JEE Chemistry Electrochemistry MCQs Set 10

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Review structured MCQ sets for JEE Chemistry Electrochemistry. Built according to official JEE guidelines, these downloadable questions support daily revision and core concept reinforcement.

Chapter-wise Objective Questions: Electrochemistry

Navigate directly to the 50 objective questions for Electrochemistry using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question 1: If the limiting molar conductances of Ca²⁺ and Cl⁻ ions are 119 and 76.3 S cm² mol⁻¹ respectively, the limiting molar conductivity of CaCl₂ is :
(a) 347 S cm² mol⁻¹
(b) 271.6 S cm² mol⁻¹
(c) 42.7 S cm² mol⁻¹
(d) 195.3 S cm² mol⁻¹
Show Answer & Explanation

Answer: (b) 271.6 S cm² mol⁻¹

Explanation:
1. CaCl₂ gives one Ca²⁺ and two Cl⁻.
2. Λ°m = 119 + 2 × 76.3 = 271.6 S cm² mol⁻¹.

Teacher's Note:
1. Multiply each ionic value by the number of ions in the formula.
2. Kohlrausch's law for molar conductivity.

Question 2: Consider the change in oxidation state of bromine corresponding to different EMF values (reduction potentials) as shown : BrO₄⁻ → BrO₃⁻ (1.82 V), BrO₃⁻ → HBrO (1.5 V), HBrO → Br₂ (1.595 V), Br₂ → Br⁻ (1.0652 V). Then the species undergoing disproportionation is :
(a) BrO₃⁻
(b) BrO₄⁻
(c) Br₂
(d) HBrO
Show Answer & Explanation

Answer: (d) HBrO

Explanation:
1. A species disproportionates if its reduction potential to the lower state is greater than the potential for the oxidation to the upper state.
2. For HBrO, E° for HBrO → Br₂ is 1.595 V, which is greater than 1.5 V for BrO₃⁻ → HBrO, so HBrO disproportionates.

Teacher's Note:
1. The right-hand potential must exceed the left-hand potential.
2. Check each intermediate species in the same way.

Question 3: For an electrolytic solution of concentration 0.1 M, the resistance is 1 kΩ. If the distance between the electrodes is 100 cm and the area is 1 cm², calculate Λm in S cm² mol⁻¹.
(a) 100
(b) 1000
(c) 500
(d) 50
Show Answer & Explanation

Answer: (b) 1000

Explanation:
1. κ = (1/R)(ℓ/A) = (1/1000) × 100 = 0.1 S cm⁻¹.
2. Λm = κ × 1000 / M = 0.1 × 1000 / 0.1 = 1000 S cm² mol⁻¹.

Teacher's Note:
1. Cell constant is ℓ/A = 100 cm⁻¹.
2. Resistance 1 kΩ is 1000 Ω.

Question 4: What will be the value of Ecell for the reaction Mg(s) + 2Ag⁺ (10⁻³ M) ⇌ Mg²⁺ (0.02 M) + 2Ag(s), given E°cell is 3.17 V ?
(a) 2.67
(b) 2.13
(c) 3.27
(d) 3.041
Show Answer & Explanation

Answer: (d) 3.041

Explanation:
1. E = 3.17 − (0.0591/2) log(0.02/(10⁻³)²) = 3.17 − 0.02955 × log(2 × 10⁴).
2. log(2 × 10⁴) = 4.301, so E = 3.17 − 0.127 = 3.043 V ≈ 3.04 V.

Teacher's Note:
1. Substitute the ion concentrations into the reaction quotient.
2. The Ag⁺ concentration is squared.

Question 5: The specific conductance of 0.1 M nitric acid is 6.3 × 10⁻² ohm⁻¹ cm⁻¹. The molar conductance of the solution is :
(a) 630 ohm⁻¹ cm² mol⁻¹
(b) 315 ohm⁻¹ cm² mol⁻¹
(c) 100 ohm⁻¹ cm² mol⁻¹
(d) 6.300 ohm⁻¹ cm² mol⁻¹
Show Answer & Explanation

Answer: (a) 630 ohm⁻¹ cm² mol⁻¹

Explanation:
1. Λm = κ × 1000 / M = 6.3 × 10⁻² × 1000 / 0.1.
2. = 630 ohm⁻¹ cm² mol⁻¹.

