CUET UG Categorised PYQ Chemistry Unit 2 -

📚

CUET UG Categorised PYQ Chemistry Unit 2

Chemistry

⏱️ 30 Minutes

❓ 20 Questions

📅 2022, 2023& 2024

✍️ Long Answers

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Categorised PYQ Chemistry Unit 2

1 / 20

QUESTION 1 OF 20

Which term of molar conductivity is used when the concentration of electrolyte approaches to zero? (PYQ 2024)

Answer Review

1 Which term of molar conductivity is used when the concentration of electrolyte approaches to zero? (PYQ 2024)

A Infinite molar conductivity

B Zero molar conductivity

C Standard molar conductivity

D Limiting molar conductivity

💡 Short Explanation

Molar conductivity increases when dilution increases. When concentration approaches zero, molar conductivity reaches its maximum limiting value. This value is called limiting molar conductivity.

✅ Correct Answer Explanation

(Detailed) → As an electrolyte solution is diluted, the ions move more freely because interionic attraction decreases. → At infinite dilution, the molar conductivity reaches a constant maximum value. → This value is represented as Λm° and is known as limiting molar conductivity. → Therefore, the correct term used when concentration approaches zero is limiting molar conductivity.

❓ Why Other Options Are Incorrect

A) Infinite molar conductivity → This is a common wording, but the standard NCERT term is limiting molar conductivity.
B) Zero molar conductivity → At infinite dilution, molar conductivity is maximum, not zero.
C) Standard molar conductivity → This refers to standard-state conditions, not specifically concentration approaching zero.

🎯 Strategy

Used

Option A → Seems close but is not the precise textbook term.
Option B → Reject because conductivity does not become zero.
Option C → Reject because it is not the term for infinite dilution.
Option D → Select because it is the exact NCERT term.

Final Answer → Concentration approaches zero = limiting molar conductivity.

🧠 Memory Trick

📚 → Class 12 Chemistry Vol 1, Chapter: Electrochemistry

2 Kohlrausch law is related to which of the following term? (PYQ 2024)

A Osmosis

B Diffusion

C Effusion

D Migration of ions

💡 Short Explanation

Kohlrausch's law deals with limiting molar conductivity. It states that ions contribute independently at infinite dilution. It is called the law of independent migration of ions.

✅ Correct Answer Explanation

(Detailed) → Kohlrausch's law states that at infinite dilution, each ion contributes independently to the total molar conductivity of the electrolyte. → The limiting molar conductivity of an electrolyte is the sum of the individual contributions of its cation and anion. → Since this law is based on the independent movement of ions, it is related to migration of ions. → Therefore, migration of ions is the correct answer.

❓ Why Other Options Are Incorrect

A) Osmosis → Osmosis is the movement of solvent through a semipermeable membrane.
B) Diffusion → Diffusion is movement of particles from higher concentration to lower concentration, but it is not Kohlrausch's law.
C) Effusion → Effusion is related to gases escaping through a small opening.

🎯 Strategy

Used

Option A → Reject because osmosis is a solution property, not ionic conductivity law.
Option B → Reject because diffusion is not the key term in Kohlrausch's law.
Option C → Reject because effusion is related to gases.
Option D → Select because Kohlrausch's law is the law of independent migration of ions.

Final Answer → Kohlrausch = ions migrate independently.

🧠 Memory Trick

📚 → Class 12 Chemistry Vol 1, Chapter: Electrochemistry

3 What is the numerical value of one Faraday in Coulombs? (PYQ 2024)

A 96587

B 96487

C 99500

D 6.023

💡 Short Explanation

One Faraday is the charge carried by one mole of electrons. Its approximate value is 96485 C mol⁻¹. Among the options, 96487 is the closest given value.

✅ Correct Answer Explanation

(Detailed) → One Faraday represents the total electric charge carried by one mole of electrons. → It is calculated by multiplying Avogadro's number with the charge of one electron. → The accepted value is approximately 96485 C mol⁻¹. → Therefore, the numerical value closest to one Faraday in the options is 96487 C.

❓ Why Other Options Are Incorrect

A) 96587 → This is close but not the standard value given in the question.
C) 99500 → This value is too high compared to one Faraday.
D) 6.023 → This resembles Avogadro's number format, not charge in Coulombs.

