25 questions in 25 minutes on the NTA computer-based test interface, all from Electrochemistry. +4 / -1 marking, instant score, full solutions. Free, no login.
A chapter test is the fastest honest check on whether a chapter has actually landed. This one draws 25 questions from Electrochemistry, of which 20 are real NEET previous-year questions and 6 sit at the harder end of the bank. Sit it in one 25-minute block, the way you would in the hall.
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Duration: 180 minutes · 180 questions · 720 marks maximum
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How many questions are in this Electrochemistry mock test?
25 questions in 25 minutes, marked +4 for a correct answer and -1 for a wrong one, exactly like the real paper. 20 of them are NEET previous-year questions.
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Yes. On-screen countdown, colour-coded question palette, Save & Next, Clear Response and Mark for Review, then an instant score with full solutions.
Should I take this before or after revising Electrochemistry?
Both, and that is the point. Take it cold to find out what you actually do not know, revise from the chapter notes, then retake it to confirm the gap closed.
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The complete paper in text form, with the correct option and a worked explanation for every question. Sit the timed test above first: the score is only worth having if you earn it. Then open this to revise, or to re-read a question you got wrong without replaying the whole paper.
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Chemistry
- Q1For the cell Zn|Zn²⁺(1 M)||Cu²⁺(c)|Cu, the cell EMF will be maximum when [Cu²⁺] is:
- Equal to [Zn²⁺]
- Very high
- Zero
- Very low
Answer: (B) Very high
Ecell = E°cell − (0.0591/2)log([Zn²⁺]/[Cu²⁺]). Increasing [Cu²⁺] (the reactant) makes the log term more negative, so −(…) raises EMF. Highest [Cu²⁺] gives maximum EMF.
- Q2Assertion (A): For a Daniell cell Zn/Zn²⁺||Cu²⁺|Cu with E_(cell) = 1.1 V, the application of opposite potential greater than 1.1 V results into the flow of electrons from cathode to anode. Reason (R): Zn is deposited at zinc electrode and Cu is dissolved at copper electrode.
- Both (A) and (R) are true and (R) is the correct explanation of (A).
- Both (A) and (R) are true, but (R) is not the correct explanation of (A).
- (A) is true and (R) is false.
- Both (A) and (R) are false.
Answer: (A) Both (A) and (R) are true and (R) is the correct explanation of (A).
A. Tests: what happens to a Daniell cell when an opposing external voltage is pushed past its emf. Why A: With Eₑₓₜ > 1.1 V the external source overpowers the cell, so electrons are driven backwards, from the copper electrode (the cathode of the spontaneous cell) towards the zinc electrode, and the assertion is true. Reversing the electron flow reverses both electrode reactions: Zn²⁺ + 2e⁻ arrow Zn now plates zinc onto the zinc electrode, while Cu arrow Cu²⁺ + 2e⁻ dissolves the copper electrode, so the reason is true as well. The reason is not a separate fact, it is the chemical statement of the very reversal the assertion describes, so it explains it. Why not B: the reason is not an unrelated truth, depositing Zn and dissolving Cu is exactly what reversed electron flow means at each electrode [misconception B: “two true statements about the same cell must be independent”]. Why not C: the reason is true, and this reversal is the standard NCERT description of a Daniell cell driven above 1.1 V [misconception C: “electrode reactions cannot be reversed once the cell is built”]. Why not D: both statements are true, the assertion following directly from Eₑₓₜ > E_(cell) [misconception D: “an external voltage can only stop a cell, never reverse it”]. Remember: push a Daniell cell past 1.1 V from outside and it runs backwards, plating Zn and eating Cu.
- Q3The efficiency of a fuel cell is given by:
- ΔG/ΔS
- ΔG/ΔH
- ΔH/ΔG
- ΔS/ΔG
Answer: (B) ΔG/ΔH
The thermodynamic efficiency of a fuel cell is the fraction of the fuel’s enthalpy converted to useful electrical (free-energy) work: η = ΔG/ΔH. Fuel cells can approach 100% efficiency.
