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

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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

queen_honey_blossom answered 1 year ago

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