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
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