CBSE Class 12 Chemistry Electrochemistry Case Study Questions Set 01

Answer: (a) Infinite dilution refers to a hypothetical situation when the ions are infinitely far apart.

Explanation:
1. Infinite dilution is a hypothetical limiting situation (concentration → 0) at which the ions are assumed to be infinitely far apart and exert no interionic attraction.

Teacher's Note:
1. For weak electrolytes, this limit is only ever reached by extrapolation (Kohlrausch's law), never by direct measurement – that's the whole reason the law is needed.

Answer: (d) HCl

Explanation:
1. HCl is a strong acid and dissociates almost completely in water, making it a strong electrolyte; HNO₂, HCN and NH₃ are all weak electrolytes.

Teacher's Note:
1. The common strong acids/bases/salts (HCl, HNO3, H2SO4, NaOH, KCl, etc.) are worth memorising as a quick reference set for these identification questions.

Answer: (d) H₂SO₃

Explanation:
1. Weak electrolytes do not dissociate completely in aqueous solution. H₂SO₃ (sulphurous acid) is a weak acid, unlike the soluble salts and strong base listed in the other options.

Teacher's Note:
1. Weak electrolytes set up a dissociation equilibrium rather than dissociating fully, which is exactly why they need a separate treatment (Ostwald's dilution law) from strong electrolytes.

Answer: (a) 25.96 S cm² mol⁻¹

Explanation:
1. By Kohlrausch's law: Λ°(MgI₂) = Λ°[(CH₃COO)₂Mg] + 2Λ°(NaI) − 2Λ°(CH₃COONa)
2. = 18.78 + 2(12.69) − 2(9.10) = 25.96 S cm² mol⁻¹

Teacher's Note:
1. Build the target salt from pieces you have data for: here, swapping acetate for iodide across two sodium salts and one magnesium salt isolates Mg²⁺ and I⁻ together.

Answer: (a) Li⁺ < Na⁺ < K⁺ < Rb⁺

Explanation:
1. Ionic conductivity increases down the alkali metal group because the smaller, more heavily hydrated Li⁺ ion moves more slowly than the larger, less hydrated Rb⁺ ion. So Li⁺ < Na⁺ < K⁺ < Rb⁺.

Teacher's Note:
1. Counterintuitively, the smallest bare ion (Li+) is the slowest in solution because it carries the largest hydration shell, which effectively increases its size while moving.

Answer: (b) 4 × 10⁻¹⁵

Explanation:
1. 0.059 = \( \frac{0.059}{2} \) log \( \frac{0.001}{[M^{2+}]} \)
2. log \( \frac{0.001}{[M^{2+}]} \) = 2, so [M²⁺] = 10⁻⁵ M
3. Let the solubility of the salt be s mol/litre: MX₂ → M²⁺ + 2X⁻, so [M²⁺] = s, [X⁻] = 2s
4. K_sp = 4s³ = 4 × (10⁻⁵)³ = 4 × 10⁻¹⁵

Teacher's Note:
1. The saturated electrode of a concentration cell is a neat way to work backwards from EMF to a solubility product – a pattern worth recognising on sight.

Answer: (d) −11.4

Explanation:
1. ΔG = −nFE = −2 × 96500 × 0.059 = −11387 J mol⁻¹ = −11.4 kJ mol⁻¹

Teacher's Note:
1. n = 2 here because M²⁺/M exchanges two electrons, matching the MX2 stoichiometry given in the passage.

Answer: (a) 7.6 × 10¹²

Explanation:
1. E°_cell = E°_{Ce⁴⁺/Ce³⁺} − E°_{Fe³⁺/Fe²⁺} = 1.44 − 0.68 = 0.76 V
2. E°_cell = \( \frac{0.059}{1} \) log K_c
3. log₁₀ K_c = \( \frac{0.76}{0.059} \) = 12.88
4. K_c = 7.6 × 10¹²

Teacher's Note:
1. For a one-electron reaction, log K = E°_cell/0.059, so a cell potential of 0.76 V already gives K of the order of 10¹².
2. A large K means the reaction goes almost to completion, which is why Ce⁴⁺ is used to titrate Fe²⁺.

Answer: (b) 2.287 × 10⁻¹² mol³ L⁻³

Explanation:
1. E_cell = E°_cell − \( \frac{0.059}{1} \) log \( \frac{[Ag^{+}]}{[Ag^{+}]_{satd.\,Ag_2CrO_4}} \)
2. 0.164 = \( \frac{0.059}{1} \) log \( \frac{0.1}{[Ag^{+}]_{satd.\,Ag_2CrO_4}} \) ⇒ [Ag⁺]_satd. = 1.66 × 10⁻⁴ M
3. [CrO₄²⁻] = \( \frac{1.66\times10^{-4}}{2} \)
4. K_sp(Ag₂CrO₄) = [Ag⁺]²[CrO₄²⁻] = (1.66 × 10⁻⁴)² × (1.66 × 10⁻⁴/2) = 2.287 × 10⁻¹² mol³ L⁻³

Teacher's Note:
1. Same concentration-cell-to-Ksp pattern as the earlier MX2 question, just with the 2:1 stoichiometry of Ag2CrO4 instead of 1:2.

