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JEE Chemistry

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

225 Q 5 Topics Take Test
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Difficulty: All Easy Medium Hard 51–60 of 225
Topics in JEE Chemistry
Q.51 Medium Electrochemistry
For the electrode reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, the number of electrons transferred in the redox reaction is:
A 2
B 3
C 5
D 8
Correct Answer:  C. 5
EXPLANATION

The equation explicitly shows that 5 electrons (5e⁻) are transferred in this reduction half-reaction where Mn goes from +7 to +2 oxidation state.

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Q.52 Medium Electrochemistry
What is the purpose of a salt bridge in an electrochemical cell?
A To complete the external circuit
B To maintain electrical neutrality by ion migration
C To increase cell potential
D To prevent mixing of solutions
Correct Answer:  B. To maintain electrical neutrality by ion migration
EXPLANATION

A salt bridge completes the internal circuit and allows ion migration to maintain electrical neutrality as electrons flow through the external circuit. This prevents charge accumulation.

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Q.53 Medium Electrochemistry
The molar conductivity of a solution decreases with dilution for:
A Strong electrolytes only
B Weak electrolytes only
C Both strong and weak electrolytes
D Neither strong nor weak electrolytes
Correct Answer:  C. Both strong and weak electrolytes
EXPLANATION

Molar conductivity (Λm) decreases with dilution for both strong and weak electrolytes, but the decrease is greater for weak electrolytes due to increased ionization upon dilution.

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Q.54 Medium Electrochemistry
In a concentration cell with two zinc electrodes in different concentrations of Zn²⁺, which statement is true?
A E°cell = 0 V
B Both electrodes are identical in composition
C Cell potential depends only on the concentration difference
D All of the above
Correct Answer:  D. All of the above
EXPLANATION

In a concentration cell: E°cell = 0 (identical electrodes), electrodes are chemically identical but in different concentrations, and cell potential depends on concentration difference via Nernst equation.

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Q.55 Medium Electrochemistry
Which metal cannot be obtained by electrolysis of its aqueous salt solution?
A Copper
B Silver
C Sodium
D Gold
Correct Answer:  C. Sodium
EXPLANATION

Sodium has a very negative reduction potential (-2.71 V). Water is preferentially reduced instead. Sodium is obtained by electrolysis of molten NaCl. Copper, silver, and gold can be obtained from aqueous solutions.

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Q.56 Medium Electrochemistry
The cell potential at 25°C for the reaction: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) is 0.46 V. Calculate ΔG°:
A -44.6 kJ/mol
B -88.7 kJ/mol
C -223 kJ/mol
D -356 kJ/mol
Correct Answer:  B. -88.7 kJ/mol
EXPLANATION

ΔG° = -nFE°cell = -2 × 96485 × 0.46 = -88,766 J/mol ≈ -88.7 kJ/mol. The negative value confirms spontaneity.

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Q.57 Medium Electrochemistry
In the electrolysis of aqueous KCl solution with inert electrodes, which gas is produced at the cathode?
A Cl₂
B H₂
C O₂
D HCl
Correct Answer:  B. H₂
EXPLANATION

At the cathode in KCl solution, H⁺ ions (from water) are preferentially reduced to H₂ gas because water reduction potential (-0.83 V) is higher than K⁺ reduction potential (-2.93 V).

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Q.58 Medium Electrochemistry
The Nernst equation relates cell potential to which of the following?
A Temperature only
B Concentration only
C Temperature and concentration
D Pressure only
Correct Answer:  C. Temperature and concentration
EXPLANATION

The Nernst equation: E = E° - (RT/nF)ln(Q) shows that cell potential depends on both temperature (T) and concentration (through Q, the reaction quotient).

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Q.59 Medium Electrochemistry
In the electrolysis of dilute H₂SO₄ with inert electrodes, the cathode reaction is:
A 2H⁺ + 2e⁻ → H₂
B SO₄²⁻ + 2e⁻ → S + 4O²⁻
C 2H₂O + 2e⁻ → H₂ + 2OH⁻
D H⁺ + e⁻ → H
Correct Answer:  A. 2H⁺ + 2e⁻ → H₂
EXPLANATION

In dilute H₂SO₄, H⁺ is preferentially reduced over water at the cathode (less negative reduction potential), producing H₂ gas.

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Q.60 Medium Electrochemistry
Calculate the number of Faradays required to deposit 3.2 g of Cu (atomic mass 64) from CuSO₄ solution:
A 0.1 F
B 0.05 F
C 0.2 F
D 0.5 F
Correct Answer:  A. 0.1 F
EXPLANATION

Cu²⁺ + 2e⁻ → Cu. Moles of Cu = 3.2/64 = 0.05 mol. Charge = 0.05 × 2 = 0.1 F (since 1 F = 1 mole of electrons).

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