JEE Main Previous Year Questions (2025): Electrochemistry | Chemistry for JEE Main & Advanced PDF Download

 Page 1


JEE Main Previous Year Questions 
(2025): Electrochemistry 
Q1: A solution of aluminium chloride is electrolysed for ???? minutes using a current of 
2 A . The amount of the aluminium deposited at the cathode is ____ . 
[Given : molar mass of aluminium and chlorine are ???? ?? ?????? -?? and ???? . ?? ?? ?????? -?? 
respectively. Faraday constant = ???????????? ?????? -?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 1.660 g 
B. 1.007 g 
C. 0.336 g 
D. 0.441 g 
Ans: C 
Solution: 
Al
3+
+ 3?? -
? Al 
The mass of aluminium deposited can be calculated using Faraday's laws of electrolysis. The 
steps are as follows: 
Compute the total charge passed: 
?? = ?? · ?? = 2 A · (30 × 60 s) = 3600C 
Determine the moles of aluminium deposited. Since three electrons are required to deposit one 
mole of aluminium, the number of moles is given by: 
?? (Al ) =
?? 3?? =
3600
3 × 96500
˜ 0.01242 mol 
Finally, calculate the mass of aluminium using its molar mass ( ?? = 27 g/mol ): 
?? (Al ) = ?? (Al ) × ?? = 0.01242× 27 ˜ 0.335 g 
 
Q2: Which of the following electrolyte can be used to obtain ?? ?? ?? ?? ?? ?? by the process 
of electrolysis ? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Concentrated solution of sulphuric acid 
B. Dilute solution of sodium sulphate. 
C. Acidified dilute solution of sodium sulphate. 
D. Dilute solution of sulphuric acid 
Ans: A 
Solution: 
Page 2


JEE Main Previous Year Questions 
(2025): Electrochemistry 
Q1: A solution of aluminium chloride is electrolysed for ???? minutes using a current of 
2 A . The amount of the aluminium deposited at the cathode is ____ . 
[Given : molar mass of aluminium and chlorine are ???? ?? ?????? -?? and ???? . ?? ?? ?????? -?? 
respectively. Faraday constant = ???????????? ?????? -?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 1.660 g 
B. 1.007 g 
C. 0.336 g 
D. 0.441 g 
Ans: C 
Solution: 
Al
3+
+ 3?? -
? Al 
The mass of aluminium deposited can be calculated using Faraday's laws of electrolysis. The 
steps are as follows: 
Compute the total charge passed: 
?? = ?? · ?? = 2 A · (30 × 60 s) = 3600C 
Determine the moles of aluminium deposited. Since three electrons are required to deposit one 
mole of aluminium, the number of moles is given by: 
?? (Al ) =
?? 3?? =
3600
3 × 96500
˜ 0.01242 mol 
Finally, calculate the mass of aluminium using its molar mass ( ?? = 27 g/mol ): 
?? (Al ) = ?? (Al ) × ?? = 0.01242× 27 ˜ 0.335 g 
 
Q2: Which of the following electrolyte can be used to obtain ?? ?? ?? ?? ?? ?? by the process 
of electrolysis ? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Concentrated solution of sulphuric acid 
B. Dilute solution of sodium sulphate. 
C. Acidified dilute solution of sodium sulphate. 
D. Dilute solution of sulphuric acid 
Ans: A 
Solution: 
H
2
 S
2
O
8
 is obtained by electrolysis of concentrated solution of sulphuric acid. 
At anode : 
2HSO
4
-
? H
2
 S
2
O
8
+ 2e
-
 
Q3: Given below are two statements : 
Statement (I): Corrosion is an electrochemical phenomenon in which pure metal acts 
as an anode and impure metal as a cathode. 
Statement (II): The rate of corrosion is more in alkaline medium than in acidic 
medium. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true 
Ans: C 
Solution: 
Let's examine both statements one by one and see whether they hold true for typical corrosion 
processes (e.g., iron rusting): 
Statement (I) 
"Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure 
metal as a cathode." 
Corrosion is electrochemical 
True. Corrosion, especially rusting of iron, involves oxidation at one region (the anode) and 
reduction at another region (the cathode) on the metal's surface. 
Pure metal as anode, impurity as cathode 
In many practical cases (e.g., iron containing small amounts of carbon or other impurities), the 
relatively pure region of the iron is more active (less noble) and tends to oxidize (lose electrons) 
- that is, it serves as the anode. 
The impurity (e.g., carbon-rich region) often behaves more nobly and thus becomes the cathode 
region where reduction (e.g., oxygen reduction) occurs. 
Page 3


