# Question c7688

Jun 11, 2017

Here's what I got.

#### Explanation:

The trick here is to realize that peroxydisulfuric acid contains the peroxydisulfate anion, ${\text{S"_2"O}}_{8}^{-}$, which has a particularly interesting Lewis structure.

Notice that two oxygen atoms are bonded via a single bond--this is known as a peroxide linkage. This will influence the average oxidation state of oxygen in this anion.

More specifically, the two oxygen atoms that are bonded to each other will have a $\textcolor{b l u e}{- 1}$ oxidation state, the same oxidation state that oxygen has in peroxides (hence the name peroxide linkage).

The other six oxygen atoms will have a $\textcolor{b l u e}{- 2}$ oxidation state, which implies that the average oxidation state of oxygen in the peroxydisulfate anion is

$\frac{6 \times \textcolor{b l u e}{- 2} + 2 \times \textcolor{b l u e}{- 1}}{8} = - \frac{7}{4}$

This means that instead of using two different oxidation states for the oxygen atoms, you can use an average oxidation state of $\textcolor{b l u e}{- \frac{7}{4}}$ for all $8$ atoms of oxygen present in the anion.

This implies that you have--keep in mind that hydrogen has a $\textcolor{b l u e}{+ 1}$ oxidation state in this compound

${\stackrel{\textcolor{b l u e}{+ 1}}{\text{H")_ 2 stackrel(color(blue)(?))("S") _ 2 stackrel(color(blue)(-7/4))("O}}}_{8}$

Now, peroxydisulfuric acid is a neutral compound, which implies that the sum of the oxidation states of all the atoms that make up a molecule of peroxydisulfuric acid must be equal to $0$.

This means that you have

${\overbrace{2 \times \textcolor{b l u e}{\left(+ 1\right)}}}^{\textcolor{red}{\text{2 atoms of H")) + overbrace( 2 xx ?)^(color(red)("2 atoms of S")) + overbrace(8 xx color(blue)((-7/4)))^(color(red)("8 atoms of O}}} = 0$

Solve to find the oxidation state of sulfur

2 * 2 * ? - 14 = 0

2 * ? = 12 implies ? = +6

Therefore, you can say that sulfur has a $\textcolor{b l u e}{+ 6}$ oxidation state in peroxydisulfuric acid

${\stackrel{\textcolor{b l u e}{+ 1}}{\text{H")_ 2 stackrel(color(blue)(+6))("S")_ 2stackrel(color(blue)(-7/4))("O}}}_{8}$

$\textcolor{w h i t e}{\frac{a}{a}}$
Now, here's an example of how NOT to answer this question.

$\textcolor{w h i t e}{a}$
$\downarrow \downarrow \textcolor{red}{\text{Incorrect solution below}} \downarrow \downarrow$

$\textcolor{w h i t e}{\frac{a}{a}}$
Your ultimate goal here is to figure out the oxidation state of sulfur in peroxydisulfuric acid, ${\text{H"_2"S"_2"O}}_{8}$, because you should already know the oxidation states for hydrogen and oxygen in this compound.

More specifically, hydrogen has a $\textcolor{b l u e}{+ 1}$ oxidation state and oxygen has a $\textcolor{b l u e}{- 2}$ oxidation state.

This means that you have

${\stackrel{\textcolor{b l u e}{+ 1}}{\text{H")_ 2 stackrel(color(blue)(?))("S")_ 2stackrel(color(blue)(-2))("O}}}_{8}$

Now, peroxydisulfuric acid is a neutral compound, which implies that the sum of the oxidation states of all the atoms that make up a molecule of peroxydisulfuric acid must be equal to $0$.

This means that you have

${\overbrace{2 \times \textcolor{b l u e}{\left(+ 1\right)}}}^{\textcolor{red}{\text{2 atoms of H")) + overbrace( 2 xx ?)^(color(red)("2 atoms of S")) + overbrace(8 xx color(blue)((-2)))^(color(red)("8 atoms of O}}} = 0$

Solve to find the oxidation state of sulfur

2 + 2 * ? - 16 = 0

2 * ? = 14 implies ? = 14/2 = 7#

Therefore, sulfur is in a $\textcolor{b l u e}{+ 7}$ oxidation state in peroxydisulfuric acid.

${\stackrel{\textcolor{b l u e}{+ 1}}{\text{H")_ 2 stackrel(color(blue)(+7))("S")_ 2stackrel(color(blue)(-2))("O}}}_{8}$