JEE Chemistry Electrochemistry MCQs Set 27

Multiple Choice Questions (MCQs) for JEE Chemistry: Electrochemistry

Explore reliable objective questions for Electrochemistry tailored for JEE learners. Utilizing these Chemistry multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Practice Electrochemistry MCQs for JEE Chemistry

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Multiple Choice Questions

Question 1: The standard Gibbs energy for the cell reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s), with E° = 2 V at 298 K (F = 96000 C mol⁻¹), in kJ mol⁻¹ is:
(a) 384
(b) 192
(c) −384
(d) −192
Show Answer & Explanation

Answer: (c) −384

Explanation:
1. \( \DeltaG^{\circ} \) = −nFE° with \( n = 2 \).
2. = \( -2 \times 96000 \times 2 = -384000 J = -384 \) kJ mol⁻¹.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 2: A solution of Ni(NO₃)₂ is electrolysed between platinum electrodes using 0.1 Faraday of electricity. The number of moles of Ni deposited at the cathode is:
(a) 0.20
(b) 0.10
(c) 0.15
(d) 0.05
Show Answer & Explanation

Answer: (d) 0.05

Explanation:
1. Ni²⁺ \( + 2e^{-} \) → Ni, so \( 2 F \) deposits 1 mol.
2. \( 0.1 F \) deposits 0.05 mol.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 3: The decreasing order of electrical conductivity of 0.1 M formic acid (A), 0.1 M acetic acid (B) and 0.1 M benzoic acid (C) is:
(a) A > C > B
(b) C > B > A
(c) A > B > C
(d) C > A > B
Show Answer & Explanation

Answer: (a) A > C > B

Explanation:
1. Conductivity follows the degree of ionisation, so it follows acid strength.
2. HCOOH > C₆H₅COOH > CH₃COOH.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 4: The standard cell potential of Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is 2 V at 300 K. Taking R = 8 J K⁻¹ mol⁻¹ and F = 96000 C mol⁻¹, the equilibrium constant is approximately:
(a) e⁻¹⁶⁰
(b) e¹⁶⁰
(c) e⁻⁸⁰
(d) e³²⁰
Show Answer & Explanation

Answer: (b) e¹⁶⁰

Explanation:
1. \( \ln K \) = nFE°/RT.
2. = \( 2 \times 96000 \times 2/(8 \times 300) = 160 \), so \( K = e^{160} \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 5: Reducing power of the metals Zn (E° = −0.76 V), Ca (−2.87 V), Mg (−2.36 V) and Ni (−0.25 V) increases in the order:
(a) Zn < Mg < Ni < Ca
(b) Ni < Zn < Mg < Ca
(c) Ca < Zn < Mg < Ni
(d) Ca < Mg < Zn < Ni
Show Answer & Explanation

Answer: (b) Ni < Zn < Mg < Ca

Explanation:
1. More negative \( E^{\circ} \) means a stronger reducing agent.
2. Ni (−0.25) < Zn (−0.76) < Mg (−2.36) < Ca (−2.87).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 6: For the cell Zn|Zn²⁺||Mˣ⁺|M with E°(Zn²⁺/Zn) = −0.76 V, which cathode gives the maximum E°cell per electron transferred? Data: Au³⁺/Au = 1.40 V, Ag⁺/Ag = 0.80 V, Fe³⁺/Fe²⁺ = 0.77 V, Fe²⁺/Fe = −0.44 V.
(a) Ag⁺/Ag
(b) Fe²⁺/Fe
(c) Au³⁺/Au
(d) Fe³⁺/Fe²⁺
Show Answer & Explanation

Answer: (a) Ag⁺/Ag

Explanation:
1. E°cell divided by \( n \): Ag \( 1.56/2 = 0.78 V \), Fe²⁺ \( 0.32/2 = 0.16 V \), Au \( 2.16/6 = 0.36 V \), Fe³⁺ \( 1.53/2 = 0.765 V \).
2. Ag⁺/Ag gives the highest value, \( 0.78 V \) per electron.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 7: Λ°m for NaCl, HCl and NaA are 126.4, 425.9 and 100.5 S cm² mol⁻¹. If the conductivity of 0.001 M HA is 5 × 10⁻⁵ S cm⁻¹, the degree of dissociation of HA is:
(a) 0.25
(b) 0.50
(c) 0.75
(d) 0.125
Show Answer & Explanation

