What you are proposing is basically and net ionically
2Al
3+ + 3SO
32- + 3H
2O -> 2Al + 3SO
42- + 6H
+You may check half potentials to see whether this reaction is possible, but it is enough to think for a moment. Where is Al on the reactivity series? Will it survive in the presence of H
+? No, it will react:
2Al + 6H
+ -> 2Al
3+ + 3H
2If so, what you propose to happen looks like
2Al
3+ + 3SO
32- + 3H
2O -> 2Al + 3SO
42- + 6H
+ -> 2Al
3+ + 3H
2 + 3SO
42-or just
SO
32- + H
2O -> H
2 + SO
42-Obvious now?
(did I ever told you how I love to use EBAS when preparing such equations? UBBC codes are generated automatically and I have balanced and correctly displayed equation in no time - and everyone thinks "gee, Borek knows how to post"
)