REDOX REACTION CHEMISTRY

EXERCISE-I OBJECTIVE PROBLEMS (JEE MAIN)

1. The equivalent weight of MnSO4 is half its 9. The number of moles of oxalate KHC 2 O 4 .
molecular weight when it is converted to : H 2 C 2 O 4 .2H 2 O oxidised by one mole of
(A) Mn2O3 (B) MnO2 permanganate ion is:

(C) MnO –
(D) MnO42– (A) 3/4 (B) 1
4

2. Given the equation S2O82– + 2e–  2SO42– , Mn2+ (C) 5/4 (D) 6/4
+ 4H2O  MnO4– + 8H+ + 5e– . How many moles 10. How many equivalents are there per mol of H2S
of S2O82– ions are require to oxiDes 1 mole of in its oxidation to SO2 ?
Mn2+:
(A) 2 (B) 4
(A) 0.4 (B) 0.5
(C) 6 (D) 8
(C) 2.5 (D) 1.0
11. 3 mol of a mixture of FeSO 4 and Fe 2(SO4)3
3. A solution of 10 ml 0.1 M FeSO4 was titrated required 100 mL of 2M KMnO4 solution is acidic
with KMnO4 solution in acidic medium. The medium .
amount of KMnO4 used will be:
Hence mol fraction of FeSO4 in the mixture is:
(A) 5 ml of 0.1 M (B) 10 ml of 0.1 M
(A) 1/3 (B) 2/3
(C) 10 ml of 0.5 M (D) 10 ml of 0.02 M
(C) 2/5 (D) 3/5
4. NH2OH reacts with ferric sulphate as follows :
2NH2OH + 4 Fe3+  N2O + H2O + 4 Fe2+ 4H+. 12. In a reaction 4 moles of electron are transferred
The eq. wt. of NH2OH in this reaction is : to one mole of HNO3. When acted as an oxidant.
(A) (mol. wt.)/1 (B) (mol.wt.)/2 The possible reduction product is :

(C) (mol.wt.)/3 (D) (mol.wt.)/4 (A) 1/2 mole of N2 (B) 1/2 mole of N2O
(C) 1 mole of NO2 (D) 1 mole of NH3
5. 20 ml of 0.1 M solution of metal ion reaCteD
with 20 ml of 0.1 M SO2 solution. SO2 reacted 13. One mole of N2H4 loses ten moles of electrons
according to the equation. SO2 + 2H2O  SO42– to from a new Compound y. Assuming that all
+ 4H+_ + 2e–. If the oxidation no. of metal ion the nitrogen appears in the new compound. What
was +3, the new oxidation number of the metal is the oxidation state of nitrogen in y. There is
would be: no change in the oxidation state of hydrogen:
(A) 0 (B) +1 (A) –1 (B) –3
(C) +2 (D) None of these (C) + 3 (D) +5
6. How many ml of 0.150 M Na2CrO4 will be required 14. 1 g equiv of a substance is the weight of that
to oxidize 40 ml of 0.5 M Na2S2O3. amount of a substance which is equivalent to :
CrO42– + S2O32– Cr(OH)4– + SO42–. (A) 0.25 mol of O2 (B) 0.50 mol of O2
(A) 225 ml (B) 355 ml (C) 1 mol of O2 (D) 8 mol of O2
(C) 455 ml (D) 555 ml 15. Which of the following changes requires reducing
7. Number of moles of electrons take up when 1 agent?
mole of NO3– ions is reduced to 1 mole of NH2OH (A) CrO42–  Cr2O72– (B) BrO3–  BrO–
is: (C) H2AO3  HAsO42– (D) Al(OH)3 Al(OH)4–
(A) 2 (B) 4 16. Which of the following is a disproportonation
(C) 5 (D) 6 reaction?

8. The number of moles of thiosulphate (S2O32–) (A) CaCO3 + 2H+  Ca2+ + H2O + CO2
that will required to react completely with one (B) 2CrO42– + 2H+  Cr2O72+ + H2O
mole I2 in alkaline medium (where it gets oxidised to (C) Cr2O42– + 2OH–  2CrO42– + H2O
SO42–) is :
(D) Cu2O + 2H+  Cu + Cu2+ + H2O
(A) 1/4 (B) 4
(C) 8 (D) 1/8

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CHEMISTRY REDOX REACTION

17. If equal volumes of 1M KMnO4 and 1M K2Cr2O7 24. When one gm mole of KMnO4 reacts with HCl,
solutions are allowed to oxidise Fe2+ in acidic the volume of chlorine liberated at NTP will be:
medium. (A) 11.2 litre (B) 22.4 litre
The amount of iron oxidised will Be : (C) 44.8 litre (D) 56.0 litre
(A) more by KMnO4 solution
25. 10.78 g of H3PO4 in 550 mL solution is 0.40 N.
(B) more by K2Cr2O7 solution Thus this acid:
(C) equal in both the cases (A) has been neutralised to HPO42–
(D) cannot be determined (B) has been neutralised to PO43–
18. In the following reaction (unbalanced), (C) has been reduced to HPO32–
equivalent wt. of As2S3 is related to molecular
wt. M By: (D) has been neutralised to H2PO4–