Teacher's Note:
1. Multiply κ by 1000 and divide by the molarity.
2. Check that κ is in ohm⁻¹ cm⁻¹.

Question 6: The highest electrical conductivity of the following aqueous solutions is of :
(a) 0.1 M fluoroacetic acid
(b) 0.1 M difluoroacetic acid
(c) 0.1 M acetic acid
(d) 0.1 M chloroacetic acid
Show Answer & Explanation

Answer: (b) 0.1 M difluoroacetic acid

Explanation:
1. Fluorine is strongly electron withdrawing, and two fluorine atoms make difluoroacetic acid the strongest acid of the four.
2. It ionises most, giving the highest ion concentration and the highest conductivity.

Teacher's Note:
1. Conductivity of a weak electrolyte depends on its degree of ionisation.
2. More electron-withdrawing groups give a stronger acid.

Question 7: The resistance of a conductivity cell filled with a solution of an electrolyte of concentration 0.1 M is 100 Ω. The conductivity of this solution is 1.29 S m⁻¹. The resistance of the same cell when filled with 0.02 M of the same solution is 520 Ω. The molar conductivity of the 0.02 M solution of the electrolyte will be :
(a) 124 × 10⁻⁴ S m² mol⁻¹
(b) 1240 × 10⁻⁴ S m² mol⁻¹
(c) 1.24 × 10⁻⁴ S m² mol⁻¹
(d) 12.4 × 10⁻⁴ S m² mol⁻¹
Show Answer & Explanation

Answer: (a) 124 × 10⁻⁴ S m² mol⁻¹

Explanation:
1. Cell constant = κR = 1.29 × 100 = 129 m⁻¹.
2. For 0.02 M, κ = 129/520 = 0.2481 S m⁻¹, and Λm = κ / C = 0.2481 / 20 mol m⁻³ = 1.24 × 10⁻² S m² mol⁻¹ = 124 × 10⁻⁴ S m² mol⁻¹.

Teacher's Note:
1. In SI units the concentration must be in mol m⁻³ (0.02 M = 20 mol m⁻³).
2. Find the cell constant from the first solution.

Question 8: The equivalent conductance of NaCl at concentration C and at infinite dilution are λC and λ∞ respectively. The correct relationship between λC and λ∞ is given by (where the constant B is positive) :
(a) λC = λ∞ − B√C
(b) λC = λ∞ + B√C
(c) λC = λ∞ + B C
(d) λC = λ∞ − B C
Show Answer & Explanation

Answer: (a) λC = λ∞ − B√C

Explanation:
1. The Debye-Hückel-Onsager equation for strong electrolytes is Λ = Λ° − b√C.
2. So λC = λ∞ − B√C.

Teacher's Note:
1. A graph of Λ against √C is a straight line with negative slope for strong electrolytes.
2. The intercept gives the limiting value.

Question 9: Which of the following statements is TRUE for the electrochemical Daniell cell ?
(a) Electrons flow from the copper electrode to the zinc electrode
(b) Current flows from the zinc electrode to the copper electrode
(c) Cations move towards the copper electrode
(d) Cations move towards the zinc electrode
Show Answer & Explanation

Answer: (c) Cations move towards the copper electrode

Explanation:
1. Zn is the anode and Cu is the cathode.
2. Electrons flow from Zn to Cu and the current flows from Cu to Zn in the external wire, and cations move towards the cathode (Cu), so (c) is true.

Teacher's Note:
1. Cations go to the cathode and anions go to the anode in any cell.
2. Check electrons and current directions separately.

Question 10: Aluminium displaces hydrogen from dilute HCl whereas silver does not. The e.m.f. of a cell made by combining Al | Al³⁺ and Ag | Ag⁺ is 2.46 V. The reduction potential of silver electrode is +0.80 V. The reduction potential of the aluminium electrode is :
(a) −3.26 V
(b) +1.66 V
(c) −1.66 V
(d) +3.26 V
Show Answer & Explanation

Answer: (c) −1.66 V

Explanation:
1. Al is the anode and Ag is the cathode.
2. E°cell = 0.80 − E°(Al) = 2.46, so E°(Al³⁺/Al) = −1.66 V.

Teacher's Note:
1. Al displaces hydrogen, so its reduction potential is negative.
2. Solve for the unknown anode potential.

