Which one of the following pairs of substances on reaction will not not evolve `H_2` gas?

Options

A. Iron and `H_(2)SO_(4)` (aqueous)
B. Iron and copper
C. Copper and HCl (aqueous)
D. Sodium and ethyl alcohol

Answer

C

Explanation

According to elctrochemical series `E_(Cu^(2+)//Cu)^(@)` is positive while `E_(Fe^(2+)//Fe)^(@)` and `E_(Na^(2+)//Na)^(@)`are negative. A positvie `E^(@)` means that the redox couple is a weaker reducing agent than the `H^(+)/H_(2)` couple while a negative `E^(@)` means that the redox couple is a stronger reducing agent thatn the `H^(+)/H_(2)` couple. Thus, copper metal will not be able to reduce `H^(+)` ions of an acid to realease `H_(2)` gas while other meatls can do that:
`Cu+HCl(aq.) rarr` No reaction
`Fe+H_(2)SO_(4)(aq.) rarr FeSO_(4)(aq.)+H_(2)(g)`
`3Fe+underset(steam)(3H_(2)O_(g)) rarr Fe_(3)O_(4)(s)+4H_(2)(g)`
`2Na(s)+c_(2)H_(5)OH(l) rarr (2C_(2)H_(5)ONa+H_(2)(g)`