As2S3 + H+ NO3–  NO + H2O + AsO43– + SO42– 26. When 0.75 gm of a substance was kjeldalised,
it produced NH3. Which neutralizes 30 ml of 0.25
(A) M/2 (B) M/4 N sulphuric acid. The percentage of nitrogen in
(C) M/28 (D) M/24 the organic compound is :
19. Mass of KHC 2O 4 (potassium acid oxalate) (A) 14 (B) 11
required to reduce 100 mL of 0.02 M KMnO4 in (C) 1 (D) None
acidic medium (to Mn2+) is x g, and to neutralise
100 mL of 0.05 M Ca(OH)2 is y g then : 27. 15 mol of KMnO4 are treated with excess H2C2O4
in H2SO4 medium. How many moles of CO2 will
(A) x = y (B) 2x = y be formed and how many moles of H2C2O4 will
(C) x = 2y (D) None is CorreCt be consumed?
20. 100 mL of 1 M KMnO4 oxidised 100 mL of H2O2 in (A) 75, 37, 5. (B) 3, 15
acidic medium (when MnO4– is reduced to Mn2+)
(C) 3, 6 (D) 75, 150
; volume of same KMnO4 required to oxidise 100
mL of H2O2 in basic medium (when MnO4– is 28. An equimolar mixture of NaHC2O4 and H2C2O4
reduced to MnO2) will be: consumes 20 ml 0.3 M NaOH solution for complete
neutralization. The same mixture requires V ml.
(A) (100/3) mL (B) (500/3) mL
0.05 M KMnO4 solution in acidic medium for
(C) (300/5 ) mL (D) 100 mL oxidation.
21. 1 mol of ferric oxalate is oxidised by x mol of The value of V is :
MnO4– and also 1 mol of ferrous oxalate is oxidised
By y mol of MnO4– in acidic medium. The ratio (A) 160 ml (B) 32 ml
(x/y) is: (C) 24 ml (D) None of these
(A) 2 : 1 (B) 1 : 2 29. Oxidation number of Fe in Fe3O4 is fractional
(C) 3 : 1 (D) 1 : 3 because-
22. 0.7 gm of Na2CO3.xH2O is dissolved in 100 ml, 20 (A) It is a mixed [Fe(+2) – Fe(+4)] oxide
ml of which required 19.8 ml of 0.1 N HCl. The (B) It is a non-stoichiometric compound
value of x is:
(C) It is a mixed [Fe(+2) – Fe(+3)] oxide
(A) 4 (B) 3
(D) None of the above
(C) 2 (D) 1
30. The oxidation state of Oxygen atom in
23. A metal is burnt in oxygen and all the products potassium superoxide is-
of combustion are weighed. It is found that the
wt. of the metal seems to have increased by 1
24%. The equivalent wt. of the above metal. (A) Zero (B) –
2
(A) 25 (B) 24
(C) – 1 (D) – 2
(C) 33.34 (D) 76

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REDOX REACTION CHEMISTRY

31. The oxidation state of tungsten in Codes :–

Na2W4O13.10H2O is – A B C D
(A) + 7 (B) + 6 (A) c d b a
(C) + 4 (D) + 4.5 (B) d c b a
32. Amongst the following identify the species (C) c d a b
with an atom in +6 oxidation state –
(D) d c a b
(A) MnO4¯ (B) Cr(CN)63–
38. In the reaction,
(C) NiF62– (D) CrO2Cl2
3Br2 + 6NaOH  NaBrO3 + 5NaBr + 3H2O
33. In [Cr(O2)(NH3)4 (H2O)] Cl2 oxidation number which element loses as w ell as gains
of Cr is + 3, then O2 will be in the form : electrons–
(A) dioxide (B) peroxide (A) Na (B) Br
(C) superoxide (D) oxide (C) O (D) H
34. An example of redox process is – 39. Oxidation number of S in H2S2O7 is –
(A) CuSO4 + Fe  FeSO4 + Cu (A) +4 (B) –6
(B) Ca(OH)2 + 2HCl  CaCl2 + 2H2O (C) –5 (D) +6
(C) CaO + 2HCl  CaCl2 + H2O 40. Oxidation number of S in H2SO5 is 6. This is
(D) CaCO3
heat
  CaO + CO2 observed, because –

35. In the reduction of dichromate by Fe(II), the (A) There are five oxygen atoms in the
number of electrons involved per chromium molecule
atom is – (B) The hydrogen atom is directly linked with
(A) 3 (B) 1 non-metal
(C) 2 (D) 4 (C) There is peroxide linkage in the molecule
36. Consider the following statement in the (D) The sulphur atom shows co-ordinate
reaction linkage
KIO3 + 5KI + 6HCl = 3I2 + 6KCl + 3H2O 41. The oxidation number of S in Na2S4O6 is -
(a) KI is oxidised to I2 (A) + 2.5
(b) KIO3 is oxidised to I2 (B) + 2 and + 3 (two S have + 2 and other two
(c) KIO3 is reduced to I2 have + 3)

(d) Oxidation number of I increases from (C) + 2 and + 3 (three S have + 2 and one S
(– ) in KI to zero in I2 of these statements has + 3)

(A) a, c and d are correct (D) + 5 and 0 (two S have + 5 and the other
(B) a, b and d are correct two have 0)

(C) b and d are correct 42. The oxidation state of molybdenum in its
(D) a alone is correct oxocomplex species [Mo2O4(C2H4)2(H2O)2]2–
is
37. Match list – I (compounds) with list – II
(A) +2 (B) +3
(Oxidation state of nitrogen) and select (C) +4 (D) +5
the correct answer using the codes given
below the lists – 43. Which element will have the maximum
oxidation number in K2Cr2O7 and KMnO4 –
List – I List – II
(A) Mn (B) Cr
(A) NaN3 (a) +5 (C) O (D) K
(B) N2H4 (b) +2
(C) NO (c) –1/3
(D) N2O5 (d) –2

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CHEMISTRY REDOX REACTION

44. Select the pair of oxidation processes, 47. Oxygen shows oxidation state of –1 in the
(a) 2Cu2+  Cu2 2+ compound –
(b) MnO4¯  Mn2+ (A) NO2 (B) MnO2
(c) [Fe(CN)6]–4  [Fe(CN)6]–3 (C) PbO2 (D) Na2O2
(d) 2I¯ I2 48. The oxidation number of Pt in [Pt(C2H4)Cl3]¯
(A) a, b (B) c, d is –
(C) a, d (D) b, c (A) +1 (B) +2
45. Carbon is in the lowest oxidation state in – (C) +3 (D) +4
(A) CH4 (B) CCl4 49. Which of the following reactions does not
(C) CF4 (D) CO2 involve either oxidation or reduction –
(A) VO2+V2O3 (B) Na  Na+
46. AB4¯ + C+2  C+3 + A+2
(C) Zn+2Zn (D) CrO4–2Cr 2O7–2
If the O.N. of B is –2. Choose the true
50.Which one is correctly matched:
statement for the above change –
Substance O.N. of S
(A) O.N. of A decreases by –5
(A) H2S +2
(B) O.N. of C decreases by +1
(B) H2SO5 +4
(C) O.N. of A decreases by + 5 and that of
C increases by +1 (C) H2SO4 +4