Answer: (a) EMF = E°_cathode − E°_anode

Explanation:
1. By definition, the standard EMF of a cell is EMF = E°_cathode − E°_anode, using reduction potentials for both electrodes.

Teacher's Note:
1. This formula recurs in almost every question in the chapter – it's worth being completely automatic with it.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. log(C₁/C₂) < 0 for spontaneity. ∴ C₁ < C₂.
2. Both statements are correct, and the reason correctly explains the assertion.

Teacher's Note:
1. For a concentration cell, spontaneity simply means ions flow (net) from the more concentrated half-cell toward the more dilute one, which is exactly what C1 < C2 encodes here.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both statements are correct, and the reason correctly explains why the voltmeter reads zero: with no net concentration change, there is no further driving force and so no EMF left.

Teacher's Note:
1. A dead battery is simply a cell that has reached this equilibrium – EMF = 0, even though both electrodes are still physically present.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Both statements are correct, but the direction of current flow does not explain why EMF depends on concentration.

Teacher's Note:
1. Use (a) only when the reason actually explains the assertion. Two true statements that are unrelated give (b).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. Both statements are correct, and the reason gives the Nernst equation that correctly explains why raising [Cu²⁺] raises E_cell.

Teacher's Note:
1. Raising a reactant-side (here, cathode-side) concentration always pushes Ecell up, by Le Chatelier intuition as much as by the equation itself.

Answer: (d) A is false, but R is true.

Explanation:
1. The 0.059/n coefficient in the Nernst equation is valid at 298 K, not at STP (273 K), so the assertion is false.
2. The reason – that STP conditions require 273 K – is itself a correct statement, giving answer (d).

Teacher's Note:
1. A frequently confused pair: STP (0°C / 273 K) is not the same as the 298 K (25°C) at which the standard 0.059 V coefficient is derived.

Answer: (b) E_cell > 0; ΔG < 0

Explanation:
1. M_(s) → M⁺_(aq)(0.05 M) + e⁻ (anode); M⁺_(aq)(1.0 M) + e⁻ → M_(s) (cathode)
2. For the concentration cell, E_cell = − \( \frac{0.059}{1} \) log \( \frac{0.05}{1} \) = − \( \frac{0.059}{1} \) log(5 × 10⁻²) = −0.059[(−2) + log 5] = −0.059(−1.302) = 0.0768 V
3. Since E_cell is positive, ΔG = −nFE_cell is negative.

Teacher's Note:
1. The cell is spontaneous as written because the net process moves M+ from the more concentrated side to the less concentrated side, which is always how a concentration cell discharges.

Answer: (c) 154 mV

Explanation:
1. With [M⁺] = 0.05 M, E₁ = 0.059 log(1/0.05) = 0.0768 V (from the previous question).
2. With [M⁺] = 0.0025 M, E₂ = 0.059 log(1/0.0025); comparing the two, \( \frac{E_1}{E_2} \) = \( \frac{\log(1/0.05)}{\log(1/0.0025)} \) = \( \frac{1.301}{2.602} \) = \( \frac{1}{2} \), so E₂ = 2E₁ ≈ 154 mV.

Teacher's Note:
1. A ten-fold dilution step doesn't simply double the cell potential in general – it's the particular log ratio here that happens to come out close to 2×.

Answer: (c) > 1

Explanation:
1. K = antilog(nE°/0.0591). For a feasible cell, E° is positive, hence K > 1 for a feasible cell reaction.

Teacher's Note:
1. A positive E°cell, a negative ΔG°, and K > 1 are three equivalent ways of saying the same thing: the reaction is thermodynamically favoured as written.

Answer: (b) 0

Explanation:
1. At equilibrium, there is no further net driving force for the reaction, so the EMF of the cell is zero (this is also when ΔG = 0).

Teacher's Note:
1. This is the same idea tested earlier with the voltmeter reading zero for the Zn/Cu cell at equilibrium – here it is restated in general terms.

Answer: (a) concentration of ions in solution

Explanation:
1. Electrode potential, by the Nernst equation, depends on the concentration of the relevant ions in solution – not on the physical position of the electrodes or the overall cell voltage.

Teacher's Note:
1. This is exactly the dependence the Nernst equation quantifies, which is why it's central to so many questions in this chapter.

Answer: (b) 1.0

Explanation:
1. n_NaCl = \( \frac{4\times500}{1000} \) = 2 mol
2. ∴ n_Cl2 = 1 mol (since 2 mol Cl⁻ gives 1 mol Cl₂)

Teacher's Note:
1. With 500 mL (half a litre) of a 4 M solution, start by finding total moles of NaCl present (= 2) before working out how much Cl2 that can give.

