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The basic steps involved in the conversion of $\ce{FeCr2O4 -> Cr}$ involves,

$$\ce{FeCr2O4 ->[NaOH / air, \Delta] Na_2CrO4 ->[C, \Delta] CrO3 ->[Al, \Delta] Cr}$$

Basically, I'm having problems in learning so large amount of inorganic reactions, so I thought it would be easy to learn the mechanism behind these reactions, just like we do in organic chemistry.

Therefore, I'm keen to know the mechanism behind first conversion (i.e., from $\ce{FeCr2O4 -> Na_2CrO4}$). I've already tried to search for this, but found nothing related.


My attempt

I know some basic sets of inorganic mechanisms, as follows

  • Ion-exchange
  • Decomposition
  • Water-addition
  • Redox
  • Complex reaction

From the above listed, I don't think it would be ion-exchange (bcoz, $\ce{FeCr2O4 -> FeCr+ + CrO4-}$ is very unlikely to happen), water-addition or complex reaction. I guess it could be decomposition or redox, but I'm not sure on the "bond-movement" mechanism part.

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  • $\begingroup$ This paper may help $\endgroup$ – Safdar Jul 12 at 10:16
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    $\begingroup$ @SafdarFaisal: From the paper you've linked, it seems that inorganic mechanisms don't have bond movement visualizations / e- or nu- attacks / addition or subs, unlike organic ones. Is it true? $\endgroup$ – Rahul Verma Jul 12 at 10:45
  • $\begingroup$ unfortunately feels so.. There are mechanisms for some reactions like hydrolysis.. $\endgroup$ – Safdar Jul 12 at 10:58
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    $\begingroup$ @RahulVerma The majority of the reactions you quoted occur either completely in solid state, or have at least one solid reactant, which suggests the main limiting factor is likely a diffusion. For the same reason one doesn't use arrow pushing schemes in IC as there are usually no molecules at all, the change occurs to a network solid. For molecular inorganic species, however, one can use similar schematics from OC, and several examples are covered, for instance, in the textbook Arrow pushing in inorganic chemistry by Ghosh and Berg. $\endgroup$ – andselisk Jul 12 at 11:35
  • $\begingroup$ This answer by @Maurice may be of interest. If so, please consider upvoting that answer. $\endgroup$ – Ed V Jul 12 at 12:40
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The first reaction is a double oxidation. Both Fe and Cr are oxidized by O2 from the air. The reaction is made of two half-reactions. The first half-reaction is not easy to establish, because two elements (Fe, Cr) are oxidized simultaneously, Fe from +II to +III, and Cr from +III to +VI. Sorry to say it : It is one of the most difficult half-equations I ever had to state. It is : $$\ce{FeCr_2O_4 + 11 OH^- -> Fe(OH)_3 + 2 CrO_4^{2-} + 4 H_2O + 7 e-}$$ The second half-equation is easier to write. $$\ce{O_2 + 2H_2O + 4 e- -> 4 OH^-}$$ The overall equation is obtained by multiplying the first equation by $4$ and the second by $7$ and adding the whole. After simplification, it yields : $$\ce{4 FeCr_2O_4 + 16 OH^- + 7 O_2 -> 4 Fe(OH)_3 + 8 CrO_4^{2-} + 2 H_2O}$$ or, without ions : $$\ce{4 FeCr_2O_4 + 16 NaOH + 7 O_2 -> 4 Fe(OH)_3 + 8 Na_2CrO_4 + 2 H_2O}$$ This was the first step, defining how to pass from the mineral $\ce{FeCr_2O_4}$ to $\ce{Na_2CrO_4}$, at high temperature, and in the presence of air and $\ce{NaOH}$. The final mixture can be washed with water, which dissolves easily $\ce{Na_2CrO_4}$, as $\ce{Fe(OH)_3}$ is insoluble in water.

