JEE Advanced Previous Year Questions (2021-2023): Redox Reactions | Chemistry for JEE Main & Advanced PDF Download

 Page 1


JEE Advanced Previous Year Questions 
(2021-2023): Redox Reactions 
2021 
Numerical Type 
JEE Advanced 2021 Paper 2 Online 
Q1. A sample (?? . ?? ?? ) containing iron is completely dissolved in cold dilute ?????? to prepare a 
?????? ???? of solution. Titration of ???? . ?? ???? of this solution requires ???? . ?? ???? of ?? . ?????????????? 
?? 
solution to reach the end point. Number of moles of ????
?? +
 present in ?????? ???? solution is ?? × ????
-?? 
(consider complete dissolution of ????????
?? ). The amount of iron present in the sample is ?? % by 
weight. 
(Assume : ???????? ?? reacts only with ????
?? +
 in the solution 
Use : Molar mass of iron as ???? ?? ?????? -?? ) 
The value of ?? is 
Ans: 1.875 
Fe
(?? ×10
-2
?????? )
+ 2HCl ? FeCl
2
(?? ×10
-2
 mol)
+ H
2
 
Concentration of Fe
2+
, ?? 1
=
?? ×10
-2
250
× 1000 
=
10?? 250
?? =
?? 25
?? 
Volume of Fe
2+
 solution titrated, V
1
= 25.0 mL 
Concentration of MnO
4
-
, M
2
= 0.03M 
Volume of MnO
4
-
used, V
2
= 12.5 mL 
5Fe
2+
+ MnO
4
-
? Mn
2+
+ 5Fe
3+
 
Using relation, 
?? 1
?? 1
?? 1
=
?? 2
?? 2
?? 2
 
?? 2
?? 1
?? 1
= ?? 1
?? 2
?? 2
 [ ?? 1
 and ?? 2
 are number of moles of Fe
2+
 and MnO
4
-
reacting] 
1 ×
?? 25
× 25 = 5 × 0.03 × 12.5 ? ?? = 1.88 
The volume of x is 1.88 . 
JEE Advanced 2021 Paper 2 Online 
Q2. A sample (?? . ?? ?? ) containing iron is completely dissolved in cold dilute ?????? to prepare a 
?????? ?? ?? of solution. Titration of ???? . ?? ???? of this solution requires ???? . ?? ???? of ?? . ?????????????? ?? ?? 
solution to reach the end point. Number of moles of ????
?? +
 present in ?????? ???? solution is ?? × ????
-?? 
Page 2


JEE Advanced Previous Year Questions 
(2021-2023): Redox Reactions 
2021 
Numerical Type 
JEE Advanced 2021 Paper 2 Online 
Q1. A sample (?? . ?? ?? ) containing iron is completely dissolved in cold dilute ?????? to prepare a 
?????? ???? of solution. Titration of ???? . ?? ???? of this solution requires ???? . ?? ???? of ?? . ?????????????? 
?? 
solution to reach the end point. Number of moles of ????
?? +
 present in ?????? ???? solution is ?? × ????
-?? 
(consider complete dissolution of ????????
?? ). The amount of iron present in the sample is ?? % by 
weight. 
(Assume : ???????? ?? reacts only with ????
?? +
 in the solution 
Use : Molar mass of iron as ???? ?? ?????? -?? ) 
The value of ?? is 
Ans: 1.875 
Fe
(?? ×10
-2
?????? )
+ 2HCl ? FeCl
2
(?? ×10
-2
 mol)
+ H
2
 
Concentration of Fe
2+
, ?? 1
=
?? ×10
-2
250
× 1000 
=
10?? 250
?? =
?? 25
?? 
Volume of Fe
2+
 solution titrated, V
1
= 25.0 mL 
Concentration of MnO
4
-
, M
2
= 0.03M 
Volume of MnO
4
-
used, V
2
= 12.5 mL 
5Fe
2+
+ MnO
4
-
? Mn
2+
+ 5Fe
3+
 
Using relation, 
?? 1
?? 1
?? 1
=
?? 2
?? 2
?? 2
 
?? 2
?? 1
?? 1
= ?? 1
?? 2
?? 2
 [ ?? 1
 and ?? 2
 are number of moles of Fe
2+
 and MnO
4
-
reacting] 
1 ×
?? 25
× 25 = 5 × 0.03 × 12.5 ? ?? = 1.88 
The volume of x is 1.88 . 
JEE Advanced 2021 Paper 2 Online 
Q2. A sample (?? . ?? ?? ) containing iron is completely dissolved in cold dilute ?????? to prepare a 
?????? ?? ?? of solution. Titration of ???? . ?? ???? of this solution requires ???? . ?? ???? of ?? . ?????????????? ?? ?? 
solution to reach the end point. Number of moles of ????
?? +
 present in ?????? ???? solution is ?? × ????
-?? 
(consider complete dissolution of ????????
?? ). The amount of iron present in the sample is ?? % by 
weight. 
(Assume : ???????? ?? reacts only with ????
?? +
 in the solution 
Use : Molar mass of iron as ???? ?? ?????? -?? ) 
The value of ?? is 
Ans: 18.75 
Explanation 
Mass of sample = 5.6 g 
Number of moles of Fe = 1.88 × 10
-2
 mol 
Molar mass of Fe = 56 g/mol 
Mass of Fe = Moles of Fe × Molar mass 
= 1.88 × 10
-2
× 56 = 1.05 g 
%Fe =
 Mass of Fe 
 Mass of sample 
× 100 =
1.05
5.6
× 100 
? y = 18.75% 
The value of y is 18.75 . 
JEE Advanced 2021 Paper 2 Online 
Q3. Ozonolysis of ?????? ?? produces an oxide of chlorine. The average oxidation state of chlorine 
in this oxide is 
Ans: 6 
2ClO
2
 Chlorine dioxide 
+ 2O
3
? Cl
2
O
6
 Chloral perchlorate 
+ 2O
2
 
Oxidation state of Cl
in 
Cl
2
O
6
= 2x + 6(-2) = 0 
2?? - 12 = 0 
2?? = 12 ? ?? = +6 
Average oxidation state of Cl in Cl
2
O
6
 is 6 .