- Q4The conductivity of a solution is 0.025 S cm⁻¹ and its molar conductivity is 250 S cm² mol⁻¹. The molar concentration is:
- 0.1 mol L⁻¹
- 0.001 mol L⁻¹
- 1 mol L⁻¹
- 0.01 mol L⁻¹
Answer: (A) 0.1 mol L⁻¹
Λm = κ×1000/c ⇒ c = κ×1000/Λm = 0.025×1000/250 = 0.1 mol L⁻¹.
- Q5In a galvanic cell, the anode is the electrode where:
- Reduction occurs and it is positive
- Oxidation occurs and it is negative
- Reduction occurs and it is negative
- Oxidation occurs and it is positive
Answer: (B) Oxidation occurs and it is negative
In a galvanic cell, oxidation always occurs at the anode, which is the negative terminal (electrons leave it to the external circuit). The cathode is positive and is the reduction site.
- Q6In the electrochemical cell Zn | ZnSO₄(0.01 M) || CuSO₄(1.0 M) | Cu, the EMF is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ to 0.01 M, the EMF becomes E₂. The relationship between E₁ and E₂ is (RT/F = 0.059):
- E₁ = E₂
- E₂ = 0 ≠ E₁
- E₁ > E₂
- E₁ < E₂
Answer: (C) E₁ > E₂
E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]). For E₁: log(0.01/1) = −2, so E₁ = E° + 0.059. For E₂: log(1/0.01) = +2, so E₂ = E° − 0.059. Therefore E₁ > E₂.
- Q7The specific conductance of a 0.1 N KCl solution at 23 °C is 0.012 S cm⁻¹. The resistance of the cell containing this solution at the same temperature is 55 Ω. The cell constant is:
- 0.142 cm⁻¹
- 0.66 cm⁻¹
- 0.918 cm⁻¹
- 1.12 cm⁻¹
Answer: (B) 0.66 cm⁻¹
κ = (1/R) × cell constant ⇒ cell constant = κ × R = 0.012 × 55 = 0.66 cm⁻¹.
- Q8The reaction; (1)/(2) H₂(g) + AgCl(s) arrow H⁺(aq) + Cl⁻(aq) + Ag(s) occurs in which of the following galvanic cell:
- Pt | H₂(g) | KCl(soln) | AgCl(s) | Ag
- Pt | H₂(g) | HCl(soln) | AgCl(s) | Ag
- Ag | AgCl(s) | KCl(soln) | AgNO₃(aq) | Ag
- Pt | H₂(g) | HCl(soln) | AgNO₃(aq) | Ag
Answer: (B) Pt | H₂(g) | HCl(soln) | AgCl(s) | Ag
D. Tests: translating a written cell reaction into correct cell notation, including the choice of silver electrode. Why D: the reaction consumes H₂ and makes H⁺, so the left electrode must be a hydrogen electrode, Pt | H₂(g) | H⁺. It consumes solid AgCl and makes Ag and Cl⁻, so the right electrode is the silver silver chloride electrode, AgCl(s) | Ag, dipping in a chloride solution. Both products, H⁺ and Cl⁻, appear in the same solution, so one HCl solution serves both electrodes and no salt bridge is needed. That is Pt | H₂(g) | HCl(soln) | AgCl(s) | Ag. Why not A: its right electrode is Ag⁺ | Ag from AgNO₃, so the cathode reaction is Ag⁺ + e⁻ arrow Ag and solid AgCl is never consumed [misconception A: “any silver electrode reduces silver in the same way”]. Why not B: it contains no hydrogen electrode at all, so H₂ cannot be oxidised and H⁺ cannot be produced anywhere in that cell. Why not C: with KCl as the electrolyte the solution holds no H⁺, so the hydrogen electrode has no H⁺ to be reversible to and the written product H⁺(aq) does not belong to that solution. Remember: split the given reaction into its two half reactions first, then pick electrodes that create exactly those species in the solution shown.