Answer: (b) 446 g

Explanation:
1. n_Na deposited = 2 mol
2. ∴ n_Na–Hg formed = 2 mol
3. ∴ Mass of amalgam formed = 2 × 223 = 446 g (using Na + Hg ≈ 223 u per unit of amalgam)

Teacher's Note:
1. An amalgam here is just the sodium deposited combined 1:1 with mercury at the cathode – the mole count of Na from the previous question carries straight over.

Answer: (a) H₂ gas

Explanation:
1. With an inert Pt electrode, water is preferentially reduced at the cathode over Na⁺, liberating H₂ gas.

Teacher's Note:
1. Na+ is very hard to reduce (it's near the top of the reactivity series), so water always wins at the cathode in aqueous NaCl electrolysis, regardless of electrode material.

Answer: (c) 0.614 S m⁻¹

Explanation:
1. Cell constant = κ × R = 1.29 S m⁻¹ × 200 Ω = 258 m⁻¹
2. Conductivity of 0.02 mol L⁻¹ KCl solution = Cell constant/resistance, i.e., κ = \( \frac{G^{*}}{R} \) = \( \frac{258}{420} \) = 0.614 S m⁻¹

Teacher's Note:
1. Always find the cell constant from the standard KCl solution first; the same G* is then used for the unknown solution in the same cell.

Answer: (d) Decreases

Explanation:
1. The conductivity decreases with dilution because on dilution number of ions per unit volume that are charge carriers decreases.

Teacher's Note:
1. Here the data agree: the more dilute 0.02 M solution has the larger resistance (420 Ω), i.e. lower conductivity.

Answer: (d) Remains constant for a cell.

Explanation:
1. The cell constant of a conductivity cell remains constant for a cell as it only depends on distance between the electrodes and their cross-section area.

Teacher's Note:
1. G* = l/A has units of length⁻¹ (cm⁻¹ or m⁻¹) – a quick way to recognise it in a numerical.

Answer: (a) S m⁻¹

Explanation:
1. SI unit for conductivity of a solution is S m⁻¹ as conductivity is reciprocal of resistivity, where κ = C \( \frac{l}{A} \)
2. unit of κ = S \( \frac{(m)}{(m^{2})} \), κ = S m⁻¹

Teacher's Note:
1. S m² mol⁻¹ is the SI unit of molar conductivity, not conductivity – a common mix-up.

Answer: (d) is independent of ions movement under a potential gradient

Explanation:
1. The conductivity of solutions of different electrolytes in the same solvent and at a given temperature differs due to size and charge of the ions in which they dissociate, concentration of ions, ease with which the ions move under a potential gradient.

Teacher's Note:
1. Conduction in a solution is ion movement under an applied potential, so it can never be independent of how easily ions move.

Answer: (a) Zn(s) | Zn²⁺(aq) || Pb²⁺(aq) | Pb(s)

Explanation:
1. In galvanic cells, redox reaction that start from left corner follow:
2. Reduction: Pb²⁺ + 2e⁻ → Pb; Oxidation: Zn → Zn²⁺ + 2e⁻
3. Complete cell reaction: Zn + Pb²⁺ → Zn²⁺ + Pb
4. Zn(s) | Zn²⁺(aq) || Pb²⁺(aq) | Pb(s)

Teacher's Note:
1. Zn has the lower (more negative) E°, so it is the anode and goes on the left.

Answer: (c) 0.609 V

Explanation:
1. According to Nernst equation: E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[Zn^{2+}]}{[Pb^{2+}]} \)
2. E_cell = [−0.13 − (−0.76)] − \( \frac{0.0591}{2} \) log \( \frac{0.1}{0.02} \)
3. = 0.63 − 0.02955 × log 5 = 0.63 − 0.02955 × 0.6990
4. = 0.63 − 0.0206 = 0.6094 V

Teacher's Note:
1. Students often forget the E° values of Pb²⁺/Pb and Zn²⁺/Zn; E°_cell = 0.63 V is worth remembering for this common cell.

Answer: (b) Pb

Explanation:
1. Anode reaction: Zn(s) → Zn²⁺(aq) + 2e⁻
2. Cathode reaction: Pb²⁺(aq) + 2e⁻ → Pb(s)

Teacher's Note:
1. Reduction always takes place at the cathode, so the product there is the reduced form – metallic Pb.

Answer: (d) It provides surface for redox reaction.

Explanation:
1. Inert electrode does not participate in redox reaction and acts only as source or sink for electrons. It provides surface either for oxidation or for reduction reaction.

Teacher's Note:
1. An inert electrode serves only one half-reaction – oxidation if it is the anode, reduction if it is the cathode – never the whole redox reaction.