CBSE Class 12 Chemistry Electrochemistry MCQs Set 09

Answer: (b) Ag < Cr < Mg < K

Explanation:
1. Higher the oxidation potential, more easily it is oxidised and hence greater is the reducing power.
2. Hence, the increasing order of reducing power is Ag < Cr < Mg < K.

Teacher's Note:
1. Reducing power runs opposite to reduction potential – the metal with the lowest (most negative) E° is oxidised most easily.
2. Read the half-cell values as reduction potentials first before ranking; a common slip is ranking by oxidation potential by mistake.

Answer: (b) −2.13 × 10⁵ J mol⁻¹

Explanation:
1. E°_cell = E°_Ag2O/Ag − E°_Zn²⁺/Zn = 0.344 − (−0.76) = 1.104 V
2. ΔG° = −nFE°_cell = −2 × 96500 × 1.104 = −2.13 × 10⁵ J mol⁻¹

Teacher's Note:
1. n is the number of electrons in the balanced cell reaction – here 2, from the 2Ag and 2OH⁻ in the equation.
2. A negative ΔG° confirms the reaction as written is spontaneous under standard conditions.

Answer: (a) E(Mⁿ⁺/M) = E°(Mⁿ⁺/M) − \( \frac{RT}{nF} \) ln \( \frac{1}{[M^{n+}]} \)

Explanation:
1. E(Mⁿ⁺/M) = E°(Mⁿ⁺/M) − \( \frac{RT}{nF} \) ln \( \frac{[M]}{[M^{n+}]} \)
2. Since the concentration of a solid is taken as unity, [M] = 1, so
3. E(Mⁿ⁺/M) = E°(Mⁿ⁺/M) − \( \frac{RT}{nF} \) ln \( \frac{1}{[M^{n+}]} \)

Teacher's Note:
1. Pure solids and liquids are always assigned activity 1 in the Nernst equation, which is why [M] drops out here.
2. Note this uses natural log (ln) with RT/nF, not the 0.0591/n log form – the two are equivalent only after converting ln to log₁₀.

Answer: (c) 271.6, 91.0 and 126.4 S cm² mol⁻¹

Explanation:
1. Λ°_m CaCl₂ = λ°_Ca²⁺ + 2λ°_Cl⁻ = 119.0 + 2 × 76.3 = 271.6 S cm² mol⁻¹
2. Λ°_m CH₃COONa = λ°_CH3COO⁻ + λ°_Na⁺ = 40.9 + 50.1 = 91 S cm² mol⁻¹
3. Λ°_m NaCl = λ°_Na⁺ + λ°_Cl⁻ = 50.1 + 76.3 = 126.4 S cm² mol⁻¹

Teacher's Note:
1. This is a direct application of Kohlrausch's law: multiply each ion's contribution by its number of ions in the formula unit, here 2 for Cl⁻ in CaCl₂.
2. A frequent slip is forgetting the stoichiometric multiplier for divalent ions like Ca²⁺.

Answer: (d) all of these

Explanation:
1. All three factors – the nature and structure of the metal, the number of valence electrons per atom, and temperature – affect electronic conductance.

Teacher's Note:
1. Metallic conductance decreases with rising temperature (more lattice vibration scatters electrons), unlike electrolytic conductance, which increases.

Answer: (a) The reaction stops and no current flows through the cell.

Explanation:
1. If an external potential exactly equal to 1.1 V is applied to the cell, the reaction stops and no current flows through the cell.
2. Any further increase in the external potential starts the reaction again, but in the opposite direction, and the cell then functions as an electrolytic cell.

Teacher's Note:
1. This balance point is exactly how the Nernst equation is used to measure a cell's EMF – by finding the opposing potential at which current just stops.

Answer: (a) the chemical energy liberated during the redox reaction is converted to electrical energy

Explanation:
1. A Daniell cell is a galvanic cell: it converts the chemical energy liberated during the spontaneous redox reaction into electrical energy, and has an electrode potential of 1.1 V.

Teacher's Note:
1. Keep the direction straight: a galvanic (voltaic) cell turns chemical energy into electrical energy; an electrolytic cell does the reverse.

