Nernst Equation + Cell EMF
Electrochemistry
48
JEE Qs
21%
Hard
75
min
Master the accurate determination of 'n' and the correct formulation of 'Q' for various types of electrochemical cells and reactions to avoid common calculation errors.
🧮 Key Formulas
✅ Key Points for JEE
- 1The Nernst equation allows calculation of cell potential (EMF) under non-standard conditions, i.e., when concentrations of species or partial pressures of gases are not unity.
- 2The term 'n' represents the total number of electrons transferred in the balanced redox reaction for the complete cell.
- 3The reaction quotient 'Q' must be correctly formulated using activities (approximated by concentrations for solutes and partial pressures for gases), excluding pure solids and liquids.
- 4At equilibrium, the cell potential (E_cell) becomes zero, and the reaction quotient (Q) becomes equal to the equilibrium constant (K_eq); this allows calculation of K_eq from standard cell potential.
- 5Nernst equation establishes a fundamental connection between cell potential (E), Gibbs free energy (ΔG), and the equilibrium constant (K_eq).
⚠️ Common Mistakes
- ✕Incorrectly determining the value of 'n' (number of electrons transferred) for the balanced cell reaction.
- ✕Making errors in constructing the reaction quotient 'Q', especially by including pure solids/liquids or by incorrectly assigning exponents.
- ✕Confusing 'ln' (natural logarithm) with 'log' (base 10 logarithm) and misusing the constants (RT/nF vs 0.0592/n).
📝 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