Electrochemistry Practice

Take timed practice tests on Electrochemistry for JEE Main and JEE Advanced with session-wise drills, score review, and explanation-led revision.

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Electrochemistry Practice Tests

Six 20-question timed sessions plus a 60-question chapter module. Each item is original and reframed for copyright safety.

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JEE Main Session 1

5Questions

60s/QTimer

+4 / -1Scoring

Preview all 18 questions in JEE Main Session 1 (no login required)

Electrolytic cells
1. In an electrolytic cell, oxidation occurs at the:
- A. Anode (Correct)
- B. Cathode
- C. Both electrodes equally
- D. Electrolyte
Explanation: In both electrolytic and galvanic cells: Anode = Oxidation, Cathode = Reduction. Mnemonic: A-O-C-R (Anode-Oxidation, Cathode-Reduction).
Faraday's first law
2. The mass deposited in electrolysis is proportional to:
- A. Charge passed (Correct)
- B. Voltage only
- C. Resistance only
- D. Temperature
Explanation: Faraday's First Law: m = ZIt = (M/nF)Q. Mass is proportional to the charge Q = It passed through the electrolyte.
Standard electrode potential
3. The standard hydrogen electrode (SHE) has a potential of:
- A. 0 V by convention (Correct)
- B. +1 V
- C. −1 V
- D. +0.5 V
Explanation: The standard hydrogen electrode (SHE) is the reference electrode with E° = 0.00 V by convention. All other electrode potentials are measured relative to SHE.
Cell EMF
4. E°cell = E°cathode − E°anode. For Zn-Cu cell, E°Zn/Zn²⁺ = −0.76 V and E°Cu²⁺/Cu = +0.34 V. E°cell is:
- A. +1.10 V (Correct)
- B. −1.10 V
- C. +0.42 V
- D. −0.42 V
Explanation: Zn is the anode (oxidised, more negative E°) and Cu is cathode. E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V.
Conductance
5. The unit of molar conductivity is:
- A. S cm² mol⁻¹ (Correct)
- B. S m⁻¹
- C. Ω cm
- D. S
Explanation: Molar conductivity Λm = κ/c where κ is specific conductivity (S cm⁻¹) and c is concentration (mol cm⁻³). Units: S cm⁻¹/(mol cm⁻³) = S cm² mol⁻¹.
Nernst equation
6. The Nernst equation for a cell reaction at 298 K is E = E° − (0.0592/n) log Q. For n = 2 and Q = 100, E − E° is:
- A. −0.0592 V (Correct)
- B. +0.0592 V
- C. −0.1184 V
- D. +0.1184 V
Explanation: E − E° = −(0.0592/2) log 100 = −(0.0296)(2) = −0.0592 V.
Kohlrausch's law
7. Kohlrausch's law states that at infinite dilution, the molar conductivity equals:
- A. Sum of ionic conductivities of cation and anion (Correct)
- B. Conductivity of cation alone
- C. Specific conductivity
- D. Resistance × area
Explanation: Kohlrausch's Law: Λ°m = λ°cation + λ°anion. At infinite dilution, each ion contributes independently to molar conductivity.
ΔG and EMF
8. The relationship between ΔG° and E°cell is:
- A. ΔG° = −nFE°cell (Correct)
- B. ΔG° = nFE°cell
- C. ΔG° = −RTlnK
- D. ΔG° = −E°cell/nF
Explanation: ΔG° = −nFE°cell, where n = moles of electrons transferred, F = Faraday constant (96485 C/mol). Combining with ΔG° = −RTlnK gives E°cell = (RT/nF)lnK.
Faraday's second law
