# What does *oxidizing* have to do with *hydrogen*? [closed]

An oxidizing acid is a Brønsted acid that is a strong oxidizing agent. All Brønsted acids can act as oxidizing agents, because the acidic proton can be reduced to hydrogen gas.

(source: Wikipedia)

But what does oxidizing have to do with hydrogen? Isn't it about oxygen?

Like how iron becomes iron oxide after prolonged exposure to oxygen in the atmosphere?

So why is hydrogen relevant?

• Do you know the chemical definition of oxidation (and reduction)? – orthocresol Oct 2 at 18:20
• Moreover, generally speaking, oxidation has nothing to do with oxygen either. – Ivan Neretin Oct 2 at 18:22
• i believed so. perhaps not. the definition seems to vary with the source. i assumed the acquisition of an oxygen atom was the crux of the concept. – voices Oct 2 at 18:33
• is oxidation not contingent on the synthesis of an oxide? – voices Oct 2 at 18:58
• @voices Well, reaction of fluorine with oxygen is fluorine reduction, not oxidation. Free oxygen is one of many potentially oxidation agens. Some of them contain oxygen, some do not. One of them is hydronium ion H3O+, which can e.g. oxidize metallic zinc to Zn^2+ ions. Zn + 2 H2O+ -> Zn^2+ + H2 + 2 H2O – Poutnik Oct 2 at 19:00

An oxidizing acid is a Brønsted acid that is a strong oxidizing agent. All Brønsted acids can act as oxidizing agents, because the acidic proton can be reduced to hydrogen gas."

The whole sentence is written in a very convoluted fashion, and it is partially wrong as well. Someone should improve this Wikipedia section.

All they are saying is that strong acids do react with many metals and oxidize them. In turn their acidic proton is reduced to hydrogen.

For example, metallic Zn reacts with dilute sulfuric acid to form zinc sulfate and hydrogen gas. Zinc (0) has been oxidized to Zn(II). Consequently, Zn(0) also reduced the hydrogen ions to hydrogen gas.

Why this statement is wrong? The reason is that it says ALL Bronsted acids can do this, which is simply an over-generalization. For example, benzoic acid would not produce hydrogen with zinc. It is sparingly soluble in water.

Independent of this point, sulfuric and nitric acid are strong oxidizers in concentrated form. Here it is not behaving like an acid. Sulfuric acid, if concentrated will oxidize copper to Cu(II), and itself get reduced to sulfur dioxide.

The definition of oxidation and reduction are

Oxidation is the loss of electrons or an increase in the oxidation state of an atom, an ion, or of certain atoms in a molecule.

Reduction is the gain of electrons or a decrease in the oxidation state of an atom, an ion, or of certain atoms in a molecule.

By this definition a $$\ce{H+}$$ ion can gain an electron by oxidizing a species, in this process since it has gained an electron we say it has been reduced.

All acids are oxidizing. They are all able to oxidize metals $$\ce{M}$$ whose redox potentials are negative with respect to hydrogen, like zinc $$\ce{Zn} ~(E° = - 0.76~\mathrm V$$), iron $$\ce{Fe}~ (E° = -0.41~\mathrm V)$$, and magnesium $$\ce{Mg} ~(E° = -2.37~\mathrm V)$$. The reaction produces some hydrogen gas $$\ce{H2}$$ and the metallic cation $$\ce{M^{z+}}$$. But usual acids like $$\ce{HCl}$$ or $$\ce{H2SO4}$$ cannot oxidize metals whose redox potentials are positive, like copper $$\ce{Cu}$$ or silver $$\ce{Ag}$$, at least at room temperature. That is why $$\ce{HCl}$$ and $$\ce{H2SO4}$$ are not called "strong oxidizing acids". For example : $$\ce{2 HCl + Fe ->Fe^{2+} + 2 Cl- + H2}$$ The $$\ce{Fe}$$ atom is oxidized from $$0$$ to $$+2$$, and the $$\ce{H}$$ atom is reduced from $$+1$$ to $$0$$. $$\ce{HCl + Cu -> no reaction}$$

On the other hand, nitric acid $$\ce{HNO3}$$ is able to oxidize metals whose redox potentials are positive, like copper $$\ce{Cu} ~(E° = +0.34~\mathrm V)$$ and silver $$\ce{Ag}~ (E° = +0.80~\mathrm V)$$. That is why nitric acid is called a "strong oxidizing acid". It is worth observing that the reaction does not produce hydrogen gas $$\ce{H2}$$. It produces nitrogen oxides like $$\ce{NO}$$ or $$\ce{NO2}$$. For example with copper $$\ce{Cu}$$: $$\ce{3 Cu + 8 HNO3 -> 3 Cu^{2+} + 2 NO + 4 H2O + 6 NO3-}$$ In this reaction the $$\ce{Cu}$$ atom is oxidized from $$0$$ to $$+2$$, but the $$\ce{H}$$ atom is neither reduced nor oxidized. It is the $$\ce{N}$$ atom that is reduced (from $$+5$$ to $$+2$$).