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The aim of my experiment was to make a Cr(II) aquacomplex in situ, and for that I have the reaction:

$$\ce{CrCl3·6H2O + Zn}$$

and I add concentrated sulfuric acid to it. I know that chromium(III) reduces to chromium(II), but I'm not sure about the reaction. Would it be

$$\ce{CrCl3·6H2O + Zn -> Cr(H2O)6Cl2 + ZnCl + H2}?$$

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    $\begingroup$ 6H2O is irrelevant, since we are dealing with water solutions anyway. As for the rest, the reduction can be performed by Zn, or it can be done by the nascent hydrogen. I strongly suspect the latter (otherwise you won't need acid around). $\endgroup$
    – Ivan Neretin
    Commented Jun 18, 2019 at 17:22
  • $\begingroup$ @IvanNeretin The instructions were to use fine Zn powder so we didn't use nascent hydrogen $\endgroup$
    – bobsburger
    Commented Jun 18, 2019 at 17:31
  • $\begingroup$ But you added acid, didn't you? That was my point. $\endgroup$
    – Ivan Neretin
    Commented Jun 18, 2019 at 17:32
  • $\begingroup$ The equation you have given is not balanced. $\endgroup$
    – Nilay Ghosh
    Commented Jun 18, 2019 at 18:09
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    $\begingroup$ ZnCl is not a product. It does not exist at normal conditions. $\endgroup$
    – Nilay Ghosh
    Commented Jun 19, 2019 at 4:30

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I know that chromium(III) reduces to chromium(II), but I'm not sure about the reaction. Would it be $$\ce{CrCl3·6H2O + Zn -> Cr(H2O)6Cl2 + ZnCl + H2}?$$

Your concept is right and I am glad nobody taught you the concept of nascent hydrogen. It is an obsolete idea of the 1920-40s. Please don't use it anywhere. Wikipedia has a nice summary https://en.wikipedia.org/wiki/Nascent_hydrogen and of course there are several articles which negate this very concept.

You can just simply write the ionic equation involving Cr(III), Zn, resulting in Cr(II) and Zn(II). Hydrogen is just a side-product, which is not participating in the reaction. Write the conditions over the arrow.

Alternatively, write the aquated forms as a complex. Your current equation has incorrect formula of zinc salt. Check what it should be!

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  • $\begingroup$ so the full formula would be: $\ce{3CrCl3 + 3Zn + 12H_2O-> 2[Cr(H2O)6]^2+ + 3ZnCl_2}$. I kept the water molecules to be able to see the metal-aqua complex that builds in the end $\endgroup$
    – bobsburger
    Commented Jun 19, 2019 at 13:09
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    $\begingroup$ No need to add water, the starting product is most likely [Cr(H2O)6]^3+ . Chloride is just a spectator ion. Show the acid over the arrow. I would personally write, Zn and acid over the arrow, [Cr(H2O)6]^3+ -> [Cr(H2O)6]^2+, because nobody knows the exact mechanism. $\endgroup$
    – ACR
    Commented Jun 19, 2019 at 13:28

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