(D) O.N. of A decreases by +5 and that of (D) Na2S4O6 +2.5

C decreases by +1

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REDOX REACTION CHEMISTRY

EXERCISE-II MULTIPLE CHOICE PROBLEMS (JEE ADVANCED)

Multiple choice Question. (A) 11.2 ml of N2 and 0.015 g of nitric oxide
1. In the titration of K2Cr2O7 and ferrous sulphate, (B) 22.4 L of nitrous oxide and 22.4 L of CO2
following data is obtained: V1 ml of 1.0 M1 (C) 1 mole of HCl and 0.5 mole of H2S
K2Cr2O7 requires V2 ml of 1.0 M2 FeSO4. Which (D) 1 mole of H2O2 and 1 mole of H2SO4
of the following relations is/are true for the above 8. When 100 ml of 0.1 M KNO3, 400 ml of 0.2 M
titration ? HCl and 500 ml of 0.3 M H2SO4 are mixed. Then
(A) 6 M1V1 = M2V2 (B) M1V1 = 6 M2V2 in the resulting solution
(C) N1V1 = N2V2 (D) M1V1 = M2V2 (A) The molarity of K+ = 0.01 M

2. Cr2O72 is reduced to Cr3+ by Fe2+. Identify (B) The molarity of SO24 = 0.15 M
the correct statement from the following : (C) The molarity of H+ = 0.38 M
(A) 6 moles of Fe2+are oxidised to Fe3+ ions. (D) The molarity of NO3– = 0.08 M and Cl– = 0.01 M
(B) The solution becomes yellow 9. A 110 % sample of oleum contains
(C) The solution becomes green (A) 44.4% of SO3
(D) 3 moles of Fe2+ get oxidised to Fe3+ (B) 55.6% of sulphuric acid
3. 15 g of KMnO4 in acidic medium equal to (C) 55.6% of SO3
(A) 0.095 moles (D) 44.4% of sulphuric acid
(B) 0.477 g equivalents 10. A mixture of 1 mole each of FeSO4 and FeC2O4
(C) 9.54 L of 0.05 N KMnO4 are taken then :
(D) 10 ml of 0.05 M KMnO4 (A) nmoles of KMnO4 required to oxidise Fe2+ to
4. Which of the following statements are correct Fe3+ in acidic medium is 0.4
?
(B) nmoles of KMnO4 required to oxidise SO24 is 6/
(A) One mole of Cl2 means 8 equivalents of chlorine
5
when Cl2  Cl– + ClO 4
(C) nmoles of KMnO4 required to oxidise oxalate
(B) One mole of Cl2 means 6 equivalents of chlorine ion is 0.4

when Cl2  Cl– + ClO 4 (D) The total number of moles of KMnO 4
required to completely oxidise the mixture is
(C) When one mole of As2S3 is oxidised to As2O5 + 0.8
SO2 then it mean 22 equivalents of As2S3 are
11. 100 ml of 0.06 M Ca(NO3)2 is added to 50 ml of
oxidised.
0.06 M Na2C2O4. After the reaction is complete
(D) The equivalent weight of As2S3 is always = mol.
(A) 0.003 moles of calcium oxalate will get
wt./22.
precipitated
5. 27 g of Al will react completely with
(B) 0.003 M of excess of Ca2+ will remain in
(A) 24 g of O2 (B) 0.75 mole of O2 excess
(C) 16.8 L of O2 at STP (D) 1 mole of O2
(C) Na2C2O4 is limiting reagent.
6. 10.78 g of H3PO4 in 550 ml solution is 0.40 N.
(D) Ca(NO3)2 is excess reagent.
Thus this acid
12. To 25 ml of H2O2 solution, excess of acidified
(A) has been neutralized to HPO42–
solution of KI was added. The iodine liberated
(B) has been neutralized to PO43–
required 20 ml of 0.1 N Na2S2O3 solution.
(C) has been reduced to HPO32– Calculate the % of H2O2 solution.
(D) has been neutralized to H2PO4–
(A) 0.0136 % (B) 0.136 %
7. Which of the following contains equal number (C) 0.0068 % (D) 0.068 %
of atoms ?

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CHEMISTRY REDOX REACTION

13. 10 g mixture of KI and NaClO3 treated with 200 19. What is the value of n
cc HCl gives a gas which absorbes in 40 ml of (A) 4 (B) 3
N (C) 1 (D) 2
sodium thiosulphate solution. Then the
10
correct statement is/are Question No. 20 to 22 (3 questions)
(A) equivalent weight of iodine in the reaction is A steel sample is to be analysed for Cr and Mn
M/2 simultaneously. By suitable treatment Cr is
oxidized as Cr2O72– and the Mn to MnO4–.
(B) percentage of KI in the mixture is 66.4%
Cr  Cr2O72–
(C) sodium thiosulphate converted into
Na2S4O6. Mn  MnO4–
(D) percentage of NaClO3 is 66.4% A 10 gm sample of steel is used to produce
250.0 mL of a solution containing Cr2O72– and
14. H2C2O4 and NaHC2O4 behave as acids as well
MnO4–. A 10 mL portion of this solution is added
as reducing agents. Which is/are correct
to a BaCl2 solution and by proper adjustment of
statement(s) ?
the acidity, the chromium is completely
(A) Equivalent wt. of H2C2O4 and NaHC2O4 are precipitated as BaCrO4 ; 0.0549 g is obtained.
equal to their molecular weights when behaving
2– H
as reducing agent Cr2 O 7 
 BaCrO 4
(B) 100 ml of 1 N solution of each is neutralised A second 10 mL portion of this solution requires
by equal volumes of 1 M Ca(OH)2 exactly 15.95 mL of 0.0750 M standard Fe2+
(C) 100 ml of 1 N solution of each is neutralised solution for its titration (in acid solution)
by equal volumes of 1 N Ca(OH)2 20. % of chromium in the steel sample
(D) 100 ml of 1 M solution of each is oxidised (A) 1.496 (B) 2.82
by equal volumes of 1 M KMnO4 (C) 1.96 (D) 5
15. 100 ml of 0.15 N H2O2 is completely oxidized by 21. Equivalent of Fe2+ required for reduction of MnO4– is
(A) 150 ml of 0.1 N KMnO4 solution (A) 5.44 × 10–4 (B) 0.544 × 10–2
(B) 2.5 × 10–3 moles of K2Cr 2O7 in acidic
medium (C) 1.196 × 10–3 (D) 11.96 × 10–4

