Physical Chemistry MCQs

Correct: A — Chemical to electrical energy. In a galvanic (voltaic) cell, a spontaneous redox reaction produces electrical energy. Example: Daniell cell (Zn–Cu). Oxidation at anode (negative terminal in galvanic cell): Zn → Zn²⁺ + 2e⁻. Reduction at cathode (positive terminal): Cu²⁺ + 2e⁻ → Cu. Electrons flow from anode to cathode through external circuit. Contrast: electrolytic cell uses electrical energy to drive non-spontaneous reactions.

Correct: B — Anode, negative terminal. Mnemonic: OIL RIG at anode (Oxidation Is Loss), reduction at cathode. In a galvanic cell: anode is negative (electrons leave it), cathode is positive (electrons arrive). In electrolytic cell: anode is positive (connected to +ve battery terminal), cathode is negative. Oxidation always at anode and reduction always at cathode — this rule applies to both types of cells.

Correct: C — 1 atm H₂, 1 M H⁺, Pt electrode. SHE: Pt electrode immersed in 1 M HCl (a_H⁺ = 1), H₂ gas bubbled at 1 atm. By convention, E°(H⁺/H₂) = 0.00 V. All other standard electrode potentials are measured relative to SHE. E°_cell = E°_cathode − E°_anode. The Pt electrode is used because it is inert (does not react) but conducts electricity.

Correct: D — E°. When [Zn²⁺] = [Cu²⁺], Q = [Zn²⁺]/[Cu²⁺] = 1. log(1) = 0. So E = E° − (0.0592/2) × 0 = E°. The Daniell cell has E° = E°_Cu²⁺/Cu − E°_Zn²⁺/Zn = +0.34 − (−0.76) = +1.10 V. The Nernst equation shows E = E° only when all concentrations are at standard state (1 M) or when they are equal and cancel.

Correct: B — ΔG° = −nFE°_cell. For a spontaneous reaction, ΔG° < 0 and E°_cell > 0 (they have opposite signs). F = Faraday constant = 96,500 C/mol. n = number of electrons transferred. This relationship links thermodynamics (ΔG°) to electrochemistry (E°). Also: ΔG° = −RT ln K_eq = −nFE°, so E° = (RT/nF) ln K_eq. A cell with E°_cell > 0 has K > 1 (products favoured).

Correct: A — w = MIt/nF. Faraday's first law: mass deposited is directly proportional to the quantity of electricity passed (Q = I × t). w = (M/nF) × Q = (M × I × t)/(n × F). Where M = molar mass, n = number of electrons in electrode reaction, F = 96500 C/mol. Example: depositing Cu (M=64, n=2): w = (64 × I × t)/(2 × 96500). 96500 C deposits 1 equivalent (M/n grams) of any substance.

Correct: C — Sum of individual ion conductances. Kohlrausch's law of independent migration of ions: Λ°_m = ν₊λ°₊ + ν₋λ°₋. At infinite dilution, each ion makes its own independent contribution to conductance regardless of what other ions are present. Used to calculate Λ°_m for weak electrolytes (like acetic acid) by combining data from strong electrolytes: Λ°_m(CH₃COOH) = Λ°_m(CH₃COONa) + Λ°_m(HCl) − Λ°_m(NaCl).

Correct: D — Pb → PbSO₄ at anode. Lead storage battery reactions during discharge: Anode (−): Pb + SO₄²⁻ → PbSO₄ + 2e⁻ (oxidation). Cathode (+): PbO₂ + 4H⁺ + SO₄²⁻ + 2e⁻ → PbSO₄ + 2H₂O (reduction). Overall: Pb + PbO₂ + 2H₂SO₄ → 2PbSO₄ + 2H₂O. Both electrodes become coated with PbSO₄ during discharge; H₂SO₄ is consumed (density decreases — used to check charge level).

Correct: B — Increases with dilution for both. Molar conductance (Λ_m = κ × 1000/M) increases with dilution: for strong electrolytes — interionic attractions decrease as concentration drops (Λ_m = Λ°_m − A√C, Debye-Hückel-Onsager equation); for weak electrolytes — degree of dissociation (α) increases dramatically with dilution, releasing more ions. Both approach Λ°_m at infinite dilution, but weak electrolytes rise much more steeply.

