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JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction PDF Download

2021

Q1: A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the end point. Number of moles of Fe2+ present in 250 mL solution is x  10-2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.
(Assume: KMnO4 reacts only with Fe2+ in the solution
Use : Molar mass of iron as 56 g mol-1)    [JEE Advanced 2021 Paper 2]
The value of x is ______.
Ans:
1.875
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
Using relation,
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
n2M1V1 = n1M2V2 [n1 and n2 are number of moles of Fe2+ and MnO-4 reacting]
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
The volume of x is 1.88.

Q2: A sample (5.6 g) containing iron is completely dissolved in cold dilute HCl to prepare a 250 mL of solution. Titration of 25.0 mL of this solution requires 12.5 mL of 0.03 M KMnO4 solution to reach the end point. Number of moles of Fe2+ present in 250 mL solution is x  10-2 (consider complete dissolution of FeCl2). The amount of iron present in the sample is y% by weight.
(Assume : KMnO4 reacts only with Fe2+ in the solution
Use : Molar mass of iron as 56 g mol-1)
The value of y is ______.     [JEE Advanced 2021 Paper 2]
Ans: 
18.75
Mass of sample = 5.6 g
Number of moles of Fe = 1.88  x 10-2 mol
Molar mass of Fe = 56 g/mol
Mass of Fe = Moles of Fe  x Molar mass
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
⇒ y = 18.75%
The value of y is 18.75.

Q3: Ozonolysis of ClO2 produces an oxide of chlorine. The average oxidation state of chlorine in this oxide is __________.   [JEE Advanced 2021 Paper 2]
Ans:
6
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
Oxidation state of Cl in Cl2O6 = 2x + 6(2) = 0
2x  - 12 = 0
2x = 12 ⇒  x = + 6
Average oxidation state of Cl in Cl2O6 is 6.

2020

Q1: Choose the correct statement(s) among the following:   [JEE Advanced 2020 Paper 2]
(a) SnCl2 . 2H2O is a reducing agent.
(b) SnO2 reacts with KOH to form K2[Sn(OH)6].
(c) A solution of PbCl2 in HCl contains Pb2+ and Cl ions.
(d) The reaction of Pb3O4 with hot dilute nitric acid to give PbO2 is a redox reaction.
Ans:
(a, b)
Sn2+ of stannous chloride dihydrate (SnCl2 . 2H2O) tends to convert into Sn4+.
Hence, statement (a) is correct.
(b) SnO2 reacts with KOH and gives K2SnO3 . 3H2O or K2[Sn(OH)6] because it is amphoteric in nature.
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
Hence, statement (b) is correct.

(c) In conc. HCl, PbCl2 exists as chloroplumbous acid, H2[PbCl4]
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
Hence, statement (c) is incorrect.
(d) Pb3O4 is a mixture of JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
JEE Advanced Previous Year Questions (2020 - 2021): Redox Reaction
It is not a redox reaction. Thus, the statement (d) is incorrect.

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