9. When the same charge passes through AgNO₃ and CuSO₄ solutions, the ratio of Ag:Cu deposited (by moles) is:
- A. 2:1 (Correct)
- B. 1:2
- C. 1:1
- D. 3:1
Explanation: Faraday's Second Law: masses deposited are proportional to equivalent weights. n-factor of Ag⁺ = 1, n-factor of Cu²⁺ = 2. For same charge: moles Ag = 2 × moles Cu. Ratio Ag:Cu = 2:1.
Electrolysis calculation
10. 0.965 A current passes for 100 s through CuSO₄ solution. Mass of Cu deposited (M = 64, n = 2) is:
- A. 0.032 g (Correct)
- B. 0.64 g
- C. 0.128 g
- D. 0.016 g
Explanation: Q = It = 0.965 × 100 = 96.5 C. m = (M/nF) × Q = (64/(2×96500)) × 96.5 = (64/193000) × 96.5 = 0.032 g.
Battery chemistry
11. In a lead storage battery, which reaction occurs at the anode during discharge?
- A. Pb → PbSO₄ (Correct)
- B. PbO₂ → PbSO₄
- C. H₂SO₄ → H₂O
- D. Pb → Pb²⁺
Explanation: During discharge: Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ (oxidation). Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O (reduction).
JEE Main
12. At equilibrium, E_cell is:
- A. 0 (Correct)
- B. E°cell
- C. ∞
- D. −E°cell
Explanation: At equilibrium, ΔG = 0 → Ecell = 0. The Nernst equation: E = E° − (RT/nF)lnK = 0 at equilibrium because E° = (RT/nF)lnK.
Conductance variation
13. For a strong electrolyte, as concentration decreases (dilution increases), molar conductivity:
- A. Increases, approaching Λ°m (Correct)
- B. Decreases
- C. Remains constant
- D. First increases then decreases
Explanation: Molar conductivity Λm increases on dilution and approaches the limiting molar conductivity Λ°m at infinite dilution (Debye-Hückel-Onsager equation: Λm = Λ°m − b√c).
JEE Advanced
14. E°cell for the reaction 2Ag⁺ + H₂ → 2Ag + 2H⁺ at 25°C if E°Ag⁺/Ag = +0.80 V is:
- A. +0.80 V (Correct)
- B. +1.60 V
- C. −0.80 V
- D. 0 V
Explanation: Cathode: Ag⁺ + e⁻ → Ag (E° = +0.80 V). Anode: H₂ → 2H⁺ + 2e⁻ (E° = 0 V). E°cell = 0.80 − 0 = 0.80 V. Note: E° doesn't change when the equation is multiplied.
JEE Advanced
15. The degree of dissociation of an electrolyte can be determined from conductance by:
- A. α = Λm/Λ°m (Correct)
- B. α = Λ°m/Λm
- C. α = Λm × c
- D. α = 1/Λm
Explanation: For weak electrolytes, the degree of dissociation at concentration c: α = Λm/Λ°m. As dilution increases, α increases from near 0 towards 1.
Mixed
16. Which of the following is used as electrolyte in the hydrogen fuel cell?
- A. Aqueous KOH or H₃PO₄ (Correct)
- B. NaCl solution
- C. H₂SO₄ only
- D. Pure water
Explanation: Hydrogen fuel cells use alkaline (KOH) or acidic (H₃PO₄ or PEM membrane) electrolytes. Both H₂ (fuel) and O₂ (oxidant) are supplied externally.
Mixed
17. The EMF of the cell: Zn(s)|Zn²⁺(0.01 M)||Cu²⁺(1 M)|Cu(s) compared to E°cell (+1.10 V) at 298 K is:
- A. Greater than 1.10 V (Correct)
- B. Equal to 1.10 V
- C. Less than 1.10 V
- D. Zero
Explanation: Q = [Zn²⁺]/[Cu²⁺] = 0.01/1 = 0.01 1.10 V.
Mixed
18. 1 Faraday of electricity deposits 108 g of Ag (n=1). If 0.5 Faraday is passed, mass of Ag deposited is:
- A. 54 g (Correct)
- B. 108 g
- C. 27 g
- D. 216 g
Explanation: By Faraday's law: m ∝ charge. 1 F deposits 108 g. 0.5 F deposits 108 × 0.5 = 54 g.

