$$\ce{2Cr(OH)3 + 4NaOH + 3H2O2 -> 2Na2CrO4 + 8H2O}$$
How does this reaction occur? We suddenly get sodium chromate in the place of chromium hydroxide. I can't understand it. We mix two hydroxides and get a salt?
What role does the peroxide play?
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asked Jun 9 '16 at 19:14
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2$\begingroup$ The peroxide oxidizes the Cr (III) to Cr (VI). $\endgroup$ – f'' Jun 9 '16 at 22:48
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$\begingroup$ @f'' - why isn't it CrO3 then or Cr(OH)6? $\endgroup$ – CowperKettle Jun 10 '16 at 5:10
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1$\begingroup$ Because there is NaOH around. $\endgroup$ – Ivan Neretin Jun 10 '16 at 14:11
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The hydrogen peroxide is a strong oxidiser. In basic conditions, this is partly due to $$\ce{H2O2->H2O + O\ \ }[*].$$ The monooxygen is very eager to gain electrons. Therefore, $$\ce{Cr^{III}->[\ce{[O]}]Cr^{VI} +3e-}.$$Chromium(III) hydroxide$^{[**]}$ is amphoteric, it reacts with bases, too.
$$\ce{Cr(OH)3 + 3NaOH <=>Na3[Cr(OH)6]}$$
This complex is then oxidised. For common chromium(VI) compounds, one would expect to see the equilibrium $$\ce{2CrO4^{2}^- <=>[\ \ \ce{H3O+}][\ \ \ce{OH-}]Cr2O7^{2}^{-}} \ \ \text{[***].}$$
Thus, in (strongly) basic conditions, the chromate ion $\ce{CrO4^{2}^-}$ would be favoured.
Extra
One thing that might be confusing are the equations themselves. In other words, they simply show
- the most likely process (not always the case),
- the end result (there are exceptions),
- the formulae are not true to the actual structure (cannot be sure).
To make inorganic chemistry more tolerable, it is best to accept that there are multiple reactions occurring concurrently. Then it is easier to arrive at the "correct" end result by taking into account reaction conditions and stability.
[*] This is a simplified explanation taken only as a model for basic understanding. The actual decomposition mechanism is more complicated, depending on pH, other ions in solution, etc. For more information on specific situations, look at Kinetic Studies and Mechanism of Hydrogen Peroxide Catalytic Decomposition [---]. There you can find many other great references as well.
[**] Chromium(III) hydroxide might actually be $\ce{CrOOH*H2O}$, or $\ce{Cr2O3*}$ $n$$\ce{H2O}$. Even if we were to be dealing with the hydroxide, it actually comprises of $\ce{Cr(OH)6}$ octahedrons, and is claimed to be inactive (active form is the oxide hydrate). [N. S. Ahmetov's Inorganic chemistry]
I would make the claim that the hydroxide does not actually exist at all, and is, like for iron, mistaken for the oxidehydroxide hydrate. Furthermore, the reaction might not be direct, it could be a case of Le Châtelier – Braun principle.
[***] The overall transition is $$\ce{2CrO4^{2}^{-} +2H3O+ <=>Cr2O7^{2}^{-} +3H2O}$$
from where it is easy to see that adding hydroxide ions tilts the equilibrium to the left. Also, you might wish to look up different aqueous complexes of chromium and the structures of the ions above. That also helps to visualise the process.
Suggestion: Chromium complexes tend to polymerise ;).
Even if $\ce{CrO3}$ would form, as you suggested, it reacts rather well with water via
$$\ce{CrO3 + H2O \rightarrow H2CrO4, \\ 2CrO3 + H2O \rightarrow H2Cr2O7,}$$
and hydroxides:
$$\ce{CrO3 + 2NaOH \rightarrow Na2CrO4 + H2O}.$$
answered Jun 10 '16 at 13:40