Correct: C — Iron is the anode. In rusting, iron forms a tiny galvanic cell with an impurity (e.g., carbon). Fe (anode): Fe → Fe²⁺ + 2e⁻ (oxidation). At cathode (impurity): O₂ + 4H⁺ + 4e⁻ → 2H₂O (in acidic medium) or O₂ + 2H₂O + 4e⁻ → 4OH⁻ (neutral/basic). Overall: 4Fe + 3O₂ + xH₂O → 2Fe₂O₃·xH₂O (rust). Requires both O₂ and water. Prevention: galvanisation (Zn coating), painting, cathodic protection.

Correct: A — Sacrificial anode. E°(Zn²⁺/Zn) = −0.76 V, E°(Fe²⁺/Fe) = −0.44 V. Zinc has a more negative (lower) reduction potential → stronger tendency to oxidise (corrode). In a galvanic cell, Zn acts as anode and Fe as cathode. Zn corrodes preferentially, protecting Fe even if the coating is scratched. This is called cathodic protection (Fe is cathodically protected). Used in galvanised iron sheets (GI pipes, roofing sheets).

Correct: D — Cathode. In Down's cell (molten NaCl): Cathode (−): Na⁺ + e⁻ → Na (reduction, liquid Na floats to top). Anode (+): 2Cl⁻ → Cl₂ + 2e⁻ (oxidation, chlorine gas). Overall: 2NaCl → 2Na + Cl₂. This is how metallic sodium is commercially produced. In the chlor-alkali process (electrolysis of brine/NaCl solution): cathode gives NaOH + H₂, anode gives Cl₂.

Correct: B — Increased ionic mobility. Unlike metals (where resistance increases with temperature due to lattice vibrations), electrolyte solutions conduct better at higher temperatures. As temperature rises, viscosity of the solvent decreases and ions can move faster (higher mobility). This increases conductance. This is the opposite of metallic conduction and is a key distinguishing feature of ionic/electrolytic conduction.

Correct: C — 2H₂ + O₂ → 2H₂O. In a hydrogen fuel cell: Cathode: O₂ + 4H⁺ + 4e⁻ → 2H₂O. Anode: 2H₂ → 4H⁺ + 4e⁻. Overall: 2H₂ + O₂ → 2H₂O + electrical energy. The only product is water — zero carbon emissions. High efficiency (~70% vs ~40% for combustion engines). Used in Apollo space missions, submarines, electric vehicles. Fuel is supplied continuously (unlike batteries which store chemicals internally).

Correct: A — 3 Faradays. To deposit 1 mole of Al, 3 moles of electrons are needed (n = 3). 1 Faraday = charge of 1 mole of electrons = 96,500 C. So 3 F = 3 × 96,500 = 289,500 C. For Cu (n=2): 2 F per mole. For Ag (n=1): 1 F per mole. This follows from Faraday's second law: different substances are deposited in proportion to their equivalent masses (M/n) by the same charge.

Correct: B — κ = G × G*. Cell constant G* = l/A (length between electrodes / area). Specific conductance κ = G × (l/A) = G × G*. Units: G in siemens (S), G* in cm⁻¹, κ in S·cm⁻¹. Molar conductance Λ_m = κ × 1000/M (in S·cm²·mol⁻¹, where M is molarity in mol/L). The cell constant is determined experimentally using a solution of known conductance (e.g., KCl).

Correct: D — E°_cell > 0 and ΔG° < 0. For a spontaneous reaction: ΔG° < 0 (thermodynamics). Since ΔG° = −nFE°_cell, if ΔG° < 0 then E°_cell > 0 (positive EMF). The equilibrium constant K > 1 for a spontaneous reaction. At equilibrium, ΔG = 0 and EMF = 0. These three conditions are equivalent: ΔG° < 0 ⟺ E°_cell > 0 ⟺ K > 1.

Correct: C — Cathode. In electroplating: cathode = object to be plated (reduction deposits the metal). Anode = pure copper strip (which slowly dissolves: Cu → Cu²⁺ + 2e⁻). Electrolyte = CuSO₄ solution. As copper deposits on cathode, the copper anode dissolves, maintaining Cu²⁺ concentration. Current flows: battery positive → anode → (through solution) → cathode → battery negative.

Correct: B — Zero. E = E° − (0.0592/n)log Q. At equilibrium, Q = K_eq, and E = 0. So 0 = E° − (0.0592/n)log K_eq, giving E° = (0.0592/n)log K_eq. This is why a discharged battery has EMF = 0 — the cell has reached equilibrium. The relationship E° = (0.0592/n)log K allows calculation of equilibrium constants from standard electrode potentials.