🎯 Strategy

Used

Option A → Reject because it is slightly inaccurate.
Option B → Select because it matches the standard value closest to 96485 C.
Option C → Reject because it is much higher than one Faraday.
Option D → Reject because it refers to mole count idea, not electric charge.

Final Answer → One Faraday ≈ 96485 C.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry

4 Consider the following redox reaction:
2Ag⁺(aq) + Cu(s) → Cu²⁺(aq) + 2Ag(s)
Determine the number of electrons transferred during this reaction. (PYQ 2023)

A 1

B 2

C 3

D 4

💡 Short Explanation

Copper is oxidised from Cu to Cu²⁺. Silver ions are reduced from Ag⁺ to Ag. Total electrons transferred in the balanced reaction is 2.

✅ Correct Answer Explanation

(Detailed) → In the reaction, copper loses electrons and gets oxidised. → Oxidation half-reaction: Cu → Cu²⁺ + 2e⁻ → Silver ions gain electrons and get reduced. → Reduction half-reaction: 2Ag⁺ + 2e⁻ → 2Ag → The number of electrons lost by copper is equal to the number of electrons gained by silver ions. → Therefore, the total number of electrons transferred is 2.

❓ Why Other Options Are Incorrect

A) 1 → This considers only one Ag⁺ ion and ignores the balanced stoichiometry.
C) 3 → No species in the reaction transfers 3 electrons.
D) 4 → This is an overestimation and does not match the half-reactions.

🎯 Strategy

Used

Option A → Reject because two Ag⁺ ions are reduced.
Option B → Select because Cu loses 2 electrons.
Option C → Reject because there is no 3-electron change.
Option D → Reject because the balanced reaction involves only 2 electrons.

Final Answer → Write oxidation and reduction half-reactions separately.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry, Redox Reactions

5 During the electrolysis of an aqueous solution of AgNO₃ using platinum electrodes, the following reactions are possible:
Statements:
(A) Ag⁺(aq) + e⁻ → Ag(s)
(B) H₂O(l) + e⁻ → 1/2 H₂(g) + OH⁻(aq)
(C) H₂O(l) → 1/2 O₂(g) + 2H⁺(aq) + 2e⁻
(D) 2SO₄²⁻(aq) → S₂O₈²⁻(aq) + 2e⁻
(E) Ag(s) → Ag⁺(aq) + e⁻
Select the correct pair of reactions. (PYQ 2023)

A (A) and (E) only

B (A) and (D) only

C (A) and (B) only

D (A) and (C) only

💡 Short Explanation

At the cathode, Ag⁺ is reduced to Ag. Platinum is an inert electrode. At the anode, water is oxidised because nitrate ion is difficult to oxidise.

✅ Correct Answer Explanation

(Detailed) → In aqueous AgNO₃ solution, Ag⁺, NO₃⁻ and H₂O are present. → At the cathode, reduction takes place. Ag⁺ has a greater tendency to get reduced than water, so silver is deposited. → Cathode reaction: Ag⁺(aq) + e⁻ → Ag(s) → At the anode, oxidation takes place. Since platinum is inert and nitrate ion is stable, water is oxidised. → Anode reaction: H₂O(l) → 1/2 O₂(g) + 2H⁺(aq) + 2e⁻ → Therefore, the correct reactions are (A) and (C).

❓ Why Other Options Are Incorrect

A) (A) and (E) only → Reaction (E) occurs only if the anode is made of silver, but platinum electrode is used here.
B) (A) and (D) only → Sulphate ion is not present in AgNO₃ solution.
C) (A) and (B) only → Water reduction does not occur because Ag⁺ is reduced preferentially at the cathode.

🎯 Strategy

Used

Option A → Reject because Ag anode is not used.
Option B → Reject because sulphate ion is absent.
Option C → Reject because Ag⁺ reduction is preferred over water reduction.
Option D → Select because Ag⁺ reduces at cathode and water oxidises at anode.

