Electrochemistry NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Electrochemistry question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Electrochemistry previous year questions
44 questions. Tap an option to check your answer.
Q1NEET 2021
The molar conductance of NaCl, HCl and CH₃COONa at infinite dilution are 126.45, 426.16 and 91.0 S cm² mol⁻¹ respectively. The molar conductance of CH₃COOH at infinite dilution is. Choose the right option for your answer.
Not quite — the answer is B.
Kohlrausch's law: Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl) = 91.0 + 426.16 − 126.45 = 390.71 S cm² mol⁻¹. Simply adding all three gives 643.61, which is not the right combination.
Q2NEET 2021
The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. What is the dissociation constant of acetic acid ? Choose the correct option. [Λ°(H⁺) = 350 S cm² mol⁻¹, Λ°(CH₃COO⁻) = 50 S cm² mol⁻¹]
Not quite — the answer is C.
Λ°m = 350 + 50 = 400 S cm² mol⁻¹, so α = Λm/Λ°m = 20/400 = 0.05. Ka = cα²/(1 − α) ≈ 0.007 × 0.0025 = 1.75 × 10⁻⁵ mol L⁻¹ (exactly 1.84 × 10⁻⁵ without approximation, still nearest).
Q3NEET 2020 · Q99
On electrolysis of dil. sulphuric acid using Platinum (Pt) electrode, the product obtained at anode will be :
Not quite — the answer is B.
In dilute H₂SO₄, oxidation of water (2H₂O → O₂ + 4H⁺ + 4e⁻) is easier than oxidation of sulphate, so O₂ is liberated at the inert Pt anode. Hydrogen is formed at the cathode, not the anode.
Q4NEET 2020 · Q106
The number of Faradays (F) required to produce 20 g of calcium from molten CaCl₂ (Atomic mass of Ca = 40 g mol⁻¹) is :
Not quite — the answer is A.
Ca²⁺ + 2e⁻ → Ca, so each mole of Ca needs 2 F. 20 g is 20/40 = 0.5 mol, which needs 0.5 × 2 = 1 F. Using 2 F straight from the equation ignores that only half a mole is produced.
Q5NEET 2019 · Q53
For the cell reaction 2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq), E°cell = 0.24 V at 298 K. The standard Gibbs energy (ΔrG°) of the cell reaction is: [Given that Faraday constant F = 96500 C mol⁻¹]
Not quite — the answer is A.
Two electrons are transferred, so ΔG° = −nFE° = −2 × 96500 × 0.24 = −46320 J = −46.32 kJ mol⁻¹. Using n = 1 gives −23.16 kJ mol⁻¹, and a positive E° must give a negative ΔG°.
Q6NEET 2019 · Q82
For a cell involving one electron E°cell = 0.59 V at 298 K, the equilibrium constant for the cell reaction is: [Given that 2.303RT/F = 0.059 V at T = 298 K]
Not quite — the answer is C.
log K = nE°/0.059 = 1 × 0.59/0.059 = 10, so K = 1.0 × 10¹⁰. 10³⁰ would correspond to three electrons.
Q7NEET 2017 · Q25
In the electrochemical cell: Zn|ZnSO₄ (0.01 M)||CuSO₄ (1.0 M)|Cu, the emf of this Daniel cell is E₁. When the concentration of ZnSO₄ is changed to 1.0 M and that of CuSO₄ changed to 0.01 M, the emf changes to E₂. From the followings, which one is the relationship between E₁ and E₂? (Given, RT/F = 0.059)
Not quite — the answer is C.
E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]). For E₁ the ratio is 0.01/1 so E₁ = E° + 0.059 V; for E₂ the ratio is 1/0.01 so E₂ = E° − 0.059 V. Hence E₁ > E₂; reversing the concentrations does not make the emf zero.
Q8NEET 2017 · Q37
Ionic mobility of which of the following alkali metal ions is lowest when aqueous solution of their salts are put under an electric field?
Not quite — the answer is D.
