The standard reduction potential for Zn²⁺/Zn is -0.76 V and for Cu²⁺/Cu is +0.34 V. For the cell Zn-Cu, E°cell is:
Answer: B
E°cell = E°cathode - E°anode = (+0.34) - (-0.76) = +1.10 V. Cu²⁺ is reduced (cathode), Zn is oxidized (anode).
Q.62Hard
An electrochemical cell requires 193,700 C of charge to deposit 19.6 g of a metal X. The valency of metal X is:
Answer: B
Charge = 193,700 C; moles of electrons = 193,96700,500 = 2. If 19.6 g = ? mol; then valency n = (moles of e⁻)/(moles of metal). Assuming atomic mass from calculation gives valency = 3 (like Al).
Q.63Hard
At 25°C, for a cell reaction: Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s), if [Zn²⁺] = 1 M and [Ag⁺] = 0.1 M, and E°cell = 1.56 V, the Ecell is approximately:
The conductivity of a solution containing 0.1 M NaCl is 1.29 S·m⁻¹. The molar conductivity is:
Answer: A
Λm = κ/C, where κ = 1.29 S·m⁻¹ = 0.0129 S·cm⁻¹ and C = 0.1 M. Λm = 0.00129.1 = 0.129 S·cm²·mol⁻¹ = 12.9 S·cm²·mol⁻¹.
Q.65Hard
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.
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Q.66Hard
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.67Hard
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.68Hard
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.69Hard
Two cells with the same emf but different internal resistances are connected. Which cell will deliver more current in an external circuit?
Answer: B
Current I = E/(R + r). For the same emf E, lower internal resistance r means higher current. The cell with lower internal resistance delivers more current.
Q.70Hard
A galvanic cell has E°cell = +1.2 V at 25°C with n = 2. At what concentration ratio [Zn²⁺]/[Ag⁺] will the cell potential equal zero?
Answer: B
At E = 0: 0 = 1.2 - (0.2059) log([Zn²⁺]/[Ag⁺]). Solving: log([Zn²⁺]/[Ag⁺]) = (1.2 × 2)/0.059 ≈ 40.68. So [Zn²⁺]/[Ag⁺] ≈ 10⁴⁰.
Q.71Hard
In electrochemistry, overpotential is important because:
Answer: B
Overpotential (η) is the excess potential needed to overcome kinetic barriers and drive the reaction at significant rates. It depends on current density and the nature of the electrode.
Q.72Hard
According to the latest JEE chemistry pattern (2024-25), the conductivity of a strong electrolyte solution depends primarily on:
Answer: B
Conductivity (κ) depends on multiple factors: concentration of ions, mobility of ions (which depends on electrolyte nature, temperature, and solvent nature). These are the primary factors affecting conductivity measurements.
Q.73Hard
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:
For a reaction with mechanism: A ⇌ B (fast equilibrium), B + C → D (slow), the rate law is:
Answer: C
From fast equilibrium: K = [B]/[A], so [B] = K[A]. The slow step rate law is rate = k'[B][C] = k'K[A][C] = k[A]^(21)[C] where k combines constants.
Q.75Hard
The rate constant for a reaction increases 4 times when temperature increases from 27°C to 47°C. What is the activation energy? (R = 8.314 J mol⁻¹ K⁻¹)
In a reaction, the rate increases by a factor of 8 when [A] doubles and by a factor of 2 when [B] doubles. What is the overall order of the reaction?
Answer: C
When [A] doubles, rate increases by 8 = 2³, so order w.r.t. A = 3. When [B] doubles, rate increases by 2 = 2¹, so order w.r.t. B = 1. Overall order = 3 + 1 = 4.
Q.77Hard
For the consecutive reaction A → B → C, if the rate constants are k₁ = 0.1 s⁻¹ and k₂ = 0.05 s⁻¹, and k₁ > k₂, which statement is true?
Answer: B
Since k₁ > k₂, A converts to B faster than B converts to C, so B accumulates initially and then decreases as it slowly converts to C.
Q.78Hard
In the Lindemann mechanism for unimolecular reactions, A* represents an activated molecule. The rate-determining step is:
Answer: B
In the Lindemann mechanism: Step 1 (fast equilibrium): A + A ⇌ A* + A, Step 2 (slow): A* → products. The slow step is rate-determining.
Q.79Hard
For a pseudo-first-order reaction where [B]₀ >> [A]₀, the rate law simplifies to first-order even though the actual order is higher. This is because:
Answer: A
When [B]₀ >> [A]₀, the concentration of B doesn't change significantly during the reaction, so it can be incorporated into the rate constant, making the reaction appear first-order in A only.
Q.80Hard
The rate constant for a reaction at 298 K is 2 × 10⁻⁵ s⁻¹ with Ea = 80 kJ/mol. What is the frequency factor (A) if rate = Ae^(-Ea/RT)?
