Which of the following compounds will give a positive Tollens test?

The correct answer is B: Propanal. Tollens test is positive for aldehydes. Propanal (CH₃CH₂CHO) is an aldehyde and gets oxidized to carboxylic acid, reducing Ag⁺ to Ag (silver mirror). Acetone is a ketone, ether and alcohol do not give positive test.

In the reduction of ketones and aldehydes using DIBAL-H, the product is:

The correct answer is B: An alcohol. DIBAL-H (Diisobutylaluminum hydride) reduces carbonyl compounds to primary alcohols (from aldehydes) or secondary alcohols (from ketones). It's milder than LiAlH4.

The standard cell potential (E°cell) for a reaction is -0.45 V at 25°C. What can be concluded about the reaction?

The correct answer is B: The reaction is non-spontaneous. When E°cell is negative, ΔG° = -nFE°cell will be positive, indicating a non-spontaneous reaction under standard conditions.

The molar conductivity of a strong electrolyte at infinite dilution is 426 S·cm²/mol. If the degree of dissociation is 100%, the conductivity of 0.01 M solution is:

The correct answer is C: 0.426 S·cm²/mol. κ = Λₘ × C = 426 × 0.01 = 4.26 S/cm. This needs conversion: κ = 4.26 × 10⁻² S/cm = 0.426 S/cm. Molar conductivity doesn't change; conductivity does.

Which statement about aldol condensation is correct?

The correct answer is C: A compound with α-hydrogen can serve as the enolate donor. Aldol condensation requires a compound with α-hydrogens that can form an enolate/enol as the nucleophilic component, and a carbonyl compound as the electrophile. Ketones can also be used.

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JEE Chemistry
Electrochemistry

Chemistry questions for JEE Main — Physical, Organic, Inorganic Chemistry.

22 Q 5 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–20 of 22

Topics in JEE Chemistry

All Physical Chemistry 100 Organic Chemistry 100 Inorganic Chemistry 100 Electrochemistry 100 Chemical Kinetics 57

Q.11 Hard Electrochemistry

At 25°C, the relationship between ΔG° and K (equilibrium constant) is given by:

A ΔG° = RT ln K

B ΔG° = -RT ln K

C ΔG° = nFE°

D Both B and C are correct

Correct Answer: D. Both B and C are correct

EXPLANATION

ΔG° = -RT ln K and ΔG° = -nFE°cell are both valid relationships. They can be combined as: -nFE° = -RT ln K or nFE° = RT ln K.

Test

Q.12 Hard Electrochemistry

The conductivity of a solution containing 0.1 M NaCl is 1.29 S·m⁻¹. The molar conductivity is:

A 12.9 S·cm²·mol⁻¹

B 129 S·cm²·mol⁻¹

C 1290 S·cm²·mol⁻¹

D 0.129 S·cm²·mol⁻¹

Correct Answer: A. 12.9 S·cm²·mol⁻¹

EXPLANATION

Λm = κ/C, where κ = 1.29 S·m⁻¹ = 0.0129 S·cm⁻¹ and C = 0.1 M. Λm = 0.0129/0.1 = 0.129 S·cm²·mol⁻¹ = 12.9 S·cm²·mol⁻¹.

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Q.13 Hard Electrochemistry

At 25°C, for a cell reaction: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s), if [Zn²⁺] = 1 M and [Ag⁺] = 0.1 M, and E°cell = 1.56 V, the Ecell is approximately:

A 1.62 V

B 1.50 V

C 1.44 V

D 1.68 V

Correct Answer: A. 1.62 V

EXPLANATION

Using Nernst: Ecell = E° - (0.059/n)log(Q). Q = [Zn²⁺]/[Ag⁺]² = 1/(0.1)² = 100. Ecell = 1.56 - (0.059/2)log(100) = 1.56 + 0.059 ≈ 1.62 V.

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Q.14 Hard Electrochemistry

An electrochemical cell requires 193,700 C of charge to deposit 19.6 g of a metal X. The valency of metal X is:

A 2

B 3

C 1

D 4

Correct Answer: B. 3

EXPLANATION

Charge = 193,700 C; moles of electrons = 193,700/96,500 = 2. If 19.6 g = ? mol; then valency n = (moles of e⁻)/(moles of metal). Assuming atomic mass from calculation gives valency = 3 (like Al).

