Electrolysis — Faraday's first and second laws
Electrochemistry
9
JEE Qs
8%
Hard
75
min
Master the precise stoichiometry of electrode reactions and maintain unit consistency to correctly apply Faraday's quantitative laws.
🧮 Key Formulas
✅ Key Points for JEE
- 1Faraday's first law relates the mass of substance deposited/liberated to the quantity of charge passed (Q=It).
- 2The 'n_factor' (or valency factor) is crucial for calculating equivalent weight (E = M/n_factor) and determining moles of electrons involved per mole of substance.
- 3One Faraday (F) is the charge of one mole of electrons (approx. 96485 C/mol), serving as a bridge between electrical charge and chemical stoichiometry.
- 4Faraday's second law states that when the same quantity of electricity is passed through different electrolytic cells connected in series, the masses of substances deposited/liberated are proportional to their equivalent weights.
- 5Always identify the correct half-reaction at the electrode to determine the stoichiometry (moles of electrons per mole of product) accurately.
⚠️ Common Mistakes
- ✕Incorrectly determining the 'n_factor' for a given reduction or oxidation reaction at the electrode.
- ✕Failing to convert units consistently (e.g., using minutes instead of seconds for time, or not converting current from mA to A).
- ✕Confusing molar mass (M) with equivalent weight (E) in calculations.
- ✕Not understanding that Faraday's second law strictly applies to cells connected in series, implying the same charge passes through each.
📝 Practice Questions
See allQ54.The element that does not belong to the same period of the remaining elements (modern periodic table) is: 2025 (23 Jan Shift 1) JEE Main Previous Year Paper (1) Iridium (2) Platinum (3) Osmium (4) Palladium
Q67.For the given cell Fe2+ (aq) + Ag+ (aq) →Fe3+(aq) + Ag(s) The standard cell potential of the above reaction is Ag+ + e−→Ag Eθ = xV Given: Fe2+ + 2e−→Fe Eθ = yV Fe3+ + 3e−→Fe Eθ = zV (1) x + y −z (2) x + 2y (3) x + 2y −3z (4) y −2x
Q58.Which of the following electrolyte can be used to obtain H2 S2O8 by the process of electrolysis? (1) Dilute solution of sodium sulphate. (2) Acidified dilute solution of sodium sulphate. (3) Dilute solution of sulphuric acid (4) Concentrated solution of sulphuric acid
Q66.A solution of aluminium chloride is electrolysed for 30 minutes using a current of 2 A . The amount of the aluminium deposited at the cathode is [Given : molar mass of aluminium and chlorine are 27 g mol−1 and 35.5 g mol−1 respectively. Faraday constant = 96500Cmol−1] (1) 1.660 g (2) 0.336 g (3) 0.441 g (4) 1.007 g 2025 (22 Jan Shift 1) JEE Main Previous Year Paper
Q58.Standard electrode potentials for a few half cells are mentioned below : E∘ = 0.34 V, E∘ = −0.76 V Cu2+/Cu Zn2+/Zn Which one of the following cells gives the most negative value of E∘Ag+/Ag = 0.80 V, E∘Mg2+/Mg = −2.37 V ΔG∘ ? (1) Zn Zn2+(1M) Ag+(1M) Ag (2) Zn Zn2+(1M) Mg2+(1M) Mg (3) Ag Ag+(1M) Mg2+(1M) Mg (4) Cu Cu2+(1M)∥Ag+(1M) Ag
Q57. E∘Cr2O2−7 /Cr3+ = 1.33 V E∘Cl2/Cl(−) = 1.36 V Based on the data given below : the strongest reducing agent is : E∘ = 1.51 V E∘ = −0.74 V MnO−4 /Mn2+ Cr3+/Cr (1) Cr (2) Cl− (3) MnO−4 (4) Mn2+
NCERT Chapters
- Class 12 Chemistry Part 1 Ch 3: Electrochemistry