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JEE Main Session 1

5Questions

60s/QTimer

+4 / -1Scoring

Preview all 18 questions in JEE Main Session 1 (no login required)

Electrolytic cells
1. In an electrolytic cell, oxidation occurs at the:
- A. Anode (Correct)
- B. Cathode
- C. Both electrodes equally
- D. Electrolyte
Explanation: In both electrolytic and galvanic cells: Anode = Oxidation, Cathode = Reduction. Mnemonic: A-O-C-R (Anode-Oxidation, Cathode-Reduction).
Faraday's first law
2. The mass deposited in electrolysis is proportional to:
- A. Charge passed (Correct)
- B. Voltage only
- C. Resistance only
- D. Temperature
Explanation: Faraday's First Law: m = ZIt = (M/nF)Q. Mass is proportional to the charge Q = It passed through the electrolyte.
Standard electrode potential
3. The standard hydrogen electrode (SHE) has a potential of:
- A. 0 V by convention (Correct)
- B. +1 V
- C. −1 V
- D. +0.5 V
Explanation: The standard hydrogen electrode (SHE) is the reference electrode with E° = 0.00 V by convention. All other electrode potentials are measured relative to SHE.
Cell EMF
4. E°cell = E°cathode − E°anode. For Zn-Cu cell, E°Zn/Zn²⁺ = −0.76 V and E°Cu²⁺/Cu = +0.34 V. E°cell is:
- A. +1.10 V (Correct)
- B. −1.10 V
- C. +0.42 V
- D. −0.42 V
Explanation: Zn is the anode (oxidised, more negative E°) and Cu is cathode. E°cell = E°cathode − E°anode = 0.34 − (−0.76) = 1.10 V.
Conductance
5. The unit of molar conductivity is:
- A. S cm² mol⁻¹ (Correct)
- B. S m⁻¹
- C. Ω cm
- D. S
Explanation: Molar conductivity Λm = κ/c where κ is specific conductivity (S cm⁻¹) and c is concentration (mol cm⁻³). Units: S cm⁻¹/(mol cm⁻³) = S cm² mol⁻¹.
Nernst equation
6. The Nernst equation for a cell reaction at 298 K is E = E° − (0.0592/n) log Q. For n = 2 and Q = 100, E − E° is:
- A. −0.0592 V (Correct)
- B. +0.0592 V
- C. −0.1184 V
- D. +0.1184 V
Explanation: E − E° = −(0.0592/2) log 100 = −(0.0296)(2) = −0.0592 V.
Kohlrausch's law
7. Kohlrausch's law states that at infinite dilution, the molar conductivity equals:
- A. Sum of ionic conductivities of cation and anion (Correct)
- B. Conductivity of cation alone
- C. Specific conductivity
- D. Resistance × area
Explanation: Kohlrausch's Law: Λ°m = λ°cation + λ°anion. At infinite dilution, each ion contributes independently to molar conductivity.
ΔG and EMF
8. The relationship between ΔG° and E°cell is:
- A. ΔG° = −nFE°cell (Correct)
- B. ΔG° = nFE°cell
- C. ΔG° = −RTlnK
- D. ΔG° = −E°cell/nF
Explanation: ΔG° = −nFE°cell, where n = moles of electrons transferred, F = Faraday constant (96485 C/mol). Combining with ΔG° = −RTlnK gives E°cell = (RT/nF)lnK.
Faraday's second law
9. When the same charge passes through AgNO₃ and CuSO₄ solutions, the ratio of Ag:Cu deposited (by moles) is:
- A. 2:1 (Correct)
- B. 1:2
- C. 1:1
- D. 3:1