(C) 15 × 10–3 moles of KMnO4 in basic medium 22. Amount of BaCl2 required for conversion of
(D) 15 moles of O3 in acidic medium Cr2O72– to BaCrO4 in steel sample

Question No. 16 to 19 (4 questions) (A) 0.045 (B) 0.0549

30cc of a solution containing 9.15 gm of salt (C) 1.125 (D) 2.82

KxHY (C2O4)z. nH2O per litre required 27cc of 0.12 Question No. 23 to 25 (3 questions)
N NaOH for neutralization. The same quantity of 25 ml from a stock solution containing NaHCO3
solution was also found to require 36cc of 0.12 and Na2CO3 was diluted to 250 ml with CO2 free
N KMnO4 solution for complete oxidation. distilled water. 25 ml of the diluted solution when
16. What is the value of X titrated with 0.12 M HCl required 8 ml., when
(A) 1 (B) 2 phenolphthalein was used as an indicator.

(C) 3 (D) 4 HPh

Na2CO3 + HCl   NaHCO3
17. What is the value of Y When 20 ml of diluted solution was titrated with
(A) 4 (B) 3 same acid it required 18 ml when methlyorange
(C) 2 (D) 1 was used as an indicator.

18. What is the value of Z Na2CO3 + 2HCl MeOH

 2NaCl + H2O + CO2
(A) 4 (B) 1 MeOH
NaHCO3 + HCl  
 NaCl + H2O + CO2
(C) 2 (D) 3

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REDOX REACTION CHEMISTRY

23. Concentration of NaHCO3 in gm/lit. (A) 5 (B) 4

(A) 0.312 (B) 2.62 (C) 3 (D) 2
(C) 3.12 (D) 26.208 32. The oxidation state of molybdenum in its oxo
24. Amount of NaOH that should be added to convert complex species [Mo2O4(C2H4)2 (H2O2)]2–
all bicarbonate into carbonate in 100 ml stock [IIT-Screening 1991]
solution (A) 2 (B) 3
(A) 1.248 gm (B) 0.312 gm (C) 4 (D) 5
(C) 3.12 × 10–2 gm (D) 7.8 × 10–3 gm 33. The oxidation states of the most electronegative
25. Millimoles of NaHCO3 present in stock solution element in the products of the reaction, BaO2
(A) 0.624 (B) 2.16 with dilute H2SO4 are - [IIT- 1991]
(C) 1.536 (D) 7.8 (A) 0 and –1 (B) –1 and –2
Question No. 26 to 28 (3 questions) (C) –2 and 0 (D) –2 and +1
1.16 g CH3(CH2)n COOH was burnt in excess air 34. For the readox reaction,
and the resultant gases (CO2 and H2O) were
Mn O 4 + C2 O 24 + H+  Mn+2+CO2 + H2O
passed through excess NaOH solution. The
resulting solution was divided in two equal parts. the correct coefficients of the reactants for
One part requires 50 mL of 1 N HCl for the balanced reaction are– [IIT- 1992]
neutralization using phenolphthalein as indicator.
Another part required 80 mL of 1 N HCl for Mn O 4 C2 O 24 H+
neutralization using methyl orange as indicator.
(A) 2 5 16
26. Produced mole of the CO2
(B) 16 5 2
(A) 0.1 (B) 0.01
(C) 5 16 2
(C) 0.06 (D) None of these
(D) 2 16 5
27. What is the value of n
35. What mass of N2H4 can be oxidised to N2 by 24
(A) 4 (B) 3 gm of K2CrO4 which is reduced to [Cr(OH)4]–.
(C) 2 (D) 1 36. The ion An+ is oxidised to AO3– by MnO4– changing
28. Amount of excess NaOH solution taken initially. to Mn2+ in acid medium. Given that 2.68 ×10–3
mole of An+ required 1.61 ×10–3 mole of MnO4–.
(A) 3.2 gm (B) 6.4 gm
What is the value of n.
(C) 1.2 gm (D) None of these
37. 2.180 gram of a sample contains a mixture of
29. The oxidation number of phosphorus in XO and X2O3 which are completely oxidised to
Ba(H2PO2)2 is - [IIT-1990] XO4– by 0.015 mole of K2Cr2O7. Calculate the
(A) +3 (B) +2 atomic weight of X, if 0.0187 mole XO4– is formed.
(C) +1 (D) –1 38. The number of moles of KMnO4 that will be
30. The compound which could act both as oxidising needed to react completely with one mole of
as well as reducing agent is - ferrous oxalate in acid solution is [JEE 1996]
[IIT-Screening 1991] (A) 3/5 (B) 2/5
(C) 4/5 (D) 1
(A) SO2 (B) KMnO4
39. The number of moles of KMnO4 that will be
(C) Al2O3 (D) CrO3
needed to react with one mole of sulphite ions
31. The number of electrons to balance the following in acidic solution is [JEE 1997]
equation, the value of x is-
(A) 2/5 (B) 3/5
[IIT-Screening 1991] (C) 4/5 (D) 1

