CBSE EXAMINATION PAPER-2022

Answer: The cell reaction is: Ni(s) + 2Ag⁺(aq) → Ni²⁺(aq) + 2Ag(s). Given data: Standard cell potential, E°cell = 1.05 V; Faraday's constant, F = 96,500 C mol^-1; Number of electrons transferred, n = 2 (because Ni goes from 0 to +2 and 2Ag⁺ gain 2 electrons).
(a) Calculate ΔrGº:
Use the formula ΔrGº = - n F E°cell
ΔrGº = - 2 * 96,500 * 1.05 = -202,650 J mol^-1 = -202.65 kJ mol^-1
(b) Calculate log K_e:
Use the relation ΔrGº = - R T ln K_e, where R = 8.314 J mol^-1 K^-1, T = 298 K.
ln K_e = - ΔrGº / (R T) = 202,650 / (8.314 * 298) = 202,650 / 2477.572 ≈ 81.74
Therefore, log K_e = ln K_e / 2.303 = 81.74 / 2.303 ≈ 35.5
Final answers:
ΔrGº = -202.65 kJ mol^-1, log K_e = 35.5.
This shows the reaction is spontaneous and favors product formation at standard conditions.

Answer: To calculate the e.m.f. of the given cell, we can use the Nernst equation, which is: E = E° - (RT/nF) ln Q. Here, E° = 0.44 V and T = 298 K. The standard conditions give us R = 8.314 J/(mol·K) and F = 96500 C/mol. For the cell reaction: Fe(s) → Fe²⁺(0.001 M) + 2e⁻, n = 2. First, calculate Q = [Fe²⁺]/[H⁺]² = 0.001/(0.01)² = 0.01. Substituting E°, Q, R, T, and n into the Nernst equation: E = 0.44 - (8.314 × 298)/(2 × 96500) ln(0.01). Calculating: E = 0.44 - 0.00412 × 4.605 = 0.44 - 0.01896 ≈ 0.421 V. Thus, the e.m.f. of the cell at 298 K is approximately 0.421 V.

Answer: Transition metals are defined by IUPAC as metallic elements with an incomplete d subshell in either their neutral state or in their ions. These metals exhibit multiple oxidation states, paramagnetism, and the ability to form colored ions. Examples of transition metals include iron, copper, and nickel. However, zinc (Zn), cadmium (Cd), and mercury (Hg) possess a d10 electron configuration in their ground and common oxidation states, meaning they have fully filled d orbitals. Thus, they do not exhibit the variability in oxidation states typically associated with transition metals. Instead, their oxidation states are more stable and generally lower than that of transition metals. Moreover, the oxidation state variability of transition metals differs from that of p-block elements where oxidation states typically change by increments of two, due to the differing electron configurations; transition metals change by one unit owing to their partially filled d orbitals allowing for the loss of electrons from both s and d orbitals.

Answer: In the reactions involving the generation of free radicals, the primary product formed typically stems from the propagation steps that occur after the initial chain initiation. For instance, when chlorine reacts with methyl radicals under UV light, the major product formed is chloromethane (CH3Cl) as free radicals interact with one another to create stable compounds. The chain process involves methyl and chlorine radicals engaging through multiple steps, eventually leading to the formation of various products. However, chloromethane is the principal product due to its stability and lower energy compared to other possible products. This reaction exemplifies the process of halogenation where the free radical mechanism plays a crucial role in determining the major product. It is also important to note that besides the main product, minor side reactions can yield additional products, but these are often negligible in organic reaction results due to their lower yields. The presence of free radicals and subsequent termination steps contribute to the overall outcome of the reactions as radicals can further react or combine to produce other species.