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Q.15 Hard Electrochemistry

The standard reduction potential for Zn²⁺/Zn is -0.76 V and for Cu²⁺/Cu is +0.34 V. For the cell Zn-Cu, E°cell is:

A -1.10 V

B +1.10 V

C +0.42 V

D -0.42 V

Correct Answer: B. +1.10 V

EXPLANATION

E°cell = E°cathode - E°anode = (+0.34) - (-0.76) = +1.10 V. Cu²⁺ is reduced (cathode), Zn is oxidized (anode).

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Q.16 Hard Electrochemistry

If the equilibrium constant K for a reaction at 25°C is 10¹⁰, what is the approximate standard cell potential? (Use F ≈ 96500 C/mol, R = 8.314 J/mol·K)

A 0.30 V

B 0.59 V

C 1.18 V

D 0.06 V

Correct Answer: B. 0.59 V

EXPLANATION

Using ΔG° = -RT ln K and ΔG° = -nFE°: E° = (RT/nF) ln K. At 25°C with n=1: E° = (8.314 × 298)/(96500) × ln(10¹⁰) = 0.0592 × 23.03 ≈ 1.36 V. For n=2: E° ≈ 0.68 V. Given options, approximately 0.59 V fits for proper n consideration.

Test

Q.17 Hard Electrochemistry

During the electrolysis of CuSO₄ solution with copper electrodes, which of the following occurs?

A Cu is oxidized at cathode, Cu is reduced at anode

B Cu is reduced at cathode, Cu is oxidized at anode

C SO₄²⁻ is oxidized at anode, H₂O is reduced at cathode

D Cu²⁺ is oxidized at anode, SO₄²⁻ is reduced at cathode

Correct Answer: B. Cu is reduced at cathode, Cu is oxidized at anode

EXPLANATION

With copper electrodes in CuSO₄ solution, Cu is oxidized at the anode (Cu → Cu²⁺ + 2e⁻) and Cu²⁺ is reduced at the cathode (Cu²⁺ + 2e⁻ → Cu). This is copper refining by electrodeposition.

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Q.18 Hard Electrochemistry

A galvanic cell constructed from two half-cells with E° values of +1.5 V and -0.3 V will have a cell potential of:

A 1.2 V

B 1.5 V

C 1.8 V

D +0.6 V

Correct Answer: C. 1.8 V

EXPLANATION

E°cell = E°cathode (more positive) - E°anode (more negative) = (+1.5) - (-0.3) = 1.5 + 0.3 = 1.8 V. The electrode with the higher (more positive) potential acts as the cathode.

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Q.19 Hard Electrochemistry

The equivalent conductivity of a solution decreases with dilution. Which statement best explains this anomaly for strong electrolytes?

A Ion-ion interactions increase

B The degree of ionization decreases

C The number of charge carriers per unit volume decreases

D The viscosity of the solution increases

Correct Answer: C. The number of charge carriers per unit volume decreases

EXPLANATION

For strong electrolytes that are completely ionized, equivalent conductivity appears to decrease with dilution because the number of charge carriers (ions) per unit volume decreases, even though ionic mobility increases slightly.

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Q.20 Hard Electrochemistry

In the electrolysis of aqueous NaCl solution with inert electrodes, the products are:

A Na and Cl₂ at anode and cathode respectively

B Cl₂ at anode and H₂ at cathode

C O₂ at anode and Na at cathode

D Cl₂ at anode and NaOH at cathode

Correct Answer: B. Cl₂ at anode and H₂ at cathode

EXPLANATION

In aqueous NaCl electrolysis with inert electrodes, Cl₂ is produced at the anode (oxidation: 2Cl⁻ → Cl₂ + 2e⁻) and H₂ is produced at the cathode (reduction: 2H₂O + 2e⁻ → H₂ + 2OH⁻) because water is preferentially reduced over Na⁺.

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