Question
A solution is prepared in which a trace or small amount of Fe2+ is added to a much larger amount of solution in which the [OH-] is 1.0 x 10^-2 M. Some Fe(OH)2 precipitates. The value of Ksp for Fe(OH)2 = 8.0 x10^-10.
A. Assuming that the hydroxide is 1.0 x 10^-2 M, calculate the concentration of Fe2+ ions in the solution.
B. A battery is prepared using the above solution with an iron wire dipping into it as one half-cell. The other half-cell is the standard nickel electrode. Write the balanced net ionic equation for the cell reaction.
c. use the nerst equation to calculate the potential of the above cell.
I don't know how to do B & C
For (A) i got 8.0*10^-6M. if not please help
Thank You!
A. Assuming that the hydroxide is 1.0 x 10^-2 M, calculate the concentration of Fe2+ ions in the solution.
B. A battery is prepared using the above solution with an iron wire dipping into it as one half-cell. The other half-cell is the standard nickel electrode. Write the balanced net ionic equation for the cell reaction.
c. use the nerst equation to calculate the potential of the above cell.
I don't know how to do B & C
For (A) i got 8.0*10^-6M. if not please help
Thank You!
Answers
A is right.
What values are you using for Eo Ni and Eo Fe? My best table for those is over 50 years old.
What values are you using for Eo Ni and Eo Fe? My best table for those is over 50 years old.
For Ni its -0.28V
& for Fe its -.440V
Is it right?
& for Fe its -.440V
Is it right?
EFe = EoFe -(0.0592/2)log(Fe/Fe^2+)
Substitute 1 for Fe and 8E-6 for Fe^2+, solve for EFe. Then reverse the sign and the half equation and add to the Ni reduction equation.
Fe + Ni^2+ ==> Fe^2+(8E-6M) + Ni
Substitute 1 for Fe and 8E-6 for Fe^2+, solve for EFe. Then reverse the sign and the half equation and add to the Ni reduction equation.
Fe + Ni^2+ ==> Fe^2+(8E-6M) + Ni
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