its lowest possible oxidation
state in a compound or in free state. It can function as a powerful
reducing agent.
e.g. I
–
10. (b) In iodometry, K
2
Cr
2
O
7
 liberates I
2
 from iodides (NaI or KI). Which
is titrated with Na
2
S
2
O
3
 solution.
Here, one mole of K
2
Cr
2
O
7
 accepts 6 mole of electrons.
11. (c) 
Hence H
2
SO
3
 is the reducing agent because it undergoes oxidation.
12. (d) 
All undergo disproportionation
13. (d) H
2
S, the oxidation state of S is – 2. So it cannot accept more
electrons because on accepting 2 electrons S accquires a noble gas
configuration. So, it can acts only as a reducing agent by loosing
electron. On the other hand, the oxidation state of S in SO
2
 is + 4
which is an intermediate oxidation state of sulphur so it can 
reduce as well oxidise.
14. (b) The compound undergo oxidation itself and reduces others is
known as reducing agent. In this reaction O. N. of Ni changes from
0 to + 2 and hence Ni acts as a reducing agent.
15. (c) A reaction, in which a substance undergoes simultaneous oxidation
and reduction, is called disproportionation reaction. In these
reactions, the same substance simultaneously acts as an oxidising
agent and as a reducing agent. Here Cl undergoes simultaneous
oxidation and reduction.
16. (b) Na
2
S
4
O
6
 has the structure :
O.N. of two S* atoms are +5 each and that of other two S atoms is zero
each.
17. (a)
Hence, bromine is a stronger oxidising agent than I
2
, as it oxidises S of 
 whereas I
2
 oxidises it only into 
18. (a) Higher the value of reduction potential higher will be the oxidising
power whereas the lower the value of reduction potential higher
will be the reducing power.
19. (c) 
i.e. maximum change in oxidation number is observed in Cl (+5 to –1).
20. (d) Zinc gives H
2
 gas with dil H
2
SO
4
/HCl but not with HNO
3
 because
in HNO
3
, NO
3
? –
 ion is reduced and give NH
4
NO
3
, N
2
O, NO and NO
2
(based upon the concentration of HNO
3
)
Zn is on the top position of hydrogen in electrochemical series. So Zn
displaces H
2
 from dilute H
2
SO
4
 and HCl with liberation of H
2
.
Zn + H
2
SO
4
  ZnSO
4
 + H
2
21. (23)
First half reaction
.... (i)
On balancing
.... (ii)
Second half reaction
.... (iii)
On balancing
.... (iv)
On multiplying eqn. (ii) by 5 and (iv) by 2 and then adding we get
 
? X + Y + Z = 2 + 5 + 16 = 23
22. (+ 6) Xe = 53.5 % ? F = 46.5%
Relative number of atoms Xe
Simple ratio Xe = 1 and F = 6 ; Molecular formula is XeF
6
O.N.of Xe is +6
23. (3) The reaction given is
Cr
2
O
7
? 2–
 + Fe
2+
 + C
2
O
4
? 2–
 Cr
3+
 + Fe
3+
 + CO
2
? Cr
2
O
7
? 2–
  2Cr
3+
On balancing
14H
+
 + Cr
2
O
7
? 2–
 + 6e
–
 2Cr
3+
 + 7H
2
O ......(i)
Fe
2+
  Fe
3+
 + e
–
......(ii)
C
2
O
4
? 2–
  2CO
2
 + 2e
–
.......(iii)
On adding all the three equations.
Cr
2
O
7
? 2–
 + Fe
2+
 + C
2
O
4
? 2–
 + 14H
+
 + 3e
–
   2Cr
3+
 + Fe
3+ 
+ 2CO
2
 + 7H
2
O
Hence the total no. of electrons involved in the reaction = 3
24. (0.67) The balanced equation is
Ratio of the coefficients of CO
2
 and H
2
O