Posted by DAN on Wednesday, April 11, 2012 at 5:57pm.
A is right.
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?
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
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