Answer: (c) \( \frac{[Fe^{2+}][H_2]}{[Fe][H^{+}]^{2}} \) form: E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[Fe^{2+}][H_2]}{[Fe][H^{+}]^{2}} \)

Explanation:
1. At anode: Fe → Fe²⁺(0.001 M) + 2e⁻
2. At cathode: 2H⁺(1 M) + 2e⁻ → H₂(1 bar)
3. Net reaction: Fe + 2H⁺ → Fe²⁺ + H₂
4. Nernst equation: E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[Fe^{2+}][H_2]}{[Fe][H^{+}]^{2}} \)

Teacher's Note:
1. Write the half-reactions first; the reaction quotient in the log term always goes products over reactants, raised to their stoichiometric coefficients.

Answer: (a) silver is deposited at cathode and O₂ is liberated at anode

Explanation:
1. At cathode: Ag⁺_(aq) + e⁻ → Ag_(s)
2. At anode: 2OH⁻_(aq) → ½O_2(g) + H₂O_(l) + 2e⁻

Teacher's Note:
1. Ag⁺ is preferentially reduced over H⁺ at the cathode because its reduction potential is much higher, so silver plates out rather than hydrogen gas forming.

Answer: (c) the electrode potential is assumed to be zero

Explanation:
1. By convention, the standard hydrogen electrode is assigned a zero potential at all temperatures, and all other electrode potentials are measured relative to it.

Teacher's Note:
1. This is a defined reference point, not a measured property of hydrogen itself – every other E° value in the chapter is relative to this zero.

Answer: (a) E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[ion]_{RHS}}{[ion]_{LHS}} \)

Explanation:
1. At 25°C, E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[ion]_{RHS}}{[ion]_{LHS}} \)
2. where LHS and RHS refer to the two half-cells as written in the cell notation.

Teacher's Note:
1. Remember the cell-notation convention: the left electrode is the anode (oxidation) and the right is the cathode (reduction), so the ratio is RHS over LHS.

Answer: (a) Br⁻ < Fe²⁺ < Al

Explanation:
1. Lower the reduction potential, more is the reducing power.
2. E° values: Br₂/Br⁻ = +1.09 V > Fe³⁺/Fe²⁺ = +0.77 V > Al³⁺/Al = −1.66 V.
3. Thus, the increasing order of reducing power is Br⁻ < Fe²⁺ < Al.

Teacher's Note:
1. Reducing power runs opposite to reduction potential: the species on the right of the half-reaction with the lowest E° is the strongest reducing agent.
2. Here Al (E° = −1.66 V) is the strongest reducer and Br⁻ (E° = +1.09 V) the weakest.

Answer: (c) Nernst equation is E_cell = E°_cell − \( \frac{0.0591}{n} \) log \( \frac{[Products]}{[Reactants]} \)

Explanation:
1. (a) The correct relation is log K = \( \frac{nFE_{cell}^{\circ}}{2.303RT} \), not with E in place of E°cell alone as loosely written.
2. (b) The correct form is E_cell = E°_cell − \( \frac{0.0591}{2} \) log \( \frac{[a_1]}{[a_2]} \).
3. (d) The 0.0591/n (or 0.059/n) coefficient is only valid at 298 K, not 273 K.
4. Option (c), the Nernst equation written as products over reactants, is the correctly matched expression.

Teacher's Note:
1. Treat this as an error-spotting question: three of the four relationships have a small but real mistake (wrong ratio, wrong temperature), and only one is written correctly.

Answer: (b) 1.166 cm⁻¹

Explanation:
1. κ = G × \( \frac{l}{A} \)
2. \( \frac{l}{A} \) = κ × \( \frac{1}{G} \) = κ × R = 0.0212 × 55 = 1.166 cm⁻¹

Teacher's Note:
1. Cell constant = κ × R always, whatever the solution – it's a fixed property of the physical cell, not of what's inside it.

Answer: (a) Zn_(s)|Zn²⁺_(aq)||Ag⁺_(aq)|Ag_(s)

Explanation:
1. Zn + 2Ag⁺ → Zn²⁺ + 2Ag can be represented as Zn_(s)|Zn²⁺_(aq)||Ag⁺_(aq)|Ag_(s).

Teacher's Note:
1. Cell notation convention: anode (oxidation, solid first) on the left, salt bridge as ||, cathode (reduction) on the right, with phases always shown.

Answer: (a) 189 S cm² mol⁻¹

Explanation:
1. Λ°_Al2(SO4)3 = 2λ°_Al³⁺ + 3λ°_SO4²⁻
2. λ°_Al³⁺ = \( \frac{858 - (3\times160)}{2} \) = \( \frac{858 - 480}{2} \) = 189 S cm² mol⁻¹

Teacher's Note:
1. Always match the subscript on λ° to how many of that ion are in one formula unit – 2 Al and 3 SO₄ here.

