Chemistry carries the highest scoring potential in JEE for anyone who keeps the three branches separate in revision. This set covers physical chemistry numericals, organic reaction mechanisms and named reactions, and inorganic chemistry including periodic trends, chemical bonding and coordination compounds. Organic questions show the mechanism arrow by arrow, so the reasoning transfers to reactions you have not seen before.
In the electrolysis of aqueous KCl solution with inert electrodes, which gas is produced at the cathode?
Answer: B
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).
Q.342Medium
The cell potential at 25°C for the reaction: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) is 0.46 V. Calculate ΔG°:
Answer: B
ΔG° = -nFE°cell = -2 × 96485 × 0.46 = -88,766 J/mol ≈ -88.7 kJ/mol. The negative value confirms spontaneity.
Q.343Medium
Which metal cannot be obtained by electrolysis of its aqueous salt solution?
Answer: C
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.
Q.344Medium
In a concentration cell with two zinc electrodes in different concentrations of Zn²⁺, which statement is true?
Answer: D
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.
Q.345Medium
The molar conductivity of a solution decreases with dilution for:
Answer: C
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.
Q.346Medium
What is the purpose of a salt bridge in an electrochemical cell?
Answer: B
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.
Q.347Medium
For the electrode reaction: MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O, the number of electrons transferred in the redox reaction is:
Answer: C
The equation explicitly shows that 5 electrons (5e⁻) are transferred in this reduction half-reaction where Mn goes from +7 to +2 oxidation state.
Q.348Medium
Using the data: E°(Cu²⁺/Cu) = +0.34 V, E°(Zn²⁺/Zn) = -0.76 V, what is E°cell for the Daniell cell?
Answer: A
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V. Zinc is oxidized (anode) and copper is reduced (cathode).
Q.349Medium
In electroplating of iron with copper, which electrode should be made of copper?
Answer: B
The anode is made of copper (the plating metal) which gets oxidized and dissolved, while the cathode (iron object) receives Cu²⁺ ions that are reduced to form the copper coating.
Q.350Hard
At 25°C, the relationship between ΔG° and K (equilibrium constant) is given by:
Answer: D
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.
Q.351Hard
For the cell: Pt | H₂(1 atm) | H⁺(0.1 M) || Ag⁺(0.1 M) | Ag, calculate E at 25°C if E°cell = 0.80 V and log(0.1) = -1:
Answer: A
Using Nernst: E = E° - (0.059/n)log Q. For this cell, n = 1, Q = [H⁺]/[Ag⁺] = 0.01.1 = 1, log Q = 0. At different concentrations: Q = [H⁺]²/[Ag⁺] = 0.001.1 = 0.1, so E = 0.80 - (0.059)(-1) = 0.859 ≈ 0.82 V
Q.352Hard
The conductivity of a solution is 2.0 × 10⁻⁴ S/cm and the cell constant is 1.0 cm⁻¹. What is the molar conductivity of 0.01 M solution?
Answer: A
Molar conductivity = (κ × 1000)/C = (2.0 × 10⁻⁴ × 1.0 × 1000)/0.01 = 20 S·cm²/mol. where κ is conductivity and C is molarity.
Q.353Hard
In the galvanic cell using Pb-PbSO₄ electrode and Hg-Hg₂Cl₂ electrode (calomel), which is the cathode if E°(Hg₂Cl₂/Hg) = 0.27 V and E°(PbSO₄/Pb) = -0.36 V?
Answer: B
The electrode with higher reduction potential acts as cathode. Since E°(Hg₂Cl₂/Hg) = 0.27 V > E°(PbSO₄/Pb) = -0.36 V, the calomel electrode (Hg-Hg₂Cl₂) is the cathode.
Q.354Medium
What mass of Cu is deposited when 2 Faradays of charge pass through a CuSO₄ solution? (Atomic mass of Cu = 64)
Answer: C
For Cu²⁺ + 2e⁻ → Cu, 2 Faradays deposit 1 mole of Cu. Therefore, 2 Faradays deposit 2 moles of Cu = 2 × 64 = 128 g.
Q.355Easy
In a galvanic cell, the electrode where oxidation occurs is called:
Answer: A
In a galvanic cell, oxidation occurs at the anode (negative electrode), while reduction occurs at the cathode (positive electrode).
Q.356Easy
The standard cell potential (E°cell) for a reaction is -0.45 V at 25°C. What can be concluded about the reaction?
Answer: B
When E°cell is negative, ΔG° = -nFE°cell will be positive, indicating a non-spontaneous reaction under standard conditions.
Q.357Easy
Which of the following is NOT an application of electrochemistry?
Answer: C
Rusting of iron is a corrosion process (galvanic corrosion) and not a direct application like electroplating, electrolysis, or electrorefining.
Q.358Easy
The Faraday constant (F) is numerically equal to:
Answer: A
F = N_A × e = 6.022 × 10²³ × 1.602 × 10⁻¹⁹ C = 96485 C/mol, which equals Avogadro's number times the charge of one electron.
Q.359Easy
In the electrochemical series, which metal is most easily oxidized?
Answer: C
Potassium (K) has the most negative standard reduction potential and is most easily oxidized among the given options. The order of ease of oxidation is K > Cu > Ag > Au.
Q.360Medium
The molar conductivity of a solution increases with dilution because:
Answer: B
As dilution increases, ions are farther apart, reducing interionic attractions and increasing ionic mobility, thus increasing molar conductivity.