Li⁺ is the smallest alkali metal ion, so it has the highest charge density and is the most heavily hydrated. Its large hydrated radius makes it move most slowly in solution, so Li⁺ has the lowest ionic mobility even though the bare ion is the smallest.
Q9NEET 2017 (Gujarati paper) · Q28
Given that Λ°ₘ = 133.4 (AgNO₃); Λ°ₘ = 149.9 (KCl); Λ°ₘ = 144.9 S cm² mol⁻¹ (KNO₃), the molar conductivity at infinite dilution for AgCl is:
Not quite — the answer is B.
By Kohlrausch's law, Λ°(AgCl) = Λ°(AgNO₃) + Λ°(KCl) − Λ°(KNO₃) = 133.4 + 149.9 − 144.9 = 138.4 ≈ 138 S cm² mol⁻¹. Adding all three, or subtracting the wrong salt, gives none of the ion combinations needed.
Q10NEET 2016 Phase 2 · Q47
The 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
Not quite — the answer is B.
Λm = κ × 1000/M = 5.76 × 10⁻³ × 1000/0.5 = 11.52 S cm² mol⁻¹. Multiplying by the concentration instead of dividing gives 2.88 S cm²/mol.
Q11NEET 2016 Phase 2 · Q56
If the E°cell for a given reaction has a negative value, which of the following gives the correct relationships for the values of ΔG° and K_eq?
Not quite — the answer is A.
ΔG° = −nFE°cell, so a negative E° gives a positive ΔG°. Also ΔG° = −RT ln K, so a positive ΔG° means ln K < 0, i.e. K < 1. A negative ΔG° and K > 1 belong to a positive E°.
Q12NEET 2016 Phase 2 · Q60
The number of electrons delivered at the cathode during electrolysis by a current of 1 ampere in 60 seconds is (charge on electron = 1.60 × 10⁻¹⁹ C)
Not quite — the answer is C.
Charge passed Q = It = 1 × 60 = 60 C. Number of electrons = 60/(1.60 × 10⁻¹⁹) = 3.75 × 10²⁰. 6 × 10²³ is close to Avogadro's number and would need about one faraday of charge.
Q13NEET 2016 Phase 2 · Q63
Zinc can be coated on iron to produce galvanized iron but the reverse is not possible. It is because
Not quite — the answer is D.
E°(Zn²⁺/Zn) = −0.76 V is more negative than E°(Fe²⁺/Fe) = −0.44 V, so zinc is oxidised in preference and protects iron sacrificially. An iron coating on zinc would not protect the zinc, since iron is less easily oxidised. Melting point and density are not the reason.
Q14NEET 2016 Phase 1 · Q75
The pressure of H₂ required to make the potential of H₂-electrode zero in pure water at 298 K is:
Not quite — the answer is A.
For 2H⁺ + 2e⁻ → H₂, E = 0 − (0.059/2) log(p_H₂/[H⁺]²). In pure water [H⁺] = 10⁻⁷ M, so E = 0 requires p_H₂ = [H⁺]² = 10⁻¹⁴ atm. Using [H⁺] instead of [H⁺]² would give 10⁻⁷ atm, which is not even offered.
Q15AIPMT 2015
A device that converts energy of combustion of fuels like hydrogen and methane, directly into electrical energy is known as:
Not quite — the answer is B.
A fuel cell converts the free energy of combustion of fuels like H₂ or CH₄ directly into electricity. A Ni–Cd cell is a rechargeable storage cell, and a dynamo converts mechanical energy, not chemical.
Q16AIPMT 2014
The weight of silver (at. wt. = 108) displaced by a quantity of electricity which displaces 5600 mL of O₂ at STP will be:
Not quite — the answer is C.
5600 mL O₂ at STP = 0.25 mol; since 2H₂O → O₂ + 4H⁺ + 4e⁻, this requires 1 mol electrons = 1 F. One faraday deposits 1 mol Ag = 108 g.