Answer: (a) Since E° value for the cell will be positive, the redox reaction is feasible.

Explanation:
1. Cell reaction: Ni_(s) + 2Ag⁺_(aq) → Ni²⁺_(aq) + 2Ag_(s)
2. E°_cell = E°_cathode − E°_anode = 0.80 − (−0.25) = +1.05 V
3. ΔG° = −nFE°_cell; as E°_cell is +ve, ΔG° is −ve, hence the reaction is feasible.

Teacher's Note:
1. The electrode with the higher (more positive) reduction potential is always the cathode, so Ag⁺/Ag reduces and Ni/Ni²⁺ oxidises here.

Answer: (a) Reduction takes place at the magnesium electrode and oxidation at the SHE.

Explanation:
1. Mg²⁺_(aq) + 2e⁻ → Mg_(s); E° = −2.36 V
2. 2H⁺_(aq) + 2e⁻ → H_2(g); E° = 0.00 V
3. Thus, oxidation takes place at the magnesium electrode and reduction at the hydrogen electrode, so statement (a) is the incorrect one.

Teacher's Note:
1. Magnesium has the lower reduction potential, so it is oxidised (anode); hydrogen, with the higher potential, is reduced (cathode).

Answer: (b) Λ°_mNaBr = Λ°_mNaCl + Λ°_mKBr − Λ°_mKCl

Explanation:
1. By Kohlrausch's law of independent migration, combining NaCl and KBr double-counts K⁺ and Cl⁻, so KCl must be subtracted:
2. Λ°_mNaBr = Λ°_mNaCl + Λ°_mKBr − Λ°_mKCl

Teacher's Note:
1. This "add two salts, subtract the third" trick is standard for building an unknown electrolyte's Λ° from salts you do have data for.

Answer: (a) least, first

Explanation:
1. The ion which requires the least energy will be liberated first at a given electrode.

Teacher's Note:
1. This is why, say, Cu²⁺ is discharged before H⁺ during electrolysis of CuSO₄ – it needs less energy to be reduced.

Answer: (a) 1.2 × 10⁶

Explanation:
1. log K_c = \( \frac{nE_{cell}^{\circ}}{0.0591} \). For the given reaction, n = 1.
2. log K_c = \( \frac{1\times0.36}{0.0591} \) = 6.09
3. K_c = antilog 6.09 = 1.2 × 10⁶

Teacher's Note:
1. Disproportionation reactions like this one (Cu⁺ going to both Cu and Cu²⁺) still follow the same log K = nE°/0.0591 relation.

Answer: (c) (i), (ii) and (iii)

Explanation:
1. Iron coated with zinc does not get rusted even if cracks appear on the surface, because Zn will take part in the redox reaction, not Fe, as Zn is more reactive than Fe – statement (i) is correct.
2. If iron is coated with tin and cracks appear on the surface, Fe will take part in the redox reaction because Sn is less reactive than Fe – statement (ii) is correct.
3. In the reactivity series, Zn is above Fe and Sn is below Fe, so statement (iii) is also correct. All three statements hold.

Teacher's Note:
1. A metal coating protects the base metal only if it is more reactive (sacrificial, like Zn on steel); a less reactive coating (like Sn) actually accelerates corrosion once it's breached.

Answer: (c) Mg, Al, Zn, Fe, Cu

Explanation:
1. In the reactivity series, Mg > Al > Zn > Fe > Cu.
2. Hence, Mg can displace Al, Al can displace Zn, and so on, giving the order Mg, Al, Zn, Fe, Cu.

Teacher's Note:
1. A more reactive (higher-up) metal always displaces a less reactive one from its salt solution – this is just the activity series read left to right.

Answer: (a) Cd + NiO₂ + 2H₂O → Cd(OH)₂ + Ni(OH)₂

Explanation:
1. Nickel-cadmium battery: Anode – Cd; Cathode – NiO₂; Electrolyte – KOH.
2. At anode: Cd_(s) + 2OH⁻_(aq) → Cd(OH)_2(s) + 2e⁻
3. At cathode: NiO_2(s) + 2H₂O_(l) + 2e⁻ → Ni(OH)_2(s) + 2OH⁻_(aq)
4. Overall: Cd_(s) + NiO_2(s) + 2H₂O_(l) → Cd(OH)_2(s) + Ni(OH)_2(s)

Teacher's Note:
1. Ni-Cd is a secondary (rechargeable) cell – the reverse of this reaction regenerates Cd and NiO₂ on charging.