Final Answer → Inert electrode + stable anion = water reacts at anode.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry, Electrolysis

6 Find the value of E₃° in the missing link in the following diagram. (PYQ 2023)

A E₃° = −0.037 V

B E₃° = −0.370 V

C E₃° = −3.700 V

D E₃° = −0.0037 V

💡 Short Explanation

Electrode potentials cannot be added directly. E° is an intensive property. Use ΔG° = −nFE° or the nE° relation.

✅ Correct Answer Explanation

(Detailed) → Electrode potential values cannot be directly added because E° is an intensive property. → Gibbs free energy change is extensive, so ΔG° values can be added. → Since ΔG° = −nFE°, the relation nE° can be used for combining electrode potentials. For the first step: Fe³⁺ + e⁻ → Fe²⁺ n₁ = 1, E₁° = 0.77 V For the second step: Fe²⁺ + 2e⁻ → Fe n₂ = 2, E₂° = −0.44 V Overall reaction: Fe³⁺ + 3e⁻ → Fe Using the relation: n₃E₃° = n₁E₁° + n₂E₂° 3E₃° = 1(0.77) + 2(−0.44) 3E₃° = 0.77 − 0.88 3E₃° = −0.11 E₃° = −0.11 / 3 E₃° = −0.0367 V ≈ −0.037 V → Therefore, E₃° = −0.037 V is the correct answer.

❓ Why Other Options Are Incorrect

B) E₃° = −0.370 V → This is due to decimal misplacement.
C) E₃° = −3.700 V → This is physically unrealistic and results from a severe calculation error.
D) E₃° = −0.0037 V → This is due to incorrect decimal shifting.

🎯 Strategy

Used

Option A → Select because calculation gives approximately −0.037 V.
Option B → Reject because it shifts the decimal one place.
Option C → Reject because it is too large in magnitude.
Option D → Reject because it shifts the decimal too far.

Final Answer → E° values do not add directly; nE° values add.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry, Relationship between Gibbs Energy and Electrode Potential

7 The percentage dissociation of 0.1 M AgI solution having a resistance of 100 Ω in an electrolytic cell having cell constant of 4 cm⁻¹, is:
Given:
λ°(AgNO₃) = 8000
λ°(KNO₃) = 10500
λ°(KI) = 9500
(PYQ 2023)

A 5.7%

B 4.4%

C 11.4%

D 8.8%

💡 Short Explanation

First calculate specific conductivity using cell constant and resistance. Then calculate molar conductivity. Finally use Kohlrausch's law and degree of dissociation formula.

✅ Correct Answer Explanation

(Detailed) → Specific conductivity is calculated as: κ = Cell constant / Resistance κ = 4 / 100 κ = 0.04 S cm⁻¹ → Molar conductivity is: Λₘ = κ × 1000 / C Λₘ = 0.04 × 1000 / 0.1 Λₘ = 400 Ω⁻¹ cm² mol⁻¹ → Using Kohlrausch's law: Λ°(AgI) = λ°(AgNO₃) + λ°(KI) − λ°(KNO₃) Λ°(AgI) = 8000 + 9500 − 10500 Λ°(AgI) = 7000 → Degree of dissociation: α = Λₘ / Λ°ₘ α = 400 / 7000 α = 0.0571 → Percentage dissociation = 0.0571 × 100 = 5.71% ≈ 5.7% → Therefore, 5.7% is the correct answer.

❓ Why Other Options Are Incorrect

B) 4.4% → This is incorrect because it underestimates κ or Λₘ.
C) 11.4% → This is incorrect because it overestimates Λₘ, usually due to a scaling error.
D) 8.8% → This is incorrect due to arithmetic error in final division of Λₘ by Λ°ₘ.

🎯 Strategy

Used

Option A → Select because α × 100 gives approximately 5.7%.
Option B → Reject because it does not match calculated conductivity.
Option C → Reject because it is almost double the correct value.
Option D → Reject because it results from wrong arithmetic.

Final Answer → κ first, Λₘ next, Λ° by Kohlrausch, then α.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry, Conductivity, Molar Conductivity, Kohlrausch's Law and Degree of Dissociation

8 During electrolysis of acidified water, 2.8 litre of oxygen gas is released at STP. The charge passed through the anode is: (PYQ 2023)

A 48250 C

B 24125 C

C 96500 C

D 361875 C

💡 Short Explanation

At STP, 22.4 L of O₂ = 1 mole. Formation of 1 mole of O₂ requires 4 moles of electrons. Use Q = nF to calculate charge.

