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For the given cell; Cu s Cu 2 + C 1 M Cu 2 + C 2 M Cu s change in Gibbs energy ∆ G is negative, it :

Options

  1. AC 1 = C 2
  2. BC 2 = C 1 2
  3. CC 1 = 2 C 2
  4. DC 2 = 2   C 1

Correct answer

D. C 2 = 2   C 1

Step-by-step solution

For concentration cell E cell o ¯ = 0 Anode : Cu s → Cu 2 + aq A Cathode : Cu 2 + aq c → Cu s Overall : Cu 2 + aq c → Cu 2 + aq ¯ A As ΔG = − nF   E cell If ΔG = − ve then E cell is positive. E cell = E cell o ¯ − 0 .059 2 log C 1 C 2 E cell = − 0 .059 2 log C 1 C 2 E cell > 0 ⇒ C 2 > C 1

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