# What is the oxidation state of "H"_2"S"_2"O"_2"?

Jul 22, 2017

The oxidation state of the molecule is $0$.

Refer to the explanation below.

#### Explanation:

The overall oxidation state for a neutral compound always equals zero.

The oxidation numbers for each element are indicated below.

Hydrogen

The oxidation number of hydrogen in this case is $+ 1$. Since there are two hydrogen atoms the total oxidation number is $2 \times + 1 = + 2$

${\stackrel{+ 2}{\text{H}}}_{2}$

Oxygen

The oxidation number of oxygen is $- 2$. Since there are two oxygen atoms, the total oxidation number for oxygen is $2 \times - 2 = - 4$.

${\stackrel{- 4}{\text{O}}}_{2}$

Sulfur

The oxidation number of sulfur can be determined from the oxidation numbers for hydrogen and oxygen. The combined oxidation numbers for hydrogen and oxygen are $+ 2 + \left(- 4\right) = - 2$.

Since the overall oxidation number of a neutral compound must be zero, the total oxidation number for sulfur in this compound must be $+ 2$, which means that each sulfur atom in the compound is $+ 1$, because $2 \times + 1 = + 2$.

${\stackrel{+ 2}{\text{S}}}_{2}$

The oxidation numbers for each element can be written above each element in the compound.

${\stackrel{+ 2}{\text{H"_2""stackrel(+2)"S"_2""stackrel(-4)"O}}}_{2}$