Correct: A — Reverse reaction restores original electrodes. During charging (electrolytic cell driven by external power): at anode (+), PbSO₄ → PbO₂ (oxidation); at cathode (−), PbSO₄ → Pb (reduction); H₂O is consumed and H₂SO₄ is regenerated. The density of H₂SO₄ increases during charging (back to ~1.28 g/mL when fully charged). A fully charged battery has well-separated Pb and PbO₂ electrodes and concentrated H₂SO₄.

Correct: C — mol L⁻¹ s⁻¹. For rate = k[A]⁰ = k (zero order), rate has units mol L⁻¹ s⁻¹, so k has units mol L⁻¹ s⁻¹. General formula for units of k: (mol L⁻¹)^(1−n) s⁻¹, where n = order. Zero order (n=0): mol L⁻¹ s⁻¹. First order (n=1): s⁻¹. Second order (n=2): L mol⁻¹ s⁻¹. Third order (n=3): L² mol⁻² s⁻¹.

Correct: B — t½ = 0.693/k. For first order: [A] = [A]₀e^(−kt). At t = t½, [A] = [A]₀/2. Solving: t½ = ln2/k = 0.693/k. The half-life is constant and independent of initial concentration — a unique property of first order reactions. This is why radioactive decay (always first order) has a constant half-life. Compare: zero order t½ = [A]₀/2k (depends on [A]₀).

Correct: D — Key distinction. Molecularity: number of molecules/ions that collide simultaneously in an elementary reaction step. Always a positive integer (1=unimolecular, 2=bimolecular, 3=termolecular max — higher are statistically improbable). It is a theoretical/mechanistic concept. Order: sum of exponents in the experimental rate law (rate = k[A]^m[B]^n). Determined from experiment. Can be 0, fractional, or even negative. For complex multi-step reactions, order ≠ molecularity.

Correct: A — Excess water (pseudo constant). True rate: rate = k[ester][H₂O]. Since [H₂O] >> [ester] and barely changes, it is absorbed into a pseudo rate constant: k' = k[H₂O]. Rate = k'[ester]. This appears first order — a pseudo first order reaction. Other examples: inversion of cane sugar (sucrose hydrolysis), acid hydrolysis of methyl acetate. The prefix "pseudo" means "false" — it looks like first order but is actually second order.

Correct: C — Slope = −Ea/R. ln k = ln A − Ea/(RT). Comparing with y = c + mx: y = ln k, x = 1/T, slope m = −Ea/R, intercept = ln A. The slope is negative (as 1/T increases, ln k decreases → k decreases). Ea = −R × slope. This is the standard method to determine activation energy experimentally from a series of rate constant measurements at different temperatures.

Correct: B — Lowers Ea. A catalyst provides an alternative mechanism with lower activation energy. More molecules can cross the energy barrier → faster reaction. Key points: (1) catalyst is NOT consumed — regenerated in each cycle; (2) does not change ΔH, ΔG, or equilibrium constant; (3) increases BOTH forward and reverse rates equally; (4) does not change the equilibrium position. The transition state is different (lower energy) but the reactants and products are unchanged.

Correct: D — 3rd order. Overall order = sum of all powers of concentration in rate law = 2 (for NO) + 1 (for O₂) = 3. Order w.r.t. NO = 2 (second order in NO); order w.r.t. O₂ = 1 (first order in O₂). Note: the stoichiometric coefficients in the balanced equation (2 for NO and 1 for O₂) do NOT necessarily equal the orders in the rate law. Rate law is always determined experimentally.

Correct: A — k = 0.693/t½. From t½ = 0.693/k → k = 0.693/t½ = 0.693/5730 ≈ 1.21 × 10⁻⁴ yr⁻¹. This k is used in carbon dating: age of sample = (2.303/k) × log([N]₀/[N]). Since living organisms have constant C-14 (equilibrium with atmosphere), and dead organisms decay at this known rate, the ratio of C-14 to stable C-12 gives the age. Used for samples up to ~50,000 years old.

Correct: C — Remains the same. t½ = 0.693/k for first order. The half-life is independent of [A]₀ — it depends only on k, which is a constant at given temperature. This is the distinguishing feature of first order kinetics: equal time intervals lead to equal fractional decreases. After 1 half-life: 50% remains. After 2 half-lives: 25%. After n half-lives: (1/2)ⁿ fraction remains. Zero order t½ = [A]₀/2k does depend on initial concentration.

Correct: B — Exponential increase in fraction with E ≥ Ea. In the Boltzmann distribution, the fraction of molecules with energy ≥ Ea = e^(−Ea/RT). A 10°C rise causes this fraction to roughly double for typical reactions with Ea ≈ 50–80 kJ/mol. Collision frequency increases only ~2% per 10°C rise (proportional to √T), which alone cannot explain the large rate increase. The exponential dependence in the Arrhenius equation accounts for this temperature sensitivity.