NO 3 + 4H+ + xe–  2H2O + NO is -

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CHEMISTRY REDOX REACTION

Each of the questions given below con- Column Match

sists of Statement – I and Statement – II. 45. Column I Column II
Use the following Key to choose the ap-
(1) Increase in oxidation (a) Loss of electrons
propriate answer.
number
(A) If both Statement-I and Statement-II
are true, and Statement-II is the cor- (2) Decrease in oxidation (b) Redox reaction
rect explanation of Statement–I. number
(B) If both Statement-I and Statement-II (3) Oxidising agent (c) Fractional oxidation
are true but Statement-II is not the number
correct explanation of Statement–I. (4) Reducing agent (d) Zero oxidation
(C) If Statement-I is true but Statement-II number
is false. (5) 2Cu+  Cu2+ + Cu (e)Simple neutralisation
(D) If Statement-I is false but Statement-II reaction
is true. (6) MnO2+4HClMnCl2 (f )Gain of electrons
40. Statement I: NaOH+H3PO4NaH2PO4+H2O +Cl2+ 2H2O
(7) Mn3O4 (g) Disproportionation
in given reaction equivalent weight of
(8) CH2Cl2 (h) Oxidation
M (9) NaOH+HClNaCl (i) Reduction
H3PO4 is .
3 +H2O
Statement II : H3PO4 is tribasic acid. 46. The normality of 0.3 M phosphorus acid (H3PO3) is
41. Statement I : In CuO equivalent weight (A) 0.1 (B) 0.9
of Cu is 63.5 & in Cu2O equivalent weight (C) 0.3 (D) 0.6
of Cu is 31.8. [JEE 1999]
Statement II : Equivalent weight of any 47. One mole of calcium phosphide on reaction with
metal is the gm quantity of metal which excess of water gives [JEE 1999]
is combined with 8 gm of oxygen in the (A) one mole of phosphine
formation of metal oxide. (B) Two moles of phsophoric acid
(C) Two moles of phosphine
8
42. Statement I : In Pb3O4 all Pb has + (D) One mole of phosphorus pentoxide
3
oxidation number. 48. An aqueous solution of 6.3 gm of oxalic acid
dihydrate is made upto 250 ml. The volume of
Statement II : Pb3O4 is mixed oxide of 0.1 N NaOH required to completely neutralize
PbO & PbO2 10 ml of this solution is [JEE 2001]
43. Statement I : In estimation of FeSO 4 (A) 40 ml (B) 20 ml
KMnO4 along with HCl can be used as (C) 10 ml (D) 4 ml
oxidising agent. 49. In the standarization of Na2S2O3 using K2Cr2O7
Statement II : In acidic medium KMnO4 is by iodometry the equivalent mass of K2Cr2O7 is
more powerful oxidant compared to neutral/ [JEE 2001]
basic medium.
M.Mass M.Mass
44. Column - I Column- II (A) (B)
2 6
M
(A) Cr2O7–2  Cr+3 (P) M.Mass
7 (C) (D) Same as M. Mass.
3
M
(B) Fe(NO3)3  Fe (NO2)2 (Q) 50. How many litres of a 0.5 N solution of an
6 oxidising agent are reduced by 2 litres of a
(C) H2O2  O2 (R) M 2.0 N solution of a reducing agent ?
M (A) 8 (B) 4
(D) PH3 + H+  PH4+ (S)
2 (C) 6 (D) 7 litres

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REDOX REACTION CHEMISTRY

EXERCISE-III (JEE ADVANCED)

Balancing of reactions 14. Balance following equations in proper medium
1. Complete and balance the following equations P + OH– + H2O  H2PO4– + PH3
(a) KMnO4 + H2SO4 + H2O2  15. Balance following equations in proper medium
K2SO4 + MnSO4 + H2O + ......... S + OH–  S2– + S2O32–
(b) Cu2+ + I–  Cu+ + I2 16. Balance following equations in proper medium
2. Balance the following in basic medium Na2S2O3 + KMnO4 + H2O 
(i) CrI3 + H2O2 + OH  CrO
–
4
2–
+ IO 4
–
+ H2O Na2S4O6 + MnO2 + KOH + NaOH
(ii) KOH + K4Fe(CN)6 + Ce(NO3)4  17. Balance following equations in proper medium
Fe(OH)3 + Ce(OH)3 + K2CO3 + KNO3 + H2O FeC2O4 + KMnO4 + H2SO4 
3. Balance the following equations using desired Fe2(SO4)3 + CO2 + MnSO4 + K2SO4 + H2O
medium:
18. Balance following equations in proper medium
(a) SbCl3 + KIO3 + HCl  SbCl5 + ICl + H2O + KCl
C2H5OH + I2+ OH–  CHI3 + HCOO– + H2O + I–
(b) FeC2O4 + KMnO4 + H2SO4 
Acid Base Titration
Fe2(SO4)3 + CO2 + MnSO4 + K2SO4 + H2O
19. A solution containing 4.2 g of KOH and Ca(OH) 2
4. Balance the following equations using desired is neutralized by an acid. It consumes 0.1
medium: equivalent of acid, calculate the percentage
FeCr2O4 + K2CO3 + KClO3  composition of the sample.

Fe2O3 + K2CrO4 + KCl + CO2 20. How many ml of 0.1 N HCl are required to react
completely with 1 g mixture of Na2CO3 and
5. Balance the following equations using desired
NaHCO3 containing equimolar amounts of two ?
medium:
21. 0.5 g of fuming H2SO4 (oleum) is diluted with
Pb(N3)2 + Co(MnO4)3
water. The solution requires 26.7 ml of 0.4 N
CoO + MnO2 + Pb3O4 + NO NaOH for complete neutralization. Find the % of
6. Balance the following equations in acidic medium free SO3 in the sample of oleum.
KClO3 + H2SO4  KHSO4 + HClO4 + ClO2 + H2O 22. H3PO4 is a tri basic acid and one of its salt is
NaH2PO4. What volume of 1 M NaOH solution
7. Balance the following equations in acidic medium
should be added to 12 g of NaH2PO4 to convert
Br– + BrO3– + H+  Br2 + H2O
it into Na3PO4 ?
8. Balance the following equations in acidic medium
Double titration
H2S + Cr2O72– + H+  Cr2O3 + S8 + H2O
23. A solution contains Na2CO3 and NaHCO3. 20 ml
9. Balance the following equations in acidic medium
of this solution required 4 ml of 1 N HCl for
H2S + Cr2O72– + H+  Cr2O3 + S8 + H2O titration with Ph indicator. The titration was
10. Balance the following equations in acidic medium repeated with the same volume of the solution
but with MeOH. 10.5 ml of 1 N HCl was required
Cu2O + H+ + NO3–  Cu2+ + NO + H2O
this time. Calculate the amount of Na2CO3 &
11. Balance the following equations in acidic medium NaHCO3.
[Fe(CN)6]4– + MnO4–  24. A solution contains a mixture of Na2CO3 and
Fe3+ + CO2 + NO3– + Mn2+ NaOH. Using Ph as indicator 25 ml of mix required
19.5 ml of 1 N HCl for the end point. With MeOH,
12. Balance following equations in proper medium
25 ml of the solution required 25 ml of the same
13. Balance following equations in proper medium HCl for the end point. Calculate gms/L of each
Cr2O72– + C2H4O + H+  C2H4O2 + Cr3+ substance in the mixture.