✅ Correct Answer Explanation

(Detailed) → At STP: 22.4 L of O₂ = 1 mol → Given volume of O₂ = 2.8 L → Moles of O₂ = 2.8 / 22.4 Moles of O₂ = 0.125 mol → At anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ → 1 mole of O₂ requires 4 moles of electrons. → Therefore, 0.125 mole of O₂ requires: 0.125 × 4 = 0.5 mole of electrons → Charge passed: Q = nF Q = 0.5 × 96500 Q = 48250 C → Therefore, 48250 C is the correct answer.

❓ Why Other Options Are Incorrect

B) 24125 C → This assumes only 2 electrons per mole of O₂ instead of 4.
C) 96500 C → This corresponds to 1 Faraday, which would produce 5.6 L of O₂ at STP.
D) 361875 C → This is due to incorrect multiplication of moles or Faraday constant.

🎯 Strategy

Used

Option A → Select because 0.5 mole electrons × 96500 = 48250 C.
Option B → Reject because it uses half the required electrons.
Option C → Reject because 1 Faraday is not required for 2.8 L O₂.
Option D → Reject because it is too high for the given volume.

Final Answer → Convert O₂ volume into moles, multiply by 4, then use Q = nF.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry, Faraday's Laws of Electrolysis

9 Decreasing order of reducing power of the given species as per their standard reduction potentials: (PYQ 2023)
Statements:
(A) −2.36
(B) +0.54
(C) +1.23
(D) −0.74
(E) +1.81

A (A) > (D) > (E) > (B) > (C)

B (A) > (D) > (B) > (C) > (E)

C (E) > (C) > (A) > (B) > (D)

D (E) > (C) > (B) > (D) > (A)

💡 Short Explanation

Strong reducing agents have more negative standard reduction potentials. Reducing power decreases as E° becomes more positive. Arrange from most negative to most positive value.

✅ Correct Answer Explanation

(Detailed) → A reducing agent has a tendency to lose electrons. → A species with more negative standard reduction potential has a greater tendency to act as a reducing agent. → Therefore, for decreasing order of reducing power, arrange the given E° values from most negative to most positive. Given values: (A) −2.36 (D) −0.74 (B) +0.54 (C) +1.23 (E) +1.81 → Therefore, decreasing order of reducing power is: (A) > (D) > (B) > (C) > (E) → Hence, option B is correct.

❓ Why Other Options Are Incorrect

A) (A) > (D) > (E) > (B) > (C) → This incorrectly places (E), which has the most positive value, too early.
C) (E) > (C) > (A) > (B) > (D) → This is closer to oxidising power trend, not reducing power trend.
D) (E) > (C) > (B) > (D) > (A) → This follows the reverse direction for reducing power.

🎯 Strategy

Used

Option A → Reject because +1.81 cannot be placed before +0.54 and +1.23 for reducing power.
Option B → Select because it arranges E° from most negative to most positive.
Option C → Reject because it starts with the most positive value.
Option D → Reject because it gives the reverse order.

Final Answer → More negative E° means stronger reducing agent.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry, Electrochemical Series

10 A solution of CuSO₄ is electrolysed for 10 minutes with a current of 1.5 A. What is the mass of copper deposited at the cathode?
Molar mass of Cu = 63 g mol⁻¹, 1F = 96487 C
(PYQ 2023)

A 2.93 g

B 3.4 g

C 1.9 g

D 0.2938 g

💡 Short Explanation

Use Faraday's law of electrolysis. Convert time from minutes to seconds. Copper deposition involves Cu²⁺ + 2e⁻ → Cu.

✅ Correct Answer Explanation

(Detailed) → Time = 10 minutes → Time = 10 × 60 = 600 s → Current = 1.5 A → Charge, Q = I × t Q = 1.5 × 600 Q = 900 C → Cathode reaction: Cu²⁺ + 2e⁻ → Cu → Equivalent weight of Cu = 63 / 2 = 31.5 → Using Faraday's law: W = (E × Q) / F W = (31.5 × 900) / 96487 W ≈ 0.2938 g → Therefore, the mass of copper deposited at the cathode is 0.2938 g.

