What would you use for the 2 half reactions for the following redox reaction?? Al + NO3- + OH- + H2O --> Al(OH)4- + NH3
2006-07-26 14:11:50 · 3 answers · asked by Me 1 in Science & Mathematics ➔ Chemistry
Well, aluminum goes from a valence state of 0 on the left to +3 in aluminum hydroxide on the right (loss of 3 electrons) and nitrogen goes from +5 in the nitrate on the left to -3 in the ammonia on the right (a gain of 8 electrons).
2006-07-26 14:21:49 · answer #1 · answered by rb42redsuns 6 · 0⤊ 0⤋
5 OH- + 8Al + 18H2O + 3NO3- --> 8Al(OH)4 - + 3NH3
is what I get. Rb42redsuns gives a good explanation as to how to solve for this answer... make two half reaction an oxidation and a reduction reaction (Al --> Al3+) and (NO3- --> NH3) and then balance for OH-, then H+, and then electrons.
2006-07-26 19:00:11 · answer #2 · answered by kbarnus 2 · 0⤊ 0⤋
There is something wrong with this equation. For one, there is no charge balance, in the products or in the reagents (negative has to equal positive for both)
What's the starting oxidation state of the aluminium?
2006-07-26 17:15:35 · answer #3 · answered by niuchemist 6 · 0⤊ 0⤋