Gibbs Energy & EMF — ΔG = -nFE
Electrochemistry
9
JEE Qs
8%
Hard
75
min
Always correctly determine 'n' (moles of electrons transferred) for the balanced redox reaction and meticulously apply sign conventions for ΔG and E_cell to assess spontaneity.
🧮 Key Formulas
✅ Key Points for JEE
- 1ΔG represents the maximum useful electrical work obtainable from an electrochemical cell reaction under specific conditions.
- 2For a spontaneous cell reaction, ΔG must be negative, which directly implies that the cell potential (E_cell) must be positive.
- 3'n' is the number of moles of electrons transferred in the balanced redox reaction; its accurate determination is crucial for all calculations.
- 4The formulas connect thermodynamic spontaneity (ΔG) with electrical potential (E) and chemical equilibrium (K_eq), linking three fundamental areas of physical chemistry.
- 5The relationship applies to both standard (ΔG°, E°) and non-standard (ΔG, E) conditions, with the reaction quotient (Q) connecting them via the Nernst equation and ΔG = ΔG° + RTlnQ.
⚠️ Common Mistakes
- ✕Incorrectly determining 'n' (number of moles of electrons transferred) from the balanced redox reaction.
- ✕Confusing standard state conditions (ΔG°, E°) with non-standard conditions (ΔG, E) and failing to apply the Nernst equation or ΔG = ΔG° + RTlnQ when necessary.
- ✕Sign errors, particularly in interpreting spontaneity (ΔG < 0 for spontaneous, E > 0 for spontaneous) or using the wrong sign for the work done.
- ✕Using an incorrect value or units for the gas constant (R) or Faraday constant (F).
📝 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+
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