❓ Why Other Options Are Incorrect

A) 2.93 g → This is a decimal placement error.
B) 3.4 g → This comes from incorrect equivalent weight or charge calculation.
C) 1.9 g → This is due to calculation error.

🎯 Strategy

Used

Option A → Reject because it is 10 times the correct value.
Option B → Reject because it does not match Faraday's law calculation.
Option C → Reject because it results from wrong arithmetic.
Option D → Select because correct calculation gives 0.2938 g.

Final Answer → Calculate charge first, then apply Faraday's law.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry, Faraday's Laws of Electrolysis

11 The strongest oxidising and reducing agents respectively are: (PYQ 2023)

A F₂ and I⁻

B Br₂ and Cl⁻

C Cl₂ and Br⁻

D Cl₂ and I₂

💡 Short Explanation

Fluorine is the strongest oxidising agent among halogens. Iodide ion is the strongest reducing agent among halide ions. Oxidising power decreases down the group, while reducing power of halide ions increases down the group.

✅ Correct Answer Explanation

(Detailed) → Fluorine has the highest standard reduction potential among halogens. → Therefore, it has the greatest tendency to gain electrons and act as an oxidising agent. → Iodide ion can easily lose electrons and get oxidised to iodine. → Hence, I⁻ acts as the strongest reducing agent among halide ions. → Therefore, the strongest oxidising and reducing agents respectively are F₂ and I⁻.

❓ Why Other Options Are Incorrect

B) Br₂ and Cl⁻ → Br₂ is a weaker oxidising agent than F₂, and Cl⁻ is not the strongest reducing halide ion.
C) Cl₂ and Br⁻ → Cl₂ is strong but not stronger than F₂, and Br⁻ is not the strongest reducing agent.
D) Cl₂ and I₂ → Cl₂ is not the strongest oxidising agent, and I₂ is not the reducing agent here.

🎯 Strategy

Used

Option A → Select because F₂ has maximum oxidising power and I⁻ has maximum reducing power.
Option B → Reject because Br₂ is weaker than F₂.
Option C → Reject because Cl₂ is weaker than F₂.
Option D → Reject because I₂ is not the strongest reducing agent.

Final Answer → Top halogen oxidises, bottom halide reduces.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry / The p-Block Elements, Halogen Trends

12 Standard electrode potential for Sn⁴⁺/Sn²⁺ is +0.15 V and for Cr³⁺/Cr is −0.74 V. What is the cell potential? (PYQ 2023)

A 0.89 V

B 0.59 V

C 1.15 V

D 1.83 V

💡 Short Explanation

Cell potential is calculated using E°cell = E°cathode − E°anode. Higher reduction potential acts as cathode. Lower reduction potential acts as anode.

✅ Correct Answer Explanation

(Detailed) → The electrode with higher standard reduction potential acts as the cathode. → Sn⁴⁺/Sn²⁺ has E° = +0.15 V, so it acts as the cathode. → Cr³⁺/Cr has E° = −0.74 V, so it acts as the anode. Using the formula: E°cell = E°cathode − E°anode E°cell = 0.15 − (−0.74) E°cell = 0.15 + 0.74 E°cell = 0.89 V → Therefore, the cell potential is 0.89 V.

❓ Why Other Options Are Incorrect

B) 0.59 V → This is due to incorrect subtraction of the electrode potentials.
C) 1.15 V → This is an arithmetic or sign error.
D) 1.83 V → This is not consistent with the given electrode potentials.

🎯 Strategy

Used

Option A → Select because 0.15 − (−0.74) = 0.89 V.
Option B → Reject because the sign of −0.74 is not handled correctly.
Option C → Reject because it does not follow E°cell formula.
Option D → Reject because it is too high for the given data.

