Question

Justify that the following reaction is redox reaction: $2K(s)+F_{2}(g)\longrightarrow ]2K^{+}F^{-}(s)$

Answer 1

Step 1: Write the reaction along with oxidation number of each element that is present in the reactants as well as the products of the reaction.

$\begin{matrix} 0\quad \quad \quad \quad 0 &  & +1\quad \quad -1 \\ 2K(s)+{ F }_{ 2 }(g) & \longrightarrow  & 2{ K }^{ + }{ F }^{ - }(s) \end{matrix}$

Step 2: Notice that the oxidation number of potassium (K) is 0 in the reactants but it is +1 in the products. So, there is an increase in the oxidation number of potassium during the chemical reaction. Hence it is evident that potassium is oxidized to potassium fluoride.

Step 3: Notice that the oxidation number of fluorine(F) is 0 in the reactants but it is -1 in the products. So, there is a decrease in the oxidation number of fluorine during the chemical reaction. Hence it is evident that ${ F }_{ 2 }$ is reduced to KF.

Thus, it is proved that the given reaction is a redox reaction .