Question

Justify that the following reaction is redox reaction: $4BCl_{3}(g)+3LiAlH_{4}(s)\longrightarrow 2B_{2}H_{6}+3LiCl(s)+3AlCl_{3}(s)$

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.

$\begin{matrix} \quad +3-1 \quad +1+3-1\quad \quad -3+1\quad \quad \quad \quad +1-1\quad \quad +3\quad -1\quad  \\ 4B{ Cl }_{ 3 }(g)+3LiAl{ H }_{ 4 }(s) \longrightarrow 2{ B }_{ 2 }{ H }_{ 6 }(g)+3LiCl(s)+3Al{ Cl }_{ 3 }(s) \end{matrix}$

Step 2: Notice that the oxidation number of boron (B) is +3 in the reactants but it is -3 in the products. So there is a decrease in the oxidation number of boron during the chemical reaction. Hence it is evident that $4B{ Cl }_{ 3 }$ is reduced to ${ B }_{ 2 }{ H }_{ 6 }$.

Step 3: Notice that the oxidation number of hydrogen (H) is -1 in the reactants but it is +1 in the products. So there is an increase in the oxidation number of hydrogen during the chemical reaction. Hence it is evident that $LiAl{ H }_{ 4 }$  is oxidized to ${ B }_{ 2 }{ H }_{ 6 }$.

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