# What is an oxidizer?

When talking about gaseous fuel / oxidizer mixture what does one mean by oxidizer if there is no oxygen involved?

Is it correct to think of oxidizer as the more electronegative specie, because it is pulling the electron? Or is this definition only valid for solutions? The Wikipedia article is not clear whether the definition applies to gaseous reactants or only valid for solutions.

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@Brian thanks for the fix, that was rather stupid of me :) I went to capitalize the title and botched the rest of it. –  ManishEarth Feb 11 '13 at 12:05

Oxidation: losing electron(s), Oxidizer / oxidizing agent: a chemical that can oxidize another reagent. Reduction: gaining electron(s) - think of it as reduction of charge! Reducing agent: A chemical that can reduce another reagent.

And oxidation can never occur without a reduction, meaning if you use an oxidizing agent to oxidize a substance, when the oxidizing agent is in turn being reduced itself. It gains the electrons of the oxidized substance.

To describe how well a substance functions as an oxidizer, we use reduction potentials, measured in volts! A substance with a large, positive reduction potential is easy to reduce, which in turn means it is a good oxidizing agent. Similarly, if a substance has a large, negative reduction potential, it means that it is hard to reduce this substance. This is because it's good at reducing other substances itself - it's a reducing agent.

To figure out if a redox reaction is spontaneous (if it can potentially proceed on its own), we use standard reduction potentials, $E_0$. This is for concentrations of 1 M and a temperature of 25 degrees celcius. You look these up in your textbook or online.

An example: Is the following reaction spontaneous? $$\ce{Cu^{2+}(aq) + 2Ag(s) -> Cu(s) + 2Ag+(aq)}$$

To find out we first find the standard reduction potentials for each participating species:

$\ce{Cu^{2+}(aq) + 2e- -> Cu(s)}, E_0=+0.34V$

$\ce{Ag^{+}(aq) + e- -> Ag(s)}, E_0=+0.80V$

This describes how much they want to be reduced (GAIN electrons). The silver ion wants it the most! However, notice that in our reaction we're asking for it to LOSE electrons (be oxidized). THEREFORE: No, the reaction is not spontaneous under the aforementioned standard conditions. However, the REVERSE reaction is spontaneous. To calculate the potential for our reaction, we simply say:

$$E_0 = (+0.34V) - (+0.80V) = -0.46V$$

The reduction potential for the substance that we want to gain electrons (reduced), minus the potential for the substance we want to lose electrons (be oxidized). The fact that the result is negative, is what tells us the reaction will not proceed - and that it will in fact proceed in the OPPOSITE direction.

Take notice that these reduction potentials are not given the unit Volt for no reason! They are true electric potentials. You can think of electric potentials as "electron pressures". The pole with the highest electron pressure will be (-), because electrons are negatively charged, and the other will be (+) relative to it.

Remember that the reduction potential (the willingness to take electrons) for the silver ion is higher than that of the copper ion. Silver wants to take up electrons more than copper, meaning that electrons will travel from the copper to the silver. you could just as well say that the copper wants to get rid of its electrons more than the silver - copper has the higher electron pressure.

What about electronegativity? Electronegativity does indeed correlate somewhat with reduction potentials, at least for simple reactions involving pure elements. And you can see that Ag does indeed have a higher electronegativity than Cu, which makes sense. Electronegativity doesn't take into account the oxidation states of different species etc., and quickly it becomes a better idea to just leave electronegativity out of the picture when dealing with redox reactions.

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Brian's answer is very good and thorough, but there is one rather important empirical fact you need to consider with reduction potentials. While the difference in $\ce{E_0}$ values can tell you if a reaction can occur, it cannot tell you if a reaction will occur. There are other factors, like reaction rate and activation energy, that can interfere with what $\ce{E_0}$ values indicate.

There are two terrific CHEM Study videos that demonstrate these points. The first, Bromine: Element from the Sea, shows a procedure for oxidizing bromide ion in seawater to elemental bromine. At about the 9:30 mark, they discuss looking up $\ce{E_0}$ values to find an oxidizer for bromine. They first try oxygen, which indicates a spontaneous reaction by $\ce{E_0}$, but in fact the reaction does not occur, likely because the rate is too slow. They then try chlorine, which works to oxidize the bromine. In this example, the difference in $\ce{E_0}$ values is higher between $\ce{Br_2}$ and $\ce{Cl_2}$ than between $\ce{Br_2}$ and $\ce{O_2}$, so you might reasonably conclude that a higher difference in $\ce{E_0}$ values indicates a faster reaction.

Unfortunately, there is more to it, as the second video shows. The CHEM Study video on Nitric Acid shows that nitric acid--$\ce{HNO_3}$--can be used as a strong oxidizing agent, owing to nitrogen in the $\ce{+5}$ state. At around the 10:30 mark, they discuss the potential reduction products that can be produced from nitric acid (mostly nitrogen-oxygen gases) by consulting $\ce{E_0}$ values. The highest potential is for nitrogen gas, $\ce{N_2}$. But when an experiment to oxidize metals is conducted, poisonous nitrogen dioxide $\ce{(NO_2)}$ is the product, not $\ce{N_2}$, even though $\ce{NO_2}$ has a smaller $\ce{E_0}$ value. The answer is that reduction to $\ce{N_2}$ requires a higher activation energy, and at room temperature this reaction mostly does not occur, but reduction to $\ce{NO_2}$ dominates. Not realizing this possibility could literally be the difference between life $\ce{(N_2)}$ and death $\ce{(NO_2)}$!

As always in science, experimentation is the ultimate determinant of what works. Theory such as reduction potentials helps to determine what could work, but it is only the beginning of the process.

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Important indeed :-) –  Brian Feb 11 '13 at 14:32