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Nernst equation and Gibbs-energy relation: NEET MCQs with answers

89 questions on this topic of Redox Reactions and Electrochemistry (NEET 2027 syllabus). Here are 12; tap one to see its answer and explanation.

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easy

1. Given the following statements: Statement I: Electrical work done by a galvanic cell in one second is equal to electrical potential multiplied by total charge passed. Statement II: Cell potential Ecell is an extensive property, whereas ΔrG is an intensive property. In light of the above statements, choose the correct answer:

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

Statement I is correct as electrical work = potential × charge. Statement II is incorrect because cell potential (Ecell) is an intensive property (does not depend on the size of the system), whereas Gibbs energy change (ΔrG) is an extensive property.

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2. Consider the reduction half-reaction: Cr₂O₇²⁻(aq) + 14H⁺(aq) + 6e⁻ → 2Cr³⁺(aq) + 7H₂O(l). If the pH of the solution is increased from 0 to 1 at 298 K, by how much does the reduction potential change, assuming all other species remain at unit activity?

  1. Increases by 0.138 V
  2. Decreases by 0.138 V
  3. Decreases by 0.059 V
  4. Increases by 0.029 V
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Answer: (B) Decreases by 0.138 V

By Nernst equation: E = E° - (0.059/6) log([Cr³⁺]² / ([Cr₂O₇²⁻][H⁺]¹⁴)) = E° + (0.059 × 14 / 6) log[H⁺] = E° - (0.059 × 14 / 6) pH. Change ΔE = - (0.059 × 14 / 6) × ΔpH = - 0.1377 V ≈ - 0.138 V. Thus, the potential decreases by 0.138 V.

hard

3. For the cell Zn | Zn²⁺(0.1 M) || Cu²⁺(10 M) | Cu, E°cell = 1.10 V. Using E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]) at 298 K, the cell emf is:

  1. 1.1 V
  2. 1.04 V
  3. 1.22 V
  4. 1.16 V
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Answer: (D) 1.16 V

log([Zn²⁺]/[Cu²⁺]) = -2. E = 1.10 − 0.0295 × (-2) = 1.16 V.

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4. For the half-cell reaction at 298 K: O₂(g) + 4H⁺(aq) + 4e⁻ → 2H₂O(l), what is the reduction potential in neutral water (pH = 7) under standard oxygen pressure (p(O₂) = 1 bar), given E° = +1.23 V? (Take 2.303 RT/F = 0.059 V)

  1. +0.41 V
  2. +1.23 V
  3. +0.82 V
  4. +1.64 V
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Answer: (C) +0.82 V

E = E° - (0.059/4) log(1 / (p(O₂) · [H⁺]⁴)) = E° - 0.059 pH. At pH = 7, E = 1.23 - 0.059 × 7 = 1.23 - 0.413 = +0.817 V ≈ +0.82 V.

easy

5. For a spontaneous cell reaction involving the transfer of 2 moles of electrons (n=2), the standard cell EMF is measured as E⁰ = 0.2 V. Its standard Gibbs free energy change ΔG⁰ in kJ/mol is:

  1. -19 kJ/mol
  2. -39 kJ/mol
  3. -50 kJ/mol
  4. 39 kJ/mol
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Answer: (B) -39 kJ/mol

Using the thermodynamic relationship ΔG⁰ = -nFE⁰. Here, n = 2, F = 96500 C/mol = 96.5 kJ/(V·mol), and E⁰ = 0.2 V. Thus, ΔG⁰ = -2 × 96.5 × 0.2 = -39 kJ/mol.

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

hard

6. For the cell Zn | Zn²⁺(1 M) || Cu²⁺(0.1 M) | Cu, E°cell = 1.10 V. Using E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]) at 298 K, the cell emf is:

  1. 1.1 V
  2. 1.04 V
  3. 1.13 V
  4. 1.07 V
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Answer: (D) 1.07 V

log([Zn²⁺]/[Cu²⁺]) = 1. E = 1.10 − 0.0295 × (1) = 1.07 V.

hard

7. For the cell Zn | Zn²⁺(0.001 M) || Cu²⁺(1 M) | Cu, E°cell = 1.10 V. Using E = E° − (0.059/2) log([Zn²⁺]/[Cu²⁺]) at 298 K, the cell emf is:

  1. 1.1 V
  2. 1.28 V
  3. 1.19 V
  4. 1.01 V
Show answer and explanation

Answer: (C) 1.19 V

log([Zn²⁺]/[Cu²⁺]) = -3. E = 1.10 − 0.0295 × (-3) = 1.19 V.

