# Which element is being reduced in this reaction?

## $\text{H"_ 2"S"+ "HNO"_3 -> "S" + "NO" + "H"_ 2"O}$

Aug 31, 2016

Nitrogen is being reduced.

#### Explanation:

You can actually answer this question without balancing the chemical equation given to you.

All you need to notice here is that sulfur is paired with hydrogen on the reactants' side in hydrogen sulfide, $\text{H"_2"S}$, so right from the start it will have a negative oxidation number, $- 2$ to be precise, on this side of the equation.

On the products' side, sulfur is present in its elemental form as $\text{S}$. This implies that its oxidation state is equal to zero.

$\textcolor{w h i t e}{a}$
SIDE NOTE You'll often see elemental sulfur written as ${\text{S}}_{8}$. This is actually the most common allotrope of sulfur called octasulfur.
$\textcolor{w h i t e}{a}$

So sulfur goes from a negative oxidation number to an oxidation number equal to zero $\to$ sulfur is being oxidized.

Now, hydrogen and oxygen almost always have oxidation numbers of $+ 1$ and $- 2$, respectively. This can only mean that nitrogen is being reduced.

Therefore, option (C) is correct.

$\textcolor{w h i t e}{a}$
!! ANOTHER QUICK EXPLANATION !!

It's also very helpful to keep in mind that nitric acid, ${\text{HNO}}_{3}$, is a very powerful oxidizing agent.

This lets you know that in most reactions that feature nitric acid, you can expect this compound to oxidize another compound.

In this case, nitric acid oxidizes hydrogen sulfide to elemental sulfur, $\text{S}$ or ${\text{S}}_{8}$, while being reduced to nitric oxide, $\text{NO}$.

Therefore, the answer will once again be (C) $\text{N}$ is reduced.