JEE Main Previous Year Questions 
(2025): Electrochemistry 
Q1: A solution of aluminium chloride is electrolysed for ???? minutes using a current of 
2 A . The amount of the aluminium deposited at the cathode is ____ . 
[Given : molar mass of aluminium and chlorine are ???? ?? ?????? -?? and ???? . ?? ?? ?????? -?? 
respectively. Faraday constant = ???????????? ?????? -?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 1.660 g 
B. 1.007 g 
C. 0.336 g 
D. 0.441 g 
Ans: C 
Solution: 
Al
3+
+ 3?? -
? Al 
The mass of aluminium deposited can be calculated using Faraday's laws of electrolysis. The 
steps are as follows: 
Compute the total charge passed: 
?? = ?? · ?? = 2 A · (30 × 60 s) = 3600C 
Determine the moles of aluminium deposited. Since three electrons are required to deposit one 
mole of aluminium, the number of moles is given by: 
?? (Al ) =
?? 3?? =
3600
3 × 96500
˜ 0.01242 mol 
Finally, calculate the mass of aluminium using its molar mass ( ?? = 27 g/mol ): 
?? (Al ) = ?? (Al ) × ?? = 0.01242× 27 ˜ 0.335 g 
 
Q2: Which of the following electrolyte can be used to obtain ?? ?? ?? ?? ?? ?? by the process 
of electrolysis ? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Concentrated solution of sulphuric acid 
B. Dilute solution of sodium sulphate. 
C. Acidified dilute solution of sodium sulphate. 
D. Dilute solution of sulphuric acid 
Ans: A 
Solution: 
H
2
 S
2
O
8
 is obtained by electrolysis of concentrated solution of sulphuric acid. 
At anode : 
2HSO
4
-
? H
2
 S
2
O
8
+ 2e
-
 
Q3: Given below are two statements : 
Statement (I): Corrosion is an electrochemical phenomenon in which pure metal acts 
as an anode and impure metal as a cathode. 
Statement (II): The rate of corrosion is more in alkaline medium than in acidic 
medium. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true 
Ans: C 
Solution: 
Let's examine both statements one by one and see whether they hold true for typical corrosion 
processes (e.g., iron rusting): 
Statement (I) 
"Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure 
metal as a cathode." 
Corrosion is electrochemical 
True. Corrosion, especially rusting of iron, involves oxidation at one region (the anode) and 
reduction at another region (the cathode) on the metal's surface. 
Pure metal as anode, impurity as cathode 
In many practical cases (e.g., iron containing small amounts of carbon or other impurities), the 
relatively pure region of the iron is more active (less noble) and tends to oxidize (lose electrons) 
- that is, it serves as the anode. 
The impurity (e.g., carbon-rich region) often behaves more nobly and thus becomes the cathode 
region where reduction (e.g., oxygen reduction) occurs. 
This difference in electrode potentials between the pure region and the impurity region drives 
the corrosion cell. 
Hence, Statement (I) is generally considered true in the usual context of corrosion (like rusting of 
iron). 
Statement (II) 
"The rate of corrosion is more in alkaline medium than in acidic medium." 
For most common metals (like iron), acidic media typically enhance corrosion because abundant 
H
+
ions (and possibly other acidic species) can accelerate the oxidation/dissolution of the metal. 
In mildly alkaline or neutral environments, metals often form protective oxide or hydroxide 
layers that can slow down further corrosion. 
While certain strong bases can attack specific metals (e.g., Al in strong NaOH), in general for iron 
and many other metals, corrosion is more severe in acidic environments than in alkaline ones. 
Thus, Statement (II) is false under normal corrosion scenarios (e.g., rusting of iron). 
Conclusion 
Statement (I): True 
Statement (II): False 
Therefore, the correct choice (matching these truth values) is: 
(C) Statement I is true but Statement II is false. 
 
