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 $\ce{P}$?
Answer
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:
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.
The affinity of phosphorus for oxygen is greater than that of nitrogen; as a result, phosphonic acid ($\ce{H3PO3}$) is good reducing agent.
$\ce{H3PO4 + 2H+ + 2e-> H3PO3 + H2O}$; $E=-0.276 \ce{V}$
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