Nernst Equation — Concentration effect on EMF
Electrochemistry
9
JEE Qs
8%
Hard
60
min
Master the accurate determination of 'n' and the correct construction of the reaction quotient 'Q' for any given cell reaction; these are the primary sources of errors.
🧮 Key Formulas
✅ Key Points for JEE
- 1The Nernst equation quantifies how non-standard concentrations/pressures of reactants and products affect the electrode potentials and the overall cell EMF.
- 2The reaction quotient (Q) is critical; correctly identify products, reactants, and their stoichiometric coefficients from the *balanced cell reaction*. Pure solids, pure liquids, and the solvent in dilute solutions are omitted from Q.
- 3'n' represents the total number of electrons transferred in the balanced overall cell reaction, not per half-reaction.
- 4For concentration cells, E°_cell is always zero, and the cell potential (E_cell) arises solely from the difference in concentrations/pressures between the anode and cathode compartments.
- 5At equilibrium, the cell potential (E_cell) is zero, and the reaction quotient (Q) becomes the equilibrium constant (K_eq), leading to a direct relationship between E°_cell and K_eq.
⚠️ Common Mistakes
- ✕Incorrectly determining the value of 'n' (number of electrons transferred) for the overall balanced redox reaction.
- ✕Errors in setting up the reaction quotient 'Q', such as including pure solids/liquids, incorrect stoichiometric coefficients as exponents, or swapping product and reactant terms.
- ✕Using 'ln' instead of 'log' (base 10) with the 0.0591 constant, or vice-versa, or using the wrong constant with the correct log base.
- ✕Sign errors when calculating E_cell, especially when dealing with half-cell potentials or the concentration terms in log(Q).
- ✕Not recognizing a concentration cell and attempting to find a non-zero E°_cell, when it should be zero.
📝 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