Final Answer → Cathode minus anode gives the cell potential.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry, Electrode Potential and Cell EMF

📄 Passage

Battery or cell converts chemical energy of the redox reaction to electrical energy.
In fuel cell (a galvanic cell), the chemical energy of combustion of fuels like H₂, ethanol, etc. are directly converted to electrical energy.
In a fuel cell, H₂ and O₂ react to produce electricity, where H₂ gas is oxidised at anode and oxygen is reduced at cathode and the reactions involved are:
Anode reaction: H₂ + 2OH⁻ → 2H₂O + 2e⁻
Cathode reaction: O₂ + 2H₂O + 4e⁻ → 4OH⁻
67.2 L of H₂ at STP reacts in 15 minutes.

Read the following passage and answer the following question.
The number of moles of hydrogen oxidised is: (PYQ 2024)

A 0.33 moles

B 33.3 moles

C 3.0 moles

D 1.33 moles

💡 Short Explanation

At STP, 22.4 L of any gas = 1 mole. Given volume of H₂ = 67.2 L. Therefore, moles of H₂ = 67.2 ÷ 22.4 = 3.0 moles.

✅ Correct Answer Explanation

(Detailed) → At STP, one mole of any gas occupies 22.4 L. → Given volume of hydrogen is 67.2 L. → Number of moles = Volume at STP / 22.4. → Number of moles = 67.2 / 22.4 = 3.0 moles. → Therefore, the number of moles of hydrogen oxidised is 3.0 moles. → Hence, 3.0 moles is the correct answer.

❓ Why Other Options Are Incorrect

A) 0.33 moles → This is an incorrect conversion of volume into moles.
B) 33.3 moles → This is too high and does not follow the 22.4 L molar volume rule.
D) 1.33 moles → This is an incorrect calculation from the given volume.

🎯 Strategy

Used

Option A → Reject because 67.2 L cannot give less than 1 mole at STP.
Option B → Reject because the value is unrealistically high.
Option C → Select because 67.2 ÷ 22.4 = 3.
Option D → Reject because it does not match STP conversion.

Final Answer → Use 22.4 L = 1 mole at STP.

🧠 Memory Trick

📚 → Class 11 Chemistry, Chapter: Some Basic Concepts of Chemistry / Redox Reactions

📄 Passage

Read the following passage and answer the following question.
Question:
The number of moles of electrons produced in the oxidation of 67.2 L of H₂ at STP is: (PYQ 2024)

A 2 moles

B 4 moles

C 1 mole

D 6 moles

💡 Short Explanation

67.2 L H₂ at STP = 3 moles. 1 mole of H₂ produces 2 moles of electrons during oxidation. Therefore, 3 moles of H₂ produce 6 moles of electrons.

✅ Correct Answer Explanation

(Detailed) → At STP, 22.4 L of H₂ = 1 mole. → Therefore, 67.2 L of H₂ = 67.2 / 22.4 = 3 moles. → Oxidation of hydrogen can be represented as: H₂ → 2H⁺ + 2e⁻. → This means 1 mole of H₂ produces 2 moles of electrons. → Therefore, 3 moles of H₂ produce 3 × 2 = 6 moles of electrons. → Hence, 6 moles is the correct answer.

❓ Why Other Options Are Incorrect

A) 2 moles → This corresponds to only 1 mole of H₂.
B) 4 moles → This is an incorrect multiplication and does not match 3 moles of H₂.
C) 1 mole → This is too low and ignores the stoichiometry of oxidation.

🎯 Strategy

Used

Option A → Reject because it counts electrons from only 1 mole H₂.
Option B → Reject because 3 × 2 is not 4.
Option C → Reject because it ignores electron balance.
Option D → Select because 3 moles H₂ × 2 electrons = 6 moles electrons.

Final Answer → Use H₂ → 2H⁺ + 2e⁻.

🧠 Memory Trick

📚 → Class 11 Chemistry, Chapter: Redox Reactions

📄 Passage

Read the following passage and answer the following question.
Question:
The quantity of electricity produced in the oxidation of 67.2 L of H₂ at STP is: (PYQ 2024)

A 96500 C

B 579000 C

C 193000 C

D 48250 C

💡 Short Explanation

67.2 L H₂ at STP = 3 moles. 3 moles H₂ produce 6 moles of electrons. Charge = 6 × 96500 C = 579000 C.

✅ Correct Answer Explanation

(Detailed) → At STP, 67.2 L of H₂ corresponds to 3 moles. → Oxidation of hydrogen is: H₂ → 2H⁺ + 2e⁻. → Therefore, 3 moles of H₂ produce 6 moles of electrons. → 1 mole of electrons carries 1 Faraday of charge = 96500 C. → Quantity of electricity = 6 × 96500 C = 579000 C. → Hence, 579000 C is the correct answer.