Answer: (d) 0.125

Explanation:
1. Λ°(HA) = \( 425.9 + 100.5 - 126.4 = 400 \).
2. Λm = \( 1000 \times 5 \times 10^{-5}/10^{-3} = 50 \), so \( \alpha = 50/400 = 0.125 \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 8: Galvanisation is applying a coating of:
(a) Cr
(b) Cu
(c) Zn
(d) Pb
Show Answer & Explanation

Answer: (c) Zn

Explanation:
1. Iron is coated with zinc to protect it from rusting.
2. Zinc is more reactive, so it corrodes in preference to iron.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 9: The resistance of a 0.2 M solution of an electrolyte is 50 ohm and its specific conductance is 1.4 S m⁻¹. The resistance of a 0.5 M solution of the same electrolyte in the same cell is 280 ohm. The molar conductivity of the 0.5 M solution in S m² mol⁻¹ is:
(a) 5 × 10⁻⁴
(b) 5 × 10⁻³
(c) 5 × 10³
(d) 5 × 10²
Show Answer & Explanation

Answer: (a) 5 × 10⁻⁴

Explanation:
1. Cell constant = κR = \( 1.4 \times 50 = 70 m^{-1} \).
2. κ \( (0.5 M) = 70/280 = 0.25 S m^{-1} \); Λm = 0.25/500 mol \( m^{-3} = 5 \times 10^{-4} S m^{2} \) mol⁻¹.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 10: The relation between molar conductivity Λc at concentration C and limiting molar conductivity Λ∞ for a strong electrolyte such as NaCl (Debye-Huckel-Onsager equation) is:
(a) Λc = Λ∞ + B C
(b) Λc = Λ∞ − B √C
(c) Λc = Λ∞ + B √C
(d) Λc = Λ∞ − B C
Show Answer & Explanation

Answer: (b) Λc = Λ∞ − B √C

Explanation:
1. Strong electrolytes follow Λc = Λ∞ \( - B\sqrt{C} \).
2. Hence molar conductivity falls linearly with the square root of concentration.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 11: Mn²⁺ + 2e⁻ → Mn, E° = −1.18 V; Mn³⁺ + e⁻ → Mn²⁺, E° = +1.51 V. The E° for 3Mn²⁺ → Mn + 2Mn³⁺ is:
(a) −2.69 V, the reaction will not occur
(b) −2.69 V, the reaction will occur
(c) −0.33 V, the reaction will not occur
(d) −0.33 V, the reaction will occur
Show Answer & Explanation

Answer: (a) −2.69 V, the reaction will not occur

Explanation:
1. \( E^{\circ} \) = E°(reduction of Mn²⁺) + E°(oxidation of Mn²⁺ to Mn³⁺) = \( -1.18 + (-1.51) = -2.69 V \).
2. Negative \( E^{\circ} \) means non-spontaneous.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 12: The reduction potential of a hydrogen half-cell will be negative if:
(a) p(H₂) = 1 atm and [H⁺] = 2.0 M
(b) p(H₂) = 1 atm and [H⁺] = 1.0 M
(c) p(H₂) = 2 atm and [H⁺] = 1.0 M
(d) p(H₂) = 2 atm and [H⁺] = 2.0 M
Show Answer & Explanation

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

Explanation:
1. \( E = -0.0591 \log (p^\frac{1}{2}/[H^{+}]) \).
2. It is negative only when \( p^\frac{1}{2} \gt [H^{+}] \): \( 2^\frac{1}{2} = 1.41 \gt 1 \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 13: The Gibbs energy for the decomposition of Al₂O₃ at 500 °C is ΔrG = +966 kJ mol⁻¹ for 2/3 Al₂O₃ → 4/3 Al + O₂. The minimum potential difference needed for electrolytic reduction is:
(a) 4.5 V
(b) 3.0 V
(c) 2.5 V
(d) 5.0 V
Show Answer & Explanation

