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ExamsJEE AdvancedChemistry

Determine the EMF of the cell: Cr | Cr3+ (0.1 M) || Fe2+ (0.01 M) | Fe. (Given E°(Cr3+/Cr) = -0.75 V, E°(Fe2+/Fe) = -0.45 V)

  1. +0.26 V
  2. +0.30 V
  3. +0.34 V
  4. -0.26 V

Correct answer: +0.26 V

Solution

Standard cell potential E°cell = E°cathode - E°anode = (-0.45) - (-0.75) = +0.30 V. The balanced reaction transfers 6 electrons: 2Cr + 3Fe2+ -> 2Cr3+ + 3Fe. The reaction quotient Q = [Cr3+]² / [Fe2+]³ = (0.1)² / (0.01)³ = 0.01 / 1e-6 = 1e4. Nernst: Ecell = E° - (0.059/6) log Q = 0.30 - (0.059/6)(4) = 0.30 - 0.0393 = 0.26 V.

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