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#