Compute the equilibrium constant for the spontaneous reaction between Sn2+(aq) and Fe(s). It is not 1.39...

Compute the equilibrium constant for the spontaneous reaction between Sn2+(aq) and Fe(s). It is not 1.39 x 1-^10 or 1.4 x 10^10

Solutions

Expert Solution

Fe(s) -------------------> Fe^2+ (aq) + 2e^- E0 = 0.41v

Sn^2+ (aq) + 2e^- --------> Sn(s) E0 = -0.14v

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

Fe(s) + Sn^2+(aq) ----------------> Fe^2+(aq) + Sn(s) E0cell = 0.27v

n = 2

G = -nE0cell*F

= -2*0.27*96500 = -52110J

G = -RTlnK

G = -RT*2.303logK

-52110 = -8.314*298*2.303logK

logK = -52110/-5705.8483

logK = 9.132

K = 10^9.132 = 1.35*10^9 >>>>>answer

queen_honey_blossom answered 1 year ago

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