Answer: (c) 2.5 V

Explanation:
1. For this equation \( n = 4 \).
2. \( E = 966000/(4 \times 96500) = 2.5 V \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 14: Given E°(Fe³⁺/Fe) = −0.036 V and E°(Fe²⁺/Fe) = −0.439 V, the standard electrode potential for Fe³⁺ + e⁻ → Fe²⁺ is:
(a) −0.072 V
(b) 0.385 V
(c) 0.770 V
(d) −0.270 V
Show Answer & Explanation

Answer: (c) 0.770 V

Explanation:
1. Potentials are not additive, but free energies are: \( \DeltaG = \DeltaG(1) - \DeltaG(2) \).
2. \( \DeltaG_{3} = \DeltaG_{1} - \DeltaG_{2} \) gives \( -E = -3(-0.036) + 2(-0.439) \), so \( E = 3(-0.036) - 2(-0.439) = +0.770 V \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 15: Λ° for CH₃COONa and HCl at infinite dilution are 91.0 and 426.2 S cm² equiv⁻¹. What additional quantity is needed to calculate Λ° of acetic acid?
(a) Λ° of NaCl
(b) Λ° of CH₃COOK
(c) Λ° of H⁺
(d) Λ° of chloroacetic acid
Show Answer & Explanation

Answer: (a) Λ° of NaCl

Explanation:
1. Λ°(CH₃COOH) = Λ°(CH₃COONa) + Λ°(HCl) − Λ°(NaCl).
2. So Λ°(NaCl) is required.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 16: The resistance of a conductivity cell containing 0.1 M electrolyte solution is 100 ohm and its conductivity is 1.29 S m⁻¹. When filled with 0.2 M solution of the same electrolyte, the resistance is 520 ohm. The molar conductivity of the 0.2 M solution is:
(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: (d) 12.4 × 10⁻⁴ S m² mol⁻¹

Explanation:
1. Cell constant = \( 1.29 \times 100 = 129 m^{-1} \); κ = \( 129/520 = 0.248 S m^{-1} \).
2. Λm = \( 0.248/200 = 1.24 \times 10^{-3} = 12.4 \times 10^{-4} S m^{2} \) mol⁻¹.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 17: The highest electrical conductivity is shown by which 0.1 M aqueous solution?
(a) difluoroacetic acid
(b) fluoroacetic acid
(c) chloroacetic acid
(d) acetic acid
Show Answer & Explanation

Answer: (a) difluoroacetic acid

Explanation:
1. Two fluorines give the strongest \( -I \) effect and hence the greatest ionisation.
2. More ions means higher conductivity.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 18: Λ∞ values (S cm² mol⁻¹): KCl 149.9, KNO₃ 145.0, HCl 426.2, NaOAc 91.0, NaCl 126.5. Λ∞ for acetic acid in S cm² mol⁻¹ is:
(a) 217.5
(b) 390.7
(c) 552.7
(d) 517.2
Show Answer & Explanation

Answer: (b) 390.7

Explanation:
1. Λ∞(HOAc) = Λ∞(NaOAc) + Λ∞(HCl) − Λ∞(NaCl).
2. = \( 91.0 + 426.2 - 126.5 = 390.7 \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 19: In a hydrogen-oxygen fuel cell, combustion of hydrogen occurs to:
(a) generate heat
(b) create a potential difference between the two electrodes
(c) produce high purity water
(d) remove adsorbed oxygen from electrode surfaces
Show Answer & Explanation

Answer: (b) create a potential difference between the two electrodes

Explanation:
1. A fuel cell converts chemical energy directly into electrical energy.
2. The redox reactions at the electrodes set up a potential difference.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 20: Given E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(Sn²⁺/Sn) = −0.14 V, the standard potential for Sn(s) + 2Fe³⁺(aq) → 2Fe²⁺(aq) + Sn²⁺(aq) is:
(a) 1.68 V
(b) 1.40 V
(c) 0.91 V
(d) 0.63 V
Show Answer & Explanation

