• # question_answer13) Justify that following reactions are redox reactions: (a) $CuO(s)+{{H}_{2}}(g)\to Cu(s)+{{H}_{2}}O(g)$ (b) $F{{e}_{2}}{{O}_{3}}(s)+3CO(g)\to 2Fe(s)+3C{{O}_{2}}(g)$ (c) $4BC{{l}_{3}}(g)+3LiAl{{H}_{4}}(s)\to 2{{B}_{2}}{{H}_{6}}(g)+LiCl(s)+3AlC{{l}_{3}}(s)$(d) $2K(s)+{{F}_{2}}(g)\to 2{{K}^{+}}{{F}^{-}}(s)$ (e) $4N{{H}_{3}}(g)+5{{O}_{2}}(g)\to 4NO(g)+6{{H}_{2}}O(g)$

Process Reason (a)   $CuO$is reduced to $Cu(s)$ Loss of oxygen     ${{H}_{2}}(g)$oxidised to ${{H}_{2}}O(g)$ Combination with oxygen (b)   $F{{e}_{2}}{{O}_{3}}$(s) is reduced to $Fe(s).$ Loss of oxygen     $CO(g)$is oxidised to $C{{O}_{2}}(g)$ Combination with Oxygen (c)   $\overset{+3-3}{\mathop{BC{{l}_{3}}}}\,$is reduced to $\overset{-6+6}{\mathop{{{B}_{2}}{{H}_{6}}}}\,$ Oxidation state of boron changes from +3 to -3 (gain of electron)   $\overset{+1+3-4}{\mathop{LiAl{{H}_{4}}}}\,$is oxidized. Oxidation state of hydrogen changes from -1 to +1 (loss of electron) (d)     $K(s)$is oxidised to ${{K}^{+}}$. Loss of electron       ${{F}_{2}}$ is reduced to ${{F}^{-}}$. Gain of electron (e) $\overset{-3}{\mathop{N}}\,\overset{+3}{\mathop{{{H}_{3}}}}\,$ is oxidised to $\overset{+2}{\mathop{N}}\,\overset{-2}{\mathop{O}}\,.$ Loss of hydrogen and combination with oxygen (loss of electron)     ${{O}_{2}}$ is reduced to ${{O}^{2-}}$ ion found in ${{H}_{2}}O$ and $NO.$ Gain of electron