Q4: ?????? ?? ?? -
?
+?? .?? ?? 
????
?? +
?
?? .?? ?? 
????
?? +
?
-?? .?? ?? 
????
?? 
In the above diagram, the standard electrode potentials are given in volts (over the 
arrow). 
The value of ?? ?????? ?? ?? -
/????
?? + is : 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 1.2 V 
B. 2.1 V 
C. 1.4 V 
D. 1.7 V 
Ans: D 
Solution: 
Page 4


JEE Main Previous Year Questions 
(2025): Electrochemistry 
Q1: A solution of aluminium chloride is electrolysed for ???? minutes using a current of 
2 A . The amount of the aluminium deposited at the cathode is ____ . 
[Given : molar mass of aluminium and chlorine are ???? ?? ?????? -?? and ???? . ?? ?? ?????? -?? 
respectively. Faraday constant = ???????????? ?????? -?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 1.660 g 
B. 1.007 g 
C. 0.336 g 
D. 0.441 g 
Ans: C 
Solution: 
Al
3+
+ 3?? -
? Al 
The mass of aluminium deposited can be calculated using Faraday's laws of electrolysis. The 
steps are as follows: 
Compute the total charge passed: 
?? = ?? · ?? = 2 A · (30 × 60 s) = 3600C 
Determine the moles of aluminium deposited. Since three electrons are required to deposit one 
mole of aluminium, the number of moles is given by: 
?? (Al ) =
?? 3?? =
3600
3 × 96500
˜ 0.01242 mol 
Finally, calculate the mass of aluminium using its molar mass ( ?? = 27 g/mol ): 
?? (Al ) = ?? (Al ) × ?? = 0.01242× 27 ˜ 0.335 g 
 
Q2: Which of the following electrolyte can be used to obtain ?? ?? ?? ?? ?? ?? by the process 
of electrolysis ? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Concentrated solution of sulphuric acid 
B. Dilute solution of sodium sulphate. 
C. Acidified dilute solution of sodium sulphate. 
D. Dilute solution of sulphuric acid 
Ans: A 
Solution: 
H
2
 S
2
O
8
 is obtained by electrolysis of concentrated solution of sulphuric acid. 
At anode : 
2HSO
4
-
? H
2
 S
2
O
8
+ 2e
-
 
Q3: Given below are two statements : 
Statement (I): Corrosion is an electrochemical phenomenon in which pure metal acts 
as an anode and impure metal as a cathode. 
Statement (II): The rate of corrosion is more in alkaline medium than in acidic 
medium. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true 
Ans: C 
Solution: 
Let's examine both statements one by one and see whether they hold true for typical corrosion 
processes (e.g., iron rusting): 
Statement (I) 
"Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure 
metal as a cathode." 
Corrosion is electrochemical 
True. Corrosion, especially rusting of iron, involves oxidation at one region (the anode) and 
reduction at another region (the cathode) on the metal's surface. 
Pure metal as anode, impurity as cathode 
In many practical cases (e.g., iron containing small amounts of carbon or other impurities), the 
relatively pure region of the iron is more active (less noble) and tends to oxidize (lose electrons) 
- that is, it serves as the anode. 
The impurity (e.g., carbon-rich region) often behaves more nobly and thus becomes the cathode 
region where reduction (e.g., oxygen reduction) occurs. 
This difference in electrode potentials between the pure region and the impurity region drives 
the corrosion cell. 
Hence, Statement (I) is generally considered true in the usual context of corrosion (like rusting of 
iron). 
Statement (II) 
"The rate of corrosion is more in alkaline medium than in acidic medium." 
For most common metals (like iron), acidic media typically enhance corrosion because abundant 
H
+
ions (and possibly other acidic species) can accelerate the oxidation/dissolution of the metal. 
In mildly alkaline or neutral environments, metals often form protective oxide or hydroxide 
layers that can slow down further corrosion. 
While certain strong bases can attack specific metals (e.g., Al in strong NaOH), in general for iron 
and many other metals, corrosion is more severe in acidic environments than in alkaline ones. 
Thus, Statement (II) is false under normal corrosion scenarios (e.g., rusting of iron). 
Conclusion 
Statement (I): True 
Statement (II): False 
Therefore, the correct choice (matching these truth values) is: 
(C) Statement I is true but Statement II is false. 
 
Q4: ?????? ?? ?? -
?
+?? .?? ?? 
????
?? +
?
?? .?? ?? 
????
?? +
?
-?? .?? ?? 
????
?? 
In the above diagram, the standard electrode potentials are given in volts (over the 
arrow). 
The value of ?? ?????? ?? ?? -
/????
?? + is : 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 1.2 V 
B. 2.1 V 
C. 1.4 V 
D. 1.7 V 
Ans: D 
Solution: 
FeO
4
 
2
?
E 
n
1
 
,
2 V
3
Fe
-3
?
E
2
 
n
2
 
, 0.8 V
E
4
o
=?
n
4
= 4
 
?G
4
o
= ?G
1
o
+ ?G
2
o
? -n
4
FE
4
o
= -n
1
FE
1
0
- n
2
FE
2
o
? +4E
4
o
= 3 × 2 + (1 × 0.8)
?E
4
o
=
6.8
4
 V
?E
4
o
= 1.7 V
 
