Oxidation related question

1. Nov 2, 2004

Mathman23

Hi All,

$\mathrm{Fe^{3+}(aq) + SCN^{-} (aq) \rightleftharpoons FeSCN^{2+}(aq)}$

I add some $\mathrm{NaHSO_{3}(s)}$ to the above reaction !

This makes the above chemical solution sour, consequently in a sour solution $\mathrm{HSO^{-}_{3}}$ will reduce Fe^{3+} to Fe^{2+}, meanwhile $HSO_{3}^{-}$ is itself oxidated to $SO^{2-}_{4}$.

a/ The color of the original resolution changes then adding $\mathrm{NaHSO_{3}(s)}$ to yellow.

b/ By adding $\mathrm{NaHSO_{3}(s)}$ to the orginal solution the equilibrium direction changes, such that it runs from right to left.

c/ I need help writing the reaction between $HSO^{-}_{3}$ and $SO^{-}_{4}$.

d/ Which influence does the adding of $HSO^{-}_{3}$ have on the concentration of $Fe^{3+}$

I hope that there someone out there who is willing to assist me since this is my last post !

Sincerely
Fred

2. Nov 2, 2004

chem_tr

Hello,

Let me first do the redox thing in atomic level, since it's easier than dealing with a bunch of atoms.

We know that iron(III) is reduced to iron(II), and in turn, sulfite (S4+) is oxidized to sulfate (S6+):

$$S^{4+} \longrightarrow S^{6+}+2e^-$$
$$2Fe^{3+}+2e^-\longrightarrow 2Fe^{2+}$$
--------------------------------
$$S^{4+}+2Fe^{3+} \longrightarrow S^{6+}+2Fe^{2+}$$

In the real redox, you'll need to put some other correcting factors (water and hydroxide, I think) to balance additional oxygens.

Now that we've got rid of the redox reaction, we may look other things. When you add some sodium bisulfate to the medium, a redox reaction occurs; and so the thiocyanate complex changes from $\displaystyle Fe(SCN)^{2+}$ to $\displaystyle Fe(SCN)^+$; this gives a totally different color. Iron(II) sulfate is also formed from the reaction, and the color may have also come from this one, so I don't think that the equilibrium shifts to the left, but I may be wrong; I don't insist on that.

Last edited: Nov 2, 2004