Electrochemistry NEET PYQs (2005–2026) with Solutions

10 solved Electrochemistry PYQs

Real NEET and AIPMT questions from this chapter, newest first. Try each one, then open the answer and the step-by-step explanation.

1. Question 1 (NEET 2026)

   NEET 2026MediumCombining electrode potentials via Gibbs energy

   The standard electrode potential ($E^\circ$) for the half-cell reaction $\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$ at 298 K is (Given : $E^\circ(\mathrm{Fe^{3+}/Fe}) = -0.04\ \mathrm{V}$ and $E^\circ(\mathrm{Fe^{2+}/Fe}) = -0.44\ \mathrm{V}$ at 298 K)

   1. AOption A: $+0.40\ \mathrm{V}$
   2. BOption B: $+0.76\ \mathrm{V}$
   3. COption C: $-0.48\ \mathrm{V}$
   4. DOption D: $+0.92\ \mathrm{V}$

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   Correct answer: (B) $+0.76\ \mathrm{V}$

   Explanation

   Electrode potentials cannot be added directly when the number of electrons changes, but Gibbs energies can ($\Delta G^\circ = -nFE^\circ$). (i) $\mathrm{Fe^{3+} + 3e^- \rightarrow Fe}$: $\Delta G_1^\circ = -3F(-0.04) = +0.12F$ (ii) $\mathrm{Fe^{2+} + 2e^- \rightarrow Fe}$: $\Delta G_2^\circ = -2F(-0.44) = +0.88F$ The required reaction is (i) − (ii): $\mathrm{Fe^{3+} + e^- \rightarrow Fe^{2+}}$, so $\Delta G^\circ = 0.12F - 0.88F = -0.76F$. With $n = 1$: $E^\circ = -\frac{\Delta G^\circ}{1 \times F} = +0.76\ \mathrm{V}$, option (B).

2. Question 2 (NEET 2026)

   NEET 2026MediumKohlrausch law and molar conductivity

   For a salt XY, which is a strong electrolyte, the plot of $\Lambda_m$ versus $\sqrt{c}$ has a slope of $-90.0\ \mathrm{S\ cm^2\ mol^{-3/2}\ L^{1/2}}$ at 298 K. At 0.01 M concentration of XY, the value of $\Lambda_m$ is $145.0\ \mathrm{S\ cm^2\ mol^{-1}}$. The limiting molar conductivity of $\mathrm{Y^-}$ ion ($\lambda^0_{\mathrm{Y^-}}$, in $\mathrm{S\ cm^2\ mol^{-1}}$) at 298 K will be (Given : $\lambda^0_{\mathrm{X^+}} = 74.0\ \mathrm{S\ cm^2\ mol^{-1}}$)

   1. AOption A: $80.0$
   2. BOption B: $100.0$
   3. COption C: $90.0$
   4. DOption D: $76.0$

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   Correct answer: (A) $80.0$

   Explanation

   For a strong electrolyte, $\Lambda_m = \Lambda_m^\circ - A\sqrt{c}$, where the slope is $-A$. Here $A = 90.0\ \mathrm{S\ cm^2\ mol^{-3/2}\ L^{1/2}}$. At $c = 0.01\ \mathrm{M}$, $\sqrt{c} = 0.1$: $$145.0 = \Lambda_m^\circ - 90.0 \times 0.1 \Rightarrow \Lambda_m^\circ = 154.0\ \mathrm{S\ cm^2\ mol^{-1}}$$ By Kohlrausch's law, $\Lambda_m^\circ = \lambda^0_{\mathrm{X^+}} + \lambda^0_{\mathrm{Y^-}}$.$\lambda^0_{\mathrm{Y^-}} = 154.0 - 74.0 = 80.0\ \mathrm{S\ cm^2\ mol^{-1}}$, option (A).

3. Question 3 (NEET 2026)

   NEET 2026MediumElectrolysis

   A solution of copper sulphate is electrolysed for 10 minutes with a current of 1.5 amperes. The mass of copper deposited at cathode is: (Given: Molar mass of Cu = 63 g mol⁻¹, 1F = 96487 C mol⁻¹)

   1. AOption A: 1.7018 g
   2. BOption B: 0.2938 g
   3. COption C: 2.4036 g
   4. DOption D: 0.5876 g

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   Correct answer: (B) 0.2938 g

   Explanation

   Using Faraday’s law: mass=(MIt)/(nF)=63×1.5×600/(2×96487)=0.2938 g. Hence option (B) is correct.

4. Question 4 (NEET 2026)

   NEET 2026MediumElectrode Potential & Nernst Equation

   Calculate emf of the half cell:

   1. AOption A: −0.109 V
   2. BOption B: 0.035 V
   3. COption C: −0.035 V
   4. DOption D: 0.109 V

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   Correct answer: (D) 0.109 V

   Explanation

   Using Nernst equation: E=E°−(0.059/2)log([H⁺]²/PH₂). Substituting [H⁺]=0.02 M and PH₂=2 atm gives E≈0.109 V. Hence option (D) is correct.

5. Question 5 (NEET 2025)

   NEET 2025MediumConductance & Kohlrausch Law

   If the molar conductivity $(\Lambda_m)$ of a $0.050\ \mathrm{mol\ L^{-1}}$ solution of a monobasic weak acid is $90\ \mathrm{S\ cm^2\ mol^{-1}}$, its extent (degree) of dissociation will be [Assume $\Lambda_+^{\circ} = 349.6\ \mathrm{S\ cm^2\ mol^{-1}}$ and $\Lambda_-^{\circ} = 50.4\ \mathrm{S\ cm^2\ mol^{-1}}$.]

   1. AOption A: 0.115
   2. BOption B: 0.125
   3. COption C: 0.225
   4. DOption D: 0.215

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   Correct answer: (C) 0.225

   Explanation

   For a weak electrolyte, $$\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}$$ The limiting molar conductivity is: $$\Lambda_m^{\circ} = \Lambda_+^{\circ} + \Lambda_-^{\circ}$$ $$\Lambda_m^{\circ} = 349.6 + 50.4 = 400\ \mathrm{S\ cm^2\ mol^{-1}}$$ Therefore, $$\alpha = \frac{90}{400} = 0.225$$ Hence, the degree of dissociation is $0.225$.

6. Question 6 (NEET 2024)

   NEET 2024HardElectrolysis

   Match List I with List II. Choose the correct answer from the options given below:

   1. AOption A: A-II, B-III, C-I, D-IV
   2. BOption B: A-III, B-IV, C-II, D-I
   3. COption C: A-II, B-IV, C-I, D-III
   4. DOption D: A-III, B-IV, C-I, D-II

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   Correct answer: (C) A-II, B-IV, C-I, D-III

   Explanation

   A. Formation of 1 mol O₂ from water: $$2H_2O \rightarrow O_2 + 4H^+ + 4e^-$$ Thus 1 mol O₂ requires 4F, so 1 mol H₂O corresponds to 2F. Hence A-II. B. Reduction: $$MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O$$ 1 mol MnO₄⁻ requires 5F. Hence B-IV. C. Reduction: $$Ca^{2+} + 2e^- \rightarrow Ca$$ 1.5 mol Ca requires: $$1.5 \times 2 = 3F$$ Hence C-I. D. Oxidation: $$2FeO \rightarrow Fe_2O_3$$ Each Fe²⁺ changes to Fe³⁺, requiring 1 electron per Fe atom. Thus 1 mol FeO requires 1F. Hence D-III. Therefore: A-II, B-IV, C-I, D-III Hence, option (C) is correct.

7. Question 7 (NEET 2024)

   NEET 2024MediumElectrolysis

   Mass in grams of copper deposited by passing 9.6487 A current through a voltmeter containing copper sulphate solution for 1000 seconds is: (Given : Molar mass of Cu : 63 g mol⁻¹, 1F = 96487 C)

   1. AOption A: 31.5 g
   2. BOption B: 0.0315 g
   3. COption C: 3.15 g
   4. DOption D: 0.315 g

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   Correct answer: (D) 0.315 g

   Explanation

   Charge passed: $$Q = It = 9.6487 \times 1000 = 9648.7\,C$$ Reaction: $$Cu^{2+} + 2e^- \rightarrow Cu$$ Using Faraday’s law: $$m = \frac{QM}{nF}$$ Substituting values: $$m = \frac{9648.7 \times 63}{2 \times 96487}$$ $$m \approx 3.15\,g$$ Hence, option (C) is correct.

8. Question 8 (NEET 2023)

   NEET 2023MediumConductance & Kohlrausch Law

   The conductivity of centimolar solution of KCl at 25°C is 0.0210 ohm⁻¹ cm⁻¹ and resistance of cell containing the solution at 25°C is 60 ohm. The value of cell constant is -

   1. AOption A: 1.34 cm⁻¹
   2. BOption B: 3.28 cm⁻¹
   3. COption C: 1.26 cm⁻¹
   4. DOption D: 3.34 cm⁻¹

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   Correct answer: (C) 1.26 cm⁻¹

   Explanation

   Conductivity κ = 0.0210 ohm−1 cm−1 Resistance R = 60 ohm Cell constant = κ × R = 0.0210 × 60 = 1.26 cm−1 Hence, the correct answer is 1.26 cm−1.

9. Question 9 (NEET 2023)

   NEET 2023MediumElectrochemical & Galvanic Cells

   Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R : Assertion A : In equation ΔᵣG = −nFEcell, value of ΔᵣG depends on n. Reason R : Ecell is an intensive property and ΔᵣG is an extensive property. In the light of the above statements, choose the correct answer from the options given below:

   1. AOption A: Both A and R are true and R is the correct explanation of A.
   2. BOption B: Both A and R are true and R is NOT the correct explanation of A.
   3. COption C: A is true but R is false.
   4. DOption D: A is false but R is true.

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   Correct answer: (A) Both A and R are true and R is the correct explanation of A.

   Explanation

   The relation between Gibbs free energy and cell potential is: ΔᵣG = −nFE°cell Here, E°cell is an intensive property and does not depend on the amount of substance, whereas ΔᵣG is an extensive property and depends on the number of moles involved in the reaction. Therefore, ΔᵣG depends on n. Hence, both Assertion and Reason are true, and Reason correctly explains Assertion.

10. Question 10 (NEET 2022)

    NEET 2022HardElectrode Potential & Nernst Equation

    At 298 K, the standard electrode potentials of Cu²⁺/Cu, Zn²⁺/Zn, Fe²⁺/Fe and Ag⁺/Ag are 0.34 V, −0.76 V, −0.44 V and 0.80 V, respectively. On the basis of standard electrode potential, predict which of the following reaction cannot occur?

    1. AOption A: 2CuSO₄(aq) + 2Ag(s) → 2Cu(s) + Ag₂SO₄(aq)
    2. BOption B: CuSO₄(aq) + Zn(s) → ZnSO₄(aq) + Cu(s)
    3. COption C: CuSO₄(aq) + Fe(s) → FeSO₄(aq) + Cu(s)
    4. DOption D: FeSO₄(aq) + Zn(s) → ZnSO₄(aq) + Fe(s)

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    Correct answer: (A) 2CuSO₄(aq) + 2Ag(s) → 2Cu(s) + Ag₂SO₄(aq)

    Explanation

    For reaction feasibility, E°cell must be positive. In option (A): E°cell = 0.34 − 0.80 = −0.46 V Since E°cell is negative, the reaction is non-spontaneous and cannot occur. Hence option (A) is correct.