Question
Which of the following combination will produce $${H_2}$$ gas?
A.
$$Fe\,{\text{metal}}\,{\text{and}}\,{\text{conc}}{\text{.}}\,HN{O_3}$$
B.
$$Cu\,{\text{metal}}\,{\text{and}}\,{\text{conc}}{\text{.}}\,HN{O_3}$$
C.
$$Zn\,{\text{metal}}\,{\text{and}}\,{\text{NaOH}}\left( {aq} \right)$$
D.
$$Au\,{\text{metal}}\,{\text{and}}\,NaCN\left( {aq} \right)\,{\text{in}}\,{\text{the}}\,{\text{presence}}\,{\text{of}}\,{\text{air}}$$
Answer :
$$Zn\,{\text{metal}}\,{\text{and}}\,{\text{NaOH}}\left( {aq} \right)$$
Solution :
$$\left( {Fe\,{\text{becomes}}\,{\text{passive}}\,{\text{on}}\,{\text{reaction}}\,{\text{with}}\,{\text{concentrated}}\,HN{O_3}} \right).\,\,{\text{However,}}\,{\text{cold}}\,{\text{relatively}}\,{\text{conc}}{\text{.}}\,HN{O_3}\,{\text{reacts}}\,{\text{with}}\,Fe\,{\text{as}}\,{\text{below}}{\text{.}}$$
$$\eqalign{
& Fe + 6HN{O_3} \to Fe{\left( {N{O_3}} \right)_3} + 3N{O_2} + 3{H_2}O \cr
& Cu + 4HN{O_3}\,\left( {conc.} \right) \to Cu{\left( {N{O_3}} \right)_2} + 2N{O_2} + 2{H_2}O \cr
& 4Au + 8NaCN + {O_2} + 2{H_2}O \to 4Na\left[ {Au{{\left( {CN} \right)}_2}} \right] + 4NaOH \cr
& Zn + 2\,NaOH \to N{a_2}Zn{O_2} + {H_2} \cr} $$