Question

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

Short answer

3\mathrm{K}_{2}\mathrm{SiF}_{6}+4\mathrm{Al}\rightarrow3\mathrm{Si}+4\mathrm{AlF}_{3}+6\mathrm{KF}

Step-by-step solution

Identify the reactants and products

The given reactants are \(\mathrm{K}_{2}\mathrm{SiF}_{6}\) and \(\mathrm{Al}\). The products will include silicon \(\mathrm{Si}\), aluminum fluoride \(\mathrm{AlF}_{3}\), and potassium fluoride \(\mathrm{KF}\).

Analyze oxidation states

Check the oxidation states to understand what gets oxidized and what gets reduced. In \(\mathrm{K}_{2}\mathrm{SiF}_{6}\), \(\mathrm{Si}\) is in the +4 oxidation state, while \(\mathrm{Al}\) is in the 0 oxidation state. \(\mathrm{F}^{-}\) and \(\mathrm{K}^{+}\) will not change their oxidation states.

Write the balanced chemical equation

Write the equation to reflect that silicon gets reduced from \(\mathrm{K}_{2}\mathrm{SiF}_{6}\) and aluminum gets oxidized: \[\mathrm{3\,K_{2}SiF_{6}} + 4\,Al \rightarrow 3\,Si + 4\,AlF_{3} + 6\,KF\]

Key concepts

Chemical Equation

In a chemical reaction, a chemical equation represents the transformation of reactants into products. Reactants are substances that start the reaction, and products are substances formed as a result. Here, the reactants are \(\text{K}_{2}\text{SiF}_{6}\) and \(\text{Al}\).👩‍🔬 The resulting products include silicon \(\text{Si}\), aluminum fluoride \(\text{AlF}_{3}\), and potassium fluoride \(\text{KF}\).
The chemical equation for this reaction is:
\[3 \text{K}_{2}\text{SiF}_{6} + 4 \text{Al} \rightarrow 3 \text{Si} + 4 \text{AlF}_{3} + 6 \text{KF} \]
This equation shows all substances involved before and after the reaction. It is essential for understanding which elements and compounds are interacting.

Oxidation States

Oxidation states help identify the movement of electrons in a chemical reaction. An oxidation state is a number assigned to elements in a compound that reflects their charge. In this reaction:
⦁ \(\text{Si}\) in \(\text{K}_{2}\text{SiF}_{6}\) has an oxidation state of +4.
⦁ \(\text{Al}\) starts with an oxidation state of 0 (since it's in its pure form).
⦁ \(\text{F}^{-}\) and \(\text{K}^{+}\) retain their oxidation states of -1 and +1, respectively.
By identifying these states, we can see:
⦁ Silicon (Si) is reduced from +4 to 0, as it gains electrons.
⦁ Aluminum (Al) is oxidized from 0 to +3, as it loses electrons.
Understanding these changes is vital in determining which elements undergo oxidation and reduction.

Balancing Reactions

Balancing chemical reactions is necessary to follow the Law of Conservation of Mass, which states that matter cannot be created or destroyed. This means the number of atoms for each element must be the same on both sides of the chemical equation.
⦁ First, write down the unbalanced equation:
\text{K}_{2}\text{SiF}_{6} + \text{Al} \rightarrow \text{Si} + \text{AlF}_{3} + \text{KF}\.

⦁ To balance the atoms of each element:

- Balance \(\text{Si}\) atoms: 3 \(\text{K}_{2}\text{SiF}_{6}\) gives 3 \(\text{Si}\).
- Balance \(\text{Al}\) atoms: 4 \(\text{Al}\) gives 4 \(\text{AlF}_{3}\).
- Balance \(\text{F}\) atoms: 18 \(\text{F}\) from 3 \(\text{K}_{2}\text{SiF}_{6}\); this is balanced with 12 \(\text{F}\) in 4 \(\text{AlF}_{3}\) and 6 \(\text{F}\) in 6 \(\text{KF}\).
- Balance \(\text{K}\) atoms: 6 \(\text{K}\) from 3 \(\text{K}_{2}\text{SiF}_{6}\), matching 6 \(\text{KF}\).

The final balanced equation is:
\[3 \text{K}_{2}\text{SiF}_{6} + 4 \text{Al} \rightarrow 3 \text{Si} + 4 \text{AlF}_{3} + 6 \text{KF}\]
This ensures all atoms accounted for.