Why is $\ce{HNO3}$ a stronger oxidising agent than $\ce{H3PO4}$? $\ce{N}$ and $\ce{P}$ have the same oxidation number. Is it the electronegativity difference between $\ce{N}$ and $\ce{P}$?

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    $\begingroup$ Do you think the electronegativity of N helps? N: 3.04 while P: 2.19. $\endgroup$ – Huy Nguyen Jun 6 '15 at 22:15
  • $\begingroup$ Yes, may be. N atoms accept electrons more easily than P atoms, and this can affect his reduction potential. But I'm not sure, the difference in electronegativity is not that big to explain the huge difference in reduction potentials. $\endgroup$ – Ragnar Jun 6 '15 at 23:32
  • $\begingroup$ related chemistry.stackexchange.com/questions/30863/… $\endgroup$ – Mithoron Jun 7 '15 at 13:46

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.

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

  • $\begingroup$ Thank you so much, now I get it. I didn't think of the contribution of d-orbitals in P. $\endgroup$ – Ragnar Jun 7 '15 at 23:20
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    $\begingroup$ No, there is no need to bring hypothetical d-orbitals of phosphorus into the game. Electronegativity of the two elements, compared with that of oxygen is enough. $\endgroup$ – Jan Jun 8 '15 at 1:06

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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