Q22.39P - Chemistry: Molecular Nature Of Matter And Change

Question

Silicon is prepared by the reduction of ${\mathbf{\text{K}}}_{\mathbf{\text{2}}}{\mathbf{\text{SiF}}}_{\mathbf{\text{6}}}\mathbf{\text{ with Al}}$. Write the equation for this reaction. (Hint: Can ${\mathbf{\text{F}}}^{\mathbf{-}}$ be oxidized in this reaction? Can $\mathbf{K}$ be reduced?)

Step-by-Step Solution

Answer

The reaction of ${\text{K}}_{\text{2}}{\text{SiF}}_{\text{6}}$ on reduction with $\mathrm{AI}$ is,

$3 {\mathrm{K}}_2{\mathrm{SiF}}_6( \mathrm{s}) + 4\mathrm{Al}\left(\mathrm{s}\right)\to 6\mathrm{KF}\left(\mathrm{s}\right) + 4{\mathrm{AlF}}_3( \mathrm{s}) + 3\mathrm{Si}\left(\mathrm{s}\right)$

Step 1: Reduction

A reduction chemical reaction is when the electronsassociated with a single atom or a group of atoms are increased. As a result, a reduced atom takes electrons from another atom. The atom that generally provides the electrons then gets oxidized as a result. Redox reactions are another name for reduction reactions.

Step 2: Explanation

The reaction of ${\text{K}}_{\text{2}}{\text{SiF}}_{\text{6}}$ on reduction with $\mathrm{AI}$ can be written as,

${\mathrm{K}}_2{\mathrm{SiF}}_6\left(\mathrm{s}\right)+\mathrm{Al}\left(\mathrm{s}\right)\to \mathrm{KF}\left(\mathrm{s}\right)+{\mathrm{AlF}}_3\left(\mathrm{s}\right)+\mathrm{Si}\left(\mathrm{s}\right)$ 

Then, the half-reactions can be written:

Reduction: ${\text{Si}}^{\text{ + 4}}{\text{ + 4e}}^{\text{ - }}\to {\text{ Si}}^{\text{0}}$

Oxidation: ${\text{Al}}^{\text{0}}{\text{ - 3e}}^{\text{ - }}\to {\text{ Al}}^{\text{3 + }}$ 

Balancing the electron number,

Reduction: $\text{3}\left({\text{Si}}^{\text{ + 4}}{\text{ + 4e}}^{\text{ - }}\to {\text{ Si}}^{\text{0}}\right)$  

Oxidation: $\text{4}\left({\text{Al}}^{\text{0}}{\text{ - 3e}}^{\text{ - }}\to {\text{ Al}}^{\text{3 + }}\right)$ 

Thus,

Reduction: ${\text{3Si}}^{\text{ + 4}}{\text{ + 12e}}^{\text{ - }}\to {\text{ 3Si}}^{\text{0}}$ 

Oxidation: ${\text{4Al}}^{\text{0}}{\text{ - 12e}}^{\text{ - }}\to {\text{4Al}}^{\text{3 + }}$ 

Now, combining two equations,

$3{\mathrm{K}}_2{\mathrm{SiF}}_6\left(\mathrm{s}\right)+4\mathrm{Al}\left(\mathrm{s}\right)\to \mathrm{KF}\left(\mathrm{s}\right)+4{\mathrm{AlF}}_3\left(\mathrm{s}\right)+3\mathrm{Si}\left(\mathrm{s}\right)$ 

Balancing the $\text{K and F}$ , the final equation for the reaction would be,

$3 {\mathrm{K}}_2{\mathrm{SiF}}_6( \mathrm{s})+4\mathrm{Al}\left(\mathrm{s}\right)\to 6\mathrm{KF}\left(\mathrm{s}\right)+4{\mathrm{AlF}}_3( \mathrm{s})+3\mathrm{Si}\left(\mathrm{s}\right)$ 

Therefore, the equation for the reaction of ${\text{K}}_{\text{2}}{\text{SiF}}_{\text{6}}$ on reduction with $\text{Al}$ is.

$3 {\mathrm{K}}_2{\mathrm{SiF}}_6( \mathrm{s})+4\mathrm{Al}\left(\mathrm{s}\right)\to 6\mathrm{KF}\left(\mathrm{s}\right)+4{\mathrm{AlF}}_3( \mathrm{s})+3\mathrm{Si}\left(\mathrm{s}\right)$