hard

8. For the concentration cell: Ag(s) | Ag⁺(aq, 0.001 M) || Ag⁺(aq, 0.1 M) | Ag(s) at 298 K, what is the value of the Gibbs free energy change (ΔrG) for the cell reaction? (Take 2.303 RT/F = 0.059 V, F = 96500 C mol⁻¹)

  1. +5.69 kJ mol⁻¹
  2. +11.39 kJ mol⁻¹
  3. -5.69 kJ mol⁻¹
  4. -11.39 kJ mol⁻¹
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Answer: (D) -11.39 kJ mol⁻¹

For the cell, the overall reaction is Ag⁺(cathode, 0.1 M) → Ag⁺(anode, 0.001 M), with n = 1 and E°cell = 0. Ecell = 0 - 0.059 log(0.001 / 0.1) = -0.059 log(10⁻²) = +0.059 × 2 = +0.118 V. Then, ΔrG = -nFEcell = -(1)(96500 C mol⁻¹)(0.118 V) = -11387 J mol⁻¹ ≈ -11.39 kJ mol⁻¹.

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9. A galvanic cell is represented as: Zn(s) | Zn²⁺(aq) || Cu²⁺(aq) | Cu(s). A plot of Ecell versus log([Zn²⁺]/[Cu²⁺]) at 298 K yields a straight line. What are the slope and intercept of this plot respectively?

  1. -0.0295 V and E°cell
  2. -0.0591 V and E°cell
  3. +0.0295 V and -E°cell
  4. -0.0295 V and 0 V
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Answer: (A) -0.0295 V and E°cell

From the Nernst equation: Ecell = E°cell - (2.303 RT / nF) log([Zn²⁺]/[Cu²⁺]). For n = 2 at 298 K, this becomes Ecell = E°cell - 0.0295 log([Zn²⁺]/[Cu²⁺]). Comparing with y = mx + c where y = Ecell and x = log([Zn²⁺]/[Cu²⁺]), the slope is -0.0295 V and the y-intercept is E°cell.

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10. Consider the cell at 298 K: Zn(s) | Zn²⁺(aq, 0.1 M) || Cd²⁺(aq, 0.001 M) | Cd(s). If E°(Zn²⁺/Zn) = -0.76 V and E°(Cd²⁺/Cd) = -0.40 V, what is the EMF of the cell? (Take 2.303 RT/F = 0.059 V)

  1. +0.360 V
  2. +0.419 V
  3. +0.301 V
  4. +0.242 V
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Answer: (C) +0.301 V

E°cell = E°(cathode) - E°(anode) = -0.40 V - (-0.76 V) = +0.36 V. The cell reaction is Zn(s) + Cd²⁺(aq) → Zn²⁺(aq) + Cd(s), so n = 2. Applying the Nernst equation: Ecell = E°cell - (0.059/2) log([Zn²⁺]/[Cd²⁺]) = 0.36 - 0.0295 log(0.1 / 0.001) = 0.36 - 0.0295 log(100) = 0.36 - 0.0295(2) = 0.36 - 0.059 = +0.301 V.

easy

11. The standard reduction potential (E⁰) of a cell reaction is 1 V. If the temperature is 298 K, the standard Gibbs free energy change (ΔG⁰) in kJ/mol for a 2-electron transfer process (F = 96500 C/mol) is closest to:

  1. -50 kJ/mol
  2. 193 kJ/mol
  3. -96 kJ/mol
  4. -193 kJ/mol
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Answer: (D) -193 kJ/mol

Using the thermodynamic relationship ΔG⁰ = -nFE⁰. Here, n = 2, F = 96500 C/mol = 96.5 kJ/(V·mol), and E⁰ = 1 V. Thus, ΔG⁰ = -2 × 96.5 × 1 = -193 kJ/mol.

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

easy

12. Which of the following statements is INCORRECT regarding thermodynamic quantities of a reversible galvanic cell?

  1. ΔrG is an extensive property, while Ecell is an intensive property.
  2. The electrical work done in one second corresponds to electrical power.
  3. Doubling the stoichiometric coefficients of a cell reaction doubles both E°cell and ΔrG°.
  4. Maximum work obtainable from a galvanic cell is equal to the decrease in Gibbs free energy.
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Answer: (C) Doubling the stoichiometric coefficients of a cell reaction doubles both E°cell and ΔrG°.

E°cell is an intensive property and does not depend on the stoichiometric coefficients or scale of the reaction. Only ΔrG° is extensive and doubles if the stoichiometric coefficients are doubled.

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