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NEET Redox Reactions and Electrochemistry MCQs with answers

596 questions on Redox Reactions and Electrochemistry for NEET 2027, 1 of them from previous NEET papers. Below are 15 to try; tap a question to see its answer and explanation.

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15 sample questions

easy

1. What is the nominal cell potential typically delivered by a fresh Leclanché dry cell?

  1. 1.10 V
  2. 1.35 V
  3. 1.50 V
  4. 2.00 V
Show answer and explanation

Answer: (C) 1.50 V

A fresh Leclanché dry cell has an open-circuit potential of approximately 1.5 V, whereas a mercury cell delivers around 1.35 V and a single lead-acid cell delivers roughly 2.0 V.

medium

2. For a hydrogen-oxygen fuel cell operating at 298 K, the standard Gibbs free energy change ΔG° is -237.2 kJ mol⁻¹ of H₂O formed. What is the standard cell potential (E°_cell) for this fuel cell? (1 F = 96500 C mol⁻¹)

  1. +0.615 V
  2. +1.23 V
  3. +2.46 V
  4. +1.56 V
Show answer and explanation

Answer: (B) +1.23 V

For the reaction H₂(g) + 0.5 O₂(g) → H₂O(l), n = 2 electrons. Using ΔG° = -nFE°_cell, E°_cell = -(-237200 J) / (2 × 96500 C) = 237200 / 193000 ≈ +1.23 V.

hard

3. The limiting equivalent conductivities of Al³⁺ and SO₄²⁻ ions are 63.0 S cm² eq⁻¹ and 80.0 S cm² eq⁻¹ respectively. The limiting molar conductivity (Λ°m) of aluminium sulphate, Al₂(SO₄)₃, is:

  1. 858.0 S cm² mol⁻¹
  2. 429.0 S cm² mol⁻¹
  3. 143.0 S cm² mol⁻¹
  4. 286.0 S cm² mol⁻¹
Show answer and explanation

Answer: (A) 858.0 S cm² mol⁻¹

Limiting molar ionic conductivities are λ°(Al³⁺) = 3 × 63.0 = 189.0 S cm² mol⁻¹ and λ°(SO₄²⁻) = 2 × 80.0 = 160.0 S cm² mol⁻¹. Then, Λ°m(Al₂(SO₄)₃) = 2 × λ°(Al³⁺) + 3 × λ°(SO₄²⁻) = 2(189.0) + 3(160.0) = 378.0 + 480.0 = 858.0 S cm² mol⁻¹. Alternatively, Λ°eq(Al₂(SO₄)₃) = 63.0 + 80.0 = 143.0 S cm² eq⁻¹, and Λ°m = 6 × 143.0 = 858.0 S cm² mol⁻¹.

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4. During the discharging of a lead storage battery, how many moles of water (H₂O) are produced per mole of lead oxidized at the anode?

  1. 0.5 mole
  2. 1.0 mole
  3. 4.0 mole
  4. 2.0 mole
Show answer and explanation

Answer: (D) 2.0 mole

The overall discharging reaction is: Pb(s) + PbO₂(s) + 2H₂SO₄(aq) → 2PbSO₄(s) + 2H₂O(l). Here, 1 mole of Pb is oxidized at the anode, yielding exactly 2 moles of H₂O in the overall cell reaction.

easy

5. Given the following limiting ionic conductivities at 298 K: λ°(Ca²⁺) = 119.0 S cm² mol⁻¹ λ°(Cl⁻) = 76.3 S cm² mol⁻¹ λ°(Na⁺) = 50.1 S cm² mol⁻¹ λ°(SO₄²⁻) = 160.0 S cm² mol⁻¹ The limiting molar conductivity (Λm°) of CaSO₄ at 298 K is:

  1. 195.3 S cm² mol⁻¹
  2. 279.0 S cm² mol⁻¹
  3. 355.3 S cm² mol⁻¹
  4. 139.5 S cm² mol⁻¹
Show answer and explanation

