# Products of the reaction between chromium(III) chloride, zinc and sulfuric acid

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

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

• 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). – Ivan Neretin Jun 18 at 17:22
• @IvanNeretin The instructions were to use fine Zn powder so we didn't use nascent hydrogen – muhammad haider Jun 18 at 17:31
• But you added acid, didn't you? That was my point. – Ivan Neretin Jun 18 at 17:32
• The equation you have given is not balanced. – Nilay Ghosh Jun 18 at 18:09
• ZnCl is not a product. It does not exist at normal conditions. – Nilay Ghosh Jun 19 at 4:30

I know that chromium(III) reduces to chromium(II), but I'm not sure about the reaction. Would it be $$\ce{CrCl3·6H2O + Zn -> Cr(H2O)6Cl2 + ZnCl + H2}?$$
• 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 – muhammad haider Jun 19 at 13:09