A) Iron and \[{{H}_{2}}S{{O}_{4}}\] (aqueous)
B) Iron and steam
C) Copper and HCl (aqueous)
D) Sodium and ethyl alcohol
Correct Answer: C
Solution :
Fe + dil \[{{H}_{2}}S{{O}_{4}}\,\] \[\to \]\[FeS{{O}_{4}}+{{H}_{2}}\] \[\uparrow \] 3Fe+ \[\underset{Steam}{\mathop{4{{H}_{2}}O}}\,\] \[\to \] \[F{{e}_{3}}{{O}_{4}}+4{{H}_{2}}\] \[\uparrow \] Cu is below H in electrochemical series. \[Cu+dil.\,HCl\xrightarrow[{}]{{}}No\,\,\,reaction\] \[2Na+{{C}_{2}}{{H}_{5}}OH\xrightarrow[{}]{{}}2{{C}_{2}}{{H}_{5}}ONa+{{H}_{2}}\uparrow \]
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