Chemistry - Why is HNO3 a stronger oxidising agent than H3PO4?

Solution 1:

Let me create background first. Oxidizing agents are the chemical that helps something else oxidize and itself gets reduced.(reduction in charge)

$\ce{N}$ in $\ce{HNO3}$ is in the +5 oxidation state - how do we know that? $\ce{H}$ is +1, $\ce{O}$ is -2 and the overall $\ce{HNO3}$ has a zero net charge. The same goes for $\ce{H3PO4}$ resulting in +5 oxidation state of $\ce{P}$.

That +5 is the highest common oxidation state for $\ce{N}$. If $\ce{N}$ is reduced to say +4 as in $\ce{NO2}$ or +2 in $\ce{NO}$, it's charge is reduced and $\ce{N}$ is therefore an oxidizing agent for that particular reaction

So, $\ce{HNO3}$ is an strong oxidizing agent because the $\ce{N}$ can be readily reduced.

But $\ce{H3PO4}$ is a poor oxidizing agent. Here are two basic reasons:

  1. Nitrogen does not possess $d$-orbitals in valence shell and so its covalency is limited to 4. $\ce{N}$ can however achieve a formal oxidation state of +5 as in the $\ce{NO3-}$ ion. The inability of $\ce{N}$ to unpair and promote its $2s$ electron results in that $\ce{N(+5)}$ is less stable than $\ce{N(+3)}$. However, $\ce{P(+5)}$ is more stable than $\ce{P(+3)}$ and phosphoric acid shows less oxidising properties.

  2. The affinity of phosphorus for oxygen is greater than that of nitrogen; as a result, phosphonic acid ($\ce{H3PO3}$) is good reducing agent.

    $\ce{H3PO4 + 2H+ + 2e-> H3PO3 + H2O}$; $E=-0.276 \ce{V}$

Solution 2:

Frost Diagrams explain it very nicely as well. Look at the G/F = zE/V, very positive for HNO3, which makes it a potent oxidizing agent (much less stable). H3PO4 is basically the complete opposite.

As you can see, the G/F=zE/V for HNO3 is very positive, making it a potent oxidizing agent, and much less stable.

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