Electrode potentials (E ) are given below : Based on the above potentials, strongest oxidizing agent will be :
1
2
3
4
Official Solution
Correct Option: (3)
Higher the value of reduction potential stronger will be the oxidising hence based on the given values will be strongest oxidizing agent.
02
PYQ 2013
easy
chemistryID: jee-main
The standard enthalpies of formation of and glucose(.v) at 25 C- are 400 kJ/ mol. -300 kJ/mol and - 1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25 C is
1
#ERROR!
2
- 2900 kJ
3
- 16.11 kJ
4
#ERROR!
Official Solution
Correct Option: (3)
PLAN A CH (Standard heat of combustion) is the standard enthalpy change when one mole of the substance is completely oxidised. Also standard heat of formation (A fH ) can be taken as the standard of that substance Molar mass of Thus, standard heat of combustion of glucose per gram To solve such problem, students are advised to keep much importance in unit conversion. As here, value of R in must be converted into kJ by dividing the unit by 1000.
03
PYQ 2014
easy
chemistryID: jee-main
Given below are the half-cell reactions The for will be
1
; the reaction will not occur
2
; the reaction will occur
3
; the reaction will occur
4
; the reaction will occur
Official Solution
Correct Option: (1)
(1) ;
(2) ;
;
But
As value is negative reaction is non spontaneous
04
PYQ 2014
medium
chemistryID: jee-main
The equivalent conductance of at concentration and at infinite dilution are and respectively.
The correct relationship between and is given as (where, the constant is positive)
1
2
3
4
Official Solution
Correct Option: (3)
According to Debye Huckle onsager equation,
Here
05
PYQ 2014
medium
chemistryID: jee-main
Resistance of solution of an electrolyte is . The specific conductance of the solution of solution of same electrolyte is and resistance of same solution of the same electrolyte is . The molar conductivity of solution of the electrolyte in is
1
2
3
4
Official Solution
Correct Option: (1)
For solution
Now,
For M solution
Now ,
06
PYQ 2014
medium
chemistryID: jee-main
How many electrons would be required to deposit of copper at the cathode during the electrolysis of an aqueous solution of copper sulphate ? (Atomic mass of copper Avogadro's constant) :
1
2
3
4
Official Solution
Correct Option: (3)
coulomb
charge of mol of No. of
07
PYQ 2015
easy
chemistryID: jee-main
Two Faraday of electricity is passed through a solution of . The mass of copper deposited at the cathode is (At. mass of )
1
2
3
4
Official Solution
Correct Option: (2)
2 mole deposit 1 mole of mole mole of .
08
PYQ 2016
medium
chemistryID: jee-main
Galvanization is applying a coating of :
1
Pb
2
Cr
3
Cu
4
Zn
Official Solution
Correct Option: (4)
Galvanization is applying a coating of zinc.
09
PYQ 2017
medium
chemistryID: jee-main
What is the standard reduction potential for ? Given that :
1
2
3
4
Official Solution
Correct Option: (1)
Answer (a)
10
PYQ 2018
medium
chemistryID: jee-main
How long (approximate) should water be electrolysed by passing through amperes current so that the oxygen released can completely burn of diborane ? (Atomic weight of )
1
6.4 hours
2
0.8 hours
3
3.2 hours
4
1.6 hours
Official Solution
Correct Option: (3)
of mole of which requires three moles of oxygen for complete burning
(On electrolysis)
Number of faradays Amount of charge
second
hour
hours
11
PYQ 2018
medium
chemistryID: jee-main
When current was passed for into nitrobenzene in acidic medium, the amount of p-aminophenol produced is_____.
1
9.81 g
2
10.9 g
3
98.1 g
4
109.0 g
Official Solution
Correct Option: (1)
The reaction involved is
We know
From the above reaction, we can see that of electrons are required to reduce nitrobenzene to -aminophenol.
Therefore, mol of electrons will reduce mol of nitrobenzene to -aminophenol.
Mass of -aminophenol produced = Molar mass No. of moles
12
PYQ 2018
medium
chemistryID: jee-main
When an electric current is passed through acidified water, of hydrogen gas at N.T.P. was collected at the cathode in seconds. The current passed, in ampere, is :
1
1
2
0.5
3
0.1
4
2
Official Solution
Correct Option: (1)
At cathode:
At NTP, or of of of of
We know
13
PYQ 2019
medium
chemistryID: jee-main
A solution of is electrolysed between platinum electrodes using Faraday electricity. How many mole of will be deposited at the cathode?
1
0.2
2
0.05
3
0.1
4
0.15
Official Solution
Correct Option: (2)
0.1 e of will be discharged. No. of eq = (No of moles) (n-factor) 0.1 = (No. of moles) 2 No. of moles of Ni =
14
PYQ 2021
medium
chemistryID: jee-main
These are physical properties of an element (A) Sublimation enthalpy (B) Ionisation enthalpy (C) Hydration enthalpy (D) Electron gain enthalpy The total number of above properties that affect the reduction potential is _________. (Integer answer)
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: The standard electrode potential (reduction potential) of a metal electrode ( ) is determined by the enthalpy changes in a thermodynamic cycle (Born-Haber cycle for ions in solution). Step 2: Detailed Explanation: The process can be broken down into: 1. Hydration of the ion: (Reverse of hydration enthalpy) 2. Ionisation of gas atoms: (Reverse of ionisation enthalpy) 3. Atomisation/Sublimation: (Reverse of sublimation enthalpy)
Therefore, sublimation enthalpy, ionisation enthalpy, and hydration enthalpy are the three key factors that determine the reduction potential of metallic elements. Electron gain enthalpy is typically used for defining the oxidizing power of non-metals (like halogens) and is not part of the standard metal electrode potential cycle. Step 3: Final Answer: The total number of properties is 3.
15
PYQ 2021
medium
chemistryID: jee-main
Consider the following cell reaction
The value of is at . If , the standard entropy change in J K is . (Nearest integer) [Given : Faraday constant = ]
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: We use the relationship between the standard cell potential and the standard Gibbs free energy change, and then apply the thermodynamic relation involving enthalpy and entropy. Step 2: Key Formula or Approach: 1. 2. Step 3: Detailed Explanation: 1. Determine : In the reaction, is oxidised to in , and in is reduced to . Thus, . 2. Calculate :
3. Calculate : Given . Temperature .
Step 4: Final Answer: The standard entropy change to the nearest integer is 25.
16
PYQ 2021
medium
chemistryID: jee-main
For the galvanic cell,
Zn(s) + Cu (0.02 M) Zn (0.04 M) + Cu(s),
E = __________ 10 V. (Nearest integer)
[Use : E = -0.34 V, E = +0.76 V, ]
Official Solution
Correct Option: (1)
Step 1: Understanding the Question
We need to calculate the cell potential (E ) for a galvanic cell under non-standard conditions using the Nernst equation. Step 2: Key Formula or Approach
The Nernst equation for a cell is:
where is the standard cell potential, is the number of electrons transferred, and is the reaction quotient. Step 3: Detailed Calculation Calculate the Standard Cell Potential ( ): The overall reaction is Zn + Cu Zn + Cu. Oxidation (Anode): Zn Zn + 2e Reduction (Cathode): Cu + 2e Cu We need the standard reduction potentials: E : Given E = +0.76 V (standard oxidation potential), so E = -0.76 V (standard reduction potential). E : Given E = -0.34 V (standard oxidation potential), so E = +0.34 V (standard reduction potential). Now, calculate : Determine n and Q: From the half-reactions, the number of electrons transferred, , is 2. The reaction quotient, , is: Apply the Nernst Equation: Using log(2) 0.301: Express the answer in the required format: Step 4: Final Answer
Rounding to the nearest integer, the value is 109.
17
PYQ 2021
medium
chemistryID: jee-main
Given below are two statements :
Statement I : The limiting molar conductivity of KCl (strong electrolyte) is higher compared to that of (weak electrolyte).
Statement II : Molar conductivity decreases with decrease in concentration of electrolyte.
In the light of the above statements, choose the most appropriate answer from the options given below :
1
Both Statement I and Statement II are true
2
Both Statement I and Statement II are false
3
Statement I is true but Statement II is false
4
Statement I is false but Statement II is true
Official Solution
Correct Option: (2)
Step 1: Understanding the Concept:
Limiting molar conductivity ( ) depends on the individual ionic conductivities at infinite dilution. Molar conductivity ( ) is the conducting power of all ions produced by one mole of an electrolyte in a given volume of solution. Step 2: Detailed Explanation:
1. Analysis of Statement I: According to Kohlrausch's law, is the sum of limiting ionic conductivities. The limiting ionic conductivity of ( ) is much higher than that of ( ). Because of this, is actually higher than . Statement I is false.
2. Analysis of Statement II: Molar conductivity is defined as . As concentration decreases (dilution increases), the volume containing one mole of electrolyte increases. This increase in volume more than compensates for the decrease in , leading to an increase in molar conductivity upon dilution for both strong and weak electrolytes. Statement II is false. Step 3: Final Answer:
Both Statement I and Statement II are false.
18
PYQ 2021
medium
chemistryID: jee-main
Match List - I with List - II :
Choose the most appropriate answer from the options given below :
1
(a)-(ii), (b)-(i), (c)-(iii), (d)-(iv)
2
(a)-(iii), (b)-(i), (c)-(iv), (d)-(ii)
3
(a)-(i), (b)-(iv), (c)-(iii), (d)-(ii)
4
(a)-(iii), (b)-(i), (c)-(ii), (d)-(iv)
Official Solution
Correct Option: (2)
Step 1: Understanding the Concept: This question involves matching various electrochemical parameters with their respective SI or common units. Step 2: Key Formula or Approach: 1. Cell Constant ( ) = . Unit = . 2. Molar Conductivity ( ) = . Unit = . 3. Conductivity ( ) = . Unit = or . 4. Degree of dissociation ( ) = Ratio of conductivities. Being a ratio of same physical quantities, it is dimensionless. Step 3: Detailed Explanation: - (a) Cell constant is the ratio of distance between electrodes to the cross-sectional area. Its unit is . Match: (a)-(iii). - (b) Molar conductivity represents the conducting power of all ions produced by dissolving 1 mole of an electrolyte. Its unit is typically expressed as . Match: (b)-(i). - (c) Conductivity (specific conductance) is the reciprocal of resistivity. Its unit is . Match: (c)-(iv). - (d) Degree of dissociation is the fraction of total electrolyte that dissociates into ions. It is a pure number. Match: (d)-(ii). Step 4: Final Answer: The correct matching is (a)-(iii), (b)-(i), (c)-(iv), (d)-(ii).
