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JEE Main201911 Jan 2019Evening ShiftChemistryElectrochemistryActual

Given the equilibrium constant: K _ C of the reaction: Cu ( s )+2 Ag ⁺( aq ) Cu ²⁺( aq )+2 Ag ( s ) is 10 10¹⁵ calculate the E_ cell ⁰ of this reaction at 298 ~K [2.303 RT F . at .298 ~K =0.059 ~V ]

Options

  1. A0.04736 mV
  2. B0.4736 mV
  3. C0.4736 ~V
  4. D0.04736 ~V

Correct answer

C. 0.4736 ~V

Step-by-step solution

E_ cell ⁰= 2.303 RT nF K _ C or E _ cell ⁰= 0.059 ~V n K _ C = 0.059 ~V 2 10¹⁶=0.4736 ~V

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