 
Q5: Standard electrode potentials for a few half cells are mentioned below : 
?? ????
?? +
/????
°
= ?? . ???? ?? , ?? ????
?? +
/????
°
= -?? . ???? ?? 
?? ????
+
/????
°
= ?? . ???? ?? , ?? ????
?? +
/????
°
= -?? . ???? ?? 
Which one of the following cells gives the most negative value of ?? ?? °
 ? 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. Zn|Zn
2+
(1M )||Ag
+
(1M )|Ag 
B. Ag |Ag
+
(1M )||Mg
2+
(1M )|Mg 
C. Zn|Zn
2+
(1M )||Mg
2+
(1M )|Mg 
D. Cu|Cu
2+
(1M )||Ag
+
(1M )|Ag 
Ans: A 
Solution: 
? ?G
°
= -nFE
°
 
Option (1) E
°
= 0.8 + 0.76 
 = 1.56 V
? ?G
°
 = -2 × F × 1.56
 = -3.12 V
 
Option (2) E
°
= -2.37 + 0.76 
= -1.61 V
? ?G
°
= -2 × F × (-1.61)
= +3.22 V
 
Option (3) E
°
= -2.37 - 0.8 
Page 5


JEE Main Previous Year Questions 
(2025): Electrochemistry 
Q1: A solution of aluminium chloride is electrolysed for ???? minutes using a current of 
2 A . The amount of the aluminium deposited at the cathode is ____ . 
[Given : molar mass of aluminium and chlorine are ???? ?? ?????? -?? and ???? . ?? ?? ?????? -?? 
respectively. Faraday constant = ???????????? ?????? -?? ] 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. 1.660 g 
B. 1.007 g 
C. 0.336 g 
D. 0.441 g 
Ans: C 
Solution: 
Al
3+
+ 3?? -
? Al 
The mass of aluminium deposited can be calculated using Faraday's laws of electrolysis. The 
steps are as follows: 
Compute the total charge passed: 
?? = ?? · ?? = 2 A · (30 × 60 s) = 3600C 
Determine the moles of aluminium deposited. Since three electrons are required to deposit one 
mole of aluminium, the number of moles is given by: 
?? (Al ) =
?? 3?? =
3600
3 × 96500
˜ 0.01242 mol 
Finally, calculate the mass of aluminium using its molar mass ( ?? = 27 g/mol ): 
?? (Al ) = ?? (Al ) × ?? = 0.01242× 27 ˜ 0.335 g 
 
Q2: Which of the following electrolyte can be used to obtain ?? ?? ?? ?? ?? ?? by the process 
of electrolysis ? 
JEE Main 2025 (Online) 22nd January Morning Shift 
Options: 
A. Concentrated solution of sulphuric acid 
B. Dilute solution of sodium sulphate. 
C. Acidified dilute solution of sodium sulphate. 
D. Dilute solution of sulphuric acid 
Ans: A 
Solution: 
H
2
 S
2
O
8
 is obtained by electrolysis of concentrated solution of sulphuric acid. 
At anode : 
2HSO
4
-
? H
2
 S
2
O
8
+ 2e
-
 
Q3: Given below are two statements : 
Statement (I): Corrosion is an electrochemical phenomenon in which pure metal acts 
as an anode and impure metal as a cathode. 
Statement (II): The rate of corrosion is more in alkaline medium than in acidic 
medium. 
In the light of the above statements, choose the correct answer from the options 
given below : 
JEE Main 2025 (Online) 22nd January Evening Shift 
Options: 
A. Both Statement I and Statement II are true 
B. Both Statement I and Statement II are false 
C. Statement I is true but Statement II is false 
D. Statement I is false but Statement II is true 
Ans: C 
Solution: 
Let's examine both statements one by one and see whether they hold true for typical corrosion 
processes (e.g., iron rusting): 
Statement (I) 
"Corrosion is an electrochemical phenomenon in which pure metal acts as an anode and impure 
metal as a cathode." 
Corrosion is electrochemical 
True. Corrosion, especially rusting of iron, involves oxidation at one region (the anode) and 
reduction at another region (the cathode) on the metal's surface. 
Pure metal as anode, impurity as cathode 
In many practical cases (e.g., iron containing small amounts of carbon or other impurities), the 
relatively pure region of the iron is more active (less noble) and tends to oxidize (lose electrons) 
- that is, it serves as the anode. 
The impurity (e.g., carbon-rich region) often behaves more nobly and thus becomes the cathode 
region where reduction (e.g., oxygen reduction) occurs. 
This difference in electrode potentials between the pure region and the impurity region drives 
the corrosion cell. 
Hence, Statement (I) is generally considered true in the usual context of corrosion (like rusting of 
iron). 
Statement (II) 
"The rate of corrosion is more in alkaline medium than in acidic medium." 
For most common metals (like iron), acidic media typically enhance corrosion because abundant 
H
+
ions (and possibly other acidic species) can accelerate the oxidation/dissolution of the metal. 
In mildly alkaline or neutral environments, metals often form protective oxide or hydroxide 
layers that can slow down further corrosion. 
While certain strong bases can attack specific metals (e.g., Al in strong NaOH), in general for iron 
and many other metals, corrosion is more severe in acidic environments than in alkaline ones. 
Thus, Statement (II) is false under normal corrosion scenarios (e.g., rusting of iron). 
Conclusion 
Statement (I): True 
Statement (II): False 
Therefore, the correct choice (matching these truth values) is: 
(C) Statement I is true but Statement II is false. 
 