19
PYQ 2021
medium
chemistryID: jee-main
The resistance of a conductivity cell with cell constant 1.14 cm , containing 0.001 M KCl at 298 K is 1500 . The molar conductivity of 0.001 M KCl solution at 298 K in S cm mol is _________. (Integer answer)
Official Solution
Correct Option: (1)
Step 1: Understanding the Question:
We are given the resistance, cell constant, and concentration for a solution and asked to calculate its molar conductivity. Step 2: Key Formulas:
1. Conductivity ( ): It relates the resistance (R) and the cell constant ( ).
2. Molar Conductivity ( ): It relates conductivity to the molar concentration (C). (The factor of 1000 is used when is in S cm and C is in mol L to get in S cm mol ). Step 3: Calculate Conductivity ( ):
Given:
- Cell constant, cm .
- Resistance, . Step 4: Calculate Molar Conductivity ( ):
Given:
- Concentration, M. Step 5: Final Answer:
The molar conductivity of the solution is 760 S cm mol .
20
PYQ 2021
medium
chemistryID: jee-main
The molar conductivities at infinite dilution of barium chloride, sulphuric acid and hydrochloric acid are 280, 860 and 426 S cm² mol⁻¹ respectively. The molar conductivity at infinite dilution of barium sulphate is ________ S cm² mol⁻¹. (Round off to the Nearest Integer).
Official Solution
Correct Option: (1)
Step 1: .
Step 2: .
21
PYQ 2021
medium
chemistryID: jee-main
For the cell Cu(s)|Cu (aq) (0.1 M)||Ag (aq) (0.01 M)|Ag(s) the cell potential E = 0.3095 V For the cell Cu(s)|Cu (aq) (0.01 M)||Ag (aq) (0.001 M)|Ag(s) the cell potential = ________ V. (Round off to the Nearest Integer). [Use : ]
Official Solution
Correct Option: (1)
Step 1: Determine the standard cell potential (E ) using the data for the first cell.
The overall cell reaction is: Cu(s) + 2Ag (aq) Cu (aq) + 2Ag(s).
The number of electrons transferred, n = 2.
The Nernst equation for the cell potential E is:
, where . For the first cell (E ):
V, [Cu ] = 0.1 M, [Ag ] = 0.01 M.
.
.
.
V. Step 2: Calculate the cell potential for the second cell (E ) using the calculated E .
For the second cell:
[Cu ] = 0.01 M, [Ag ] = 0.001 M.
.
.
.
.
V. Step 3: Express the answer in the required format.
The question asks for the potential in units of V.
V.
Rounding to the nearest integer, the value is 28.
22
PYQ 2021
medium
chemistryID: jee-main
Consider the cell at : . The fraction of total iron present as ion at the cell potential of 1.500 V is . The value of is ________. (Nearest integer) (Given: , )
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: The cell potential is calculated using the Nernst equation, which accounts for non-standard concentrations. The total iron consists of both and ions. Step 2: Key Formula or Approach: 1. Standard Cell Potential . 2. Nernst Equation: . Step 3: Detailed Explanation: 1. Cell Reaction: Anode: . Cathode: (multiply by 2 for electron balance). Net: . ( )
2. Calculate : .
3. Apply Nernst Equation:
4. Calculate Fraction of : Let and . Total Iron = . Fraction . Step 4: Final Answer: The value of is 24.
23
PYQ 2021
medium
chemistryID: jee-main
The magnitude of the change in oxidising power of the MnO4− / Mn2+ couple is V, if the H+ concentration is decreased from 1 M to M at 25 °C. (Assume concentration of MnO4− and Mn2+ to be same on change in H+ concentration). The value of is __________ .
For the reaction 2Fe (aq) + 2I (aq) → 2Fe (aq) + I₂(s) the magnitude of the standard molar free energy change, = - _________ kJ (Round off to the Nearest Integer). [Given : = -0.440 V; = -0.036 V = 0.539 V; F = 96500 C]
Official Solution
Correct Option: (1)
Step 1: Find .
.
V.
Step 2: V.
Step 3: J kJ.
25
PYQ 2021
medium
chemistryID: jee-main
Copper reduces NO into NO and NO depending upon the concentration of HNO in solution. (Assuming fixed [Cu ] and ), the HNO concentration at which the thermodynamic tendency for reduction of NO into NO and NO by copper is same is M. The value of 2x is ________. (Rounded-off to the nearest integer)
[Given, E =0.34 V, E =0.96 V, E =0.79 V and at 298 K, (2.303)=0.059]
Official Solution
Correct Option: (1)
Copper reduces into NO and NO depending on the concentration of .
Given:
Step 1: Standard cell potentials For reduction to NO:
For reduction to NO :
Step 2: Condition for equal thermodynamic tendencyStep 3: Apply Nernst equation Let , so .
Assuming , , . Step 4: Final answer
26
PYQ 2021
easy
chemistryID: jee-main
The conductivity of a weak acid HA of concentration 0.001 mol L is S cm . If S cm mol , the ionization constant (K ) of HA is equal to _________ . (Round off to the Nearest Integer)
Official Solution
Correct Option: (1)
27
PYQ 2021
medium
chemistryID: jee-main
The product obtained from the electrolytic oxidation of acidified sulphate solutions, is :
1
HO SOSO H
2
HO SOSO H
3
HO SOOSO H
4
HSO
Official Solution
Correct Option: (3)
The electrolytic oxidation of an acidified concentrated solution of a sulphate, such as ammonium sulphate or sulfuric acid, at high current density is a method for preparing peroxodisulphates. At the anode, the bisulphate ion ( ) or sulphate ion ( ) is oxidized. The primary reaction is the coupling of two bisulphate radicals. The overall anode reaction can be represented as:
The product is the peroxodisulphate ion, . The corresponding acid is peroxodisulphuric acid, , also known as Marshall's acid. Let's examine the structure of this acid. It contains a peroxide linkage (-O-O-) between the two sulfur atoms. The structure is: \begin{center} O \hspace{0.5cm} O \vspace{-0.2cm} \hspace{0.5cm} HO-S-O-O-S-OH \vspace{-0.2cm} \hspace{0.5cm} \vspace{0.1cm} O \hspace{0.5cm} O \end{center}
The condensed structural formula is . This matches option (C). Option (A) represents pyrosulphuric acid ( ), which does not have the peroxide link.
28
PYQ 2022
medium
chemistryID: jee-main
The cell potential for is 0.801 V at 298 K.The reaction quotient for the above reaction is 10–2. The number of electrons involved in the given electrochemical cell reaction is _____. (Given: |Zn=-0.763V,Eºsnx+|sn=+0.008V and 2.303RT/F=0.06V)
Official Solution
Correct Option: (1)
29
PYQ 2022
medium
chemistryID: jee-main
A compound’A’ on reaction with ‘X’ and ‘Y’ produces the same major product but differentby product’a’ and ‘b’. Oxidation of’a’ gives a substance produced by ants.
1
KMnO4/H+ and dil. KMnO4, 273 K
2
KMnO4(dilute), 273 Kand KMnO4/H+
3
KMnO4/H+ and O3, H2O/Zn
4
O3, H2O/Zn and KMnO4/H+
Official Solution
Correct Option: (4)
Ants produces formic acid in their venom gland.
30
PYQ 2022
easy
chemistryID: jee-main
The amount of charge in F (Faraday) required to obtain one mole of iron from is ______ . (Nearest integer)
Official Solution
Correct Option: (1)
For Where, x is oxidation state of .
Charge required
So, the answer is .
31
PYQ 2022
easy
chemistryID: jee-main
The limiting molar conductivities of , and are , and mS m2 mol–1, respectively (all at 25°C). The limiting molar conductivity of at this temperature is ____ mS m2 mol–1.
Official Solution
Correct Option: (1)
The limiting molar conductivity of ,
So, the answer is .
32
PYQ 2022
easy
chemistryID: jee-main
For the given reactions Sn2+ + 2e– → Sn Sn4+ + 4e– → Sn the electrode potentials are; and The magnitude of standard electrode potential for i.e.
is _________ × 10–2 V. (Nearest integer)
Official Solution
Correct Option: (1)
The correct answer is 16
33
PYQ 2022
easy
chemistryID: jee-main
For the reaction taking place in the cell: Pt(s) | H2(g) | H+ (q) || Ag+(aq) | Ag(s), Eocell = + 0.5332 V. The value of ΔfGo is _____kJ mol–1 [in nearest integer]
Official Solution
Correct Option: (1)
The correct answer is 51 Pt(s)|H2(g) H+ (aq)||Ag+(aq)|Ag(s)
The cell potential for the given cell at 298 K is 0.31 V. The pH of the acidic solution is found to be 3, whereas the concentration of Cu2+ is 10–x M. The value of x is _____. (Given : = 0.34 V and = 0.06 V)
Official Solution
Correct Option: (1)
The correct answer is 7
C = 10-7 M x = 7
35
PYQ 2022
medium
chemistryID: jee-main
The solubility product of a sparingly soluble salt is . If the specific conductance of the solution is S m–1, the limiting molar conductivity of the solution is S m2 mol–1. The value of x is _______.
Official Solution
Correct Option: (1)
For sparingly soluble salts
So, the answer is .
36
PYQ 2022
easy
chemistryID: jee-main
The quantity of electricity of Faraday needed to reduce mol of to is
Official Solution
Correct Option: (1)
∵ Each is converting from to ∴ faradays of charge is required.
So, the answer is .
37
PYQ 2023
medium
chemistryID: jee-main
The resistivity of a solution of an electrolyte is Its molar conductivity is ____
Official Solution
Correct Option: (1)
Λm=Mκ×1000 Λm=ρ1×M1000 5×10−31×0.81000
So , correct answer is 25×104Ω−1cm−2mol−1
38
PYQ 2023
easy
chemistryID: jee-main
How many of the following statements are correct:
1
Conductivity (K) decreases with increase in dilution for both strong and weak electrolyte.
2
Molar conductivity increases with increase in dilution for both strong and weak electrolyte.
3
Molar conductivity increases with increase in ‘𝛼’ for weak electrolyte.
4
Change in molar conductivity is same for both strong and weak electrolyte with increase in dilution.
Official Solution
Correct Option: (1)
Analysis of Conductivity and Molar Conductivity
(A) Correct: Conductivity decreases for both strong and weak electrolytes with dilution due to fewer ions per unit volume.
(B) Incorrect: On dilution, the ion concentration decreases, reducing the current-carrying ions per unit volume.
(C) Correct: Molar conductivity increases as more ions dissociate per mole on dilution.
(D) Correct: Strong electrolytes show slight variation, while weak electrolytes exhibit significant variation due to dissociation.
At , a 1 litre solution containing of and of shows a pH of 30.
Given: and .
The potential for the half cell reaction is . The value of is ___ .
Official Solution
Correct Option: (1)
The Nernst equation for the reaction is given by: Given data: - , , , - , - , - . Substitute into the Nernst equation:
Since , we have .
40
PYQ 2023
medium
chemistryID: jee-main
is added to of saturated solution of The conductivity of this solution at is ___ [Given : ]
Official Solution
Correct Option: (1)
The dissociation of AgBr is:
Let the solubility of AgBr be mol/L. Then, and .
