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For an electrochemical cell, S n s S n 2 + aq , 1 M P b 2 + aq , 1 M Pb s , the ratio S n 2+ P b 2 + when this cell attains equilibrium is Given: E S n 2 + | Sn 0 = - 0.14 V, E P b 2 + | Pb 0 = - 0.13 V, 2.303 RT F = 0.06

Options

  1. A4.3
  2. B1
  3. C-2.15
  4. D2.14

Correct answer

D. 2.14

Step-by-step solution

At equilibrium E cell = 0 E cell 0 = 0.01 V Sn + P b 2+ → S n 2+ + Pb E cell = E cell 0 - 0.06 n log ⁡ Q 0 = 0.01 - 0.06 2 log ⁡ S n 2 + P b 2 + 0.01 = 0.06 2 log ⁡ S n 2 + P b 2 + 1 3 = log ⁡ S n 2 + P b 2 + Sn 2 + Pb 2 + = 10 1 3 = 2.14

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