Answer: (B) 279.0 S cm² mol⁻¹

According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of CaSO₄ is the sum of limiting molar conductivities of its constituent ions: Λm°(CaSO₄) = λ°(Ca²⁺) + λ°(SO₄²⁻) = 119.0 + 160.0 = 279.0 S cm² mol⁻¹.

hard

6. A galvanic cell consists of a standard silver electrode and a standard hydrogen electrode: Pt | H₂(g, 1 bar) | H⁺(1 M) || Ag⁺(x M) | Ag(s) If E°(Ag⁺/Ag) = +0.80 V, at what concentration of Ag⁺ will the cell potential become exactly 0.623 V at 298 K? (Take 2.303 RT/F = 0.059 V)

  1. 1.0 × 10⁻² M
  2. 1.0 × 10⁻³ M
  3. 2.0 × 10⁻³ M
  4. 1.0 × 10⁻⁴ M
Show answer and explanation

Answer: (B) 1.0 × 10⁻³ M

Ecell = E°cell - (0.059/1) log(1/[Ag⁺]) = E°cell + 0.059 log[Ag⁺]. Substituting values: 0.623 = 0.80 + 0.059 log[Ag⁺] → 0.059 log[Ag⁺] = -0.177 → log[Ag⁺] = -3. Therefore, [Ag⁺] = 10⁻³ M.

medium

7. Given below are two statements regarding commercial cells: Statement I: A Leclanché dry cell has a finite life even when not in use because the acidic NH₄Cl electrolyte slowly corrodes the zinc container. Statement II: In a lead storage battery, the positive plate consists of a lead grid packed with finely divided spongy lead.

  1. Both Statement I and Statement II are correct
  2. Statement I is correct but Statement II is incorrect
  3. Statement I is incorrect but Statement II is correct
  4. Both Statement I and Statement II are incorrect
Show answer and explanation

Answer: (B) Statement I is correct but Statement II is incorrect

Statement I is correct because NH₄Cl is acidic (salt of weak base NH₄OH and strong acid HCl) and corrodes the zinc casing over time. Statement II is incorrect because the positive plate consists of a lead grid packed with lead dioxide (PbO₂), while the negative plate contains spongy lead.

medium

8. A galvanic cell has E°cathode = 1 V and E°anode = -0.1 V, both quoted as reduction potentials. The standard cell potential is:

  1. 1.1 V
  2. 0.9 V
  3. 0.55 V
  4. -1.1 V
Show answer and explanation

Answer: (A) 1.1 V

E°cell = E°cathode − E°anode = 1 − (-0.1) = 1.1 V.

Quick tip: When both values are reduction potentials: cathode minus anode.

medium

9. A conductivity cell having two parallel electrodes of area 2.0 cm² separated by a distance of 1.2 cm is filled with an electrolytic solution. If the measured resistance of the solution is 300 Ω, what is the conductivity of this solution?

  1. 2.0 × 10⁻³ S cm⁻¹
  2. 5.0 × 10⁻³ S cm⁻¹
  3. 1.8 × 10⁻³ S cm⁻¹
  4. 3.6 × 10⁻² S cm⁻¹
Show answer and explanation

Answer: (A) 2.0 × 10⁻³ S cm⁻¹

Cell constant G* = l / A = 1.2 cm / 2.0 cm² = 0.6 cm⁻¹. Conductivity κ = G* / R = 0.6 cm⁻¹ / 300 Ω = 2.0 × 10⁻³ S cm⁻¹ (or Ω⁻¹ cm⁻¹).

medium

10. During the recharging of a lead-acid accumulator, what is the role of the electrode that was functioning as the negative electrode during discharge?

  1. It acts as the cathode and undergoes oxidation
  2. It acts as the cathode and undergoes reduction
  3. It acts as the anode and undergoes oxidation
  4. It acts as the anode and undergoes reduction
Show answer and explanation

Answer: (B) It acts as the cathode and undergoes reduction

During discharging, the negative plate is the anode where Pb is oxidized to PbSO₄. During recharging, it is connected to the negative terminal of the DC source and functions as an electrolytic cathode where PbSO₄ is reduced back to metallic lead (PbSO₄ + 2e⁻ → Pb + SO₄²⁻).

easy

11. In a standard Leclanché dry cell, manganese is reduced during discharge. What is the change in the oxidation state of manganese in this process?