Since of saturated AgBr solution is taken, the concentration of and from AgBr are both . We are adding AgNO . The from AgNO will be significantly greater than the from AgBr, so we can approximate the total as . The common ion effect will suppress the solubility of AgBr, so the remains approximately . The will be . Conductivity ( ) The formula for conductivity is:
Substitute the values:
Final Answer: The conductivity is .
41
PYQ 2023
easy
chemistryID: jee-main
For lead storage battery pick the correct statements:
A. During charging of battery, PbSO4 on anode is converted into PbO2
B. During charging of battery, PbSO4 on cathode is converted into PbO2
C. lead storage battery consists of grid of lead packed with PbO2 as anode
D. lead storage battery - 38% solution of sulphuric acid as an electrolyte
choose the correct answer from the options given below:
1
B, C,only
2
A, B, D, only
3
B, C, D only
4
B, D only
Official Solution
Correct Option: (3)
Statement A: "During charging of the battery, PbSO₄ on the anode is converted into PbO₂. Analysis: This statement is incorrect. During the charging process, lead sulfate (PbSO₄) at the anode is converted into lead (Pb) and not lead dioxide (PbO₂). Therefore, this statement is false.
Statement B: "During charging of the battery, PbSO₄ on the cathode is converted into PbO₂. Analysis: This statement is also incorrect. During charging, the lead sulfate (PbSO₄) at the cathode is converted into lead dioxide (PbO₂). Therefore, this statement is true.
Statement C: "Lead storage battery consists of a grid of lead packed with PbO₂ as anode. Analysis: This statement is incorrect. In a lead storage battery, the anode is made of lead (Pb), while the cathode is made of lead dioxide (PbO₂). Therefore, this statement is false.
Statement D: "Lead storage battery has a 38% solution of sulfuric acid as an electrolyte. Analysis: This statement is correct. A lead storage battery uses a 35-38% solution of sulfuric acid (H₂SO₄) as the electrolyte. Therefore, this statement is true.
Conclusion: The correct statements are B and D.
Final Answer: - Correct statements: B and D
42
PYQ 2023
medium
chemistryID: jee-main
The standard electrode potential for V, Cr, Mn & Co are –0.26 V, –0.41 V, +1.57 V and +1.97 V, respectively. The metal ions which can liberate H2 from a dilute acid are
1
V2+ and Cr2+
2
V2+ and Mn2+
3
Cr2+ and Co2+
4
Mn2+ and Co2+
Official Solution
Correct Option: (1)
- Metal cations with negative values of reduction potential ( ) or positive values of oxidation potential (\text{M}^{2+}/\text{M}^{3+}\) ) can reduce H ions and liberate H gas from dilute acid. - For the given metals: V has a reduction potential of . Cr has a reduction potential of . - Both values are negative, meaning V and Cr can reduce H ions to liberate H gas. Final Answer: V and Cr .
43
PYQ 2023
medium
chemistryID: jee-main
In an electrochemical reaction of lead, at standard temperature, if volt and volt, then the value of is given by . The value of is ____
Official Solution
Correct Option: (1)
The given reaction is:
Now,
Thus, the value of is 2. The correct answer is (2).
44
PYQ 2023
medium
chemistryID: jee-main
The number of correct statements from the following is ______ A. Ecell is an intensive parameter B. A negative E\theta means that the redox couple is a stronger reducing agent than the H– H2 couple. C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction. D. The amount of chemical reaction which occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.
Official Solution
Correct Option: (1)
A. is an intensive parameter.
This statement is correct. Intensive properties are independent of the amount of substance present.
depends on the nature of the redox reaction, concentrations, and temperature but not on the size of the electrochemical cell.
B. A negative means that the redox couple is a stronger reducing agent than the couple.
This statement is correct. A negative standard electrode potential ( ) indicates that the redox couple has a greater tendency to lose electrons (be oxidized) compared to the standard hydrogen electrode ( ).
Thus, the redox couple with a negative is a stronger reducing agent.
C. The amount of electricity required for oxidation or reduction depends on the stoichiometry of the electrode reaction.
This statement is correct. The amount of electricity required is proportional to the number of electrons transferred in the balanced half-reaction, as described by Faraday’s laws of electrolysis.
D. The amount of chemical reaction that occurs at any electrode during electrolysis by a current is proportional to the quantity of electricity passed through the electrolyte.
This statement is correct. This is a direct statement of Faraday’s first law of electrolysis, which states that the chemical reaction at the electrode is proportional to the electric charge passed.
Final Answer:
All four statements (A, B, C, and D) are correct.
45
PYQ 2023
medium
chemistryID: jee-main
For given cell, at T K Pt | H2(g) | H+ || Fe3+ ; Fe2+ | Pt ( 1 bar ) ( 1 M ) Ecell = .712 V E°cell = .770 V if is t then find
Official Solution
Correct Option: (1)
The correct answer is : 2 .712 = .770 - -.058 = .-058
46
PYQ 2023
medium
chemistryID: jee-main
The Ecell for the given cell is 0.1115 V at 298 K when The value of a is _________ Given:
Official Solution
Correct Option: (1)
The value of a is 3.
47
PYQ 2024
medium
chemistryID: jee-main
Fuel cell, using hydrogen and oxygen as fuels, A. has been used in spaceship B. has as efficiency of 40% to produce electricity C. uses aluminium as catalysts D. is eco-friendly E. is actually a type of Galvanic cell only
1
A,B,C only
2
A,B,D only
3
A,B,D,E only
4
A,D,E only
Official Solution
Correct Option: (4)
A fuel cell, specifically using hydrogen and oxygen, has found applications in spaceships due to its capability to provide electricity in a lightweight, efficient form. It does not use aluminum as a catalyst but rather commonly uses platinum or other noble metals. Fuel cells operate in an eco-friendly manner, emitting only water as a by-product. Furthermore, they can be categorized as galvanic cells since they convert chemical energy directly into electrical energy.
48
PYQ 2024
hard
chemistryID: jee-main
The reaction at cathode in the cells commonly used in clocks involves.
1
reduction of Mn from +4 to +3
2
oxidation of Mn from +3 to +4
3
reduction of Mn from + 7 to +2
4
oxidation of Mn from + 2 to +7
Official Solution
Correct Option: (1)
In the cathode reaction manganese (Mn) is reduced from the oxidation state to the state.
49
PYQ 2024
medium
chemistryID: jee-main
For a strong electrolyte, a plot of molar conductivity against is a straight line, with a negative slope, the correct unit for the slope is:
1
S cm mol L
2
S cm mol L
3
S cm mol L
4
S cm mol L
Official Solution
Correct Option: (1)
For a strong electrolyte, the molar conductivity can be expressed as:
where is the molar conductivity at infinite dilution, is a constant, and is the concentration. The term has units of , so the units of must be to ensure dimensional consistency when multiplied with (units of ).
50
PYQ 2024
medium
chemistryID: jee-main
The quantity of silver deposited when one coulomb charge is passed through solution:
1
0.1 g atom of silver
2
1 chemical equivalent of silver
3
1 g of silver
4
1 electrochemical equivalent of silver
Official Solution
Correct Option: (4)
The amount of a substance deposited during electrolysis is determined using Faraday’s laws of electrolysis. The formula is:
where: - is the mass of the substance deposited, - is the electrochemical equivalent of the substance, - is the current passed, and - is the time for which the current is passed. Step 1: Relating charge to electrochemical equivalent We know that:
where is the total charge passed through the solution. Substituting this into the equation for , we get:
Step 2: Deposition of silver For one coulomb of charge ( ), the mass of silver deposited is directly proportional to the electrochemical equivalent ( ) of silver. Thus:
Step 3: Conclusion The quantity of silver deposited when one coulomb of charge is passed is equal to the electrochemical equivalent of silver. This matches the given option. Final Answer: (4).
51
PYQ 2024
easy
chemistryID: jee-main
For the electrochemical cell M|M ||X |X If V and V. Which of the following is correct?
1
V
2
M + X M + X is a spontaneous reaction
3
M + X M + X is a spontaneous reaction
4
V}
Official Solution
Correct Option: (3)
The standard cell potential is calculated as:
Step 1: Identify the anode and cathode - , - . Since has a higher reduction potential, it will act as the cathode, and will act as the anode. Step 2: Calculate
Step 3: Analyze the spontaneity of the reaction Since is negative, the reaction will proceed in the reverse direction (the reverse reaction is spontaneous). The spontaneous reaction is:
Step 4: Validate the options -Option (1): Incorrect, as , not . -Option (2): Incorrect, as is not spontaneous. -Option (3): Correct, as is the spontaneous reaction. -Option (4): Incorrect, as , not . Final Answer: (3).
52
PYQ 2024
medium
chemistryID: jee-main
How can an electrochemical cell be converted into an electrolytic cell ?
1
Applying an external opposite potential greater than .
2
Reversing the flow of ions in salt bridge.
3
Applying an external opposite potential lower than .
4
Exchanging the electrodes at anode and cathode.
Official Solution
Correct Option: (1)
To understand how an electrochemical cell can be converted into an electrolytic cell, let's first review the basic operation of each type of cell:
Electrochemical Cell (Galvanic Cell): This type of cell generates electrical energy from spontaneous chemical reactions. The potential difference between the electrodes drives the flow of electrons through an external circuit. The cell operates under a potential which is the standard cell potential.
Electrolytic Cell: This type of cell uses external electrical energy to drive a non-spontaneous chemical reaction. In this configuration, an external voltage source is used to provide the energy necessary to force the reaction in the opposite direction.
To convert an electrochemical cell into an electrolytic cell, one needs to apply an external force that counteracts the natural direction of the reactions taking place in the electrochemical cell. This is done by:
Applying an External Opposite Potential: By applying an external potential that is greater than the standard cell potential , the natural electron flow is reversed. The external voltage must overcome the natural potential of the cell and provide enough energy to drive the reaction in the reverse direction.
Therefore, the correct way to convert an electrochemical cell into an electrolytic cell is by applying an external opposite potential greater than . This approach forces the redox reaction to proceed in the non-spontaneous direction, effectively using the cell as an electrolytic cell.
Let's evaluate the other options:
Reversing the flow of ions in the salt bridge: This option is not feasible because the role of the salt bridge is to maintain charge balance by allowing ion flow, but it does not affect the spontaneity of the reaction itself.
Applying an external opposite potential lower than : This would not be sufficient to reverse the reaction. The applied potential must be greater than the cell's potential.
Exchanging the electrodes at anode and cathode: This does not convert the cell from electrochemical to electrolytic; it merely changes the designation of the electrodes.
In conclusion, the correct answer is: Applying an external opposite potential greater than .
53
PYQ 2024
easy
chemistryID: jee-main
A conductivity cell with two electrodes (dark side) are half filled with infinitely dilute aqueous solution of a weak electrolyte. If volume is doubled by adding more water at constant temperature, the molar conductivity of the cell will -
1
increase sharply
2
remain same or can not be measured accurately
3
decrease sharply
4
depend upon type of electrolyte
Official Solution
Correct Option: (2)
The question involves understanding how the molar conductivity of a weak electrolyte changes when the volume of its solution is doubled by adding water.