Then $\ce{Na_2CrO_4}$ is transformed in $\ce{CrO_3}$ by adding moderately concentrated sulfuric acid : $$\ce{Na_2CrO_4 + H_2SO_4 -> CrO_3 + Na_2SO_4 + H_2O}$$ $\ce{CrO_3}$ is not soluble and can be separated by filtration. Then it is mixed with aluminum powder to get a exothermic reaction when engaged by a match : $$\ce{2 Al + CrO_3 -> Al2O3 + Cr}$$

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This is an example of industrial recovery process of $\ce{Cr}$ from toxic chromium(VI) waste, which usually use high temperature reactions. However, the other answer is acceptable since OP's main focus is on the first reaction of converting $\ce{FeCr2O4}$ (one of the products from initial reaction of chromium(VI) waste) to $\ce{Na2CrO4}$. Because these reactions are done in high temperatures (usually, $\pu{1000-1200 ^\circ C}$; Ref.1), I'd like to make few comments:

  1. $\ce{CrO3}$ would decompose to $\ce{Cr2O3}$ upon heating above $\pu{197 ^\circ C}$ (Wikipedia). The relavent reaction is: $$\ce{4CrO3 ->[$\Delta \gt 197 \ \mathrm{^\circ C}$] 2Cr2O3 + 3O2}$$ Nonetheless, $\ce{Cr2O3}$ can also be easily reduced to elemental $\ce{Cr}$.
  2. A large amount of chromic residues is discharged during the production of chromates by calcination of chromic ores. Statistics show that to produce a ton of sodium dichromate, the calcination process will discharge 2.5-3 tons of poisonous chromic residues. That chromic residues contain 3-7% of residual $\ce{Cr2O3}$, 8-11% of $\ce{Fe2O3}$, and 0.5-1.5% of water-soluble $\ce{Cr(VI)}$ (Ref.1). This reference has claimed a re-calcination of these poisonous chromic residues:
    Chronic residues are dried and crushed to 80-100 mesh, and mixed with chromic ore powder and sodium carbonate. Water is added to the the mixture to make it homogeneous (coke powder is also added as necessary), and then is heated in furnace at $\pu{1000-1200 ^\circ C}$. The main chemical reaction for this mixture is: $$\ce{4(FeO.Cr2O3) + 8Na2CO3 + 7O2 -> 8Na2CrO4 + 2Fe2O3 + 8CO2}.$$ Keep in mind that $\ce{(FeO.Cr2O3)}$ is representing $\ce{FeCr2O4}$ in this equation and $\ce{Na2CO3}$ replaces $\ce{NaOH}$ in OP's equation. Yet, the end results are the same.
  3. The OP's second reaction, $\ce{Na2CrO4 + C ->[C, \Delta] CrO3}$ does not make any sense since oxidation of chromium does not change. I'd say it may be $\ce{Na2CrO4 + C ->[C, \Delta] Cr2O3}$. For example, the following conversion has been achied directly from $\ce{FeCr2O4}$: $$\ce{FeO.Cr2O3 + C -> Fe + Cr2O3 + CO}$$ Regardless, I also found following reaction to support my suggestion: $$\ce{2Na2CrO4 + 3C -> Cr2O3 + 2Na2O + 3CO}$$ Nevertheless, the reference 1 gives a different set of reactions for the conversion of $\ce{Na2CrO4}$: $$\ce{4Na2CrO4 + 6S + 7H2O -> 4Cr(OH)3 + 3Na2S2O3 + 2NaOH}$$ $$\ce{8Na2CrO4 + 6Na2S + 23H2O -> 8Cr(OH)3 + 3Na2S2O3 + 22NaOH}$$ $$\ce{2Cr(OH)3 ->[\pu{1200 ^\circ C}] Cr2O3 + 3H2O}.$$
  4. For the 3rd reaction, $\ce{Al}$ is not the only reducing agent able to reduce $\ce{Cr2O3}$ to $\ce{Cr^\circ}$. The following reactions are listed in Ref.1: $$\ce{Cr2O3 + 3CO -> 2Cr + 3H2} \\ \ce{Cr2O3 + 3CO -> 2Cr + 3H2O} \\ \ce{Cr2O3 + 3C -> 2Cr + 3CO}$$ All of these reactions have been done at high temperatures.

Reference:

  1. Qi-Jiang Situ, Ke-Ming Xu, Pei-Nian Huang, Xing-Qin Li, De-Han Zeng, Zhi-Fa Hu, Zhi-Quan Wen, "Re-calcination and extraction process for the detoxification and comprehensive utilization of chromium residues," United States Patent 1995, 5,395,601 (PDF).
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