Question #06e3e

The tetrathionate anion, #S_4O_6^(2-)#, will gain 2 electrons and will be reduced to the thisulfate anion, #S_2O_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(color(blue)(+2.5))(S_4)# #stackrel(color(blue)(-2))(O_6^(-2)) ->2stackrel(color(blue)(+2))(S_2)# #stackrel(color(blue)(-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

![http://en.wikipedia.org/wiki/Tetrathionate]()

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

#"ON" = (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 * 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(color(blue)(+2.5))(S_4) + 2e^(-) -> 2stackrel(color(blue)(+2))(S_2)#

The net charge is equal on both sides.

#-2 + 2e^(-) = 2 * (-2)#, or
#-4 = -4#