A solution is prepared where a bit of Fe2+ is added to a solution where the [OH-] is 1E-2M. Some Fe(OH)2 precipitates. Ksp for Fe(OH)2 is 8E-10.
How do I calculate the concentration of Fe? I have solved for the half reactions and I got
Ox: 2(Fe -> Fe2+ +2e-)
Red: O2 + 2H2O +4e- -> 4OH-
Ox Volt= .44
Red Volt= .40
Total Volt= .84
I don't know if that's helpful, but I have to solve for the Fe2+ concentration and use it to find the cell potential with the Nernst equation. Any help would be greatly appreciated!
2007-01-04 08:59:15 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
A bit of Fe2+ means "almost nothing." But if any Fe(OH)2 precipitates, it has exceeded the solubility product, and therefore you have a saturated solution of Fe(OH)2.
Ksp = [Fe2+][OH-]^2 = 8 x 10^-10
Ksp = 8 x 10^-10 = (10^-2)^2[Fe+2] = (10^-4)[Fe2+]
[Fe2+] = 8 x 10^-6M
2007-01-04 09:22:52 · answer #1 · answered by steve_geo1 7 · 0⤊ 0⤋