Q17AIPMT 2014
When 0.1 mol MnO₄²⁻ is oxidised the quantity of electricity required to completely oxidise MnO₄²⁻ to MnO₄⁻ is:
Not quite — the answer is B.
MnO₄²⁻ → MnO₄⁻ + e⁻ (Mn +6 to +7), one electron per ion. 0.1 mol needs 0.1 F = 0.1 × 96 500 = 9650 C.
Q18NEET 2013
A hydrogen gas electrode is made by dipping platinum wire in a solution of HCl of pH = 10 and by passing hydrogen gas around the platinum wire at one atm pressure. The oxidation potential of electrode would be
Not quite — the answer is B.
For H₂ → 2H⁺ + 2e⁻ at 1 atm, E(oxidation) = 0 − (0.059/2) log[H⁺]² = −0.059 log[H⁺] = 0.059 × pH. With pH 10: 0.059 × 10 = 0.59 V. 0.059 V would correspond to pH 1.
Q19NEET 2013
At 25°C molar conductance of 0.1 molar aqueous solution of ammonium hydroxide is 9.54 ohm⁻¹ cm² mol⁻¹ and at infinite dilution its molar conductance is 238 ohm⁻¹ cm² mol⁻¹. The degree of ionisation of ammonium hydroxide at the same concentration and temperature is
Not quite — the answer is C.
Degree of ionisation α = Λm/Λ°m = 9.54/238 = 0.04008, i.e. 4.008%. 40.8% comes from misplacing the decimal.
Q20NEET 2013
A button cell used in watches functions as following: Zn(s) + Ag₂O(s) + H₂O(l) ⇌ 2Ag(s) + Zn²⁺(aq) + 2OH⁻(aq) If half cell potentials are Zn²⁺(aq) + 2e⁻ → Zn(s); E° = −0.76 V Ag₂O(s) + H₂O(l) + 2e⁻ → 2Ag(s) + 2OH⁻(aq); E° = 0.34 V The cell potential will be
Not quite — the answer is A.
Zinc is oxidised (anode) and Ag₂O reduced (cathode). E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V. Subtracting the magnitudes gives 0.42 V, which ignores the sign of the zinc potential.
Q21AIPMT 2012 Mains
Standard reduction potentials of the half reactions are given below: F₂(g) + 2e⁻ → 2F⁻(aq); E° = +2.85 V Cl₂(g) + 2e⁻ → 2Cl⁻(aq); E° = +1.36 V Br₂(l) + 2e⁻ → 2Br⁻(aq); E° = +1.06 V I₂(s) + 2e⁻ → 2I⁻(aq); E° = +0.53 V The strongest oxidising and reducing agents respectively are
Not quite — the answer is A.
The highest reduction potential (F₂, +2.85 V) marks the strongest oxidising agent. The lowest reduction potential (I₂/I⁻, +0.53 V) means I⁻ is most easily oxidised, so I⁻ is the strongest reducing agent; I₂ itself is an oxidant, not a reductant.
Q22AIPMT 2012 Mains
Molar conductivities (Λ°m) at infinite dilution of NaCl, HCl and CH₃COONa are 126.4, 425.9 and 91.0 S cm² mol⁻¹ respectively. Λ°m for CH₃COOH will be
Not quite — the answer is D.
By Kohlrausch's law, Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl) = 91.0 + 425.9 − 126.4 = 390.5 S cm² mol⁻¹.
Q23AIPMT 2012 Prelims
Limiting molar conductivity of NH₄OH, i.e. Λ°m(NH₄OH), is equal to
Not quite — the answer is A.
By Kohlrausch's law, Λ°m(NH₄OH) = λ°(NH₄⁺) + λ°(OH⁻) = [λ°(NH₄⁺) + λ°(Cl⁻)] + [λ°(Na⁺) + λ°(OH⁻)] − [λ°(Na⁺) + λ°(Cl⁻)] = Λ°m(NH₄Cl) + Λ°m(NaOH) − Λ°m(NaCl). The combination subtracting NaOH would cancel the OH⁻ contribution instead.