Question 11: For the redox reaction Zn(s) + Cu²⁺ (0.1 M) → Zn²⁺ (1 M) + Cu(s) taking place in a cell, E°cell is 1.10 V. Ecell for the cell will be (2.303 RT/F = 0.0591) :
(a) 1.07 V
(b) 0.82 V
(c) 2.14 V
(d) 1.80 V
Show Answer & Explanation

Answer: (a) 1.07 V

Explanation:
1. E = E° − (0.0591/2) log([Zn²⁺]/[Cu²⁺]) = 1.10 − 0.02955 × log(1/0.1).
2. = 1.10 − 0.02955 = 1.07 V.

Teacher's Note:
1. Product concentration greater than reactant concentration lowers E.
2. n = 2 for this cell.

Question 12: In a cell that utilises the reaction Zn(s) + 2H⁺(aq) ⇌ Zn²⁺(aq) + H₂(g), addition of H₂SO₄ to the cathode compartment will :
(a) increase the Ecell and shift the equilibrium to the right
(b) lower the Ecell and shift the equilibrium to the right
(c) lower the Ecell and shift the equilibrium to the left
(d) increase the Ecell and shift the equilibrium to the left
Show Answer & Explanation

Answer: (a) increase the Ecell and shift the equilibrium to the right

Explanation:
1. H⁺ is a reactant, and adding acid increases [H⁺] and so lowers Q.
2. E = E° − (0.0591/2) log Q rises, and the equilibrium shifts to the right.

Teacher's Note:
1. Raising a reactant concentration favours the forward reaction.
2. The emf rises as Q falls.

Question 13: The reduction potential of a hydrogen half cell will be negative if :
(a) p(H₂) = 2 atm and [H⁺] = 1.0 M
(b) p(H₂) = 2 atm and [H⁺] = 2.0 M
(c) p(H₂) = 1 atm and [H⁺] = 2.0 M
(d) p(H₂) = 1 atm and [H⁺] = 1.0 M
Show Answer & Explanation

Answer: (a) p(H₂) = 2 atm and [H⁺] = 1.0 M

Explanation:
1. E = −(0.0591/2) log (p(H₂)/[H⁺]²).
2. E is negative when p(H₂)/[H⁺]² > 1, which is true for p = 2 atm and [H⁺] = 1 M (ratio 2).

Teacher's Note:
1. Compare p(H₂) with [H⁺]².
2. Options (b) and (c) give a ratio below 1, option (d) gives E = 0.

Question 14: For a cell reaction involving a two-electron change, the standard e.m.f. of the cell is found to be 0.295 V at 25 °C. The equilibrium constant of the reaction at 25 °C will be :
(a) 10
(b) 1 × 10¹⁰
(c) 1 × 10⁻¹⁰
(d) 29.5 × 10⁻²
Show Answer & Explanation

Answer: (b) 1 × 10¹⁰

Explanation:
1. log K = nE° / 0.0591 = 2 × 0.295 / 0.0591 = 9.98.
2. K ≈ 10¹⁰.

Teacher's Note:
1. Use n = 2.
2. Round the logarithm to the nearest whole number.

Question 15: For a spontaneous reaction the ΔG, equilibrium constant (K) and E°cell will be respectively :
(a) −ve, < 1, −ve
(b) −ve, > 1, −ve
(c) −ve, > 1, +ve
(d) +ve, > 1, −ve
Show Answer & Explanation

Answer: (c) −ve, > 1, +ve

Explanation:
1. A spontaneous reaction has ΔG < 0.
2. Then E°cell = −ΔG°/nF > 0 and K > 1.

Teacher's Note:
1. ΔG = −nFE and ΔG° = −RT ln K.
2. All three quantities point the same way.

Question 16: On the basis of the information available from the reaction (4/3)Al + O₂ → (2/3)Al₂O₃, ΔG = −827 kJ mol⁻¹ of O₂, the minimum e.m.f. required to carry out an electrolysis of Al₂O₃ is (F = 96500 C mol⁻¹) :
(a) 2.14 V
(b) 4.28 V
(c) 6.42 V
(d) 8.56 V
Show Answer & Explanation

Answer: (a) 2.14 V

Explanation:
1. Four electrons are transferred per mole of O₂ in the reaction.
2. E = −ΔG / nF = 827000 / (4 × 96500) = 2.14 V.

Teacher's Note:
1. The minimum emf for electrolysis equals the emf of the reverse galvanic reaction.
2. n = 4 for one O₂.

