An electrochemical cell (a galvanic or voltaic cell) converts the chemical energy of a spontaneous redox reaction into electrical energy. It has two half-cells, each a metal electrode in a solution of its own ions, joined externally by a wire and internally by a salt bridge, which maintains electrical neutrality without letting the two solutions mix. Oxidation occurs at the anode, the
Electrochemistry
Class 1213 previous-year questions from this chapter every option and the correct answer, free · where the marks are
1. Electrochemical cells and Nernst equation
Exam focus: At equilibrium, E(cell) = 0, not E°(cell) = 0 — that's the condition linking E°(cell) to K through 0 = E° − (0.0591/n)log K. In the standard cell notation, oxidation (anode) is written on the left and reduction (cathode) on the right — reverse that and every EMF you calculate flips sign.
Nernst Equation and Electrochemical Cells
Physics Wallah - Alakh Pandey
Covers electrode potential, EMF of cells and the Nernst equation directly.
ChemistryNEET 2023show ▾Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: In the equation , the value of depends on .
Reason R: is an intensive property and 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 the correct explanation of A✓
- B.Both A and R are true and R is NOT the correct explanation of A
- C.A is true but R is false
- D.A is false but R is true
The worked solution is part of a paid pack.
ChemistryNEET 2022show ▾At 298 K, the standard electrode potentials of , , and 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 reactions cannot occur:
ChemistryNEET 2022show ▾Given below are half-cell reactions:
Will the permanganate ion, , liberate from water in the presence of an acid?
- A.No, because V
- B.Yes, because V✓
- C.No, because V
- D.Yes, because V
The worked solution is part of a paid pack.
ChemistryNEET 2019show ▾For a cell involving one electron transfer, V at 298 K. The equilibrium constant for the cell reaction is: (Given: V at K)
ChemistryNEET 2019show ▾For the cell reaction , V at 298 K. The standard Gibbs energy () of the cell reaction is: (Given: Faraday constant )
2. Conductance and Kohlrausch's law
Exam focus: A weak electrolyte's molar conductivity shoots up sharply on extreme dilution and its plot doesn't extrapolate to a finite value the way a strong electrolyte's does — which is exactly why the weak electrolyte's limiting molar conductivity is obtained indirectly, via Kohlrausch's law of independent migration of ions, never by extrapolating its own graph.
Conductance and Kohlrausch's Law
Physics Wallah - Alakh Pandey
Covers molar conductivity variation with concentration and Kohlrausch's law; basic conductance/cell-constant introduction is in an earlier lecture of the same series.
ChemistryNEET 2025show ▾If the molar conductivity () of a solution of a monobasic weak acid is , its extent (degree) of dissociation will be [assume and ]:
- A.0.115
- B.0.125
- C.0.225✓
- D.0.215
Solution
. Degree of dissociation .
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Save for spaced revision (free account)ChemistryNEET 2023show ▾The conductivity of a centimolar solution of KCl at C is and the resistance of the cell containing the solution at C is 60 ohm. The value of the cell constant is:
- A.1.34 cm⁻
- B.3.28 cm⁻
- C.1.26 cm⁻✓
- D.3.34 cm⁻
The worked solution is part of a paid pack.
ChemistryNEET 2021show ▾The molar conductances of NaCl, HCl and at infinite dilution are 126.45, 426.16 and respectively. The molar conductance of at infinite dilution is:
ChemistryNEET 2021show ▾The molar conductivity of 0.007 M acetic acid is . What is the dissociation constant of acetic acid? (Given and )
3. Electrolysis and Faraday's laws; batteries and corrosion
Exam focus: In aqueous NaCl electrolysis, Cl⁻ is oxidised at the anode instead of water, even though water's thermodynamic oxidation potential is lower — a real overpotential effect examiners like precisely because it contradicts the "always predict from E° alone" shortcut. Faraday's second law compares deposits by equivalent mass, not molar mass, so equal charge deposits different moles of ions of different charge.
Faraday's Laws of Electrolysis
Physics Wallah - Alakh Pandey
Covers electrolysis and Faraday's laws thoroughly with solved problems; batteries and corrosion are covered in a separate, later lecture of the same series, not this one.
ChemistryNEET 2024show ▾Match List I with List II.
List I (Conversion) List II (Number of Faraday required) A. 1 mol of to I. 3 F B. 1 mol of to II. 2 F C. 1.5 mol of Ca from molten III. 1 F D. 1 mol of FeO to IV. 5 F
Choose the correct answer from the options given below:
- A.A-III, B-IV, C-I, D-II
- B.A-II, B-III, C-I, D-IV
- C.A-III, B-IV, C-II, D-I
- D.A-II, B-IV, C-I, D-III✓
Solution
Faradays needed equal electrons transferred. : 2 e⁻ per mole (II). : 5 e⁻ (IV). : 2 e⁻ per mole, so 1.5 mol needs 3 F (I). : Fe goes +2 → +3, 1 e⁻ per Fe, 1 F (III). Hence A-II, B-IV, C-I, D-III.
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Save for spaced revision (free account)ChemistryNEET 2024show ▾The mass in grams of copper deposited by passing a current of 9.6487 A through a voltameter containing copper sulphate solution for 100 seconds is (given molar mass of Cu = , 1 F = 96487 C):
- A.0.315 g✓
- B.31.5 g
- C.0.0315 g
- D.3.15 g
Solution
Charge C F. Copper needs 2 F per mole, so mol is deposited: g.
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Save for spaced revision (free account)ChemistryNEET 2020show ▾On electrolysis of dilute sulphuric acid using platinum (Pt) electrodes, the product obtained at the anode will be:
- A.Hydrogen gas
- B.Oxygen gas✓
- C. gas
- D. gas
The worked solution is part of a paid pack.