  1. +4 to +2
  2. +7 to +2
  3. +4 to +3
  4. +6 to +3
Show answer and explanation

Answer: (C) +4 to +3

In a Leclanché cell, manganese dioxide (MnO₂) where Mn has an oxidation state of +4 is reduced at the cathode to MnO(OH), where Mn has an oxidation state of +3. The cathodic reaction is MnO₂ + NH₄⁺ + e⁻ → MnO(OH) + NH₃.

easy

12. For a spontaneous cell reaction involving the transfer of 2 moles of electrons (n=2), the standard cell EMF is measured as E⁰ = 1.1 V. Its standard Gibbs free energy change ΔG⁰ in kJ/mol is:

  1. -212 kJ/mol
  2. 212 kJ/mol
  3. -106 kJ/mol
  4. -50 kJ/mol
Show answer and explanation

Answer: (A) -212 kJ/mol

Using the thermodynamic relationship ΔG⁰ = -nFE⁰. Here, n = 2, F = 96500 C/mol = 96.5 kJ/(V·mol), and E⁰ = 1.1 V. Thus, ΔG⁰ = -2 × 96.5 × 1.1 = -212 kJ/mol.

Quick tip: Gibbs free energy change is negative for spontaneous cells with positive E⁰.

easy

13. In alkaline medium, potassium permanganate oxidises iodide ion to iodate ion and itself gets reduced to manganese dioxide: a MnO₄⁻ + b I⁻ + c H₂O → d MnO₂ + e IO₃⁻ + f OH⁻. The values of a, b, and c are:

  1. 2, 2, 1
  2. 1, 1, 2
  3. 2, 1, 1
  4. 2, 1, 2
Show answer and explanation

Answer: (C) 2, 1, 1

Reduction half-reaction: MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻ (multiply by 2). Oxidation half-reaction: I⁻ + 6OH⁻ → IO₃⁻ + 3H₂O + 6e⁻ (multiply by 1). Combining gives: 2MnO₄⁻ + I⁻ + H₂O → 2MnO₂ + IO₃⁻ + 2OH⁻. Thus, a = 2, b = 1, c = 1.

hard

14. The solubility product (Ksp) of AgCl(s) at 298 K can be determined using a galvanic cell. Given: Ag⁺(aq) + e⁻ → Ag(s), E° = +0.80 V AgCl(s) + e⁻ → Ag(s) + Cl⁻(aq), E° = +0.21 V What is the value of log₁₀ Ksp for AgCl at 298 K? (Take 2.303 RT/F = 0.059 V)

  1. -5.0
  2. -10.0
  3. -12.5
  4. -15.2
Show answer and explanation

Answer: (B) -10.0

For dissolution: AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq). Cathode: AgCl(s) + e⁻ → Ag(s) + Cl⁻, Anode: Ag(s) → Ag⁺ + e⁻. So E°cell = 0.21 - 0.80 = -0.59 V. Using E°cell = (0.059/1) log₁₀ Ksp, we get -0.59 = 0.059 log₁₀ Ksp, which gives log₁₀ Ksp = -10.0.

easy

15. Which of the following statements correctly differentiates metallic conductors from electrolytic conductors?

  1. Metallic conduction involves transfer of matter, whereas electrolytic conduction does not.
  2. Metallic conductors obey Ohm's law, whereas electrolytic conductors strictly violate it.
  3. Electrolytic conduction increases with increase in temperature, whereas metallic conduction decreases.
  4. The charge carriers in electrolytic conductors are electrons, whereas in metallic conductors they are ions.
Show answer and explanation

Answer: (C) Electrolytic conduction increases with increase in temperature, whereas metallic conduction decreases.

With rising temperature, the thermal vibrations of metal kernels increase, creating greater resistance and lowering metallic conductivity. In electrolytic solutions, higher temperature decreases viscosity and increases ionic mobility, thereby increasing electrolytic conductivity.

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