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?)

Short 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-by-step solution

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.

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.

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