- Q9An electrochemical cell is constructed using half cells in the direction of spontaneous change: Fe(OH)₂(s) + 2e⁻ arrow Fe(s) + 2OH⁻(aq), E^(θ) = -0.88 V and AgBr(s) + e⁻ arrow Ag(s) + Br⁻(aq), E^(θ) = +0.07 V Which of the following option is correct?
- Overall reaction Fe(s) + 2OH⁻(aq) + 2AgBr(s) leftharpoons Fe(OH)₂(s) + 2Ag(s) + 2Br⁻(aq)
- E^(θ)_(cell) is an extensive property
- E^(θ)_(cell) = -0.95 V
- Fe is reduced in the electrochemical cell
Answer: (A) Overall reaction Fe(s) + 2OH⁻(aq) + 2AgBr(s) leftharpoons Fe(OH)₂(s) + 2Ag(s) + 2Br⁻(aq)
A. Tests: choosing the spontaneous direction from two reduction potentials and writing the balanced overall cell reaction. Why A: The half cell with the higher (more positive) E^(θ) takes the cathode, so AgBr is reduced at +0.07 V and the Fe(OH)₂/Fe couple at -0.88 V must run in reverse as the anode: Fe + 2OH⁻ arrow Fe(OH)₂ + 2e⁻. Doubling the AgBr step to match the two electrons and adding gives Fe + 2OH⁻ + 2AgBr leftharpoons Fe(OH)₂ + 2Ag + 2Br⁻. Its E^(θ)_(cell) = 0.07 – (-0.88) = +0.95 V, positive, which confirms this is the spontaneous direction. Why not B: the magnitude is right but the sign is wrong; E^(θ)_(cell) = E^(θ)_(cathode) – E^(θ)_(anode) = 0.07 + 0.88 = +0.95 V, and a negative value would contradict the stated spontaneous direction [misconception B: “subtract the more positive potential from the more negative one”]. Why not C: iron is oxidised here, going from Fe(0) in the metal to Fe(II) in Fe(OH)₂; only the silver in AgBr is reduced [misconception C: “the metal electrode is always the species reduced”]. Why not D: E^(θ)_(cell) is intensive, because doubling every coefficient doubles Δ G^(θ) and n together and E = -Δ G/nF is unchanged [misconception D: “cell potential scales with the amount of reaction”]. Remember: the higher reduction potential wins the cathode, and E_(cell) never scales with stoichiometry.
- Q10The molar conductivity of a weak electrolyte when plotted against the square root of its concentration, which of the following is expected to be observed?
- Molar conductivity decreases sharply with increase in concentration.
- A small increase in molar conductivity is observed at infinite dilution.
- Molar conductivity increases sharply with increase in concentration.
- A small decrease in molar conductivity is observed at infinite dilution.
Answer: (A) Molar conductivity decreases sharply with increase in concentration.
B. Tests: reading the shape of the Λₘ versus c^(1/2) curve for a weak electrolyte. Why B: for a weak electrolyte Λₘ = α Λₘᵒ, where α is the degree of dissociation. As concentration rises, Ostwald’s law forces α down steeply, from near 1 in very dilute solution to a small fraction at ordinary concentrations. So Λₘ falls sharply as concentration increases, and on the plot against c^(1/2) the curve is almost vertical near the origin and then flattens out. Why not A: molar conductivity never rises with concentration; more crowding means less dissociation and more interionic drag [misconception A: “conductivity κ increases with concentration, so molar conductivity must increase too”]. Why not C: the change near infinite dilution is a very steep rise, not a small one, and that steepness is exactly why Λₘᵒ of a weak electrolyte cannot be found by extrapolating the graph. Why not D: on approaching infinite dilution Λₘ rises towards Λₘᵒ, it does not decrease, so the direction stated is wrong. Remember: on dilution κ falls and Λₘ rises, and for a weak electrolyte that rise near zero concentration is near vertical.
