In: Chemistry
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"
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
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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