Answer
Fe & Al - These two metals dissolve in HCl.
Balanced redox reaction:
$Fe(s) + 2HCl(aq) -> FeCl_{2} (aq) + H_{2} (g)$
$2Al (s) + 6HCl (aq) -> 2AlCl_{3} (aq) + 3H_{2} (g) $
Work Step by Step
In metal activity series it is seen that Fe & Al both metals are active than $H_{2}$. So, it can be concluded that these metals will be oxidized in presence of acid i.e. HCl. So, these two metals dissolve in HCl. On the other hand, Ag and Cu both metals are less active than $H_{2}$. So, they do not dissolve in HCl.
Redox reaction of Fe with HCl: $Fe (s) + H^{+} (aq) -> Fe^{2+} + H_{2}(g) $
Oxidation reaction after balancing charge: $ Fe(s) -> Fe^{2+} + 2e^{-}$
Reduction reaction after balance charge and number of element on both side we get : $2H^{+} + 2e^{-} -> H_{2}$
Now, combining both half reactions we get:
$Fe(s) + 2H^{+}(aq) -> FeCl_{2} (aq) + H_{2} (g)$
or, $Fe(s) + 2HCl(aq) -> FeCl_{2} (aq) + H_{2} (g)$
Redox reaction of Al with HCl: $Al (s) + H^{+} (aq) -> Al^{3+} + H_{2}(g) $
Oxidation reaction after balancing charge: $ Al(s) -> Al^{3+} + 3e^{-}$
Reduction reaction after balance charge and number of element on both side we get : $2H^{+} + 2e^{-} -> H_{2}$
Now, we have to combine both half reactions. But before combining we have to make sure that the electrons are eliminated from both reactions. So, we multiply oxidation reaction with co-efficient 2 and reduction reaction with co-efficient 3. Now, combining both equations we get:
$2Al(s) + 6H^{+}(aq) ->2AlCl_{3} (aq) + 3H_{2} (g)$
or, $2Al (s) + 6HCl (aq) -> 2AlCl_{3} (aq) + 3H_{2} (g) $