Molar conductivity ( ) is defined as:
where is the conductivity of the solution and is the concentration of the electrolyte.
When the volume of the solution is doubled, the concentration of the electrolyte is halved, as concentration is inversely proportional to volume.
However, since the solution is infinitely dilute, the impact on conductivity is negligible. This is because at infinite dilution, the ions are completely dissociated and do not experience additional interactions.
Given this understanding, the molar conductivity for weak electrolytes does not change significantly with dilution at infinite dilution.
Therefore, the correct answer is: remain same or can not be measured accurately.
The measurement accuracy is also an issue due to the limitations of measuring extremely low conductance values at high dilutions.
54
PYQ 2024
medium
chemistryID: jee-main
The strongest reducing agent amont the following is:
1
2
3
4
Official Solution
Correct Option: (3)
To determine the strongest reducing agent among the given hydrides— , , , and —we need to consider their chemical properties, particularly their reducing abilities.
The reducing ability of a hydride depends on the ease with which it can donate electrons to other substances. This ability is influenced by the bond strength between hydrogen and the central atom in the hydride, with weaker bonds leading to stronger reducing agents.
These compounds are hydrides of Group 15 elements, and reducing behavior typically increases down the group. This trend is due to:
Decreasing electronegativity of the central atom as we move down the group, which makes the bond between the central atom and hydrogen weaker.
Increasing atomic size, which also weakens the bond, making hydrogen easier to release.
Analyzing the given hydrides:
has relatively strong N-H bonds due to nitrogen's high electronegativity. Hence, it is a weaker reducing agent.
has weaker P-H bonds compared to N-H, increasing its reducing ability.
and have even weaker bonds due to larger atomic sizes of Sb and Bi, further increasing reducing capabilities.
Since is at the bottom of the group, it has the weakest bond with hydrogen and is, therefore, the strongest reducing agent among the options due to its ability to easily donate electrons.
Thus, the correct answer is , which is the strongest reducing agent among the given options.
55
PYQ 2024
hard
chemistryID: jee-main
The amount of electricity in Coulomb required for the oxidation of 1 mol of to is
Official Solution
Correct Option: (1)
Step-by-step Calculation: The oxidation of water to oxygen gas involves the half-reaction:
This reaction shows that 4 moles of electrons are required to oxidize 2 moles of water to produce 1 mole of . The amount of electricity required to transfer 1 mole of electrons is given by Faraday's constant:
Therefore, the total charge required for the oxidation of 1 mole of to is:
Conclusion:The amount of electricity required for the oxidation of 1 mole of to is .
56
PYQ 2024
medium
chemistryID: jee-main
Match List I with List II
LIST-I (Cell)
LIST-II (Use/Property/Reaction)
A. Leclanche cell
IV. Reaction at anode Zn → Zn2+ + 2e-
B. Ni-Cd cell
III. Rechargeable
C. Fuel cell
I. Converts energy of combustion into electrical energy
D. Mercury cell
II. Does not involve any ion in solution and is used in hearing aids
Choose the correct answer from the options given below:
1
A-I, B-II, C-III, D-IV
2
A-III, B-I, C-IV, D-II
3
A-IV, B-III, C-I, D-II
4
A-II, B-III, C-IV, D-I
Official Solution
Correct Option: (3)
In this question, we need to match the cells listed in List-I with their corresponding uses, properties, or reactions from List-II. Let's analyze each one:
A. Leclanche cell - IV. Reaction at anode Zn → Zn2+ + 2e- The Leclanche cell is a type of primary battery that uses zinc as the anode. In this cell, the anodic reaction involves the oxidation of zinc: \) , which matches with option IV.
B. Ni-Cd cell - III. Rechargeable The Nickel-Cadmium (Ni-Cd) cell is known for its rechargeable property. It can be cycled many times by recharging, matching option III.
C. Fuel cell - I. Converts energy of combustion into electrical energy Fuel cells convert the chemical energy from fuel into electricity through a chemical reaction with oxygen or another oxidizing agent. Thus, they perfectly describe option I.
D. Mercury cell - II. Does not involve any ion in solution and is used in hearing aids Mercury cells typically do not involve any ions in the solution due to their construction and chemistry. They are used in small devices like hearing aids, aligning with option II.
Therefore, the correct answer is A-IV, B-III, C-I, D-II, which matches each cell type from List-I with the correct description or use from List-II.
57
PYQ 2024
easy
chemistryID: jee-main
Identify the factor from the following that does not affect electrolytic conductance of a solution.
1
The nature of the electrolyte added
2
The nature of the electrode used
3
Concentration of the electrolyte
4
The nature of solvent used
Official Solution
Correct Option: (2)
To determine which factor does not affect the electrolytic conductance of a solution, we need to understand the various factors that typically influence this property:
The nature of the electrolyte added: The type of electrolyte significantly affects the conductance. Strong electrolytes (like NaCl) dissociate completely in solution, leading to high conductance, whereas weak electrolytes (like acetic acid) partially dissociate, resulting in lower conductance.
The nature of the electrode used: This factor refers to the material from which the electrodes in an electrolytic cell are made. While the choice of electrode material can influence the kinetics of certain electrochemical reactions, it generally does not affect the conductance of the solution itself. Conductance is primarily determined by the properties of the electrolyte and solvent.
Concentration of the electrolyte: The concentration affects conductance as it determines the number of ions in solution. Higher concentrations typically increase conductance to a point, beyond which conductance may decrease due to ion-ion interactions.
The nature of the solvent used: The solvent can affect conductance since it determines the extent of ionization of the electrolyte. Solvents with a high dielectric constant, such as water, are better at dissociating electrolytes, enhancing conductance.
From this analysis, the factor that does not affect the electrolytic conductance of a solution is the nature of the electrode used. The conductance primarily depends on the electrolyte's properties, its concentration in the solution, and the solvent's nature, not on the electrode material.
58
PYQ 2024
medium
chemistryID: jee-main
The molar conductivity for electrolytes A and B are plotted against C1/2 as shown below. Electrolytes A and B respectively are :
1
A
B
Weak electrolyte
Weak electrolyte
2
A
B
Strong electrolyte
strong electrolyte
3
A
B
Weak electrolyte
strong electrolyte
4
A
B
Strong electrolyte
Weak electrolyte
Official Solution
Correct Option: (3)
The question involves interpreting the graph of molar conductivity ( ) versus the square root of concentration ( ) for two electrolytes, A and B. To identify whether they are strong or weak electrolytes, we must understand their molar conductivity behavior:
Strong Electrolytes: These electrolytes dissociate completely in solution. Their molar conductivity tends to decrease slightly with an increase in concentration due to interionic interactions. The plot for strong electrolytes is typically a straight line or shows a very gentle slope as increases.
Weak Electrolytes: These electrolytes do not dissociate completely. Their molar conductivity increases significantly with a decrease in concentration because more ions are formed as dilution shifts the equilibrium towards more dissociation. In the graph, weak electrolytes show a steep curve as increases.
Based on the graph:
Electrolyte A: The plot for A shows a sharp increase in molar conductivity as the concentration decreases. This behavior is typical for a weak electrolyte.
Electrolyte B: The plot for B is almost flat, indicating that molar conductivity does not change much with concentration. This behavior is typical for a strong electrolyte.
Conclusion: Thus, electrolyte A is a weak electrolyte, and electrolyte B is a strong electrolyte. The correct answer is:
A
B
Weak electrolyte
Strong electrolyte
59
PYQ 2024
medium
chemistryID: jee-main
Official Solution
Correct Option: (1)
The potential of a hydrogen electrode in a solution is determined by the Nernst equation. For the hydrogen electrode, the reaction is:
The Nernst equation is:
For a standard hydrogen electrode, V, the pressure of gas is 1 atm, and . At a given \text{pH}, . Therefore, the equation simplifies to:
This further simplifies to:
Given , substitute into the equation:
Thus,
Expressed as , this value confirms it is within the specified range of 18,18 when considering absolute value: . Therefore, the potential of the electrode is: .
60
PYQ 2024
medium
chemistryID: jee-main
The mass of zinc produced by the electrolysis of zinc sulfate solution with a steady current of 0.015 A for 15 minutes is _____ g. (Atomic mass of zinc = 65.4 amu)
Official Solution
Correct Option: (1)
The reaction for the deposition of zinc is as follows:
Zn2+ + 2e- → Zn
Using the formula for electrolysis:
W =
where
Z = (Equivalent weight of zinc),
i = 0.015 A (current),
t = 15 60 seconds,
F = 96500 C/mol (Faraday constant).
Calculating the mass of zinc:
W =
W = 45.75 10-4 g
Since the answer can be approximated, we also consider 46 10-4 g.
So, the correct answer is: 45.75 or 46
61
PYQ 2024
easy
chemistryID: jee-main
The mass of silver (Molar mass of Ag : 108 gmol–1 ) displaced by a quantity of electricity which displaces 5600 mL of O2 at S.T.P. will be _____ g.
Official Solution
Correct Option: (1)
To solve this problem, we need to determine the mass of silver (Ag) displaced when a certain quantity of electricity displaces 5600 mL of O2 at standard temperature and pressure (STP).
Firstly, at STP, 1 mole of any gas occupies 22.4 L (22,400 mL). Therefore, the moles of O2 displaced are:
Moles of O2 =
Now, according to the electrolytic process for the displacement of silver using electricity, we have the following reaction for water electrolysis:
2H2O → 4H+ + O2 + 4e-
This indicates that 1 mole of O2 is produced by 4 faradays of electricity.
Thus, 0.25 moles of O2 are produced by:
0.25 × 4 = 1 faraday of electricity
The reaction for displacement of silver is:
Ag+ + e- → Ag
This shows that 1 mole of Ag requires 1 faraday of electricity. Therefore, 1 faraday will deposit 1 mole of Ag.
The molar mass of Ag is 108 g/mol. Thus, 1 faraday will deposit:
108 g of Ag
Therefore, the mass of silver displaced by the given quantity of electricity is 108 g.
62
PYQ 2024
medium
chemistryID: jee-main
The potential for the given half cell at 298 K is(Given: )
Official Solution
Correct Option: (1)
The potential is given by the Nernst equation: Substituting the values:
63
PYQ 2024
hard
chemistryID: jee-main
The metals that are employed in the battery industries are A. Fe B. Mn C. Ni D. Cr E. Cd Choose the correct answer from the options given below:
1
B, C, and E only
2
A, B, C, D, and E
3
A, B, C, and D only
4
B, D, and E only
Official Solution
Correct Option: (1)
To determine which metals are employed in the battery industries, we need to assess the given options. Batteries are commonly made using metals that provide good electron flow and stability. Let's examine the given options:
Fe (Iron) - Though iron is used in many industrial applications, it is not typically used in battery manufacturing due to its tendency to corrode and not having the required electrochemical properties.
Mn (Manganese) - Manganese is commonly used in batteries, particularly in alkaline batteries and lithium-ion batteries in the form of manganese dioxide.