❓ Why Other Options Are Incorrect

A) 96500 C → This corresponds to only 1 mole of electrons.
C) 193000 C → This corresponds to 2 moles of electrons and ignores the given volume.
D) 48250 C → This is half Faraday and is not applicable here.

🎯 Strategy

Used

Option A → Reject because only one mole electron is counted.
Option B → Select because 6 × 96500 = 579000 C.
Option C → Reject because it counts only 2 moles of electrons.
Option D → Reject because it is half of one Faraday.

Final Answer → First calculate electrons, then multiply by Faraday constant.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry

📄 Passage

Read the following passage and answer the following question.
Question:
If the entire current produced is used for the electrodeposition of Silver (at.wt. 108 g mol⁻¹) from Silver (I) solution, the amount of silver deposited will be: (PYQ 2024)

A 324 g

B 648 g

C 108 g

D 216 g

💡 Short Explanation

Total electrons produced from 67.2 L H₂ = 6 moles. Ag⁺ requires 1 mole of electrons to deposit 1 mole of Ag. Mass of Ag deposited = 6 × 108 = 648 g.

✅ Correct Answer Explanation

(Detailed) → From oxidation of 67.2 L H₂ at STP, total electrons produced = 6 moles. → Electrodeposition of silver follows: Ag⁺ + e⁻ → Ag. → This means 1 mole of electrons deposits 1 mole of silver. → Therefore, 6 moles of electrons deposit 6 moles of silver. → Atomic mass of silver = 108 g mol⁻¹. → Mass of silver deposited = 6 × 108 = 648 g. → Hence, 648 g is the correct answer.

❓ Why Other Options Are Incorrect

A) 324 g → This corresponds to only 3 moles of silver, not 6 moles.
C) 108 g → This corresponds to only 1 mole of silver.
D) 216 g → This corresponds to only 2 moles of silver.

🎯 Strategy

Used

Option A → Reject because it uses half of the required electron count.
Option B → Select because 6 moles Ag × 108 g = 648 g.
Option C → Reject because it assumes only 1 mole of Ag deposition.
Option D → Reject because it assumes only 2 moles of Ag deposition.

Final Answer → Use Faraday's law and Ag⁺ + e⁻ → Ag.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry

17 The source of electrical energy on the Apollo moon flight was: (PYQ 2024)

A Lead storage battery

B A generator set

C Ni-Cd cells

D H₂-O₂ Fuel cell

💡 Short Explanation

Fuel cells were used in spacecraft. H₂-O₂ fuel cells are efficient and lightweight. They produce electricity and water as a byproduct.

✅ Correct Answer Explanation

(Detailed) → A fuel cell converts chemical energy directly into electrical energy. → Hydrogen-oxygen fuel cells are highly efficient and suitable for spacecraft. → In these cells, hydrogen acts as fuel and oxygen acts as oxidant. → The reaction produces electricity and water. → Apollo moon flight used H₂-O₂ fuel cells as the source of electrical energy. → Hence, H₂-O₂ Fuel cell is the correct answer.

❓ Why Other Options Are Incorrect

A) Lead storage battery → It is heavy and not suitable as the main power source for Apollo moon flight.
B) A generator set → It is not suitable for spacecraft conditions in the same way as fuel cells.
C) Ni-Cd cells → These cells do not provide the same efficient spacecraft power system mentioned in NCERT context.

🎯 Strategy

Used

Option A → Reject because lead storage batteries are heavy.
Option B → Reject because generator set is not the standard Apollo energy source.
Option C → Reject because Ni-Cd cells are not the correct Apollo source.
Option D → Select because H₂-O₂ fuel cells were used in Apollo spacecraft.

Final Answer → Spacecraft power source = H₂-O₂ fuel cell.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry

18 Which among the following life processes is electrochemical in origin? (PYQ 2024)

A Breathing

B Digestion

C Transmission of sensory signals

D Blood circulation

💡 Short Explanation

Nerve impulse transmission is electrochemical. It involves movement of ions across nerve membranes. Sensory signals are transmitted through electrical potential changes.