Answer: (c) 101 Ω⁻¹ cm² mol⁻¹

Explanation:
1. Λ_m = \( \frac{\kappa\times1000}{M} \) = \( \frac{1.52\times10^{-2}\times1000}{0.15} \) = 101 Ω⁻¹ cm² mol⁻¹

Teacher's Note:
1. Keep κ in S cm⁻¹ and M in mol L⁻¹ for this formula – the ×1000 converts litres to cm³.

Answer: (b) highest reduction potential

Explanation:
1. Higher the reduction potential, stronger is the oxidising agent. Fluorine has the highest reduction potential of all elements.

Teacher's Note:
1. Don't confuse this with electron affinity – the chapter's own ranking tool is reduction potential, not electron affinity, even though the two broadly track together.

Answer: (a) 2H₂O_(l) → O_2(g) + 4H⁺_(aq) + 4e⁻

Explanation:
1. In dilute H₂SO₄, water is oxidised in preference to sulphate ions at the anode, liberating O₂: 2H₂O_(l) → O_2(g) + 4H⁺_(aq) + 4e⁻.

Teacher's Note:
1. Only concentrated (not dilute) H₂SO₄ gives S₂O₈²&supminus; at the anode – dilution changes which oxidation is preferred.

Answer: (c) A → CH₃COOH, B → CH₃COONa

Explanation:
1. For strong electrolytes, the plot of Λ_m against C^1/2 is nearly a straight line.
2. For weak electrolytes, there is a very large increase in Λ_m with dilution, especially near low concentrations.
3. So curve A (steep rise) is the weak electrolyte CH₃COOH and curve B (near-linear) is the strong electrolyte CH₃COONa.

Teacher's Note:
1. On a Λ_m vs √C plot, the curve that rises steeply at low concentration is the weak electrolyte; the gentle near-straight line is the strong one.

Answer: (a) 11.4%, 3.67 × 10⁻⁴ mol L⁻¹

Explanation:
1. λ°_HCOOH = λ°_H⁺ + λ°_HCOO⁻ = 349.6 + 54.6 = 404.2 S cm² mol⁻¹
2. α = \( \frac{\Lambda_m}{\Lambda_m^{\circ}} \) = \( \frac{46.1}{404.2} \) = 0.114 = 11.4%
3. K_a = \( \frac{C\alpha^{2}}{1-\alpha} \) = \( \frac{0.025\times(0.114)^2}{1-0.114} \) = \( \frac{0.025\times0.114\times0.114}{0.886} \) = 3.67 × 10⁻⁴ mol L⁻¹

Teacher's Note:
1. The degree of dissociation here comes straight from Λ_m/Λ°_m – always find Λ°_m first using Kohlrausch's law before computing K_a.

Answer: (b) Graph (b): straight line with positive slope and a positive intercept

Explanation:
1. E = E° + \( \frac{0.059}{2} \) log[Mg²⁺]
2. Hence, the plot of E vs log[Mg²⁺] will be linear with positive slope and intercept = E°.

Teacher's Note:
1. Rewrite log(1/x) as −log x first; the minus signs cancel and the equation becomes y = c + mx with a positive slope of 0.059/2.
2. The intercept is E° itself, so the line does not pass through the origin.

Answer: (c) +0.53 V

Explanation:
1. Cu²⁺ + e⁻ → Cu⁺; E°₁ = 0.15 V, n₁ = 1
2. Cu²⁺ + 2e⁻ → Cu; E°₂ = 0.34 V, n₂ = 2
3. Cu⁺ + e⁻ → Cu; E°₃ = ?, n₃ = 1
4. ΔG°₃ = ΔG°₂ − ΔG°₁, so −n₃FE°₃ = −n₂FE°₂ + n₁FE°₁
5. −E°₃ = −2 × 0.34 + 1 × 0.15 = −0.53
6. E°₃ = +0.53 V

Teacher's Note:
1. E° values cannot simply be added or subtracted across different n – convert to ΔG° (which is additive) first, then back to E°.

Answer: (d) In fig (Y), Zn is deposited at Cu and Cu is deposited at Zn.

Explanation:
1. In fig (Y), when the opposing potential exceeds the cell's own EMF, the cell runs in reverse: zinc is deposited at the zinc electrode and copper dissolves at the copper electrode.
2. Statement (d), which has the deposits swapped between the electrodes, is therefore incorrect.