- Q11Given Cu²⁺ + e⁻ → Cu⁺, E° = +0.15 V and Cu⁺ + e⁻ → Cu, E° = +0.50 V. The standard electrode potential E° for Cu²⁺ + 2e⁻ → Cu is:
- 0.325 V
- 0.650 V
- 0.150 V
- 0.500 V
Answer: (A) 0.325 V
Electrode potentials are not additive, but ΔG° is. ΔG°(Cu²⁺→Cu) = ΔG°₁ + ΔG°₂ ⇒ −2FE° = −1F(0.15) − 1F(0.50). So E° = (0.15 + 0.50)/2 = 0.325 V.
- Q12From the following select the one which is not an example of corrosion.
- Rusting of iron object
- Tarnishing of silver
- Development of green coating on copper and bronze ornaments
- Production of hydrogen by electrolysis of water
Answer: (D) Production of hydrogen by electrolysis of water
B. Tests: recognising corrosion as the spontaneous electrochemical attack on a metal, not just any electrochemistry. Why B: Corrosion is the spontaneous oxidation of a metal by its surroundings, running on its own with a positive cell emf. Electrolysis of water is a forced, non-spontaneous change driven by an external supply, and no metal is oxidised in it at all, so it is not corrosion. Why not A: rusting is the textbook case of corrosion, iron oxidised by moist air to hydrated Fe₂O₃ through tiny galvanic cells on the surface [misconception A: “rusting is a separate process from corrosion”]. Why not C: tarnishing of silver is corrosion too, silver oxidised to black Ag₂S by traces of sulphur compounds in air [misconception C: “corrosion only means rust”]. Why not D: the green patina on copper and bronze is basic copper carbonate formed by spontaneous oxidation in moist air [misconception D: “a coloured deposit is a coating, not corrosion”]. Remember: corrosion is metal oxidising for free, electrolysis is a reaction you have to pay for.
- Q13Alkaline oxidative fusion of MnO₂ gives “A” which on electrolytic oxidation in alkaline solution produces B. A and B respectively are
- MnO₄²⁻ and Mn₂O₇
- Mn₂O₃ and MnO₄²⁻
- Mn₂O₇ and MnO₄⁻
- MnO₄²⁻ and MnO₄⁻
Answer: (D) MnO₄²⁻ and MnO₄⁻
C. Tests: the two step preparation of permanganate, identifying the oxidation state reached at each step. Why C: alkaline oxidative fusion is 2MnO₂ + 4KOH + O₂ arrow 2K₂MnO₄ + 2H₂O, so A is the dark green manganate ion MnO₄²⁻, Mn in +6. Electrolytic oxidation in alkaline solution then strips one more electron at the anode, MnO₄²⁻ arrow MnO₄⁻ + e⁻, so B is the purple permanganate ion MnO₄⁻, Mn in +7. Why not A: Mn₂O₃ has Mn in +3, a reduction of MnO₂, and it forms on strongly heating MnO₂ alone; oxidative fusion in alkali raises the oxidation state, it does not lower it [misconception A: “heating an oxide always drives it to a lower oxide”]. Why not B: Mn₂O₇ is the oily explosive oxide obtained from KMnO₄ with concentrated H₂SO₄, not from fusion with KOH, and it cannot be the fusion product A [misconception B: “any Mn(VII) species will do”]. Why not D: it gets A right but makes the electrolysis product Mn₂O₇; anodic oxidation of manganate in alkaline solution gives the permanganate ion in solution, and the covalent heptoxide cannot survive in alkali [misconception D: “+7 manganese means Mn₂O₇ rather than MnO₄⁻“]. Remember: fuse to green MnO₄²⁻ (Mn VI), then oxidise at the anode to purple MnO₄⁻ (Mn VII).
- Q14Identify the factor from the following that does not affect electrolytic conductance of a solution.
- The nature of the electrode used.
- The nature of the electrolyte added.
- Concentration of the electrolyte.
- The nature of solvent used.
Answer: (A) The nature of the electrode used.