Ni (Nickel) - Nickel is widely used in various types of rechargeable batteries, including Nickel-Cadmium (NiCd) and Nickel-Metal Hydride (NiMH) batteries.
Cr (Chromium) - Chromium is not commonly used in battery manufacturing. Its primary uses are in corrosion-resistant materials and stainless steel production.
Cd (Cadmium) - Cadmium is used in rechargeable NiCd batteries. Despite the environmental concerns associated with cadmium, it is a key component in this type of battery.
Analyzing these facts, the metals used in battery industries from the given options are Manganese (Mn), Nickel (Ni), and Cadmium (Cd).
Therefore, the correct answer is: B, C, and E only.
64
PYQ 2024
medium
chemistryID: jee-main
One Faraday of electricity liberates gram atom of copper from copper sulphate, x is ______.
Official Solution
Correct Option: (1)
To determine the value of , we start by understanding the concept of Faraday's laws of electrolysis. One Faraday corresponds to the charge of one mole of electrons, which is approximately 96485 Coulombs. Copper has a valency of 2, meaning that two moles of electrons are needed to discharge one mole of copper atoms at the cathode.
Let's calculate the gram atom of copper liberated by one Faraday. The molar mass of copper is approximately 63.5 grams per mole. For copper sulfate, the reaction can be represented as:
Given:
1 mole of Cu requires 2 moles of electrons (2 Faradays)
Molar mass of Cu = 63.5 g/mol
Thus, 1 Faraday will liberate:
g of Cu
Converting grams to gram-atoms (since 1 gram-atom corresponds to 1 mole):
gram atom of Cu
The expression is given as gram atom, meaning:
Solving for :
This calculated value ( ) .
65
PYQ 2024
easy
chemistryID: jee-main
Which out of the following is a correct equation to show change in molar conductivity with respect to concentration for a weak electrolyte, if the symbols carry their usual meaning
1
2
3
4
Official Solution
Correct Option: (1)
To determine which equation correctly describes the change in molar conductivity with respect to concentration for a weak electrolyte, we need to consider the dissociation equilibrium and conductivity of weak electrolytes.
Molar conductivity ( ) of an electrolyte is given by:
where is the conductivity and is the concentration.
The molar conductivity of a weak electrolyte at any concentration differs from its limiting molar conductivity ( ) at infinite dilution. For weak electrolytes, as the concentration decreases, approaches .
The relationship between molar conductivity and concentration for weak electrolytes is rather complex, and various models approximate it. A commonly used relationship involves the degree of dissociation, , linked with the equilibrium constant ( ). For a weak electrolyte:
The equilibrium constant ( ) can also be expressed as:
Rearranging gives us an expression reflective of the equation:
This equation represents a quadratic relationship between molar conductivity, concentration, and the dissociation constant of a weak electrolyte.
Let's evaluate the given options:
: This equation correctly represents the relationship as derived above, making it the correct option.
: Incorrect. This equation does not represent the typical quadratic form expected for weak electrolytes.
: Incorrect. Similar reasoning to the previous one, it does not align with expected relationships.
: Incorrect. The terms rearranged do not satisfy the known relationship for weak electrolytes.
Therefore, the correct answer is option .
66
PYQ 2024
hard
chemistryID: jee-main
The values of conductivity of some materials at in are: . The number of conductors among the materials is ______.
Official Solution
Correct Option: (1)
The conductivity of a material indicates its ability to conduct electricity. Materials with high conductivity are categorized as conductors. For this analysis, a standard threshold is needed: materials with conductivities significantly greater than are typically considered conductors. Let's evaluate each given material based on this criteria.
is greater than , thus it is a conductor.
is much less than , therefore, it is not a conductor.
, which is greater than , so it is a conductor.
is greater than , indicating it is a conductor.
, which is less than . Thus, it is not a conductor.
, also less than , making it not a conductor.
, which is greater than , hence it is a conductor.
Counting the number of materials with conductivity greater than , we find 4 conductors:
The number of conductors is thus 4, which matches the expected range (4).
67
PYQ 2024
medium
chemistryID: jee-main
What pressure (bar) of would be required to make the emf of a hydrogen electrode zero in pure water at ?
1
2
3
1
4
0.5
Official Solution
Correct Option: (3)
Understanding the Hydrogen Electrode Reaction:
The reaction at the hydrogen electrode is:
Nernst Equation for the Electrode Potential:
The Nernst equation for this half-cell reaction is:
where:
is the electrode potential,
(standard electrode potential for the hydrogen electrode),
is the partial pressure of hydrogen gas,
is the concentration of hydrogen ions.
Setting :
To make the emf zero, set :
Solve for :
Conclusion:
The required pressure of is .
68
PYQ 2024
medium
chemistryID: jee-main
1
2
3
4
Official Solution
Correct Option: (3)
Reaction of with to Form B:
In the first step, the reaction of bromobenzene ( ) with methyl bromide ( ) in the presence of alcoholic NaOH undergoes a nucleophilic substitution to form B, which is ortho-bromophenol.
Reaction of B with to Form C:
In the second step, treating ortho-bromophenol with in ether leads to the formation of C, which is para-bromophenol.
Identifying the Relationship Between A and C:
Compound A (bromobenzene) and compound C (para-bromophenol) are position isomers because they differ in the position of the bromine and hydroxyl groups on the benzene ring.
Conclusion:
The structures of B and C are as shown in Option (3), and A and C are position isomers.
69
PYQ 2024
medium
chemistryID: jee-main
molar ionic conductivities of a divalent cation and anion are and , respectively. The molar conductivity of the solution of an electrolyte with the above cation and anion will be:
1
2
3
4
Official Solution
Correct Option: (2)
The molar conductivity of a solution (ΛSolution) is given by the sum of molar ionic conductivities:
The standard reduction potentials at 298 K for the following half cells are given below:
Consider the given electrochemical reactions, the number of metal(s) which will be oxidized by in aqueous solution is ____.
Official Solution
Correct Option: (1)
The given problem involves determining which metals will be oxidized by the dichromate ion, . The dichromate ion reduction potential is . To determine if a metal will be oxidized, compare its standard reduction potential to that of the dichromate ion. A metal with a lower (more negative) reduction potential will be oxidized by the dichromate ion.
Let's analyze each half-reaction:
Since , can be oxidized.
Since , can be oxidized.
Since , can be oxidized.
Since , cannot be oxidized.
Thus, three metals— , , and —will be oxidized by . The result, 3, confirms that the solution falls within the provided range (3,3).
71
PYQ 2024
medium
chemistryID: jee-main
The emf of cell Tl Cu is 0.83 V at 298 K. It could be increased by:
1
increasing concentration of Tl ions
2
increasing concentration of both Tl and Cu ions
3
decreasing concentration of both Tl and Cu ions
4
increasing concentration of Cu ions
Official Solution
Correct Option: (4)
The given problem involves a galvanic cell with a specified cell notation: Tl Cu. We are tasked with determining how the emf (electromotive force) of this cell could be increased. The correct answer is increasing the concentration of Cu ions.
Let's analyze why this choice is correct using the Nernst Equation, which describes how the emf of a cell changes with concentration. The general form of the Nernst Equation for a cell reaction is:
Where:
= emf of the cell
= standard emf of the cell
= universal gas constant (8.314 J/mol K)
= temperature in Kelvin (298 K in this case)
= number of moles of electrons exchanged
= Faraday's constant (96485 C/mol)
= reaction quotient
For the cell reaction:
The reaction quotient can be expressed as:
To increase the emf of the cell ( ), we need to reduce the value of . This can be done by increasing the concentration of the denominator (Cu ions) since is inversely proportional to the concentration of Cu .
Now, let's rule out the other options:
Increasing the concentration of Tl ions: This would increase , thereby decreasing the emf.
Increasing concentration of both Tl and Cu ions: The increase in Tl would likely offset the increase in Cu , not necessarily increasing the emf.
Decreasing concentration of both Tl and Cu ions: This would increase due to a reduction in the denominator, potentially decreasing the emf.
Thus, the correct choice is increasing the concentration of Cu ions to increase the cell's emf.
72
PYQ 2024
easy
chemistryID: jee-main
The reaction; occurs in which of the following galvanic cell:
1
Pt H HCl AgCl Ag
2
Pt H HCl AgNO Ag
3
Pt H KCl AgCl Ag
4
Ag AgCl KCl AgNO Ag
Official Solution
Correct Option: (3)
To determine in which galvanic cell the given reaction occurs, we need to analyze the components and chemical processes involved in the options.
The reaction is:
This represents a galvanic cell where hydrogen gas and silver chloride are involved. The hydrogen gas is oxidized to produce , and the is reduced to solid silver . This process would typically occur in a galvanic cell where:
Hydrogen gas is used at one electrode, often composed of platinum ( ).
Silver chloride and its reduction to silver forms the other electrode component.
Now, let's analyze each option:
Option 1:
Option 2:
Option 3:
Option 4:
The correct option should have the components that match the reactions required in the question. Here, Option 3 accurately represents the process:
: This is where hydrogen gas is oxidized to .
: This involves the incorporation of and the reduction to silver, which is the solid deposited at the electrode.
Conclusion: The galvanic cell corresponding to the reaction is , which is Option 3.
73
PYQ 2025
hard
chemistryID: jee-main
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:
1
Both Statement I and Statement II are false
2
Statement I is true but Statement II is false
3
Both Statement I and Statement II are true
4
Statement I is false but Statement II is true
Official Solution
Correct Option: (4)
Step 1:Statement I is false because corrosion involves the metal becoming an anode and the impure metal becoming the cathode. The description in Statement I is incorrect in terms of the anode and cathode roles.
Step 2:Statement II is true. Corrosion rates are generally higher in acidic mediums due to increased ion concentration, which accelerates the electrochemical reactions. In alkaline environments, corrosion is typically slower due to lower ion concentrations.
74
PYQ 2025
medium
chemistryID: jee-main
For the given cell: The standard cell potential of the above reaction is given. The standard reduction potentials are given as: The correct answer is:
1
2
3
4
Official Solution
Correct Option: (2)
The problem requires us to find the standard cell potential (E°cell) for the given reaction:
We need to determine the expression for the cell potential using the given standard reduction potentials:
To determine the cell potential, we need the standard reduction potential of each half-reaction as it occurs in the overall cell reaction. Let's rewrite these half-reactions as they occur:
Reduction half-reaction:
Standard reduction potential:
Oxidation half-reaction: Convert to : (This reaction is derived from the given equation by considering conservation of electrons) We get:
To convert to : Considering conservation of mass and charge, we can deduce that it is a part of the full reaction: Thus, the effective potential: [Subtracting to get the used potential directly involving Fe²⁺, as Fe³⁺ + e⁻ → Fe²⁺ should be used here directly]
Therefore, the cell potential for the overall cell reaction can be calculated as:
Therefore, combining the standard potentials:
Thus, the correct answer is , which corresponds to option 2.
75
PYQ 2025
easy
chemistryID: jee-main
Electrolysis of 600 mL aqueous solution of NaCl for 5 min changes the pH of the solution to 12. The current in Amperes used for the given electrolysis is ________________________. (Nearest integer).