Teacher's Note:
1. When Eext exceeds the cell's EMF, the whole cell reverses – but each electrode still only sees its own metal deposit or dissolve, never the other electrode's metal.

Answer: (d) All of these

Explanation:
1. The electrochemical series lets you compare oxidising/reducing power, predict whether a metal liberates H₂ from acid, and predict whether a redox reaction is spontaneous – all three.

Teacher's Note:
1. Keep a short list of these three uses in mind; CBSE case-based and short-answer questions draw on exactly this set repeatedly.

Answer: (b) X → NaCl, Y → CH₃COOH

Explanation:
1. Electrolyte X is a strong electrolyte: on dilution the number of ions stays the same, only interionic attraction decreases, so Λ_m rises only modestly.
2. Electrolyte Y is a weak electrolyte: dilution shifts the dissociation equilibrium forward strongly, so Λ_m rises much more sharply.
3. A 1.5× rise fits a strong electrolyte (NaCl) and a 20× rise fits a weak electrolyte (CH₃COOH).

Teacher's Note:
1. Big jumps in Λ_m on dilution are the signature of a weak electrolyte; strong electrolytes are already almost fully dissociated, so there's little room to rise.

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

Explanation:
1. The conductivity of solutions of different electrolytes in the same solvent and at a given temperature is different, so the reason is a wrong statement, while the assertion is correct.

Teacher's Note:
1. Conductivity is highly specific to the electrolyte's identity (ion size, charge, mobility), so "same solvent, same temperature" alone never guarantees equal conductivity.

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

Explanation:
1. Standard reduction potential of an electrode has a fixed, intensive value and does not scale with the stoichiometric coefficients, so the assertion is false.
2. The reason, that doubling [Ag⁺] does not change E°, is correct, giving answer (d).

Teacher's Note:
1. E° is an intensive property (per electron transferred), so multiplying an equation through by 2 never multiplies E° – only ΔG° scales with the equation.

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 the assertion: Kohlrausch's law states that Λ° of an electrolyte is the sum of the individual ionic contributions.

Teacher's Note:
1. This is the foundational equation behind every Kohlrausch's-law question in the chapter – every combination question (CH3COOH, NH4OH, MgI2, etc.) is just this idea applied with different salts.

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

Explanation:
1. When electrolytes are dissolved in a solvent, they furnish their own ions in the solution, increasing its conductivity – both statements are correct and the reason explains the assertion.

Teacher's Note:
1. Pure water conducts only very weakly (from its own slight self-ionisation), which is why dissolving even a little electrolyte raises conductivity sharply.

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

Explanation:
1. Cu²⁺ ions are deposited as metallic Cu at the cathode, which is why the blue solution turns colourless – the assertion is correct.
2. The reason, that copper is converted to Cu(OH)₂, is incorrect; copper is deposited as the metal, not as a hydroxide.

Teacher's Note:
1. The blue colour of CuSO4 solution comes from hydrated Cu²⁺ ions, so once those are plated out as Cu metal at the cathode, the colour is gone.

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

Explanation:
1. Resistance is directly proportional to length, so it increases (not decreases) with increasing length – the assertion is false.
2. Resistance is inversely proportional to area of cross-section, so it does decrease as cross-sectional area increases – the reason is correct, giving answer (d).

Teacher's Note:
1. Remember R = ρl/A: resistance rises with length and falls with cross-sectional area – opposite trends for the two geometric factors.

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 the assertion: insulators are defined precisely by their inability to let current pass.

Teacher's Note:
1. "Very low conductivity" and "does not allow current to pass" describe the same property from two angles, which is why the reason is a direct explanation here, not just a related fact.

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

Explanation:
1. In the plot of molar conductivity versus concentration for a weak electrolyte, extrapolation to zero concentration is not possible since the graph is not linear, so the assertion is correct.
2. Kohlrausch's law is indeed how Λ°_m is found instead, so the reason is also correct, but it explains how the value is obtained some other way, not why direct measurement fails – so it is not the correct explanation, giving (b).

Teacher's Note:
1. Both halves of an A-R item can be true facts about the same topic without one explaining the other – that's the distinction (a) vs (b) is testing.

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

Explanation:
1. The assertion is correct: conductance does depend on the electrolyte and its concentration.
2. The cell constant depends on the distance between the electrodes and their area of cross-section, not on the material they are made of, so the reason is incorrect, giving (c).

Teacher's Note:
1. Cell constant is a purely geometric property of the conductivity cell (electrode spacing and area) – it has nothing to do with what the electrodes are made of or what solution is inside.