Faraday's Laws of Electrolysis
Electrochemistry
45
JEE Qs
5%
Hard
45
min
Master the determination of n-factor for various ionic species in redox reactions to accurately calculate equivalent weights, which is the cornerstone for applying Faraday's laws.
🧮 Key Formulas
✅ Key Points for JEE
- 1Faraday's First Law quantifies the mass of substance (W) deposited or liberated at an electrode as directly proportional to the total charge (Q) passed.
- 2Faraday's Second Law states that when the same amount of charge is passed through different electrolytes connected in series, the masses of substances liberated at the electrodes are proportional to their respective equivalent weights.
- 3The n-factor (or valency factor) for a substance in electrolysis represents the number of electrons exchanged per mole of the substance in the electrode reaction and is crucial for calculating equivalent weight (E = M/n_factor).
- 4Always ensure time (t) is in seconds and current (I) is in Amperes to calculate charge (Q) in Coulombs, which then allows for direct application of Faraday's constant (F) in C/mol e-.
⚠️ Common Mistakes
- ✕Incorrectly determining the n-factor for the species undergoing reduction or oxidation at the electrode, leading to wrong equivalent weight calculations.
- ✕Confusing molar mass with equivalent weight in calculations, or using moles directly instead of equivalents.
- ✕Errors in unit conversion, especially using time in minutes or hours instead of seconds for charge calculations (Q = It).
📝 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 Ch 3: Electrochemistry