Q24AIPMT 2011 Prelims
The electrode potentials for Cu²⁺(aq) + e⁻ → Cu⁺(aq) and Cu⁺(aq) + e⁻ → Cu(s) are +0.15 V and +0.50 V respectively. The value of E°(Cu²⁺/Cu) will be
Not quite — the answer is C.
Potentials are not additive but free energies are: ΔG° = −nFE°. −2F·E° = −1F(0.15) − 1F(0.50), so E°(Cu²⁺/Cu) = 0.65/2 = 0.325 V. Simply adding the potentials gives 0.65 V, the trap.
Q25AIPMT 2011 Prelims
Standard electrode potential for Sn⁴⁺/Sn²⁺ couple is +0.15 V and that for the Cr³⁺/Cr couple is −0.74 V. These two couples in their standard state are connected to make a cell. The cell potential will be
Not quite — the answer is C.
The couple with higher reduction potential, Sn⁴⁺/Sn²⁺, is the cathode and Cr is the anode. E°cell = 0.15 − (−0.74) = +0.89 V. Multiplying by stoichiometric coefficients (to get 1.83 V) is wrong since E° is intensive.
Q26AIPMT 2011 Prelims
If the E°cell for a given reaction has a negative value, then which of the following gives the correct relationships for the values of ΔG° and Keq?
Not quite — the answer is A.
ΔG° = −nFE°cell, so a negative E°cell makes ΔG° positive. Since ΔG° = −RT ln K, a positive ΔG° means ln K < 0, i.e. K < 1: the reaction is non-spontaneous as written.
Q27AIPMT 2011 Prelims
Standard electrode potential of three metals X, Y and Z are –1.2 V, +0.5 V and –3.0 V respectively. The reducing power of these metals will be
Not quite — the answer is D.
The more negative the standard reduction potential, the more readily the metal is oxidised, so the stronger the reducing agent. Order: Z (−3.0 V) > X (−1.2 V) > Y (+0.5 V).
Q28AIPMT 2010 Mains · Q101
Consider the following relations for emf of an electrochemical cell: (a) emf of cell = (Oxidation potential of anode) − (Reduction potential of cathode) (b) emf of cell = (Oxidation potential of anode) + (Reduction potential of cathode) (c) emf of cell = (Reduction potential of anode) + (Reduction potential of cathode) (d) emf of cell = (Oxidation potential of anode) − (Oxidation potential of cathode) Which of the above relations are correct?
Not quite — the answer is D.
emf = E(red, cathode) − E(red, anode). Since oxidation potential = −reduction potential, this equals (oxidation potential of anode) + (reduction potential of cathode), and also (oxidation potential of anode) − (oxidation potential of cathode). The relation subtracting the reduction potential of the cathode from the anode's oxidation potential gets the sign wrong.
Q29AIPMT 2010 Prelims · Q51
An increase in equivalent conductance of a strong electrolyte with dilution is mainly due to
Not quite — the answer is B.
A strong electrolyte is already fully ionised, so dilution does not add ions. Dilution weakens interionic attractions, so the ions move faster: the increase is due to greater ionic mobility. More ions is the explanation for weak electrolytes.
Q30AIPMT 2010 Prelims · Q61
For the reduction of silver ions with copper metal, the standard cell potential was found to be +0.46 V at 25°C. The value of standard Gibbs energy, ΔG° will be (F = 96500 C mol⁻¹)
Not quite — the answer is B.
Cu + 2Ag⁺ → Cu²⁺ + 2Ag involves n = 2. ΔG° = −nFE° = −2 × 96500 × 0.46 = −88780 J ≈ −89.0 kJ. Taking n = 1 gives −44.5 kJ; −89.0 J has the wrong unit.