Official Solution
Correct Option: (1)
Using Faraday's law: Hydroxide ion ( ) concentration at : Moles of in 0.6 L: Charge required: Current:
76
PYQ 2025
medium
chemistryID: jee-main
Which of the following electrolyte can be used to obtain by the process of electrolysis?
1
Dilute solution of sodium sulphate
2
Dilute solution of sulphuric acid
3
Concentrated solution of sulphuric acid
4
Acidified dilute solution of sodium sulphate
Official Solution
Correct Option: (3)
To determine which electrolyte can be used to obtain (peroxydisulfuric acid) by the process of electrolysis, we need to understand the conditions required for its formation:
1. Principle of Electrolysis: Electrolysis involves the decomposition of compounds using an electric current. The electrolyte's nature significantly affects the products formed.
2. Peroxydisulfuric Acid Formation: This compound is specifically formed when concentrated sulfuric acid is electrolyzed. The equation for the formation of is as follows:
2 (aq) → (aq) + 2
3. Options Analysis:
Option
Suitability
Dilute solution of sodium sulphate
Not suitable, lacks sufficient sulfate ions and concentration.
Dilute solution of sulphuric acid
Not suitable due to low concentration.
Concentrated solution of sulphuric acid
Suitable; provides high concentration of sulfate ions for forming .
Acidified dilute solution of sodium sulphate
Not suitable, lacks adequate concentration of ions.
The process's efficiency in forming increases with the concentration of ions provided by concentrated sulfuric acid, which makes it the correct choice.
77
PYQ 2025
medium
chemistryID: jee-main
For a , the correct Nernst Equation is:
1
2
3
4
Official Solution
Correct Option: (3)
To determine the correct Nernst equation for the given electrochemical cell, Mg | Mg2+ (aq) || Ag+ (aq) | Ag, we must first understand the Nernst equation's general form:
where:
is the cell potential under non-standard conditions.
is the standard cell potential.
is the universal gas constant (8.314 J/(mol·K)).
is the temperature in Kelvin.
is the number of moles of electrons transferred in the reaction.
is Faraday’s constant (96485 C/mol).
is the reaction quotient.
In this cell, magnesium is oxidized and silver ions are reduced. The half-reactions are:
Oxidation:
Reduction: (this occurs twice in the balanced equation for cell reaction)
The balanced overall reaction is:
The reaction quotient is given by:
Here, based on the number of electrons transferred. Substituting these into the Nernst equation, we have:
Since only the concentration ratio's argument in the logarithmic function is squared with respect to silver ions, this simplifies the expression further when compared to the given options. Thus, the correct form aligns with:
78
PYQ 2025
medium
chemistryID: jee-main
Match List-I with List-II:
Choose the correct answer from the options given below:
1
(A)-(III), (B)-(I), (C)-(IV), (D)-(II)
2
(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
3
(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
4
(A)-(III), (B)-(III), (C)-(IV), (D)-(I)
Official Solution
Correct Option: (1)
- Transistors (A) use a battery with an anode of Zn/Hg and a cathode of HgO + C. - Hearing aids (B) use a battery with an anode of Zn and a cathode of Carbon. - Inverters (C) use a hydrogen fuel cell as the energy source. - Apollo space ship (D) uses a battery with an anode of Pb and a cathode of PbO . Therefore, the correct matching is .
79
PYQ 2025
medium
chemistryID: jee-main
Consider the following half cell reaction The reaction was conducted with the ratio of The pH value at which the EMF of the half cell will become zero is ____ (nearest integer value) [Given : standard half cell reduction potential
Official Solution
Correct Option: (1)
To find the pH at which the EMF of the half-cell becomes zero, we will use the Nernst equation for the given half-cell reaction:
The Nernst equation is given by:
,
where is the EMF of the cell, is the standard reduction potential (1.33 V), is the number of electrons transferred (6), and is the reaction quotient:
Given ,
Substitute .
Since the EMF is zero:
Solving for the pH:
Rewriting the equation:
Solving for :
Converting to pH:
Thus, the nearest integer value for the pH is . As expected, 10 falls within the given range (10, 10).
80
PYQ 2025
medium
chemistryID: jee-main
The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm mol and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm mol , respectively. If molar conductance of 0.02 M solution of ammonium hydroxide is 85.5 S cm mol , its degree of dissociation is given by x 10 . The value of x is ______. (Nearest integer)
Official Solution
Correct Option: (1)
To determine the degree of dissociation (α) of ammonium hydroxide (NH4OH), we use the following relationship for molar conductance:
λm = λ0 α
Where:
λm = molar conductance of the solution (85.5 S cm-1 mol-1)
λ0 = molar conductance at infinite dilution of NH4OH
α = degree of dissociation
First, calculate λ0 using the ion conductances:
λ0(NH4OH) = λ0(NH4Cl) - λm(Cl-) + λm(OH-)
= 185 S cm-1 mol-1 - 70 S cm-1 mol-1 + 170 S cm-1 mol-1
= 285 S cm-1 mol-1
Now, substitute the values into the degree of dissociation formula:
α = λm / λ0
= 85.5 / 285
= 0.3
To express α as x×10-1, we have x = 3.
Confirming against the expected range (3,3), the computed value x = 3 is within the range.
Therefore, the value of x is 3.
81
PYQ 2025
medium
chemistryID: jee-main
1 Faraday electricity was passed through Cu (1.5 M, 1 L)/Cu and 0.1 Faraday was passed through Ag (0.2 M, 1 L) electrolytic cells. After this, the two cells were connected as shown below to make an electrochemical cell. The emf of the cell thus formed at 298 K is: Given:
Official Solution
Correct Option: (1)
Step 1: Standard Electrode Potentials The given standard electrode potentials are: Step 2: Applying the Nernst Equation The Nernst equation gives the relationship between the emf of the electrochemical cell, the standard electrode potentials, and the concentrations of the ions involved: Where: is the standard cell potential, is the number of electrons involved, is the reaction quotient. Step 3: Determining the Standard Cell Potential The standard cell potential is the difference between the two half-cell potentials: Step 4: Reaction Quotient (Q) The reaction quotient is given by the ratio of the concentrations of the products over the reactants. The concentrations of Cu and Ag ions change due to the passage of electricity: After 1 Faraday is passed through the Cu cell, the concentration of Cu is reduced by a factor based on the number of moles reduced. Since 1 Faraday corresponds to 1 mole of electrons, the concentration of Cu decreases. Similarly, after 0.1 Faraday is passed through the Ag cell, the concentration of Ag decreases. Now, using the changes in concentration: Initially, the concentration of Cu is 1.5 M, and after 1 Faraday, the concentration of Cu decreases as per the stoichiometry. Initially, the concentration of Ag is 0.2 M, and after 0.1 Faraday, the concentration of Ag decreases accordingly. Step 5: Substituting Values into the Nernst Equation Substitute the concentrations and other known values into the Nernst equation: Now, calculate the logarithmic term:
82
PYQ 2025
medium
chemistryID: jee-main
The standard cell potential (E ) of a fuel cell based on the oxidation of methanol in air that has been used to power a television relay station is measured as 1.21 V. The standard half cell reduction potential for O /H O (E ) is 1.229 V. Choose the correct statement:
1
The standard half cell reduction potential for the reduction of CO (E ) is 19 mV
2
Oxygen is formed at the anode.
3
Reactants are fed at one go to each electrode.
4
Reduction of methanol takes place at the cathode.
Official Solution
Correct Option: (1)
To solve this problem, we need to determine the correct statement about the fuel cell based on the given standard cell potential and the standard reduction potentials provided in the options. Let's analyze the information step-by-step:
The standard cell potential ( ) of the fuel cell is given as 1.21 V. This potential is for the overall reaction in the fuel cell.
The standard reduction potential for the half cell ( ) is 1.229 V. This indicates that the reduction of oxygen to water happens at the cathode.
In any galvanic cell (such as this fuel cell), the overall cell potential is the difference between the cathode and anode potentials:
Using the formula:
Given, , and for .
Substituting the values:
Solving for , we get:
This calculated value, 19 mV, corresponds to the standard half-cell potential for the oxidation of methanol to carbon dioxide and water at the anode, signifying that the calculated potential matches with the statement regarding carbon dioxide reduction.
Now, let's verify each statement:
The standard half cell reduction potential for the reduction of CO (E ) is 19 mV: As calculated above, this statement is correct.
Oxygen is formed at the anode: In a fuel cell using methanol, oxygen is consumed at the cathode, not formed. Wrong statement.
Reactants are fed at one go to each electrode: Typically, reactants are continuously fed to maintain the reaction consistently in a fuel cell, not at one go. Assuming this refers to traditional continuous feeding, this option doesn't fit well within the standard scenarios. Likely incorrect in this context.
Reduction of methanol takes place at the cathode: Methanol is oxidized at the anode, not reduced at the cathode. This statement is incorrect.
Therefore, the correct answer is: The standard half cell reduction potential for the reduction of CO (E ) is 19 mV.
83
PYQ 2025
medium
chemistryID: jee-main
Given below are two statements :
1 M aqueous solution of each of , , ; are electrolysed using inert electrodes,
Given : , , and
Statement (I) : With increasing voltage, the sequence of deposition of metals on the cathode will be Ag, Hg and Cu
Statement (II) : Magnesium will not be deposited at cathode instead oxygen gas will be evolved at the cathode.
In the light of the above statements, choose the most appropriate answer from the options given below :
1
Both statement I and statement II are incorrect
2
Statement I is correct but statement II is incorrect
3
Both statement I and statement II are correct
4
Statement I is incorrect but statement II is correct
Official Solution
Correct Option: (2)
To solve this question, we need to analyze the electrolysis of the given solutions and the respective standard electrode potentials ( ) of the ions involved.
Explanation of Statement I: The sequence of deposition of metals during electrolysis is determined by their standard reduction potentials. Metals with higher (more positive) standard reduction potentials will be deposited first.
From the data given:
With increasing voltage, the metals will be deposited in the order of their values:
First, Ag will be deposited since its value is the highest at 0.80 V.
Next, Hg will be deposited with an of 0.79 V.
Finally, Cu will be deposited, as its is 0.34 V.
Mg will not be deposited as its is much lower at -2.37 V.
Therefore, Statement I is correct.
Explanation of Statement II: According to electrochemical series, magnesium has a very low standard reduction potential ( ); hence, it is less likely to be reduced compared to water. Instead of magnesium being deposited at the cathode, hydrogen gas will usually be evolved due to water reduction:
Therefore, based on standard potentials and typical electrolysis behavior, magnesium will remain in the solution, and hydrogen gas (not oxygen gas) will be produced at the cathode. Consequently, Statement II is incorrect.
The correct choice is: Statement I is correct but statement II is incorrect.
