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If E ClO 3 - / ClO 4 - 0 = - 0 . 3 6 V & E ClO 3 - / ClO 2 - 0 = 0 . 3 3 V at 300 K. The equilibrium concentration of perchlorate ion ClO 4 - which was initially 1.0 M in ClO 3 - when the reaction starts to attain the equilibrium, 2 ClO 3 - ⇌ ClO 2 - + ClO 4 - Given : Anti log(0.509) = 3.329

Options

  1. A0.0236 M
  2. B0.0190 M
  3. C0.123 M
  4. D0.191 M

Correct answer

D. 0.191 M

Step-by-step solution

The redox reaction can be split as Cl + 5 O 3 -1 ⟶ Cl + 7 O 4 -1 + 2 e OHR - anode Cl + 5 O 3 -1 + 2 e ⟶ Cl + 3 O 2 -1 RHR - cathode ------------------------------------------------------- 2 ClO 3 - 1 ⇄ n = 2e - ClO 4 - 1 + ClO 2 - 1 E cell o = E red cath o - E red anode o = 0.33 - 0.36 = -0.03 V as E red Anode o = E ClO 4 - 1 / ClO 3 - 1 o = + 0 . 3 6 V = - E ClO 3 - 1 / ClO 4 - 1 o at equilibrium E cell = 0 E cell o = 0 . 0 5 9 n log K eq = 0 . 0 5 9 2 log K eq Writing concentration of speces at equilibrium 2 ClO

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