C. Tests: separating bulk solution properties from interface properties in conductance measurements. Why C: electrolytic conductance is the ability of the solution itself to carry current, and it is decided by how many ions there are and how fast they move: the electrolyte, its concentration, the solvent and the temperature. What the electrodes are made of only sets what happens at the electrode surface, not the ionic transport through the bulk. Platinised platinum is used in a conductivity cell for convenience and reproducibility, and swapping it for another inert metal of the same size and spacing leaves the measured conductance unchanged. Why not A: the solvent’s dielectric constant fixes how far the electrolyte dissociates and its viscosity fixes ionic mobility, so conductance in water and in ethanol differ hugely [misconception A: “solvent is just a passive medium”]. Why not B: a strong electrolyte releases many free ions and conducts well while a weak one barely ionises, so the identity of the electrolyte is the single biggest factor [misconception B: “all dissolved salts conduct the same”]. Why not D: more concentration means more charge carriers in the same volume, so conductance rises with concentration even though molar conductivity falls [misconception D: “molar conductivity falls with concentration, so conductance must too”]. Remember: conductance depends on the ions in the solution, so electrolyte, concentration, solvent and temperature matter, but the electrode metal does not.
- Q15Using the standard electrode potential data given below, find out the most stable ion in its reduced form. Eᵒ_(Cr₂O₇²⁻/Cr³⁺) = 1.33 V Eᵒ_(Cl₂/Cl⁻) = 1.36 V Eᵒ_(MnO₄⁻/Mn²⁺) = 1.51 V Eᵒ_(Cr³⁺/Cr) = -0.74 V
- Cr
- Cr³⁺
- Cl⁻
- Mn²⁺
Answer: (D) Mn²⁺
D. Tests: connecting the size of a standard reduction potential to the stability of the reduced species. Why D: the higher the reduction potential of a couple, the harder its reduced form is to oxidise back, so the more stable that reduced form is. The largest value in the table is +1.51 V for MnO₄⁻/Mn²⁺, and its reduced form is Mn²⁺, so Mn²⁺ is the most stable reduced species here. Why not A: Cl⁻ comes from Cl₂/Cl⁻ at +1.36 V, below +1.51 V, so it is stable but not the most stable of the four. Why not B: Cr³⁺ is the reduced form of Cr₂O₇²⁻/Cr³⁺ at +1.33 V, the lowest of the three positive values, so it is the least stable of the positive set. Why not C: chromium metal is the reduced form of Cr³⁺/Cr at -0.74 V; a negative potential means it is oxidised most readily of all, making it the least stable reduced form in the list [misconception C: “the free metal is always the stable end of a couple”]. Remember: highest Eᵒ means the reduced form is hardest to push back up, so that reduced form is the most stable.
- Q16Given E°(Cu²⁺/Cu) = +0.337 V and E°(Cu²⁺/Cu⁺) = +0.153 V. The standard electrode potential E° for the half reaction Cu⁺ + e⁻ → Cu is:
- 0.38 V
- 0.52 V
- 0.90 V
- 0.30 V
Answer: (B) 0.52 V
Use ΔG° additivity. ΔG°(Cu²⁺→Cu) = ΔG°(Cu²⁺→Cu⁺) + ΔG°(Cu⁺→Cu): −2F(0.337) = −1F(0.153) − 1F(E°). So E° = 2(0.337) − 0.153 = 0.674 − 0.153 = 0.521 ≈ 0.52 V.
- Q17Which of the statements about solutions of electrolytes is not correct?
- Conductivity of solution depends upon size of ions.
- Conductivity does not depend upon solvation of ions present in solution.
- Conductivity depends upon viscosity of solution.
- Conductivity of solution increases with temperature.
Answer: (B) Conductivity does not depend upon solvation of ions present in solution.