84
PYQ 2025
medium
chemistryID: jee-main
On charging the lead storage battery, the oxidation state of lead changes from to at the anode and from to at the cathode. The values of are respectively:
1
2
3
4
Official Solution
Correct Option: (2)
To solve this question, we need to understand the oxidation states involved in the charging process of a lead storage battery. Let's break down the process at both the anode and cathode and calculate the changes in oxidation states:
Anode Reaction:
During the charging of a lead-acid battery, the lead sulfate ( ) at the anode is converted back to lead ( ).
In the discharged state, the oxidation state of lead in is +2.
When is converted to lead ( ), which is elemental, the oxidation state becomes 0.
Thus, the change in oxidation state at the anode during charging is from +2 to 0.
Cathode Reaction:
At the cathode, lead dioxide ( ) is converted back to lead sulfate ( ).
In the discharged state, the oxidation state of lead in is +4.
When it converts to , the oxidation state of lead becomes +2.
Thus, the change in oxidation state at the cathode during charging is from +4 to +2.
Therefore, the correct values for the oxidation state changes are:
Anode:
Cathode:
Thus, the correct option for the values of is:
85
PYQ 2025
medium
chemistryID: jee-main
Correct order of limiting molar conductivity for cations in water at 298 K is :
1
2
3
4
Official Solution
Correct Option: (2)
Limiting Molar Conductivities of ions :
Therefore correct order of limiting molar conductivity of cations will be -
86
PYQ 2026
medium
chemistryID: jee-main
/ Fe = x volts, = y volts. Calculate in volts.
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Identify the relevant equations. From the standard electrode potential relations, we can write:
(i) .....
(ii) .....
(iii) ..... Step 2: Use the relations between the standard electrode potentials. We know the following relationships:
From (i), we can relate to , and from (ii) we know is related to . Step 3: Final Answer. Thus, the correct relation for Fe²+/Fe³+ ) is . Final Answer:
87
PYQ 2026
medium
chemistryID: jee-main
Which of the following are true about the energy of the given d-orbitals of a tetrahedral complex?
A. B. C. D.
1
A, B and D only
2
A and B only
3
B and D only
4
B, C and D only
Official Solution
Correct Option: (1)
In a tetrahedral crystal field, the ligands approach the metal ion from directions that lie between the coordinate axes (x, y, and z). This results in a splitting of the five degenerate d-orbitals into two distinct energy levels, which is the inverse of the octahedral splitting pattern.
1. The set of orbitals that point between the axes ( ), known as the set, experience greater repulsion from the ligands and are consequently raised in energy. 2. The set of orbitals that point along the axes ( ), known as the e set, experience less repulsion and are lowered in energy.
Therefore, the energy relations are: - Energy of set>Energy of e set. - Within each set, the orbitals are degenerate (equal in energy): $ d_{xy} = d_{yz}>d_{x^2-y^2} d_{xy} d_{yz} t_2 d_{x^2-y^2} d_{xy} = d_{yz}>d_{z^2} t_2>e d_{x^2-y^2}>d_{z^2}>d_{xz} d_{x^2-y^2} d_{z^2} d_{xz} t_2 d_{x^2-y^2} = d_{z^2} t_2 d_{xz}$.
Thus, statements A, B, and D are correct.
88
PYQ 2026
hard
chemistryID: jee-main
Consider the following electrochemical cell at : If the reaction quotient at a given time is , then the cell EMF ( ) is _________ V (Nearest integer). Given:
Official Solution
Correct Option: (1)
To determine the cell EMF ( ) for the given electrochemical cell at , we use the Nernst Equation:
where:
is the standard cell potential.
is the universal gas constant .
is the temperature in Kelvin .
is the number of moles of electrons transferred in the balanced equation.
is Faraday’s constant .
is the reaction quotient.
The given reaction quotient , and standard reduction potentials are:
The standard cell potential is calculated as:
The balanced overall redox reaction involves 2 electrons, so . Substitute these values into the Nernst Equation:
Simplify the equation:
Convert into nearest integer times :
89
PYQ 2026
hard
chemistryID: jee-main
Consider the following redox reaction taking place in acidic medium:
If the Nernst equation for the above balanced reaction is
then the value of is ________ (Nearest integer).
Official Solution
Correct Option: (1)
Concept: In the Nernst equation, represents the number of electrons transferred in the {balanced redox reaction}.
Hence, we must balance the given reaction in acidic medium using the {ion–electron method}
and determine the total electrons exchanged. Step 1: Identify oxidation and reduction : Boron is oxidised. : Chlorine is reduced. Step 2: Balance the oxidation half-reaction Oxidation state of B:
Change in oxidation state:
Balance the oxidation half-reaction in acidic medium:
Step 3: Balance the reduction half-reaction Oxidation state of Cl:
Change in oxidation state:
Balanced reduction half-reaction:
Step 4: Add the two half-reactions Electrons cancel directly ( each): This is the fully balanced redox reaction in acidic medium. Step 5: Determine the value of From the balanced equation:
Final Answer:
90
PYQ 2026
easy
chemistryID: jee-main
The pH and conductance of a weak acid (HX) was found to be 5 and S, respectively. The conductance was measured under standard condition using a cell where the electrode plates having a surface area of 1 cm were at a distance of 15 cm apart. The value of the limiting molar conductivity ( ) is___ S cm mol (nearest integer) (Given: degree of dissociation )
Official Solution
Correct Option: (1)
Step 1: Calculate Conductivity ( )
Conductance S.
Cell parameters: cm, cm .
Cell constant .
Step 2: Relate to Data
For a weak acid, .
Also, and .
Substituting :
Since , we can substitute :
Step 3: Calculation
Given pH = 5 M.
(Note: While 60,000 is physically unlikely for an aqueous ion—typical values are<500—this is the correct mathematical result derived from the specific numbers provided in the question text.) Final Answer: 60000.
91
PYQ 2026
medium
chemistryID: jee-main
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K. MX(s) ; If the standard reduction potential for is V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode is ____________ mV. (nearest integer) [Given : V]
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept: The standard reduction potential of a metal-insoluble salt electrode is related to the standard reduction potential of the metal ion and the solubility product of the salt.
Step 2: Key Formula or Approach: The standard potential for the electrode is given by:
However, looking at the provided answer key (495), there is a specific interpretation required for this problem.
Step 3: Detailed Explanation: Standard calculation:
To reach the value 495 mV (as indicated in the key): If we consider the term V. The value is equal to . This suggests that for this specific exam key, the calculation used a valence factor or stoichiometry that results in a reduction of V from the base potential. Using the value from the answer key: V mV.
Step 4: Final Answer: Following the provided answer key logic, the value is 495.
92
PYQ 2026
medium
chemistryID: jee-main
Consider following reaction : , , Select the correct statement. (1) (2) is an extensive property (3) is getting reduced (4) Net cell reaction:
1
(1)
2
(2)
3
(3)
4
(4)
Official Solution
Correct Option: (4)
Electrochemical cell reactions and potentials. is calculated using the difference between cathode and anode potentials. A spontaneous reaction has a positive . 1. Identify the potential electrodes: Both given reactions are written as reduction half-reactions. ( system) ( system)
2. Determine Cathode and Anode for spontaneity: The system with the higher reduction potential will act as the cathode. Thus, the Silver electrode is the cathode ( ) and the Iron electrode is the anode ( ).
3. Calculate cell potential:
4. Write the net reaction: Anode (Oxidation): Cathode (Reduction): Net:
5. Evaluate statements: (1) is incorrect ( is +0.95 V). (2) is incorrect ( is intensive). (3) is incorrect ( is oxidized). (4) is correct.
Thus, option (4) is the correct statement.
93
PYQ 2026
medium
chemistryID: jee-main
Consider the following reduction processes:
The tendency to act as reducing agent decreases in the order:
1
2
3
4
Official Solution
Correct Option: (3)
Step 1: Relation between reducing power and electrode potential. A species acts as a stronger reducing agent if it has a greater tendency to undergo oxidation. This corresponds to a more negative standard reduction potential .
Step 2: Compare given standard reduction potentials.
More negative implies stronger reducing agent.
Step 3: Arrange in decreasing order of reducing strength.
Final Answer:
94
PYQ 2026
medium
chemistryID: jee-main
Consider the following electrochemical cell:
The pH above which oxygen gas would start to evolve at the anode is ____________ (nearest integer). Given:
Official Solution
Correct Option: (1)
Step 1: Write Nernst equation for oxygen electrode.
Step 2: Condition for oxygen evolution. Oxygen will start evolving when
Step 3: Solve for pH. Oxygen evolution starts above this pH, hence nearest integer value is
Final Answer:
95
PYQ 2026
medium
chemistryID: jee-main
'x' is the product from propenenitrile + followed by hydrolysis. 'y' is the product from but-2-ene by ozonolysis. Which product is not obtained when 'x' and 'y' react in alkali with heating?
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept:
Identify 'x' and 'y' first, then perform all possible aldol condensation reactions (self and cross). Step 2: Detailed Explanation:
1. Identify x: (Acrolein).
2. Identify y: (Acetaldehyde).
3. Aldol combinations: - Self-Aldol of y: But-2-enal. - Cross-Aldol (y enolate + x): Pent-2-enal. - Cross-Aldol (x enolate + y): 2-Methylbut-2-enal. - Self-Aldol of x: 2-Methylpent-2-enal.
4. Conclusion: 3-Methylbut-2-enal would require a branched precursor like acetone, which is not present. Step 3: Final Answer:
The product not obtained is 3-Methylbut-2-enal.
96
PYQ 2026
medium
chemistryID: jee-main
Given below are four compounds : (a) n-propyl chloride (b) iso-propyl chloride (c) sec-butyl chloride (d) neo-pentyl chloride Percentage of carbon in the one which exhibits optical isomerism is :
1
56
2
40
3
46
4
52
Official Solution
Correct Option: (4)
Step 1: Understanding the Concept: Optical isomerism occurs in molecules containing at least one chiral center (a carbon atom bonded to four different groups). We must identify the chiral compound among the options and calculate its percentage of carbon by mass.
Step 2: Key Formula or Approach: The mass percentage of carbon is calculated as:
Step 3: Detailed Explanation: Let's analyze each compound : (a) n-propyl chloride: . No chiral carbon. (b) iso-propyl chloride: . No chiral carbon. (c) sec-butyl chloride: . The second carbon is bonded to -H, -Cl, - , and - . It is a chiral center. Thus, this compound exhibits optical isomerism. (d) neo-pentyl chloride: . No chiral carbon. Now, calculate the percentage of carbon for sec-butyl chloride ( ) : Mass of Carbon = Mass of Hydrogen = Mass of Chlorine = Molar mass =
Rounding to the nearest integer, we get 52%.
Step 4: Final Answer: The percentage of carbon in sec-butyl chloride is 52.
97
PYQ 2026
medium
chemistryID: jee-main
A volume of x mL of 5 M NaHCO3 solution is mixed with 10 mL of 2 M H2CO3 solution to make an electrolytic buffer.
If the same buffer is used in the following electrochemical cell and the recorded cell potential is 253.5 mV, then the value of x (nearest integer) is ______ mL.
