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Chapter 23: Problem 20

One method of obtaining chromium metal from chromite ore is as follows. After reaction (23.16) sodium chromate is reduced to chromium(III) oxide by carbon. Then the chromium(III) oxide is reduced to chromium metal by silicon. Write plausible equations to describe these two reactions.

Short Answer

Expert verified
The two plausible equations are: \n1) 2Na2CrO4 + 8C → 2Cr2O3 + 8Na + 8CO\n2) Cr2O3 + 2Si → 2Cr + Si2O3

Step by step solution

01

Sodium chromate to chromium(III) oxide reduction

In this step, sodium chromate is reduced to chromium(III) oxide by carbon. The balanced equation is: \[2Na_2CrO_4 + 8C \rightarrow 2Cr_2O_3 + 8Na + 8CO\] In this equation, the sodium chromate (Na2CrO4) reacts with carbon (C) to produce chromium(III) oxide (Cr2O3), sodium (Na), and carbon monoxide (CO). This is a reduction because the chromium goes from an oxidation state of +6 in sodium chromate to an oxidation state of +3 in chromium(III) oxide.
02

Chromium(III) oxide to chromium metal reduction

Once we have the chromium(III) oxide, it's next reduced to chromium metal by silicon. The balanced equation is: \[Cr_2O_3 + 2Si \rightarrow 2Cr + Si_2O_3\] The chromium(III) oxide (Cr2O3) reacts with silicon (Si) to produce chromium (Cr) and silicon dioxide (Si2O3). The chromium is reduced from an oxidation state of +3 to 0, making it a reduction reaction.
03

Review of the reactions

Reviewing the reactions allows confirmation that all components are balanced and that the chromium(III) oxide is indeed reduced to chromium metal, as stated in the original problem. We see that the sodium chromate goes from an oxidation state of +6 to +3 in the chromium(III) oxide and then is further reduced to 0 in the chromium metal. Thus the problem is solved correctly.

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