JEE Chemistry - MCQ Practice Questions

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.

In electrolysis, oxidation occurs at the ANODE, not the cathode. The cathode is where reduction occurs. At the anode, electrons are removed from species.

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).

A negative E° (E°cell < 0) indicates a non-spontaneous reaction under standard conditions. ΔG° = -nFE°, so negative E° gives positive ΔG°.

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).

Fluorine has the highest reduction potential among the halogens, approximately +2.87 V. Reduction potentials decrease down the halogen group: F₂ > Cl₂ > Br₂ > I₂.

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.

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.

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.

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.

Using the Nernst equation, E = E° + (0.$\frac{0592}{2}$) 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).

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⁺.

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.

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.

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.

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.

Electrical conductivity is measured in Siemens per meter (S·m⁻¹). Ω·m is the unit of resistivity, which is the inverse of conductivity.

In a galvanic cell, the anode is the negative electrode where oxidation occurs. The cathode is positive where reduction occurs.

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.

The Nernst equation: E = E° - (RT/nF)ln(Q) calculates the cell potential when concentrations are not at standard state (1 M).