Correct: D — Slowest step. In a multi-step mechanism, the overall reaction rate cannot be faster than its slowest step. Analogy: a production line moves at the pace of its slowest worker. The rate law of the overall reaction corresponds to the molecularity of this slowest (rate-determining) step. If the slow step involves A and B, then rate = k[A][B] — the rate law can be deduced from the mechanism. Intermediates appear and disappear but do NOT appear in the overall rate law.

Correct: A — Threshold = Ea + average KE. Threshold energy is the minimum energy that colliding molecules must possess for reaction to occur. Activation energy (Ea) = Threshold energy − Average kinetic energy of reactants. Since reactant molecules already have some average kinetic energy (½mv²), they need only to acquire the additional energy (Ea) to reach the threshold. A catalyst lowers Ea (and thereby lowers the threshold energy required for the reaction to proceed).

Correct: C — Logarithmic decay. First order: −d[A]/dt = k[A] → integrating: ln[A]_t = ln[A]₀ − kt. In log form: log[A]_t = log[A]₀ − (k/2.303)t. Plot of log[A] vs t gives a straight line with slope = −k/2.303. Compare: Zero order: [A]_t = [A]₀ − kt (linear). Second order: 1/[A]_t = 1/[A]₀ + kt. The shape of the concentration–time graph identifies the order.

Correct: B — Energy AND orientation requirements. Two conditions must be met for an effective collision: (1) Energy condition: kinetic energy of colliding molecules ≥ threshold energy (Ea). (2) Orientation condition (steric factor p): molecules must approach each other in the correct geometric orientation. The rate = p × Z × e^(−Ea/RT), where Z = collision frequency and p = steric/probability factor (0 < p ≤ 1). For most reactions, only a tiny fraction (10⁻⁶ to 10⁻⁹) of all collisions are effective.

Correct: D — Experimentally determined; any value possible. Order of reaction can be: zero (e.g., thermal decomposition of HI on gold surface), fractional (e.g., decomposition of acetaldehyde is 1.5 order), negative (increasing [B] decreases rate in some chain reactions), positive integer. It is NOT necessarily equal to stoichiometric coefficients (except for elementary reactions). It is always determined by experiment, not predicted from the balanced equation alone.

Correct: C — 4 times. Rate = k[A]²[B]. If [A] → 2[A] and [B] constant: New rate = k(2[A])²[B] = k × 4[A]²[B] = 4 × original rate. Since rate is proportional to [A]², doubling [A] increases rate by 2² = 4. If [B] is also doubled: rate = k(2[A])²(2[B]) = 4 × 2 = 8 times. This type of calculation is essential for CBSE numericals on rate law.

Correct: A — Frequency factor. In k = Ae^(−Ea/RT), A (also called pre-exponential factor or frequency factor) represents the frequency of collisions with the correct orientation. It has the same units as k. Physically, A = p × Z, where p = steric factor (probability of correct orientation) and Z = total collision frequency. A is approximately independent of temperature (or weakly dependent via √T). It represents the maximum possible rate constant at infinite temperature (when all collisions are effective).

Correct: B — Straight line with negative slope. For zero order: rate = k (constant). d[A]/dt = −k → integrating: [A] = [A]₀ − kt. This is a linear equation: y = [A], x = t, slope = −k, y-intercept = [A]₀. The concentration decreases linearly with time until [A] = 0. Half-life = [A]₀/2k. Examples of zero order reactions: thermal decomposition of HI on gold surface, photochemical reactions, enzyme-catalysed reactions at saturation.

Correct: C — Two-temperature Arrhenius formula. From ln(k₂/k₁) = (Ea/R)(1/T₁ − 1/T₂), rearranging to log: log(k₂/k₁) = (Ea/2.303R) × (1/T₁ − 1/T₂). This is the standard NCERT formula for numerical problems. Note: T must be in Kelvin. If k increases when T increases (T₂ > T₁), then k₂ > k₁ and Ea > 0 (always positive). Units of Ea: J/mol or kJ/mol.

Correct: D — More collisions. Higher concentration means more reactant molecules per unit volume. By collision theory, collision frequency (Z) is proportional to concentration. More total collisions → more collisions with energy ≥ Ea → higher rate. This is captured in the rate law: rate = k[A]^m. For zero order reactions (rate = k), concentration has no effect — this occurs when molecules must first be adsorbed on a surface (surface becomes saturated, limiting the rate).