Justify that the reactions is redox reactions:$4 B C l_{3}(g)+3 \operatorname{LiAlH}_{4(s)} \rightarrow 2 B_{2} H_{6}(g)+3 \operatorname{LiCl}_{(s)}+3 A l C l_{3(s)}$

Home » Justify that the reactions is redox reactions:

Justify that the reactions is redox reactions: $4 B C l_{3}(g)+3 \operatorname{LiAlH}_{4(s)} \rightarrow 2 B_{2} H_{6}(g)+3 \operatorname{LiCl}_{(s)}+3 A l C l_{3(s)}$

Justify that the reactions is redox reactions: $4 B C l_{3}(g)+3 \operatorname{LiAlH}_{4(s)} \rightarrow 2 B_{2} H_{6}(g)+3 \operatorname{LiCl}_{(s)}+3 A l C l_{3(s)}$

Solution:

$4 \mathrm{BCl}_{3(g)}+3 \mathrm{LiAlH}_{4(s)} \rightarrow 2 \mathrm{~B}_{2} \mathrm{H}_{6}(\mathrm{~g})+3 \mathrm{LiCl}_{(s)}+3 \mathrm{AlCl}_{3}(s)$
the above response,
Oxidation no. of $\mathrm{B}$ and $\mathrm{Cl}$ in $B C l_{3}$ is $+3$ and $-1$ separately
Oxidation no. of $L i$ , Al and $H$ in $L I A l H_{4}$ is $+1,+3$ and $-1$ separately.
Oxidation no. of $B$ and $H$ in $B_{2} H_{6}$ is $-3$ and $+1$ individually
Oxidation no. of $L$ I and $\mathrm{Cl}$ in LiCl is $+1$ and $-1$ individually.
Oxidation no. of Al and $\mathrm{Cl}$ in $A l C l_{3}$ is $+3$ and $-1$ individually
The oxidation no. of $\mathrm{B}$ diminished from $+3$ in $\mathrm{BCl}_{3}$ to-3 in $B_{2} H_{6}$ . That is $B C l_{3}$ is decreased to $B_{2} H_{6}$ .
The oxidation no. of $\mathrm{H}$ expanded from $-1$ in $\mathrm{Li} \mathrm{AlH}_{4}$ to $+1$ in $B_{2} H_{6}$ . That is $\operatorname{LiAl} H_{4}$ is oxidized to $B_{2} H_{6}$
Thusly, the response is redox response.