Explanation: Faraday's Second Law: masses deposited are proportional to equivalent weights. n-factor of Ag⁺ = 1, n-factor of Cu²⁺ = 2. For same charge: moles Ag = 2 × moles Cu. Ratio Ag:Cu = 2:1.
Electrolysis calculation
10. 0.965 A current passes for 100 s through CuSO₄ solution. Mass of Cu deposited (M = 64, n = 2) is:
- A. 0.032 g (Correct)
- B. 0.64 g
- C. 0.128 g
- D. 0.016 g
Explanation: Q = It = 0.965 × 100 = 96.5 C. m = (M/nF) × Q = (64/(2×96500)) × 96.5 = (64/193000) × 96.5 = 0.032 g.
Battery chemistry
11. In a lead storage battery, which reaction occurs at the anode during discharge?
- A. Pb → PbSO₄ (Correct)
- B. PbO₂ → PbSO₄
- C. H₂SO₄ → H₂O
- D. Pb → Pb²⁺
Explanation: During discharge: Anode: Pb + SO₄²⁻ → PbSO₄ + 2e⁻ (oxidation). Cathode: PbO₂ + SO₄²⁻ + 4H⁺ + 2e⁻ → PbSO₄ + 2H₂O (reduction).
JEE Main
12. At equilibrium, E_cell is:
- A. 0 (Correct)
- B. E°cell
- C. ∞
- D. −E°cell
Explanation: At equilibrium, ΔG = 0 → Ecell = 0. The Nernst equation: E = E° − (RT/nF)lnK = 0 at equilibrium because E° = (RT/nF)lnK.
Conductance variation
13. For a strong electrolyte, as concentration decreases (dilution increases), molar conductivity:
- A. Increases, approaching Λ°m (Correct)
- B. Decreases
- C. Remains constant
- D. First increases then decreases
Explanation: Molar conductivity Λm increases on dilution and approaches the limiting molar conductivity Λ°m at infinite dilution (Debye-Hückel-Onsager equation: Λm = Λ°m − b√c).
JEE Advanced
14. E°cell for the reaction 2Ag⁺ + H₂ → 2Ag + 2H⁺ at 25°C if E°Ag⁺/Ag = +0.80 V is:
- A. +0.80 V (Correct)
- B. +1.60 V
- C. −0.80 V
- D. 0 V
Explanation: Cathode: Ag⁺ + e⁻ → Ag (E° = +0.80 V). Anode: H₂ → 2H⁺ + 2e⁻ (E° = 0 V). E°cell = 0.80 − 0 = 0.80 V. Note: E° doesn't change when the equation is multiplied.
JEE Advanced
15. The degree of dissociation of an electrolyte can be determined from conductance by:
- A. α = Λm/Λ°m (Correct)
- B. α = Λ°m/Λm
- C. α = Λm × c
- D. α = 1/Λm
Explanation: For weak electrolytes, the degree of dissociation at concentration c: α = Λm/Λ°m. As dilution increases, α increases from near 0 towards 1.
Mixed
16. Which of the following is used as electrolyte in the hydrogen fuel cell?
- A. Aqueous KOH or H₃PO₄ (Correct)
- B. NaCl solution
- C. H₂SO₄ only
- D. Pure water
Explanation: Hydrogen fuel cells use alkaline (KOH) or acidic (H₃PO₄ or PEM membrane) electrolytes. Both H₂ (fuel) and O₂ (oxidant) are supplied externally.
Mixed
17. The EMF of the cell: Zn(s)|Zn²⁺(0.01 M)||Cu²⁺(1 M)|Cu(s) compared to E°cell (+1.10 V) at 298 K is:
- A. Greater than 1.10 V (Correct)
- B. Equal to 1.10 V
- C. Less than 1.10 V
- D. Zero
Explanation: Q = [Zn²⁺]/[Cu²⁺] = 0.01/1 = 0.01 1.10 V.
Mixed
18. 1 Faraday of electricity deposits 108 g of Ag (n=1). If 0.5 Faraday is passed, mass of Ag deposited is:
- A. 54 g (Correct)
- B. 108 g
- C. 27 g
- D. 216 g
Explanation: By Faraday's law: m ∝ charge. 1 F deposits 108 g. 0.5 F deposits 108 × 0.5 = 54 g.