C. Tests: the physical factors that control the conductivity of an electrolytic solution. Why C: conductivity is the conductance of a unit cube of the solution, so it is set by how many free ions sit in that volume and how fast each one moves. A solvated ion drags a shell of solvent molecules with it, its effective radius grows, its mobility falls and the conductivity falls with it. Solvation therefore does affect conductivity, so statement C is the incorrect one and is the answer to a “not correct” question. Why not A: ion size does control conductivity, because a larger effective radius means lower mobility, so A is a true statement [misconception A: “only the charge on an ion matters, not its size”]. Why not B: viscosity does control it, since ionic mobility is inversely proportional to the drag the solvent exerts, so B is true. Why not D: heating lowers viscosity and raises both ionic mobility and the degree of dissociation, so conductivity really does increase with temperature [misconception D: “solutions conduct like metals, so heating must reduce conduction”]. Remember: anything that slows an ion down, bigger size, thicker solvation shell, higher viscosity, cuts the conductivity.
- Q18The standard cell potential (E^(ominus)_(cell)) of a fuel cell based on the oxidation of methanol in air that has been used to power television relay station is measured as 1.21 V. The standard half cell reduction potential for O₂ (E°_(O₂/H₂O)) is 1.229 V. Choose the correct statement:
- Reduction of methanol takes place at the cathode.
- The standard half cell reduction potential for the reduction of CO₂ (E°_(CO₂/CH₃OH)) is 19 mV
- Oxygen is formed at the anode.
- Reactants are fed at one go to each electrode.
Answer: (B) The standard half cell reduction potential for the reduction of CO₂ (E°_(CO₂/CH₃OH)) is 19 mV
C. Tests: extracting an unknown anode potential from E^(ominus)_(cell), together with how a fuel cell is actually run. Why C: In this cell oxygen is reduced at the cathode and methanol is oxidised at the anode, so the anode couple is CO₂/CH₃OH. From E^(ominus)_(cell) = E^(ominus)_(cathode) – E^(ominus)_(anode), E°_(CO₂/CH₃OH) = 1.229 – 1.21 = 0.019 V, which is 19 mV. Why not A: a fuel cell is fed continuously with fuel and oxidant while it runs, and that is exactly what separates it from a primary cell that carries its reactants inside [misconception A: “a fuel cell stores its reactants like a battery”]. Why not B: oxygen is the oxidant here and is consumed by reduction at the cathode, it is not a product at the anode [misconception B: “oxygen is always an anode product”]. Why not D: methanol is the fuel, so it is oxidised, and oxidation happens at the anode [misconception D: “the fuel reacts at the cathode”]. Remember: E^(ominus)_(anode) = E^(ominus)_(cathode) – E^(ominus)_(cell), and in every fuel cell the fuel is oxidised at the anode.
- Q19A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A. The amount of the aluminium deposited at the cathode is __________. [Given: molar mass of aluminium and chlorine are 27 g mol⁻¹ and 35.5 g mol⁻¹ respectively. Faraday constant = 96500 C mol⁻¹]
- 0.441 g
- 1.660 g
- 0.336 g
- 1.007 g
Answer: (C) 0.336 g
C. Tests: Faraday’s first law with the correct electron count for a trivalent cation. Why C: Q = It = 2 × 30 × 60 = 3600 C, so moles of electrons = 3600/96500 = 0.0373 mol. The cathode reaction is Al³⁺ + 3e⁻ arrow Al, so moles of Al = 0.0373/3 = 0.01243 mol and mass = 0.01243 × 27 = 0.336 g. Why not A: 1.660 g comes from using the molar mass of AlCl₃, 27 + 3(35.5) = 133.5, in place of aluminium’s: 0.0373 × 133.5/3 = 1.66 g [misconception A: “the whole formula unit is deposited, not just the metal”]. Why not B: 1.007 g is 0.0373 × 27, that is taking n = 1 and forgetting that each aluminium atom needs three electrons. Why not D: 0.441 g is 0.0373 × 35.5/3, the same calculation run with chlorine’s molar mass; chlorine is released at the anode, never deposited at the cathode. Remember: mass = (It/F) × (M/n), and n is the charge on the ion actually being discharged at that electrode.