Q31AIPMT 2009 Prelims · Q158
Given: (i) Cu²⁺ + 2e⁻ → Cu, E° = 0.337 V (ii) Cu²⁺ + e⁻ → Cu⁺, E° = 0.153 V Electrode potential, E° for the reaction, Cu⁺ + e⁻ → Cu, will be:
Not quite — the answer is D.
Combine free energies, not potentials: ΔG° for Cu⁺ + e⁻ → Cu = ΔG°(i) − ΔG°(ii) = (−2 × 0.337F) − (−1 × 0.153F) = −0.521F. With n = 1, E° = +0.52 V. Simply subtracting the potentials gives 0.18 V and adding them gives 0.49 V, both wrong.
Q32AIPMT 2009 Prelims · Q171
Al₂O₃ is reduced by electrolysis at low potentials and high currents. If 4.0 × 10⁴ amperes of current is passed through molten Al₂O₃ for 6 hours, what mass of aluminium is produced? (Assume 100% current efficiency. At. mass of Al = 27 g mol⁻¹)
Not quite — the answer is A.
Charge Q = 4.0 × 10⁴ × 6 × 3600 = 8.64 × 10⁸ C. Al³⁺ + 3e⁻ → Al, so moles of Al = 8.64 × 10⁸/(3 × 96500) ≈ 2984 mol, mass ≈ 2984 × 27 ≈ 8.1 × 10⁴ g. Forgetting the factor 3 gives 2.4 × 10⁵ g.
Q33AIPMT 2009 Prelims · Q189
The equivalent conductance of M/32 solution of a weak monobasic acid is 8.0 mhos cm² and at infinite dilution is 400 mhos cm². The dissociation constant of this acid is:
Not quite — the answer is D.
Degree of dissociation α = 8.0/400 = 0.02. Ka = Cα²/(1 − α) ≈ Cα² = (1/32)(0.02)² = 1.25 × 10⁻⁵. Forgetting to square α gives 6.25 × 10⁻⁴.
Q34AIPMT 2008 Prelims · Q55
Kohlrausch's law states that at
Not quite — the answer is A.
Kohlrausch's law of independent migration says that at infinite dilution each ion contributes a fixed amount to the limiting equivalent (or molar) conductivity, whatever the other ion is. It does not hold at finite dilution, where interionic attractions matter, and it is stated for equivalent/molar conductance, not plain conductance.
Q35AIPMT 2008 Prelims · Q67
Standard free energies of formation (in kJ/mol) at 298 K are −237.2, −394.4 and −8.2 for H₂O(l), CO₂(g) and pentane(g) respectively. The value of E°cell for the pentane-oxygen fuel cell is
Not quite — the answer is D.
C₅H₁₂ + 8O₂ → 5CO₂ + 6H₂O. ∆G° = 5(−394.4) + 6(−237.2) − (−8.2) = −3387.0 kJ. Eight O₂ molecules take 4 electrons each, so n = 32, and E° = −∆G°/nF = 3387000/(32 × 96500) ≈ 1.0968 V. Using a smaller n (for example 16) would double the voltage to about 2.19 V, which is not offered.
Q36AIPMT 2008 Prelims · Q69
The sequence of ionic mobility in aqueous solution is
Not quite — the answer is C.
Smaller ions have higher charge density and hold a larger shell of water molecules, so the hydrated Na⁺ ion is the bulkiest and moves slowest. Mobility therefore increases down the group: Cs⁺ > Rb⁺ > K⁺ > Na⁺. Na⁺ > ... > Cs⁺ follows bare ionic size and ignores hydration.
Q37AIPMT 2008 Prelims · Q76
On the basis of the following E° values, the strongest oxidizing agent is [Fe(CN)₆]⁴⁻ → [Fe(CN)₆]³⁻ + e⁻; E° = −0.35 V Fe²⁺ → Fe³⁺ + e⁻; E° = −0.77 V
Not quite — the answer is D.
