question_answer 13)

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) \[4N{{H}_{3}}(g)+5{{O}_{2}}(g)\to 4NO(g)+6{{H}_{2}}O(g)\]

Answer:

  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