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From Russian test problem 4301:

$$\ce{3NaNO3 + 8Al + 5NaOH + 18H2O -> 8Na[Al(OH)4] + 3NH3(g)}$$

How does ammonia evolve here? Is it that we get hydrogen gas evolving in the reaction between Al and NaOH and this gas reacts with the NO3 anion?

Is there a logical way to deduce this or should one just memorize that "Al and Zn reduce $\ce{NO3-}$ to ammonia in basic solutions"?

I learned that the reaction is called "a nitrate test using Devarda's alloy", but there is no specific description of the process on Wikipedia. All I learned is that the reason is certainly not the freshly-minted hydrogen:

Nascent hydrogen was supposed to be responsible for the reduction of arsenate or nitrate in arsine or ammonia respectively. Nowadays, isotopic evidence[8] has closed the nascent hydrogen debate, presently considered to be a Gedanken artifact of romanticism.

As a side question, why is NaOH needed? My thought is that it activates the aluminium by breaking through the oxide layer.

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    $\begingroup$ In the test problem linked you are not expected to know this is the case, but to guess from other data with more common knowledge. And no, neither guessing, nor remembering such corner-cases is expected and/or possible. Some should be remembered if they are part of your field of work, and that's it. $\endgroup$
    – permeakra
    Commented May 24, 2016 at 11:19
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    $\begingroup$ You should use reduction potential to deduce whether it will yield ammonia or higher nitrogen oxidation states compound. But in this case, Al and Zn are quite strong reductive metal, so both can strongly reduce nitrate ion to ammonia. $\endgroup$
    – lambda23
    Commented May 24, 2016 at 12:56
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    $\begingroup$ Does this answer your question? Testing for nitrate ions with NaNO3 in Devarda's test $\endgroup$
    – Nisarg Bhavsar
    Commented Jun 29, 2021 at 5:17
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    $\begingroup$ Assuming a redox reaction between metallic aluminium and nitrate, I'd predict that the aluminium gets attracted to the pi-system of nitrate(which contains oxygen, which aluminium has an extremely high affinity for) and donates its electrons to the antibonding N-O orbitals, splitting them and leaving off with the oxygen that it loves. What is left behind is the nitride, which immediately abstracts three protons to become ammonia. $\endgroup$
    – Kanghun Kim
    Commented Nov 10, 2021 at 2:29
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    $\begingroup$ @KanghunKim probably you don't actually get nitride ion. As 9xygen is stripped off and electron pairs beconlme available on the nitrogen they protonate. So you probably get various intermediates where nitrogen is bonded to hydrogen and oxygen, until the oxygen is gone and the bonding with hydrogen is maximized. Thus, in basic solution, NH3. $\endgroup$
    – Oscar Lanzi
    Commented Dec 24, 2023 at 11:28

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