The values given are oxidation potentials, so the reduction potentials are +0.35 V for [Fe(CN)₆]³⁻/[Fe(CN)₆]⁴⁻ and +0.77 V for Fe³⁺/Fe²⁺. The species with the higher reduction potential is the stronger oxidising agent, so Fe³⁺. [Fe(CN)₆]⁴⁻ and Fe²⁺ are the reduced forms and act as reducing agents.
Q38AIPMT 2007 Prelims · Q21
A steady current of 1.5 amp flows through a copper voltameter for 10 minutes. If the electrochemical equivalent of copper is 30 × 10⁻⁵ g coulomb⁻¹, the mass of copper deposited on the electrode will be
Not quite — the answer is C.
Faraday's first law: m = ZIt = 30 × 10⁻⁵ × 1.5 × 600 = 0.27 g. Forgetting to convert 10 minutes to 600 s gives a far smaller value.
Q39AIPMT 2007 Prelims · Q56
The efficiency of a fuel cell is given by
Not quite — the answer is B.
A fuel cell converts the free-energy change into electrical work, while the total energy released is ΔH; efficiency = ΔG/ΔH. Ratios involving ΔS are not an efficiency.
Q40AIPMT 2007 Prelims · Q91
The equilibrium constant of the reaction: Cu(s) + 2Ag⁺(aq) → Cu²⁺(aq) + 2Ag(s) ; E° = 0.46 V at 298 K is
Not quite — the answer is C.
log K = nE°/0.0591 = 2 × 0.46/0.0591 = 15.57, so K = 10¹⁵·⁵⁷ ≈ 3.7 × 10¹⁵ ≈ 4.0 × 10¹⁵. Taking n = 1 would give a value near 10⁸.
Q41AIPMT 2006 Prelims · Q1
In producing chlorine through electrolysis 100 W power at 125 V is being consumed. How much chlorine per min is liberated ? ECE of chlorine is 0.367 × 10⁻⁶ kg/C :
Not quite — the answer is A.
The current drawn is I = P/V = 100/125 = 0.8 A, so the charge passed in one minute is 0.8 × 60 = 48 C. By Faraday's first law m = ZQ = 0.367 × 10⁻⁶ × 48 ≈ 1.76 × 10⁻⁵ kg = 17.6 mg. Forgetting the 60 s factor or using V instead of I gives none of the other values.
Q42AIPMT 2006 Prelims · Q67
If E°(Fe²⁺/Fe) = −0.441 V and E°(Fe³⁺/Fe²⁺) = 0.771 V, the standard emf of the reaction : Fe + 2Fe³⁺ → 3Fe²⁺ will be :
Not quite — the answer is C.
Fe is oxidised to Fe²⁺ (anode) and Fe³⁺ is reduced to Fe²⁺ (cathode). E°cell = E°cathode − E°anode = 0.771 − (−0.441) = 1.212 V. Electrode potentials are intensive, so they are not multiplied by 2; simply adding them gives the wrong 0.33 V.
Q43AIPMT 2006 Prelims · Q85
A hypothetical electrochemical cell is shown below A | A⁺ (xM) || B⁺ (yM) | B The emf measured is +0.20 V. The cell reaction is :
Not quite — the answer is D.
In the cell notation the left electrode is the anode (oxidation) and the right the cathode (reduction). A positive emf of +0.20 V means this cell is spontaneous as written: A → A⁺ + e⁻ and B⁺ + e⁻ → B, i.e. A + B⁺ → A⁺ + B. The reverse reaction would have a negative emf.
Q44AIPMT 2005 Prelims · Q98
4.5 g of aluminium (at. mass 27 amu) is deposited at cathode from Al³⁺ solution by a certain quantity of electric charge. The volume of hydrogen produced at STP from H⁺ ions in solution by the same quantity of electric charge will be :
Not quite — the answer is C.
4.5 g Al = 1/6 mol; Al³⁺ needs 3 F per mol, so charge = 0.5 F. 0.5 F liberates 0.25 mol H₂ (2 F per mol H₂) = 0.25 × 22.4 = 5.6 L at STP. 11.2 L forgets that H₂ needs two electrons per molecule.
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