Question

Which of the following pair will not produce dihydrogen gas?

• A. $\text{Cu + HCl}$ (dil)
• B. ${\text{Fe +H}}_2{\text{SO}}_4$ (dil)
• C. $\text{Mg + steam}$
• D. $\text{Na + alcohol}$

Answer 1

The correct option is A

$\text{Cu + HCl}$ (dil)

Hydrogen cannot be prepared from metals that are below it in activity series of metals such as copper (Cu), mercury (Hg) etc, because these metals can not displace the $H^{+}$ ion from the acids.

Metals that are more reactive than hydrogen can displace it and thus liberate $H_2$ gas, as illustrated in the equations below:

$Fe\left(s\right)+H_{2}S{O}_4\left(aq\right)\to FeS{O}_4\left(aq\right)+H_2\left(g\right)\uparrow$
$Mg\left(s\right)+H_{2}O\left(g\right)\to MgO\left(aq\right)+H_2\left(g\right)\uparrow$
$Na\left(s\right)+C_2{H}_{5}OH\left(l\right)\to C_2{H}_{5}ONa\left(l\right)+H_2\left(g\right)\uparrow$