- Q20The molar conductivity of a 0.5 mol/dm³ solution of AgNO₃ with electrolytic conductivity of 5.76 × 10⁻³ S cm⁻¹ at 298 K is
- 11.52 S cm²/mol
- 2.88 S cm²/mol
- 28.8 S cm²/mol
- 0.086 S cm²/mol
Answer: (A) 11.52 S cm²/mol
B. Tests: converting molarity to mol cm⁻³ before applying Λₘ = (κ)/(c) so the units match a conductivity quoted per cm. Why B: c = 0.5 mol dm⁻³ = (0.5)/(1000) mol cm⁻³ = 5 × 10⁻⁴ mol cm⁻³. Then Λₘ = (κ)/(c) = (5.76 × 10⁻³)/(5 × 10⁻⁴) = 11.52 S cm² mol⁻¹. The same thing in the shortcut form, Λₘ = (1000κ)/(c) = (1000 × 5.76 × 10⁻³)/(0.5) = (5.76)/(0.5) = 11.52. Why not A: 1000 × 5.76 × 10⁻³ × 0.5 = 2.88, which is what you get by multiplying by the concentration instead of dividing by it [misconception A: “Λₘ = 1000κ c, because more concentrated solutions conduct more”]. Why not C: (0.5)/(1000 × 5.76 × 10⁻³) = (0.5)/(5.76) = 0.0868, the ratio taken upside down, (c)/(1000κ) instead of (1000κ)/(c) [misconception C: “the two quantities can be divided either way round since only the numbers matter, not the units”]. Why not D: no consistent substitution gives 28.8. Back-substitute it and you need c = (κ)/(Λₘ) = (5.76 × 10⁻³)/(28.8) = 2 × 10⁻⁴ mol cm⁻³, that is 0.2 mol dm⁻³, not the 0.5 given. The nearest real route is option A’s wrong product 2.88 with the decimal moved one place while juggling the 10⁻³ and the 1000 [misconception D: “a stray power of ten in the dm³ to cm³ conversion is harmless; note that actually using 10⁴ instead of 1000 would give 115.2, not 28.8″]. Remember: κ is quoted per cm, so the concentration must be per cm³; divide molarity by 1000 first, then divide κ by it.
- Q21E° for the cell Zn | Zn²⁺(aq) || Cu²⁺(aq) | Cu is 1.10 V at 25 °C. The equilibrium constant for the reaction Zn(s) + Cu²⁺(aq) ⇌ Cu(s) + Zn²⁺(aq) is of the order of:
- 10³⁷
- 10⁻¹⁸
- 10⁻²⁸
- 10¹⁷
Answer: (A) 10³⁷
log Keq = nE°/0.0591 with n = 2: log Keq = (2 × 1.10)/0.0591 = 2.20/0.0591 = 37.2. So Keq ≈ 10³⁷.
- Q22For a salt XY, which is a strong electrolyte, the plot of Λₘ versus √(c) has a slope of -90.0 S cm² mol^(-3/2) L^(1/2) at 298 K. At 0.01 M concentration of XY, the value of Λₘ is 145.0 S cm² mol⁻¹. The limiting molar conductivity of Y⁻ ion (λ⁰_(Y⁻), in S cm² mol⁻¹) at 298 K will be (Given: λ⁰_(X⁺) = 74.0 S cm² mol⁻¹)
- 76.0
- 80.0
- 90.0
- 100.0
Answer: (B) 80.0
B. Tests: reading Λ°ₘ off the Debye-Huckel-Onsager straight line, then splitting it with Kohlrausch’s law. Why B: For a strong electrolyte Λₘ = Λ°ₘ – a√(c), so the plotted slope is -a = -90.0. At c = 0.01 M, √(c) = 0.1, so the drop below the intercept is 90.0 × 0.1 = 9.0. Hence Λ°ₘ = 145.0 + 9.0 = 154.0. Kohlrausch’s law for XY gives Λ°ₘ = λ⁰_(X⁺) + λ⁰_(Y⁻), so λ⁰_(Y⁻) = 154.0 – 74.0 = 80.0 S cm² mol⁻¹. Why not A: 76.0 sits right next to the given λ⁰_(X⁺) = 74.0, which is where you land if you assume the two ions of a salt must contribute nearly the same amount; no consistent substitution of 145.0, 90.0 and 0.01 produces it [misconception A: “the cation and anion of a salt carry almost equal molar conductivities”]. Why not C: 100.0 would need Λ°ₘ = 174.0, and no combination of the slope and the concentration gives a correction of 29 units [misconception C: “the limiting molar conductivity can be rounded to a neat number instead of read off the intercept”]. Why not D: 90.0 is the slope magnitude copied straight into the answer, but the slope is a coefficient of √(c), not a conductivity of an ion [misconception D: “the slope of the plot is itself a molar conductivity”]. Remember: the intercept of the Λₘ against √(c) line is Λ°ₘ, so add back a√(c) before you use Kohlrausch’s law.
