Question

In: Chemistry

A galvanic cell consists of a iron electrode in 1 M Fe(NO3)2 and a copper electrode...

A galvanic cell consists of a iron electrode in 1 M Fe(NO3)2 and a copper electrode in 1 M Cu(NO3)2. What is the equilibrium constant for this reaction at 25oC? Enter you answer with 2 significant digits, using the syntax of "1.0x10(22)" for "1.0x1022"

Solutions

Expert Solution

Fe(s)/Fe(NO3)2 (aq) // Cu(NO3)2 (aq)/Cu

Fe------------------> Fe+2 + 2e-         E0 = 0.41V

Cu+2 +2e- --------> Cu(s)                 E0 = 0.34V

--------------------------------------------------------------------

Fe(s) + Cu+2 (aq) -------> Fe+2 (aq) + Cu(s)       E0cell = 0.75V

     n = 2

   G0   = -nE0cell*F

               = -2*0.75*96500   = -144750J

G0     = -RTlnKc

-144750 = -8.314*298*2.303logKC

logKc     = -144750/-5705.8483

logKc   = 25.37

Kc       = 10^25.37    = 2.35*10^25 >>> answer

             


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