Question

Fi + Ni^2+ = Fe^2+ Ni emf E at 25 = -0.69V ? b) what is...

Fi + Ni^2+ = Fe^2+ Ni

emf E at 25 = -0.69V ?

b) what is the value of E if the concentration of Ni^2+ is 4.17x10-2 and the concentration of Fe^2+ is 5.06x10^-3

c) If E=0.36V and Ni^2+ = 0.10M, what is the concentration

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Homework Answers

Answer #2

Oxidation : Fe (s) ---------------> Fe+2 + 2e-

reduction : Ni+2 + 2e-   -----------> Ni (s)

a)

Eocell = Eored - Eo oxid

           = - 0.26 - ( - 0.44)

Eocell = 0.18 V

b)

Ecell = Eocell - 0.05916 / 2 log [Fe+2 / Ni+2]

         = 0.18 - 0.05916 / 2 log [5.06x10^-3 /4.17x10-2 ]

Ecell = 1.096 V

c)

Ecell = Eocell - 0.05916 / 2 log [Fe+2 / Ni+2]

0.36 = 0.18 - 0.05916 / 2 log [Fe+2 / 0.10]

[Fe+2] = 8.22 x 10^-8 M

answered by: anonymous
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