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ChemistryMediumClass 12

EMF of Cell — E°cell = E°cathode - E°anode

Electrochemistry

9

JEE Qs

8%

Hard

40

min

Consistently apply the convention of using standard reduction potentials and correctly identify the cathode (reduction, higher SRP) and anode (oxidation, lower SRP) to avoid calculation errors.

🧮 Key Formulas

E°cell = E°reduction(cathode) - E°reduction(anode)
E°oxidation = -E°reduction

✅ Key Points for JEE

  • 1Always use standard reduction potentials (SRP) for both cathode and anode in the formula E°cell = E°reduction(cathode) - E°reduction(anode).
  • 2The species with the higher standard reduction potential (SRP) will act as the cathode (undergo reduction), and the species with the lower SRP will act as the anode (undergo oxidation).
  • 3Standard electrode potentials (E°) are intensive properties; they are not multiplied by the stoichiometric coefficients of the balanced redox reaction.
  • 4For a spontaneous galvanic (voltaic) cell, the calculated E°cell value must always be positive. A negative E°cell indicates a non-spontaneous process under standard conditions.
  • 5Standard conditions are 1 M concentration for all ionic species, 1 atm partial pressure for gases, and 298 K (25°C).

⚠️ Common Mistakes

  • Incorrectly identifying cathode and anode, often leading to calculating E°anode - E°cathode instead of E°cathode - E°anode.
  • Using a mix of oxidation potentials and reduction potentials without converting them to a consistent form (preferably all reduction potentials).
  • Multiplying the E° values by stoichiometric coefficients from the balanced chemical equation, which is incorrect as E° is an intensive property.

📝 Practice Questions

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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

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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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NCERT Chapters

  • Class 12 Chemistry Ch 3: Electrochemistry