# How do we represent the reaction between aluminum metal and protium ion, H^+ or H_3O^+?

Oct 5, 2016

This is formally a redox equation. Aluminum metal is sufficiently active to reduce the hydronium ion.

#### Explanation:

Aluminum is oxidized to $A {l}^{3 +}$.

And ${H}^{+}$ is reduced to ${H}_{2} \left(g\right)$.

$\text{Oxidation}$ $A l \left(s\right) \rightarrow A {l}^{3 +} + 3 {e}^{-}$

$\text{Reduction}$ ${H}^{+} + {e}^{-} \rightarrow \frac{1}{2} {H}_{2} \left(g\right)$

$\text{Overall}$ $A l \left(s\right) + 3 {H}^{+} \rightarrow A {l}^{3 +} + \frac{3}{2} {H}_{2} \left(g\right) \uparrow$

Alternatively:

$\text{Overall}$ $A l \left(s\right) + 3 {H}_{3} {O}^{+} \rightarrow A {l}^{3 +} + \frac{3}{2} {H}_{2} \left(g\right) \uparrow + 3 {H}_{2} O \left(l\right)$

else:

$A l \left(s\right) + 3 H C l \left(a q\right) \rightarrow A l C {l}_{3} \left(a q\right) + \frac{3}{2} {H}_{2} \left(g\right) \uparrow$