Question

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

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

Step-by-step solution

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.

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.

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.

Key concepts

Sodium chromate reduction

When extracting chromium metal from chromite ore, a crucial step involves the reduction of sodium chromate (Na_2CrO_4) to form chromium(III) oxide (Cr_2O_3). This process utilizes carbon (C) as a reducing agent. The balanced chemical equation for this reaction is as follows: \[2Na_2CrO_4 + 8C \rightarrow 2Cr_2O_3 + 8Na + 8CO\] In this reaction, the sodium chromate is reduced because the chromium's oxidation state decreases from +6 in sodium chromate to +3 in chromium(III) oxide.
Carbon helps facilitate this by acting as a strong electron donor, undergoing oxidation itself to form carbon monoxide (CO) as a byproduct. This reaction is energetically favorable due to carbon's strong affinity for oxygen, which aids in freeing the chromium components from their higher oxidation state.
Understanding this reduction is vital, as it sets the stage for further reduction processes needed to finally attain metallic chromium. This is the initial key step in transforming the chromium from its mineral state to a more refined form, which can then be processed further to achieve its metallic state.

Chromium(III) oxide

Chromium(III) oxide ( Cr_2O_3 ) is an important intermediary in the process of chromium extraction. It serves as the reduced form of chromium obtained from sodium chromate, and it appears as a green, stable powder. The significance of chromium(III) oxide in the extraction process lies in its role as an intermediate, which still requires further reduction to form the pure chromium metal.
This compound is highly valued for its stability and resistance to oxidation, which makes it useful in several industrial applications like pigments and refractories. However, in the context of metal extraction, its importance is in its reduction potential to chromium metal. When Cr_2O_3 is reduced further, it plays a vital part in the extraction chain leading to the elemental form of chromium. This potential is harnessed by using different elements, such as silicon, as reducing agents. Grasping this intermediate step is essential to fully understanding how raw chromium ore is processed and reduced down to its metallic state for industrial use.

Silicon reduction

Once chromium(III) oxide (Cr_2O_3) is formed, it undergoes a final reduction process where silicon (Si) plays the role of the reducing agent. The balanced chemical reaction governing this transformation is: \[Cr_2O_3 + 2Si \rightarrow 2Cr + Si_2O_3\] In this reaction, the chromium(III) oxide interacts with silicon to produce pure chromium metal (Cr) and silicon dioxide (Si_2O_3).
During this chemical reduction, the oxidation state of chromium is reduced from +3 to 0, achieving what is considered metallic chromium. This step is crucial because it converts the intermediate compound into a usable metal. The silicon dioxide formed as a byproduct is highly stable and typically harmless, allowing for an efficient separation process.
Silicon acts as an effective reducing agent because it has a high affinity for oxygen, thereby making it ideal for extracting metals from oxides. Understanding this final reduction is essential for students learning about purification processes in metallurgy, as it illustrates the step-by-step chemical transformations required to obtain pure metal from its ore.