Question 17: Given the half cell reactions Mn²⁺ + 2e⁻ → Mn, E° = −1.18 V and 2(Mn³⁺ + e⁻ → Mn²⁺), E° = +1.51 V, the E° for 3Mn²⁺ → Mn + 2Mn³⁺ will be :
(a) −0.33 V ; the reaction will not occur
(b) −0.33 V ; the reaction will occur
(c) −2.69 V ; the reaction will not occur
(d) −2.69 V ; the reaction will occur
Show Answer & Explanation

Answer: (c) −2.69 V ; the reaction will not occur

Explanation:
1. Mn²⁺ is both reduced to Mn (E° = −1.18 V) and oxidised to Mn³⁺ (reverse of +1.51 V).
2. E°cell = −1.18 − 1.51 = −2.69 V, which is negative, so the disproportionation will not occur.

Teacher's Note:
1. Add the reduction potential of the reduced step and the oxidation potential of the oxidised step.
2. A negative E° means the reaction is non-spontaneous.

Question 18: Given at 25 °C: Ag + I⁻ → AgI + e⁻, E° = 0.152 V and Ag → Ag⁺ + e⁻, E° = −0.800 V. What is the value of log Ksp for AgI ? (2.303 RT/F = 0.059 V)
(a) −8.12
(b) +8.612
(c) −37.83
(d) −16.13
Show Answer & Explanation

Answer: (d) −16.13

Explanation:
1. AgI(s) ⇌ Ag⁺ + I⁻ is obtained by adding Ag → Ag⁺ + e⁻ (−0.800 V) and AgI + e⁻ → Ag + I⁻ (−0.152 V).
2. E° = −0.952 V, so log Ksp = nE°/0.059 = −0.952 / 0.059 = −16.13.

Teacher's Note:
1. Reverse the first given reaction to get the reduction of AgI.
2. The solubility product is very small, so log Ksp is a large negative number.

Question 19: The metal that cannot be obtained by electrolysis of an aqueous solution of its salts is :
(a) Cu
(b) Cr
(c) Ag
(d) Ca
Show Answer & Explanation

Answer: (d) Ca

Explanation:
1. In an aqueous solution, water is reduced in preference to Ca²⁺ because Ca has a much lower reduction potential (−2.87 V).
2. So calcium cannot be obtained by electrolysis of aqueous solutions.

Teacher's Note:
1. Highly electropositive metals such as Na, K and Ca are obtained from molten salts.
2. Cu, Cr and Ag can be electroplated.

Question 20: Aluminium oxide may be electrolysed at 1000 °C to furnish aluminium metal (at. mass 27 amu; 1 faraday = 96500 C). The cathode reaction is Al³⁺ + 3e⁻ → Al. To prepare 5.12 kg of aluminium by this method would require :
(a) 5.49 × 10⁴ C of electric charge
(b) 5.49 × 10¹ C of electric charge
(c) 5.49 × 10⁷ C of electric charge
(d) 1.83 × 10⁷ C of electric charge
Show Answer & Explanation

Answer: (c) 5.49 × 10⁷ C of electric charge

Explanation:
1. 5.12 kg = 5120 g = 189.6 mol of Al, needing 3 F per mole, which is 568.9 F.
2. Charge = 568.9 × 96500 = 5.49 × 10⁷ C.

Teacher's Note:
1. Convert kilograms to grams before finding moles.
2. Each Al³⁺ needs 3 electrons.

Question 21: A current of i A was passed for t s through three cells P, Q and R connected in series. These contain respectively silver nitrate, mercuric nitrate and mercurous nitrate. At the cathode of cell P, 0.216 g of Ag was deposited. The weights of mercury deposited in the cathodes of Q and R respectively are (at. wt. of Hg = 200.59) :
(a) 0.4012 g and 0.8024 g
(b) 0.4012 g and 0.2006 g
(c) 0.2006 g and 0.4012 g
(d) 0.1003 g and 0.2006 g
Show Answer & Explanation

Answer: (c) 0.2006 g and 0.4012 g

Explanation:
1. 0.216 g Ag is 0.002 mol, so 0.002 F passed.
2. Hg²⁺ (cell Q) needs 2 F per mole, giving 0.001 mol = 0.2006 g, and Hg⁺ (cell R) needs 1 F per mole, giving 0.002 mol = 0.4012 g.