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CHEMISTRY REDOX REACTION

25. 200 ml of a solution of mixture of NaOH and to ferric state. Calculate the strength of
N dichromatic solution.
Na2CO 3 was first titrated with HCl with
10 34. 5 g of pyrolusite (impure MnO2) were heated
indicator Ph. 17.5 ml of HCl was required for end with conc. HCl and Cl2 evolved was passed
point. After this MeOH was added and 2.5 ml of through excess of KI solution. The iodine liberated
same HCl was again required for next end point.
Find out amounts of NaOH and Na2CO3 in the N
required 40 mL of hypo solution. Find the %
mixture. 10
Redox Titration of MnO2 in the pyrolusite.

26. It requires 40.05 ml of 1 M Ce4+ to titrate 20 ml Back Titration

of 1 M Sn2+ to Sn4+. What is the oxidation state 35. 50 gm of a sample of Ca(OH)2 is dissolved in 50
of the cerium in the product. ml of 0.5 N HCl solution. The excess of HCl was
27. A volume of 12.53 ml of 0.05093 M SeO2 reacted titrated with 0.3N NaOH. The volume of NaOH
with exactly 25.52 ml of 0.1 M CrSO4. In the used was 20cc. Calculate % purity of Ca(OH)2
reaction, Cr2+ was oxidized to Cr3+. To what 36. One gm of impure sodium carbonate is dissolved
oxidation state was selenium converted by the in water and the solution is made up to 250ml.
reaction. To 50ml of this solution, 50ml of 0.1N – HCl is
28. Potassium acid oxalate K2C2O4. 3H2C2O4.4H2O added and the mixture after shaking well
can be oxidized by MnO4– in acid medium. required 10ml of 0.16N NaOH solution for
Calculate the volume of 0.1 M KMnO4 reacting complete titration. Calculate the % purity of
in acid solution with one gram of the acid oxalate. the sample.

29. A 1 g sample of H2O2 solution containing x% 37. What amount of substance containing 60% NaCl,
H2O2 by mass requires x cm3 of a KMnO4 solution 37% KCl should be weighed out for analysis so
for complete oxidation under acidic conditions. that after the action of 25 ml of 0.1N AgNO3
Calculate the normality of KMnO4 solution. solution, excess of Ag+ is back titrated with 5
ml of NH4SCN solution? Given that 1 ml of NH4SCN
30. Metallic tin in the presence of HCl is oxidized by
= 1.1 ml of AgNO3.
K2Cr2O7 to stannic chloride, SnCl4. What volume
of deci-normal dichromate solution would be [AgNO3 + NaSCN  AgSCN + NaNO3]
reduced by 1 g of tin. 38. 10 g CaCO3 were dissolved in 250 ml of 1 M HCl.
31. 5g sample of brass was dissolved in one litre dil. What volume of 2 M KOH would be required to
H2SO4. 20 ml of this solution were mixed with KI, neutralise excess HCl ?
liberating I2 and Cu+ and the I2 required 20 ml of 39. A mixture of FeO and Fe2O3 is reacted with
0.0327 N hypo solution for complete titration. acidified KMnO4 solution having a concentration
Calculate the percentage of Cu in the alloy. of 0.2278 M, 100 ml of which was used. The
32. A 1.0 g sample of Fe2O3 solid of 55.2 % purity is solution was then titrated with Zn dust which
dissolved in acid and reduced by heating the converted Fe3+ of the solution to Fe2+. The Fe2+
solution with zinc dust. The resultant solution is required 1000 ml of 0.13 M K2 Cr2 O7 solution.
cooled and made upto 100.0 mL. An aliquot of Find the % of FeO & Fe2O3.
25.0 mL of this solution requires 17.0 mL of 40. 50 ml of a solution, containing 0.01 mole each
0.0167 M solution of an oxidant for titration. Na2CO3, NaHCO3 and NaOH was titrated with N-
Calculate the number of moles of electrons taken HCl. What will be the titre readings if
up by the oxidant in the reaction of the above
(a) only Ph is used as indicator.
titration.
(b) only MeOH is used as indicator from the
33. 0.84 g iron ore containing x percent of iron was
beginning.
taken in a solution containing all the iron in ferrous
condition. The solution required x ml of a (c) MeOH is added after the first end point with
dichromatic solution for oxidizing the iron content Ph.