- Q23The standard EMF of a galvanic cell involving a cell reaction with n = 2 is found to be 0.295 V at 25 °C. The equilibrium constant of the reaction is (use 0.0591):
- 4.0 × 10¹²
- 1.0 × 10¹⁰
- 2.0 × 10¹¹
- 1.0 × 10²
Answer: (B) 1.0 × 10¹⁰
E°cell = (0.0591/n) log K ⇒ log K = nE°cell/0.0591 = (2 × 0.295)/0.0591 = 9.98 ≈ 10. So K = 1.0 × 10¹⁰.
- Q24At 298 K, the standard electrode potentials of Cu²⁺/Cu, Zn²⁺/Zn, Fe²⁺/Fe and Ag⁺/Ag are 0.34 V, -0.76 V, -0.44 V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction cannot occur?
- FeSO₄(aq) + Zn(s) arrow ZnSO₄(aq) + Fe(s)
- CuSO₄(aq) + Zn(s) arrow ZnSO₄(aq) + Cu(s)
- CuSO₄(aq) + Fe(s) arrow FeSO₄(aq) + Cu(s)
- 2CuSO₄(aq) + 2Ag(s) arrow 2Cu(s) + Ag₂SO₄(aq)
Answer: (D) 2CuSO₄(aq) + 2Ag(s) arrow 2Cu(s) + Ag₂SO₄(aq)
D. Tests: predicting whether a metal can displace another metal ion from standard electrode potentials. Why D: a metal displaces another metal’s ion only when its own couple has the lower reduction potential, so that it is the easier of the two to oxidise. Here Eᵒ_(Ag⁺/Ag) = 0.80 V is higher than Eᵒ_(Cu²⁺/Cu) = 0.34 V, so silver holds its electrons more tightly than copper does and cannot reduce Cu²⁺. The cell potential is 0.34 – 0.80 = -0.46 V, negative, so this reaction cannot occur. Why not A: zinc reducing Cu²⁺ gives 0.34 – (-0.76) = +1.10 V, positive, so it happens; this is the Daniell cell reaction itself. Why not B: iron reducing Cu²⁺ gives 0.34 – (-0.44) = +0.78 V, positive, so it happens. Why not C: zinc reducing Fe²⁺ gives -0.44 – (-0.76) = +0.32 V, positive, so it happens too [misconception C: “two negative potentials must make the reaction impossible”]. Remember: the metal with the more negative Eᵒ sits higher in the activity series and displaces the one below it, never the other way round.
- Q25The equivalent conductance of an M/32 solution of a weak monobasic acid is 8.0 S cm² and at infinite dilution it is 400 S cm². The dissociation constant of this acid is:
- 1.25 × 10⁻⁴
- 1.25 × 10⁻⁶
- 6.25 × 10⁻⁴
- 1.25 × 10⁻⁵
Answer: (D) 1.25 × 10⁻⁵
α = Λ/Λ° = 8.0/400 = 0.02. For a weak monobasic acid Ka = Cα²/(1−α) ≈ Cα² (α≪1). C = 1/32. Ka = (1/32)(0.02)² = (1/32)(4×10⁻⁴) = 1.25×10⁻⁵.