Question

Question #c7688

Answer 1

Here's what I got.

Explanation:

The trick here is to realize that peroxydisulfuric acid contains the peroxydisulfate anion, `"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 `color(blue)(-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 `color(blue)(-2)` oxidation state, which implies that the average oxidation state of oxygen in the peroxydisulfate anion is

`(6 xx color(blue)(-2) + 2 xx color(blue)(-1))/8 = -7/4`

This means that instead of using two different oxidation states for the oxygen atoms, you can use an average oxidation state of `color(blue)(-7/4)` for all `8` atoms of oxygen present in the anion.

This implies that you have--keep in mind that hydrogen has a `color(blue)(+1)` oxidation state in this compound

`stackrel(color(blue)(+1))("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 xx color(blue)((+1)))^(color(red)("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 `color(blue)(+6)` oxidation state in peroxydisulfuric acid

`stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(+6))("S")_ 2stackrel(color(blue)(-7/4))("O")_ 8`

`color(white)(a/a)`
Now, here's an example of how NOT to answer this question.

`color(white)(a)`
`darr darrcolor(red)("Incorrect solution below")darr darr`

`color(white)(a/a)`
Your ultimate goal here is to figure out the oxidation state of sulfur in peroxydisulfuric acid, `"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 `color(blue)(+1)` oxidation state and oxygen has a `color(blue)(-2)` oxidation state.

This means that you have

`stackrel(color(blue)(+1))("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 xx color(blue)((+1)))^(color(red)("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 `color(blue)(+7)` oxidation state in peroxydisulfuric acid.

`stackrel(color(blue)(+1))("H")_ 2 stackrel(color(blue)(+7))("S")_ 2stackrel(color(blue)(-2))("O")_ 8`