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

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Answer 1 · 2016-10-05T15:36:14

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

Aluminum is oxidized to $Al^(3+)$ .

And $H^+$ is reduced to $H_2(g)$ .

$"Oxidation"$ $Al(s) rarr Al^(3+)+3e^-$

$"Reduction"$ $H^(+) + e^(-) rarr 1/2H_2(g)$

$"Overall"$ $Al(s) +3H^(+) rarr Al^(3+)+3/2H_2(g)uarr$

Alternatively:

$"Overall"$ $Al(s) +3H_3O^(+) rarr Al^(3+)+3/2H_2(g)uarr+3H_2O(l)$

else:

$Al(s) +3HCl(aq) rarr AlCl_3(aq)+3/2H_2(g)uarr$

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