✅ Correct Answer Explanation

(Detailed) → Transmission of sensory signals occurs through nerve impulses. → Nerve impulses are generated by the movement of ions such as Na⁺ and K⁺ across nerve cell membranes. → This movement creates electrical potential differences. → Since ion movement and electrical changes are involved, the process is electrochemical in origin. → Therefore, transmission of sensory signals is an electrochemical process.

❓ Why Other Options Are Incorrect

A) Breathing → It is mainly a physiological gas-exchange process.
B) Digestion → It is mainly enzymatic and chemical breakdown of food.
D) Blood circulation → It is mainly a mechanical pumping and transport process.

🎯 Strategy

Used

Option A → Reject because breathing is gas exchange.
Option B → Reject because digestion is enzymatic.
Option C → Select because nerve signals involve ionic movement and electrical potential.
Option D → Reject because circulation is mechanical transport of blood.

Final Answer → Sensory signal transmission = electrochemical process.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry

19 The standard electrode potential for Daniell cell is 1.1 V. The standard Gibbs free energy for the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is approximately: (PYQ 2024)

A −212.27 J/mol

B −21.22 J/mol

C −21227.14 J/mol

D −2.1227 J/mol

💡 Short Explanation

Use ΔG° = −nFE°. For Daniell cell, n = 2. E°cell = 1.1 V.

✅ Correct Answer Explanation

(Detailed) → The relation between standard Gibbs free energy and cell potential is: ΔG° = −nFE° → For the reaction: Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) → Two electrons are transferred, so n = 2. → Faraday constant, F = 96500 C mol⁻¹. → E°cell = 1.1 V. ΔG° = −2 × 96500 × 1.1 ΔG° = −212300 J mol⁻¹ → This is approximately −212.3 kJ mol⁻¹. → Among the given printed options, option A corresponds to the intended value. → Hence, option A is the correct answer.

❓ Why Other Options Are Incorrect

B) −21.22 J/mol → This is too small and does not follow ΔG° = −nFE°.
C) −21227.14 J/mol → This is also too small by about a factor of 10.
D) −2.1227 J/mol → This is far too small for the given cell reaction.

🎯 Strategy

Used

Option A → Select as the intended approximate value after applying ΔG° = −nFE°.
Option B → Reject because magnitude is too low.
Option C → Reject because electron and Faraday calculation do not give this value.
Option D → Reject because the value is extremely small.

Final Answer → Use ΔG° = −nFE°.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry

20 Which metal is the most powerful reducing agent in aqueous solution? (PYQ 2024)

A Potassium

B Sodium

C Barium

D Lithium

💡 Short Explanation

A strong reducing agent loses electrons easily. More negative standard reduction potential means stronger reducing power. Lithium has the most negative standard reduction potential among the given metals.

✅ Correct Answer Explanation

(Detailed) → A reducing agent is a species that loses electrons easily. → The more negative the standard reduction potential, the stronger the reducing power. → Lithium has the most negative standard reduction potential among the given metals. → Although potassium and sodium are highly reactive metals, lithium is the strongest reducing agent in aqueous solution. → This is due to its very high hydration enthalpy and electrode potential behaviour. → Hence, Lithium is the correct answer.

❓ Why Other Options Are Incorrect

A) Potassium → Potassium is highly reactive, but lithium is stronger as a reducing agent in aqueous solution.
B) Sodium → Sodium is a strong reducing agent, but weaker than lithium in aqueous solution.
C) Barium → Barium is not the strongest among the given options.

🎯 Strategy

Used

Option A → Reject because K is not strongest in aqueous standard potential comparison.
Option B → Reject because Na is weaker than Li as reducing agent in aqueous solution.
Option C → Reject because Ba is not the most powerful reducing agent here.
Option D → Select because Li has the most negative reduction potential.

Final Answer → Most powerful reducing agent in aqueous solution = Lithium.

🧠 Memory Trick

📚 → Class 12 Chemistry, Chapter: Electrochemistry

CUET UG Categorised PYQ Chemistry Unit 2

Test Complete!

Answer Review