# Question #06e3e

Apr 14, 2015

The tetrathionate anion, ${S}_{4} {O}_{6}^{2 -}$, will gain 2 electrons and will be reduced to the thisulfate anion, ${S}_{2} {O}_{3}^{2 -}$.

Here's how that happens. If you assign oxidation numbers to each atom that takes part in the reaction, you'll get

$\stackrel{\textcolor{b l u e}{+ 2.5}}{{S}_{4}}$ $\stackrel{\textcolor{b l u e}{- 2}}{{O}_{6}^{- 2}} \to 2 \stackrel{\textcolor{b l u e}{+ 2}}{{S}_{2}}$ $\stackrel{\textcolor{b l u e}{- 2}}{{O}_{3}^{- 2}}$

Notice that sulfur has a fractional oxidation number of +2.5. What that means is that not all the sulfur atoms present in the tetrathionate molecule have the same oxidation number.

In fact, +2.5 is the average of those oxidation numbers. Here's why that is

The two sulfur atoms that are bonded to each other have an oxidation number of zero. At the same time, the two sulfur atoms that are each bonded to 3 oxygen atoms will have an oxidation number of +5.

The average oxidation number of the sulfur atoms in the tetrathionate molecule will thus be

$\text{ON} = \frac{0 + 0 + 5 + 5}{4} = + 2.5$

So, on the reactants' side you have 4 sulfur atoms, ${S}_{4}$, each with an average oxidation number of +2.5. On the reactants' side, you have 4 sulfur atoms, $2 \cdot {S}_{2}$, each with an average oxidation number of +2.

You go from a total of +10 to a total of +8 by losing two electrons

$\stackrel{\textcolor{b l u e}{+ 2.5}}{{S}_{4}} + 2 {e}^{-} \to 2 \stackrel{\textcolor{b l u e}{+ 2}}{{S}_{2}}$

The net charge is equal on both sides.

$- 2 + 2 {e}^{-} = 2 \cdot \left(- 2\right)$, or
$- 4 = - 4$