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REDOX REACTION CHEMISTRY

41. 0.1 M KMnO4 solution completely reacts with 47. A mixture of H2C2O4 (oxalic acid) and NaHC2O4
0.05 M FeSO4 solution under acidic conditions. was dissolved in water and the solution made
The volume of FeSO4 used is 25 ml. What volume upto one litre. Ten milliliters of the solution
of KMnO4 was used? required 3.0 mL of 0.1 N sodium hydroxide solution
42. 20 g of a sample of Ba(OH)2 is dissolved in 50 for complete neutralization. I n another
ml. of 0.1 N HCl solution. The excess of HCl was experiment 10.0 mL of the same solution, in hot
titrated with 0.1 N NaOH. The volume of NaOH dilute sulphuric acid medium, required 4.0 mL of
used was 20cc. Calculate the % Ba(OH)2 in the 0.1 N KMnO4 solution for complete reaction.
sample. Calculate the masses of H2C2O4 and NaHC2O4 in
the mixture.
43. 1 solution contains a mixture of Na2CO3 and
NaOH. Using phenolphthalein as indicator. 25ml. 48. A solution of 0.2 g of a compound containing
of mixture required 19.5 ml. of 0.995 N HCl for Cu+2 and C2O4–2 ions an titration with 0.02 M
the end point. With methyl orange. 25ml. of KMnO4 in presence of H2SO4 consumes 22.6 mL
solution required 25 ml. of the same. HCl for the of the oxidant. The resultant solution is
point. Calculate grams per litre of each substance neutralized with Na2CO3, acidified with dilute
in the mixture. acetic acid and treated with excess KI. The
liberated iodine requires 11.3 mL of 0.05 N
44. 5.7 g of bleaching powder was suspended in Na2S2O3 solution for complete reduction. Find
500 ml. of water. 25 ml. of this solution on out the mole ratio of Cu +2 to C2O4–2 in the
treatment with KI in the presence of HCl liberated compound. Write down the balanced redox
iodine which reacted with 24.35 ml. of N/10 reactions involved in the above titration.
Na2S2O3. Calculate the % of 'available' chlorine
in the bleaching powder. 49. A 2.0 g sample of a mixture containing sodium
carbonate, sodium bicarbonate and sodium
45. A solution containing 2.68 ×10–3 moles of An+ sulphate is heated till the evolution of CO2
ions requires 1.61 ×10–3 moles of MNO4– for the ceases. The volume of CO2 at 750 mm Hg
complete oxidation of A n+ to AO 3– in acidic pressure and at 298 K is measured to be 123.9
medium. What is the value of n? ml. A 1.5 g of the sample requires 150 mL of
46. (i) What is the mas of sodium bromate and the M/10 HCl for complete neutralization. calculate
solution necessary to prepare 85.5 mL of 0.672 the percentage composition of the components
N solution when the half cell reaction is BrO3– + of the mixture.
56H+ + 6e–  Br– + 3H2O. 50. A solution contains Na2CO3 and NaHCO3 20 cm3
(ii) What would be the mass as well as molarity of this solution requires 5.0 cm3 of 0.1 M H2SO4
if the half cell reaction . solution for neutralization using phenolphthalein
2BrO3– + 12H+ + 10e–  Br2 + 6H2O as the indicator. Methylorange is then added
when a further 5.0 cm3 of 0.2 M H2SO4 was
required. Calculate the masses of Na2CO3 and
NaHCO3 in 1 L of this solution.

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CHEMISTRY REDOX REACTION

EXERCISE-IV PREVIOUS YEAR QUESTIONS LEVEL-I (JEE MAIN)

1. How many grams of H2O2 are required to convert 5. In the coordination compound, K4[Ni(CN)4], the
0.1 moles PbS to 0.1 moles PbSO4 - oxidation state of nickel is –[AIEEE-2003]
[AIEEE-2002] (A) +1 (B) +2
(A) 12.8 g (B) 13.6 g (C) –1 (D) 0
(D) 16 g (D) 3.4 g 6. What would happen when a solution of potassium
chromate is treated with an dilute nitric acid ?
2. MnO4 is good oxidising agent in different medium [AIEEE-2003]
changing to –
(A) CrO24 is reduced to +3 state of Cr
MnO 4 Mn2+ (B) CrO24 is oxidized to +7 state of Cr

MnO24 (C) Cr3+ and Cr2O72 are formed

MnO2 (D) Cr2O72 and H2O are formed

Mn2O3 7. The oxidation state of Cr in [Cr(NH3)4Cl2]+ is –

[AIEEE-2005]
Changes in oxidation number respectively
are – [AIEEE-2002] (A) +2 (B) +3
(C) 0 (D) +1
(A) 1, 3, 4, 5 (B) 5, 4, 3, 2
8. The oxidation state of chromium in the final
(C) 5, 1, 3, 4 (D) 2, 6, 4, 3 product formed by the reaction between Kl and
3. Oxidation number of Cl in CaOCl2 (bleaching acidified potassium dichromate solution is –
powder) is – [AIEEE-2002] [AIEEE-2005]

(A) Zero, since it contains Cl2 (A) +6 (B) +4

(C) +3 (D) +2
(B) –1, since it contains Cl–
9. Which of the following chemical reactions de-
(C) +1, since it contains ClO–
picts the oxidizing behaviour of H2SO4 ?
(D) +1 and –1 since it contains ClO– and Cl–
[AIEEE 2006]
4. Which of the following is a redox-reaction-
(A) Ca(OH)2 + H2SO4  CaSO4 + 2H2O
[AIEEE-2002]
(B) NaCl + H2SO4  NaHSO4 + HCl
(A) 2Na[Ag(CN)2] + Zn  Na2[Zn(CN)4]+ 2Ag
(C) 2PCl5 + H2SO4 2POCl3 + 2HCl + SO2Cl2
(B) BaO2 + H2SO4  BaSO4 + H2O2
(D) 2HI + H2SO4  I2 + SO2 + 2H2O
(C) N2O5 + H2O  2HNO3
(D) AgNO3 + K  Ag + KNO3

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REDOX REACTION CHEMISTRY

LEVEL-II (JEE ADVANCED)

1. Consider a titration of potassium dichromate 4. The reaction of white phosphorus with aqueous
solution with acidified Mohr's salt solution using NaOH gives phosphine along with another
diphenylamine as indicator. The number of moles phosphorus containing compound. The reaction
of Mohr's salt required per mole of dichromate is type; the oxidation states of phosphorus
phosphine and the other product are
[JEE' 2007]
respectively [JEE 2012]
(A) 3 (B) 4
(A) redox reaction ; –3 and –5
(C) 5 (D) 6
(B) redox reaction ; + 3 and +5
2. Match the reactions in Column I with nature of
the reactions / type of the products in Column (C) disproportionation reaction ; –3 and +5
II. [JEE' 2007]
(D) disproportionation reaction ; –3 and +3
Column I Column II

(A) O 2–  O 2  O 22 – (P) Redox reaction Paragraph for Question Nos. 5 and 6

 Bleaching powder and bleach solution are
(B) CrO 2–
4 H  (Q) One of the products
produced on a large scale and used in several
has trigonal house hold products. The effectiveness of bleach
solution is often measured by iodometry.
planar structure [JEE 2012]
(C) MnO 4–  NO 2–  H  (R) dimeric bridged 5. 25 mL of household bleach solution was mixed
with 30 mL of 0.50 M KI and 10 mL of 4 N acetic
tetrahedral metal ion acid. In the titration of the liberated iodine, 48
mL of 0.25 N Na2S2O3 was used to reach the
(D) NO 3–  H2 SO 4  Fe 2   (S) disproportionation
end point. The molarity of the household bleach
3. White phosphorus on reaction with NaOH gives solution is
PH3 as one of the products. This is a (A) 0.48 M (B) 0.96 M
[JEE 2008] (C) 0.24 M (D) 0.024 M
(A) dimerization reaction 6. Bleaching powder contains a salt of an oxoacid
(B) disproportionation reaction as one of its components. The anhydride of that
oxocacid is
(C) condensation reaction
(A) Cl2O (B) Cl2O7
(D) precipitation reaction
(C) ClO2 (D) Cl2O6

