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.
The Nernst equation relates cell potential to concentration. At 25°C, which form is correct?
Answer: C
The Nernst equation at 25°C is E = E° + (0.0592/n) log Q, where Q is the reaction quotient and n is the number of electrons transferred.
Q.302Easy
Which of the following statements about electrolysis is INCORRECT?
Answer: B
In electrolysis, oxidation occurs at the ANODE, not the cathode. The cathode is where reduction occurs. At the anode, electrons are removed from species.
Q.303Medium
A metal X is more easily oxidized than metal Y. If X and Y form a galvanic cell, which metal acts as the anode?
Answer: A
The metal that is more easily oxidized (more reactive) acts as the anode. Metal X, being more reactive, loses electrons and acts as the anode (negative electrode).
Q.304Medium
The standard cell potential for a reaction is -0.5 V. What can be concluded about this reaction?
Answer: D
A negative E° (E°cell < 0) indicates a non-spontaneous reaction under standard conditions. ΔG° = -nFE°, so negative E° gives positive ΔG°.
Q.305Medium
In the Daniel cell, the half-cell potentials are: Cu²⁺ + 2e⁻ → Cu, E° = +0.34 V and Zn²⁺ + 2e⁻ → Zn, E° = -0.76 V. Calculate E°cell:
Answer: C
E°cell = E°cathode - E°anode = 0.34 - (-0.76) = 0.34 + 0.76 = 1.10 V. Cu²⁺ is reduced (cathode) and Zn is oxidized (anode).
Q.306Medium
Which of the following has the highest reduction potential?
Answer: A
Fluorine has the highest reduction potential among the halogens, approximately +2.87 V. Reduction potentials decrease down the halogen group: F₂ > Cl₂ > Br₂ > I₂.
Q.307Medium
What is the relationship between ΔG° and E°cell?
Answer: B
The Gibbs free energy change is related to cell potential by ΔG° = -nFE°cell, where n is the number of electrons transferred, F is Faraday's constant, and E° is the standard cell potential.
Q.308Medium
In electroplating of iron with copper, the cathode is made of:
Answer: B
In electroplating iron with copper, iron (the object to be plated) acts as the cathode where Cu²⁺ ions are reduced and deposit as copper metal. The anode is made of copper.
Q.309Medium
The conductivity of a solution decreases with dilution because:
Answer: C
Upon dilution, the number of ions per unit volume decreases (concentration effect) and ionic mobility also increases slightly due to reduced ion-ion interactions, but overall conductivity decreases because the decrease in ion concentration dominates.
Q.310Hard
A cell has E°cell = 0. This means:
Answer: D
When E°cell = 0, ΔG° = -nFE°cell = 0, indicating the system is at equilibrium. Since ΔG° = -RT ln K, when ΔG° = 0, ln K = 0, so K = 1.
Q.311Hard
For the reaction: Zn + Cu²⁺ → Zn²⁺ + Cu, if the concentration of Cu²⁺ is increased at constant temperature, the cell potential will:
Answer: B
Using the Nernst equation, E = E° + (0.20592) log([Zn²⁺]/[Cu²⁺]). Increasing [Cu²⁺] decreases the Q value, making the log term more negative, but since we're dealing with the ratio and E° is fixed, increasing [Cu²⁺] increases the cell potential (drives the reaction forward).
Q.312Hard
In the electrolysis of aqueous NaCl solution with inert electrodes, the products are:
Answer: B
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⁺.
Q.313Hard
The equivalent conductivity of a solution decreases with dilution. Which statement best explains this anomaly for strong electrolytes?
Answer: C
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.
Q.314Hard
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:
Answer: C
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.
Q.315Hard
During the electrolysis of CuSO₄ solution with copper electrodes, which of the following occurs?
Answer: B
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.
Q.316Hard
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)
Answer: B
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.
Q.317Easy
What is the SI unit of electrical conductivity?
Answer: A
Electrical conductivity is measured in Siemens per meter (S·m⁻¹). Ω·m is the unit of resistivity, which is the inverse of conductivity.
Q.318Easy
In a galvanic cell, which electrode acts as the negative terminal?
Answer: B
In a galvanic cell, the anode is the negative electrode where oxidation occurs. The cathode is positive where reduction occurs.
Q.319Easy
What does Faraday's first law of electrolysis state?
Answer: A
Faraday's first law states that the mass of substance deposited/dissolved during electrolysis is directly proportional to the quantity of electricity (charge) passed through the electrolyte.
Q.320Easy
The Nernst equation is used to calculate:
Answer: A
The Nernst equation: E = E° - (RT/nF)ln(Q) calculates the cell potential when concentrations are not at standard state (1 M).
