Question

consider the following reaction and identify the substance being reduced.
si + 2 f₂ → sif₄
si
f₂
f in sif₄
si in sif₄

Explanation:

Step1: Determine oxidation states

• Oxidation state of \(Si\) in \(Si\) (elemental form) is \(0\).
• Oxidation state of \(F\) in \(F_{2}\) (elemental form) is \(0\).
• In \(SiF_{4}\), let oxidation state of \(Si=x\) and of \(F = - 1\) (since \(F\) is more electronegative). Using the formula \(x+4\times(-1)=0\) (sum of oxidation states in a neutral compound is \(0\)), we get \(x = + 4\) for \(Si\) in \(SiF_{4}\), and oxidation state of \(F\) in \(SiF_{4}\) is \(-1\).

Step2: Identify reduction

• Reduction is a decrease in oxidation state.
• For \(Si\): \(0\to+4\) (oxidation, increase in oxidation state).
• For \(F\): \(0\to - 1\) (reduction, decrease in oxidation state). So \(F_{2}\) (where \(F\) is in \(0\) oxidation state) is being reduced.

Answer:

\(F_{2}\)