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.
A current of 5 A is passed through an electrolytic cell for 1930 seconds. Calculate the number of moles of electrons transferred. (Faraday constant = 96500 C/mol)
Answer: A
Charge = I × t = 5 × 1930 = 9650 C. Moles of e⁻ = 965009650 = 0.1 mol.
Q.82Easy
In a Daniel cell, the mass of zinc electrode decreases and copper electrode increases. This indicates:
The standard reduction potential of H⁺/H₂ electrode is taken as 0.00 V because:
Answer: B
SHE (Standard Hydrogen Electrode) is the arbitrary reference against which all other reduction potentials are measured.
Q.85Medium
During the electrolysis of aqueous CuSO₄ solution with copper electrodes, which reaction occurs at the cathode?
Answer: B
At cathode with copper electrodes in CuSO₄: Cu²⁺ ions are preferentially reduced as their reduction potential (+0.34 V) is higher than H⁺ (-0.83 V).
Q.86Medium
The Gibbs free energy change for an electrochemical cell reaction is related to cell potential by:
Answer: B
The standard free energy change ΔG° = -nFE°cell, where n is moles of electrons, F is Faraday constant, and E° is standard cell potential.
Q.87Medium
In the electrorefining of copper, impure copper acts as:
Answer: B
In electrorefining, impure copper acts as anode and undergoes oxidation. Pure copper deposits at cathode. More reactive impurities go into solution.
Q.88Easy
What is the effect of temperature on the conductivity of ionic solutions?
Answer: B
Conductivity increases with temperature because ionic mobility increases due to decreased viscosity of the medium.
Q.89Easy
The electrochemical series is based on which property?
Answer: B
The electrochemical series arranges elements in order of their standard reduction potentials, with more positive values indicating stronger oxidizing agents.
Q.90Medium
In the electrolysis of dilute H₂SO₄ with inert electrodes, if 2 moles of electrons flow, what volume of gases (in liters at STP) will be produced?
Answer: A
At cathode: 2H⁺ + 2e⁻ → H₂ (1 mol H₂). At anode: 2H₂O → O₂ + 4H⁺ + 4e⁻ (0.5 mol O₂). With 2 mol e⁻: 1 mol H₂ (11.2 L) and 0.5 mol O₂ (5.6 L).
Q.91Medium
The cell potential of a galvanic cell decreases during operation because:
Answer: B
As the reaction proceeds, product concentrations increase while reactant concentrations decrease, reducing the driving force according to Nernst equation: E = E° - (0.059/n)log(Q).
Q.92Easy
Which factor does NOT significantly affect the conductance of an electrolytic solution?
Answer: C
Color of a solution is a physical property unrelated to ionic conductance. Conductance depends on nature of solute, concentration, temperature, and solvent.
Q.93Medium
For the reaction: Fe³⁺ + e⁻ → Fe²⁺ (E° = +0.77 V) and Cl₂ + 2e⁻ → 2Cl⁻ (E° = +1.36 V), which is the strongest oxidizing agent?
Answer: C
The species with highest reduction potential (+1.36 V) is Cl₂, making it the strongest oxidizing agent. Higher E° values indicate greater tendency to accept electrons.
Q.94Hard
The molar conductivity of a strong electrolyte at infinite dilution (Λ°m) can be calculated using Kohlrausch's law. For NaCl, if Λ°m(HCl) = 426, Λ°m(NaOH) = 248, and Λ°m(KCl) = 150, then Λ°m(NaCl) is:
During the electrolysis of molten NaCl using inert electrodes, if 2.3 g of Na is deposited at the cathode, what volume of Cl₂ gas (at STP) will be released at the anode?
Answer: B
At cathode: Na⁺ + e⁻ → Na; moles of Na = 2.233 = 0.1 mol. At anode: 2Cl⁻ → Cl₂ + 2e⁻; for 0.1 mol Na, electrons = 0.1 mol, so Cl₂ moles = 0.21 = 0.05 mol. Volume at STP = 0.05 × 22.4 = 1.12 L.
Q.96Medium
A galvanic cell is constructed using Zn|Zn²⁺ and Cu|Cu²⁺ half-cells. If the concentration of Zn²⁺ is increased from 1 M to 10 M at 25°C, how does this affect the cell potential? (E°cell = 1.1 V)
Answer: A
Using Nernst equation: Ecell = E°cell - (0.059/n)log(Q). Increasing [Zn²⁺] increases Q, making the log term positive, which decreases Ecell. ΔE = -(0.2059)log(10) = -0.0295 ≈ -0.0296 V.