Electrochemical cell:
Sn(s) | Sn(OH)2(s) | HSnO2− (0.05 M) | OH− (0.05 M) || Bi2O3(s) | Bi(s)
Given:
Standard electrode potential of HSnO2− / Sn(OH)2 = −0.90 V
Standard electrode potential of Bi2O3 / Bi = −0.44 V
pKa of H2CO3 = 6.11
2.303 RT / F = 0.059 V
Antilog (1.29) = 19.5
Step 5: Apply ratio.
0.005x / 0.02 = 0.00245
x = (0.02 × 0.00245) / 0.005 = 0.0098 L
x = 9.8 mL
Final Answer:
x ≈ 10 mL
98
PYQ 2026
medium
chemistryID: jee-main
500 mL of M KI solution is mixed with 500 mL of M solution in basic medium.
The liberated iodine is titrated with standard M solution in the presence of starch indicator
till the blue colour disappears.
The volume (in L) of consumed is ________ (nearest integer).
Official Solution
Correct Option: (1)
Concept: This problem involves {redox reactions} followed by {iodometric titration}. Key ideas used: In basic medium, oxidises iodide ions to iodine. Liberated iodine is titrated with sodium thiosulphate. Stoichiometry of redox reactions determines the amount of iodine formed. Step 1: Write the redox reaction in basic medium In basic medium, permanganate ion is reduced to manganese dioxide: From the equation: Step 2: Calculate moles of reactants {Moles of KI:} {Moles of :} Step 3: Identify the limiting reagent From the balanced equation: For mol , required iodide: Available iodide mol required mol Step 4: Calculate moles of iodine liberated From stoichiometry: Step 5: Titration of iodine with sodium thiosulphate Reaction: Step 6: Calculate volume of Final Answer:
99
PYQ 2026
medium
chemistryID: jee-main
Solubility product of is and . Find out .
1
0.119 V
2
-0.119 V
3
1.301 V
4
-1.301 V
Official Solution
Correct Option: (2)
Step 1: Understand the relationship between solubility product and electrode potential. The relationship between the solubility product ( ) and the electrode potential can be derived using the Nernst equation. The equation for the half-cell reaction is:
where:
- is the standard electrode potential,
- is the number of electrons involved in the reaction,
- is the reaction quotient. Step 2: Calculate the solubility from the solubility product. For the dissociation of in water, the solubility product is:
where is the solubility. Given , we can solve for :
Step 3: Apply the Nernst equation. For the reaction:
we know that the standard electrode potential for the oxidation of to is . The Nernst equation for the reverse reaction (reduction of ) is:
Substituting :
Thus, .
100
PYQ 2026
medium
chemistryID: jee-main
Consider the above electrochemical cell where a metal electrode (M) is undergoing redox reaction by forming ( ). The cation is present in two different concentrations and as shown above. Which of the following statement is correct for generating a positive cell potential?
1
If is present at anode, then .
2
If is present at cathode, then .
3
If is present at anode, then .
4
If is present at cathode, then .
Official Solution
Correct Option: (4)
The given cell is a concentration cell of the type:
For a concentration cell, the electrode with lower ion concentration
acts as the cathode, while the electrode with higher ion
concentration acts as the anode. The cell potential is given by:
For :
Therefore, if is present at the cathode, it must be
less than .
101
PYQ 2026
hard
chemistryID: jee-main
x mg of pure HCl was used to make an aqueous solution. 25.0 mL of 0.1 M Ba(OH)₂ solution is used when the HCl solution was titrated against it. The numerical value of x is_______ × 10⁻¹. (Nearest integer) Given: Molar mass of HCl and Ba(OH)₂ are 36.5 and 171.0 g mol⁻¹ respectively.}
Official Solution
Correct Option: (1)
Step 1: Understanding the Concept:
In a titration, the equivalents of acid must equal the equivalents of base at the end point. Step 2: Detailed Explanation:
Equivalents of
.
Equivalents of .
Equivalents of .
Since , Moles of . Step 3: Calculating Mass:
Mass of
Mass .
In mg: .
To express as : .
. Step 4: Final Answer:
The value of x is 1825.
102
PYQ 2026
easy
chemistryID: jee-main
For a closed circuit Daniell cell, which of the following plots is the accurate one at a given temperature ?
1
A
2
B
3
C
4
D
Official Solution
Correct Option: (2)
Step 1: Understanding the Concept: The standard cell potential, denoted as , is a fundamental thermodynamic property of an electrochemical cell. It is defined under standard conditions: all reactants and products at 1 M concentration, 1 atm pressure for gases, and a specific temperature.
Step 2: Key Formula or Approach: The instantaneous cell potential is related to concentrations by the Nernst equation:
While changes as the reaction proceeds (concentration changes), is the reference value at standard states.
Step 3: Detailed Explanation: For a given chemical reaction at a fixed temperature, is calculated as:
Since the standard reduction potentials of the electrodes are constant at a constant temperature, their difference is also a constant. As time passes in a closed circuit, the concentrations of ions change, which causes to decrease until it reaches zero at equilibrium. However, the standard reference value does not depend on time or concentration; it remains constant throughout the process. Therefore, a plot of against time will be a horizontal line.
Step 4: Final Answer: The accurate plot is the one where remains constant with respect to time, which is a horizontal line.
103
PYQ 2026
medium
chemistryID: jee-main
Find out the correct graph for the variation of with temperature.
1
1
2
2
3
3
4
4
Official Solution
Correct Option: (1)
Step 1: Understand the relationship between and temperature. The cell potential for a galvanic cell is related to the temperature through the Nernst equation. Typically, for most electrochemical reactions, the cell potential decreases as temperature increases, especially when the reaction is exothermic. This behavior is in accordance with the Gibbs free energy equation:
For exothermic reactions, an increase in temperature leads to a decrease in . Step 2: The relationship between and temperature. At constant pressure, for reactions like the one in the question, the relationship is approximately linear. The graph showing a decrease in with temperature is a characteristic behavior for most electrochemical reactions. Thus, the correct graph is the one that shows a linear decrease of with increasing temperature.
104
PYQ 2026
medium
chemistryID: jee-main
If E of the following reaction is x 10 . Find x / Sn(OH) , OH / Bi O / Bi / Pt}\) ]}\) Given ]}} \) = -0.90 V, O / Bi}} \) = -0.44 V
Official Solution
Correct Option: (1)
Step 1: Applying Nernst Equation. The Nernst equation for a cell is given by:
Where:
- is the cell potential,
- is the standard cell potential,
- is the number of moles of electrons involved in the reaction,
- is the reaction quotient. Step 2: Calculate the standard cell potential. We can calculate using the given standard electrode potentials for the two half-reactions:
Substituting the given values: Step 3: Apply the Nernst equation. Now, applying the Nernst equation:
Step 4: Conclusion. We are given that . Therefore, Thus, the correct answer is (4).
105
PYQ 2026
medium
chemistryID: jee-main
A cell representation is given: Ag/AgCl FeCl , FeCl / Pt
Which of the following can increase the EMF of the cell? (i) By increasing concentration of Fe
(ii) By increasing concentration of Fe
(iii) By decreasing concentration of Fe
(iv) By decreasing concentration of Fe
(v) By increasing concentration of Cl
1
ii, iv, v
2
iii, iv, v
3
iii, ii, v
4
i, iii, v
Official Solution
Correct Option: (2)
Step 1: Use of Nernst equation. The EMF of the cell can be calculated using the Nernst equation: Where is the standard electrode potential and is the number of electrons involved. Step 2: Effect of concentration. - Increasing the concentration of Fe would decrease the EMF because the ratio increases, leading to a more negative value of the logarithmic term. Hence, this will decrease the EMF. - Increasing the concentration of Fe would increase the EMF since the ratio decreases, increasing the value of . - Decreasing the concentration of Fe would increase the EMF. - Decreasing the concentration of Fe would decrease the EMF. - Increasing the concentration of Cl would increase the EMF by decreasing the potential difference between the electrodes. Step 3: Conclusion. The correct answer is (2) iii, iv, v, as these changes increase the EMF of the cell.
106
PYQ 2026
medium
chemistryID: jee-main
A cell is given as . For which of the following condition, is positive:
1
(If is concentration at cathode)
2
(If is concentration at anode)
3
(If is concentration at anode)
4
(If is concentration at cathode)
Official Solution
Correct Option: (4)
The cell is a concentration cell ( ). We assume . The Nernst equation is: . For to be positive, must be negative. This requires the ratio . Thus, the condition for spontaneity is . We analyze Option (D): is the cathode concentration ( ), and is the anode concentration ( ). . This condition satisfies , making positive.
107
PYQ 2026
medium
chemistryID: jee-main
Electrolysis of aqueous solution of is carried out, where of copper is deposited (atomic mass of ). After this . current is further passed for . Calculate total volume of released (in ). (Given )
1
111
2
100
3
90
4
122
Official Solution
Correct Option: (1)
1. Moles of electrons passed in Part 1 ( ): . 2. Moles of electrons passed in Part 2 ( ): . Total electrons . 3. Moles of released: . ( ). . Volume of (at ): . . .
108
PYQ 2026
medium
chemistryID: jee-main
For the given cell reaction . Cell emf 'E' is given as . Determine the value of 'n' in above equation.
Official Solution
Correct Option: (1)
Determine oxidation states and electron change.
Reduction: .
Cl state: . Change per Cl. Oxidation: (Assuming borate species like or similar).
Check H oxidation. In borohydride ( ), H is hydridic ( ). In the oxidized oxyanion product, H is protonic ( ) attached to Oxygen.
For 1 mole of , there are 4 H atoms.
Change per H atom: , loss of .
Total loss for 4 H atoms: .
Does Boron change? Usually B is in both. no.
In : B is , H is . Net: . Correct.
In Product (e.g. or ion): B is , H is .
Main redox is .
So oxidation n-factor is 8. Balance electron transfer between Oxidant (6e) and Reductant (8e).
LCM of 6 and 8 is 24.
Reaction stoichiometry: and .
Total electrons .
109
PYQ 2026
medium
chemistryID: jee-main
The relation between molar conductivity and concentration is given by
For various solution concentrations of and , the corresponding molar conductivities are and , respectively. Using the given data, determine the value of .
Official Solution
Correct Option: (1)
Concept:
For strong electrolytes, molar conductivity decreases linearly with according to:
Here, is a constant characteristic of the electrolyte. Step 1: Use Two Sets of Given Data For :
For :
Step 2: Subtract Equation (2) from (1)
110
PYQ 2026
medium
chemistryID: jee-main
of the following cell is . The cell representation is
Concentrations: Given: ion concentration is maintained by a buffer solution of mL, and mL, .
Find the value of .
Official Solution
Correct Option: (1)
Step 1: Write Cell Potential Expression Here,
Given:
Step 2: Nernst Equation Overall reaction involves electrons. Step 3: Reaction Quotient For the Sn half-cell:
Substitute given concentrations:
Step 4: Buffer Calculation Moles:
For strong basic condition:
Given high , solving gives:
