NEET Electrochemistry PYQs & Practice Questions

10 free NEET Chemistry questions on Electrochemistry, starting with previous year questions (2016–2026). Tap “Show answer” to check each one. No login needed.

Topics: Galvanic Cells · Molar Conductivity and Degree of Dissociation · Nernst Equation and Cell Potential · Intensive vs Extensive Properties of a Cell · Definition of Cell EMF · Condition for an Electrochemical Cell to Act as an Electrolytic Cell · Incorrect Statement About Electrolyte Conductivity · Charge Required to Obtain One Mole of Aluminium · Dependence of a Conductivity Cell's Cell Constant · Gibbs Energy and Cell EMF Relationship · Conductivity and Cell Constant · Standard Electrode Potential - Hess's Law

1. Q1NEET 2026Standard Electrode Potential - Hess's Law

   The standard electrode potential (E°) for the half-cell reaction Fe3+ + e⁻ → Fe2+ at 298 K is ________. (Given: E°(Fe3+/Fe) = −0.04 V and E°(Fe2+/Fe) = −0.44 V at 298 K)

   • A. +0.76 V
   • B. −0.48 V
   • C. +0.92 V
   • D. +0.40 V
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   Answer: —

   Using Hess's law with Gibbs free energies: For Fe3+ + 3e⁻ → Fe, ΔG1 = -3F(-0.04) = 0.12F. For Fe2+ + 2e⁻ → Fe, ΔG2 = -2F(-0.44) = 0.88F. The target reaction Fe3+ + e⁻ → Fe2+ = Reaction1 - Reaction2, so ΔG3 = ΔG1 - ΔG2 = 0.12F - 0.88F = -0.76F. Since ΔG3 = -nFE° with n=1: E° = 0.76 V.

2. Q2NEET 2026Kohlrausch's Law - Molar Conductivity

   For a salt XY, which is a strong electrolyte, the plot of Λm versus √c has a slope of -90.0 S cm² mol⁻³⁄² L¹⁄² at 298 K. At 0.01 M concentration of XY, the value of Λm is 145.0 S cm² mol⁻¹. The limiting molar conductivity of the other ion (in S cm² mol⁻¹) at 298 K will be (Given: limiting molar conductivity of one ion = 74.0 S cm² mol⁻¹)

   • A. 100.0
   • B. 90.0
   • C. 76.0
   • D. 80.0
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   Answer: C — 76.0

   By Kohlrausch's law for a strong electrolyte: Λm = Λm° - A√c, so Λm° = Λm + A√c = 145.0 + 90.0×√0.01 = 145.0 + 90.0×0.1 = 145.0 + 9.0 = 154.0 S cm² mol⁻¹. Since Λm° = λ°(cation) + λ°(anion), and one ion's limiting molar conductivity is given as 74.0, the other ion's value = 154.0 - 74.0 = 80.0 S cm² mol⁻¹.

3. Q3NEET 2026EMF of a half cell

   Calculate emf of the half cell given below: Pt(s) | H2(g, 2atm) | HCl(aq, 0.02M). (E°H+/H2=0V, Given: 2.303RT/F=0.059, log2=0.3010)

   • A. -0.109 V
   • B. 0.109 V
   • C. 0.035 V
   • D. -0.035 V
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   Answer: A — -0.109 V

   E = E° - 0.059 log(1/[H+]√pH2); using the Nernst equation with the given values, E ≈ 0.109 V.

4. Q4NEET 2026Mass of copper deposited during electrolysis

   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: (Molar mass of Cu=63 g mol⁻¹, 1F=96487 C mol⁻¹)

   • A. 0.2938 g
   • B. 0.5876 g
   • C. 2.4036 g
   • D. 1.7018 g
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   Answer: —

   Q=It=1.5×600=900 C. Moles of electrons=900/96487. Cu²⁺+2e⁻→Cu, moles Cu=half of that. Mass=moles×63≈0.2938 g.

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5. Q5NEET 2025Molar Conductivity and Degree of Dissociation

   If the molar conductivity (Λm) of a 0.050 mol L⁻¹ solution of a monobasic weak acid is 90 S cm² mol⁻¹, its extent (degree) of dissociation will be [Assume Λ°+ = 349.6 Scm² mol⁻¹ and Λ°− = 50.4 Scm² mol⁻¹.]

   • A. 0.115
   • B. 0.125
   • C. 0.225
   • D. 0.215
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   Answer: B — 0.125

   α = Λm/Λ°m = 90/(349.6+50.4) = 90/400 = 0.225.

6. Q6NEET 2023Gibbs Energy and Cell EMF Relationship

   Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R: Assertion A: In equation deltaG(r) = -nFE(cell), value of deltaG(r) depends on n. Reason R: E(cell) is an intensive property and deltaG(r) is an extensive property. In the light of the above statements, choose the correct answer from the options given below.

   • A. Both A and R are true and R is NOT the correct explanation of A.
   • B. A is true but R is false.
   • C. A is false but R is true.
   • D. Both A and R are true and R is the correct explanation of A.
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   Answer: C — A is false but R is true.

   Assertion A is true — deltaG depends on n (the number of moles of electrons transferred) since deltaG = -nFE(cell). Reason R is also true — E(cell) is an intensive property (independent of the amount of substance), while deltaG is an extensive property (depends on the amount, via n), and this correctly explains why deltaG depends on n while E(cell) does not. So both A and R are true and R is the correct explanation of A.

7. Q7NEET 2023Conductivity and Cell Constant

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

   • A. 3.28 cm^-1
   • B. 1.26 cm^-1
   • C. 3.34 cm^-1
   • D. 1.34 cm^-1
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   Answer: A — 3.28 cm^-1

   Cell constant G* = Conductivity (K) x Resistance (R) = 0.0210 ohm^-1cm^-1 x 60 ohm = 1.26 cm^-1.

8. Q8NEET 2022Standard Electrode Potential

   Given below are half cell reactions: MnO4- + 8H+ + 5e- -> Mn2+ + 4H2O, E° (Mn2+/MnO4-) = -1.510V 1/2 O2 + 2H+ + 2e- -> H2O, E° (O2/H2O) = +1.223V Will the permanganate ion, MnO4- liberate O2 from water in the presence of an acid?

   • A. No, because E°cell = -0.287 V
   • B. Yes, because E°cell = +2.733 V
   • C. No, because E°cell = -2.733 V
   • D. Yes, because E°cell = +0.287 V
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   Answer: C — No, because E°cell = -2.733 V

   E°cell = E°(cathode, MnO4-/Mn2+ reduction) - E°(anode, O2/H2O oxidation) = 1.510 - 1.223 = +0.287 V. Since E°cell is positive, the reaction is spontaneous, so MnO4- will liberate O2 from water in the presence of acid.

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9. Q9NEET 2022Reactivity Series

   At 298 K, the standard electrode potential 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 can not occur?

   • A. CuSO4(aq) + Fe(s) -> FeSO4(aq) + Cu(s)
   • B. FeSO4(aq) + Zn(s) -> ZnSO4(aq) + Fe(s)
   • C. 2CuSO4(aq) + 2Ag(s) -> 2Cu(s) + Ag2SO4(aq)
   • D. CuSO4(aq) + Zn(s) -> ZnSO4(aq) + Cu(s)
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   Answer: B — FeSO4(aq) + Zn(s) -> ZnSO4(aq) + Fe(s)

   Based on reactivity order Zn > Fe > Cu > Ag (from standard electrode potentials, more negative = more reactive), Ag cannot displace Cu from CuSO4 since Ag is less reactive than Cu (Ag has higher/more positive electrode potential). So this reaction cannot occur.

10. Q10NEET 2022Predicting Feasibility of Redox Reactions

    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) CuSO4(aq) + Zn(s) -> ZnSO4(aq) + Cu(s) (2) CuSO4(aq) + Fe(s) -> FeSO4(aq) + Cu(s) (3) FeSO4(aq) + Zn(s) -> ZnSO4(aq) + Fe(s) (4) 2CuSO4(aq) + 2Ag(s) -> 2Cu(s) + Ag2SO4(aq)

    • A. CuSO4(aq) + Zn(s) -> ZnSO4(aq) + Cu(s)
    • B. CuSO4(aq) + Fe(s) -> FeSO4(aq) + Cu(s)
    • C. FeSO4(aq) + Zn(s) -> ZnSO4(aq) + Fe(s)
    • D. 2CuSO4(aq) + 2Ag(s) -> 2Cu(s) + Ag2SO4(aq)
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    Answer: C — FeSO4(aq) + Zn(s) -> ZnSO4(aq) + Fe(s)

    Reaction 4 would require Ag (E=0.80V, a weaker reducing agent) to reduce Cu2+ (E=0.34V), but since Ag has a higher (more positive) reduction potential than Cu, Ag will not spontaneously reduce Cu2+ ions; rather the reverse would occur. So this reaction cannot occur spontaneously.