Q4: ?????? ?? ?? -
?
+?? .?? ?? 
????
?? +
?
?? .?? ?? 
????
?? +
?
-?? .?? ?? 
????
?? 
In the above diagram, the standard electrode potentials are given in volts (over the 
arrow). 
The value of ?? ?????? ?? ?? -
/????
?? + is : 
JEE Main 2025 (Online) 23rd January Morning Shift 
Options: 
A. 1.2 V 
B. 2.1 V 
C. 1.4 V 
D. 1.7 V 
Ans: D 
Solution: 
FeO
4
 
2
?
E 
n
1
 
,
2 V
3
Fe
-3
?
E
2
 
n
2
 
, 0.8 V
E
4
o
=?
n
4
= 4
 
?G
4
o
= ?G
1
o
+ ?G
2
o
? -n
4
FE
4
o
= -n
1
FE
1
0
- n
2
FE
2
o
? +4E
4
o
= 3 × 2 + (1 × 0.8)
?E
4
o
=
6.8
4
 V
?E
4
o
= 1.7 V
 
 
Q5: Standard electrode potentials for a few half cells are mentioned below : 
?? ????
?? +
/????
°
= ?? . ???? ?? , ?? ????
?? +
/????
°
= -?? . ???? ?? 
?? ????
+
/????
°
= ?? . ???? ?? , ?? ????
?? +
/????
°
= -?? . ???? ?? 
Which one of the following cells gives the most negative value of ?? ?? °
 ? 
JEE Main 2025 (Online) 23rd January Evening Shift 
Options: 
A. Zn|Zn
2+
(1M )||Ag
+
(1M )|Ag 
B. Ag |Ag
+
(1M )||Mg
2+
(1M )|Mg 
C. Zn|Zn
2+
(1M )||Mg
2+
(1M )|Mg 
D. Cu|Cu
2+
(1M )||Ag
+
(1M )|Ag 
Ans: A 
Solution: 
? ?G
°
= -nFE
°
 
Option (1) E
°
= 0.8 + 0.76 
 = 1.56 V
? ?G
°
 = -2 × F × 1.56
 = -3.12 V
 
Option (2) E
°
= -2.37 + 0.76 
= -1.61 V
? ?G
°
= -2 × F × (-1.61)
= +3.22 V
 
Option (3) E
°
= -2.37 - 0.8 
= -3.17 V
? ?G
°
= -2
= F × (-3.17)
= +6.34
 
Option (4) E
°
= 0.8 - 0.34 
 = 0.46 V
?G
°
 = -2 × F × 0.46
 = -0.92 V
 
 
Q6: For the given cell 
????
?? +
(???? ) + ????
(???? )
+
? ????
?? +
(???? ) + ????
(?? )
 
The standard cell potential of the above reaction is Given: 
????
+
+ ?? -
? ???? ?? ?? = ???? 
????
?? +
+ ?? ?? -
? ???? ?? ?? = ???? 
????
?? +
+ ?? ?? -
? ???? ?? ?? = ???? 
JEE Main 2025 (Online) 24th January Morning Shift 
Options: 
A. ?? + 2?? - 3?? 
B. ?? + 2?? 
C. ?? - 2?? 
D. ?? + ?? - ?? 
Ans: A 
Solution: 
Fe
2+
(aq ) + Ag
+
(aq ) ? Fe
3+
(aq ) + Ag (s) 
 
 
 
?G
3
0
= ?G
1
0
+ ?G
2
0