Teacher's Note:
1. Mercuric ion is Hg²⁺ and mercurous nitrate contains Hg⁺ (as Hg₂²⁺ in the dimer, the charge per Hg is 1).
2. The same charge passes through all cells in series.

Question 22: A galvanic cell is set up from a zinc bar weighing 100 g and 1.0 L of 1.0 M CuSO₄ solution. How long would the cell run if it is assumed to deliver a steady current of 1.0 A ? (atomic mass of Zn = 65)
(a) 1.1 hr
(b) 46 hr
(c) 53.6 hr
(d) 24.00 hr
Show Answer & Explanation

Answer: (c) 53.6 hr

Explanation:
1. Zn = 100/65 = 1.54 mol but Cu²⁺ = 1.0 mol only, so Cu²⁺ is the limiting reagent.
2. Charge available = 1.0 × 2 F = 193000 C, and t = 193000 / 1.0 A = 193000 s = 53.6 hours.

Teacher's Note:
1. Find the limiting reactant first.
2. The cell stops when either the zinc or Cu²⁺ is used up.

Question 23: The electrochemical equivalent of silver is 0.0011180 g. When an electric current of 0.5 A is passed through an aqueous silver nitrate solution for 200 s, the amount of silver deposited is :
(a) 1.1180 g
(b) 0.11180 g
(c) 5.590 g
(d) 0.5590 g
Show Answer & Explanation

Answer: (b) 0.11180 g

Explanation:
1. Charge = 0.5 × 200 = 100 C.
2. Mass = Z × Q = 0.0011180 × 100 = 0.1118 g.

Teacher's Note:
1. Electrochemical equivalent is the mass deposited per coulomb.
2. W = Z I t.

Question 24: A current of 9.65 A is passed through an aqueous solution of NaCl using suitable electrodes for 1000 s. Given that 1 faraday equals 96500 C, the amount of NaOH (mol. wt. 40.00) formed on electrolysis is :
(a) 2.0 g
(b) 8.0 g
(c) 4.0 g
(d) 1.0 g
Show Answer & Explanation

Answer: (c) 4.0 g

Explanation:
1. Charge = 9.65 × 1000 = 9650 C = 0.1 F, and one mole of OH⁻ (hence NaOH) is formed per faraday.
2. Mass of NaOH = 0.1 × 40 = 4.0 g.

Teacher's Note:
1. At the cathode, 2H₂O + 2e⁻ → H₂ + 2OH⁻.
2. One F gives one mole of OH⁻.

Question 25: An electric current is passed through a silver voltameter connected to a water voltameter. The cathode of the silver voltameter weighed 0.108 g more at the end of the electrolysis. The volume of oxygen evolved at STP is :
(a) 56 cm³
(b) 550 cm³
(c) 5.6 cm³
(d) 11.2 cm³
Show Answer & Explanation

Answer: (c) 5.6 cm³

Explanation:
1. 0.108 g Ag = 0.001 mol, so the charge is 0.001 F.
2. O₂ evolved = 0.001/4 = 0.00025 mol = 0.00025 × 22400 cm³ = 5.6 cm³.

Teacher's Note:
1. Four faradays liberate one mole of O₂.
2. Equal charge passes through both voltameters.

Question 26: The thermodynamic efficiency of a cell is given by :
(a) ΔH / ΔG
(b) nFE / ΔG
(c) −nFE / ΔH
(d) zero
Show Answer & Explanation

Answer: (c) −nFE / ΔH

Explanation:
1. Efficiency = (useful electrical work) / (enthalpy change) = ΔG / ΔH.
2. Since ΔG = −nFE, this equals −nFE / ΔH.

Teacher's Note:
1. Maximum work is −ΔG = nFE.
2. The enthalpy change is the total energy available.

Question 27: 1 L of 1 M CuSO₄ solution is electrolysed. After passing 2 F of charge, the molarity of CuSO₄ will be :
(a) M/2
(b) M/4
(c) M
(d) zero
Show Answer & Explanation

Answer: (d) zero

Explanation:
1. Cu²⁺ + 2e⁻ → Cu, so 2 F deposits 1 mol of Cu²⁺.
2. The solution contained only 1 mol, so all Cu²⁺ is used up and the molarity becomes zero.

Teacher's Note:
1. Compare the charge passed with the charge required for complete deposition.
2. Here both are equal.

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Practice MCQs for JEE Chemistry Electrochemistry

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