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CHEMISTRY REDOX REACTION

ANSWER-KEY
EXERCISE-I OBJECTIVE PROBLEMS (JEE MAIN)
1. B 2. C 3. D 4. B 5. B 6. B 7. D 8. A
9. C 10. C 11. A 12. B 13. C 14. A 15.B 16.D
17. B 18. C 19. B 20. B 21. A 22. C 23.C 24.D
25. A 26. A 27. A 28. B 29. C 30. B 31.B 32.D
33. C 34. A 35. A 36. A 37. A 38. B 39.D 40.C
41. D 42. B 43. A 44. B 45. A 46. C 47.D 48.B
49. D 50. D

EXERCISE-II MULTIPLE CHOICE PROBLEMS (JEE ADVANCED)

1. A,C 2. A,C 3. A,B,C 4. C 5. A,B,C 6. A 7. A,B
8. A,B,C 9. A,B 10. A,C,D 11. A,C,D 12. B 13. A,C 14.B,C,D
15. A,B,C 16. A 17. B 18. C 19. D 20. B 21. A
22. C 23. D 24. A 25. D 26. C 27. A 28. B
29. C 30. A 31. C 32. B 33. B 34. A 35. 3g
36. n = 2 37.100 38. A 39. A 40. D 41. D 42. D
43. D 44. AQ,BP, CS, DR 45. h,i,f,a,g,b,c,d,e 46. D
47. C 48. A 49. B 50. A
EXERCISE-III (JEE ADVANCED)
1. (a) 2KMnO4 + 3H2SO4 + 5 H2O2 

K2SO4 + 2MnSO4 + 8 H2O + 5O2

(b) 2Cu2+ + 2I–  2Cu+ + I2

2. (i) 2CrI3 + 27H2O2 + 10 OH–  2CrO42– + 6IO4– + 32H2O

(ii) 258KOH + K4Fe(CN)6 + 61Ce(NO3)4  61Ce(OH)3 + Fe(OH)3 + 36H2O + 6K2CO3 + 250KNO3

3. (a) 2SbCl3 + KIO3 + 6HCl  2SbCl5 + ICl + 3H2O + KCl

(b) 10FeC2O4 + 6KMnO4 + 24H2SO4  5Fe2(SO4)3 + 20CO2 + 6MnSO4 + 3K2SO4 + 24H2O

4. 6FeCr2O4 + 12K2CO3 + 7KClO3  3Fe2O3 + 12K2CrO4 + 7KCl + 12CO2

5. 30Pb(N3)2 + 44Co(MnO4)3 132MnO2 + 44CoO + 180NO + 10Pb3O4

6. 3KClO3 + 3H2SO4  3KHSO4 + HClO4 + 2ClO2 + H2O

7. 5Br– + BrO3– + 6H+  3Br2 + 3H2O

8. 24H2S + 8Cr2O72– + 16H+  8Cr2O3 + 3S8 + 32H2O

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REDOX REACTION CHEMISTRY

9. 2MnO4– + 5C2O42– + 16H+  2Mn2+ + 10CO2 + 8H2O

10. 3Cu2O + 14H+ + 2NO3–  6Cu2+ + 2NO + 7H2O

11. 5[Fe(CN)6]4– + 188H+ + 61MnO4–  5Fe3+ + 30CO2 + 30NO3– + 61Mn2++ 94H2O

12. 3C2H5OH + 2MnO4– + OH–  3C2H3O– + 2MnO2(s) + 5H2O

13. Cr2O72– + 3C2H4O + 8H+  3C2H4O2 + 2Cr3++ 4H2O

14. 8P + 3OH– + 9H2O  3H2PO4– + 5PH3

15. 4S + 6OH–  2S2– + S2O32–

16. 6Na2S2O3 + 2KMnO4 + 4H2O  3Na2S4O6 + 2MnO2 + 2KOH + 6NaOH

17. 10FeC2O4 + 6KMnO4 + 24H2SO4  5Fe2(SO4)3 + 20CO2 + 6MnSO4 + 3K2SO4 + 24H2O

18. C2H5OH + 4I2+ 6OH–  CHI3 + HCOO– + 5H2O + 5I–

19. KOH = 35%, Ca(OH)2 = 65% 20. V = 157.8 ml 21. 20.72 % 22. 200 mL

23. 0.424 gm; 0.21 gm 24. 23.32 gm, 22.4 gm 25. 0.06 gm; 0.0265 gm 26. +3

27. zero 28. V = 31.68 ml 29. 0.588 N 30. 337 mL

31. 41.53% 32. 6.07  6 33. 0.15 N 34. 0.174g ; 3.48 % 35. 1.406%

36. 90.1 % 37. 0.1281 g 38. V = 25 mL 39. FeO = 13.34%; Fe2O3 = 86.66%

40. 20 ml ; 40 ml; 20 ml 41. 2.5 ml 42. 1.29 % 43. 23.3, 24,4

44. 30.33 % 45. 2 46. (i) 1.446 g, 0.112 M (ii) 1.735 g, 0.1344 M

47. H2C2O4 = 0.9 g, NaHC2O4 = 1.12 g 48. 1/2

49. NaHCO3 = 42%, Na2CO3 = 26.5%, Na2SO4 = 31.5 % 50. 5.3 g, 4.2g

EXERCISE-IV PREVIOUS YEAR QUESTIONS LEVEL-I (JEE MAIN)

1. B 2. C 3. D 4. A 5. D 6. D 7. B 8. C
9. D
LEVEL-II (JEE ADVANCED)
1. D 2. (A) P, S ; (B) R ; (C) P, Q ; (D) P 3. B 4. D 5. C 6. A

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