Answer: (c) 0.91 V

Explanation:
1. E°cell = E°(cathode) − E°(anode) = 0.77 − (−0.14).
2. = \( 0.91 V \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 21: The standard emf of a cell involving one electron change is 0.591 V at 25 °C. The equilibrium constant of the reaction is (F = 96500 C mol⁻¹):
(a) 1.0 × 10¹
(b) 1.0 × 10⁵
(c) 1.0 × 10¹⁰
(d) 1.0 × 10³⁰
Show Answer & Explanation

Answer: (c) 1.0 × 10¹⁰

Explanation:
1. \( E^{\circ} = 0.0591 \log K \) for \( n = 1 \).
2. \( \log K = 10 \), so \( K = 10^{10} \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 22: When 9650 C of charge passes through an electroplating bath during electrolysis of AgNO₃, the mass of silver deposited at the cathode is:
(a) 1.08 g
(b) 10.8 g
(c) 21.6 g
(d) 108 g
Show Answer & Explanation

Answer: (b) 10.8 g

Explanation:
1. Moles of electrons = \( 9650/96500 = 0.1 \).
2. Ag⁺ \( + e^{-} \) → Ag, so 0.1 mol × \( 108 = 10.8 g \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 23: For Zn(s) + Cu²⁺(0.1 M) → Zn²⁺(1 M) + Cu(s), E°cell = 1.10 V. Ecell is (2.303RT/F = 0.0591):
(a) 2.14 V
(b) 1.80 V
(c) 1.07 V
(d) 0.82 V
Show Answer & Explanation

Answer: (c) 1.07 V

Explanation:
1. \( Q \) = [Zn²⁺]/[Cu²⁺] = \( 1/0.1 = 10 \).
2. \( E = 1.10 - (0.0591/2) \log 10 = 1.10 - 0.0295 = 1.0705 V \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 24: Conductivity S is directly proportional to area of the vessel and concentration of solution and inversely proportional to the length of the vessel. The constant of proportionality is expressed in:
(a) S m⁻¹
(b) S² m² mol⁻²
(c) S m² mol⁻¹
(d) S² m² mol
Show Answer & Explanation

Answer: (c) S m² mol⁻¹

Explanation:
1. \( S ∝ \) area \( (m^{2}) \) × conc (mol \( m^{-3} \)) / length \( (m) \).
2. So k = S × m / (m² × mol m⁻³) = S m² mol⁻¹, the unit of molar conductivity.

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 25: For a cell with hydrogen electrodes at two different pressures p₁ and p₂, Pt(H₂)p₁|H⁺(aq, 1 M)|Pt(H₂)p₂, the emf is given by:
(a) (RT/F) ln(p₁/p₂)
(b) (RT/2F) ln(p₁/p₂)
(c) (RT/F) ln(p₂/p₁)
(d) (RT/2F) ln(p₂/p₁)
Show Answer & Explanation

Answer: (b) (RT/2F) ln(p₁/p₂)

Explanation:
1. Cell reaction \( H_{2}(p_{1}) \to H_{2}(p_{2}), n = 2, E^{\circ} = 0 \).
2. \( E \) = −(RT/2F) \( \ln(p_{2}/p_{1}) \) = (RT/2F) \( \ln(p_{1}/p_{2}) \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

Question 26: The limiting molar conductivities Λ° for NaCl, KBr and KCl are 126, 152 and 150 S cm² mol⁻¹. The Λ° for NaBr is:
(a) 128 S cm² mol⁻¹
(b) 176 S cm² mol⁻¹
(c) 278 S cm² mol⁻¹
(d) 302 S cm² mol⁻¹
Show Answer & Explanation

Answer: (a) 128 S cm² mol⁻¹

Explanation:
1. Kohlrausch's law: Λ°(NaBr) = Λ°(NaCl) + Λ°(KBr) − Λ°(KCl).
2. = \( 126 + 152 - 150 = 128 \).

Teacher's Note:
1. Apply the definition carefully.
2. Check each option against the data.

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FAQs

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