A) iron
B) zinc
C) copper
D) None of these
Correct Answer: B
Solution :
Metals which are above \[{{H}_{2}}\] in electrochemical series and have negative value of \[{{E}^{o}}\] displace \[{{H}_{2}}\] from acid. \[Na,\] \[Mg,\] \[Al,\] \[Zn\] etc liberate \[{{H}_{2}}\] from acids. These metals are above than \[{{H}_{2}}\] in electrochemical series and are more reactive. \[Zn+{{H}_{2}}S{{O}_{4}}\xrightarrow{{}}ZnS{{o}_{4}}+{{H}_{2}}\] \[Zn\] metal also react with \[NaOH\] and give \[{{H}_{2}}\]. \[Zn+2NaOH\xrightarrow{{}}\underset{sodium\,\,zincate}{\mathop{N{{a}_{2}}Zn{{O}_{2}}}}\,+{{H}_{2}}\]
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