When we check the charge balance of each equation, the first is incorrect. Leaving off the spectator sodium ions, we have
Al(s) + 2OH-(aq) + 2H2O(l) --> Al(OH)4-(aq) + H2(g)
The left side has a total ion charge of -2, while the right side total is only -1.
The second reaction is charge balanced with a -2 total charge on each side:
2Al(s) + 2OH-(aq) + 6H2O(l) --> 2Al(OH)4-(aq) + 3 H2(g)
Also note that if we put in the Na+ ions (as NaOH and Na[Al(OH)4] (aq) ), the sodiums do not balance in the first reaction (two are on the left side but only one is on the right side). The second equation has the sodiums balanced with two on each side.
When balancing oxidation-reduction reactions like this, we must always be sure that the reaction is balanced in terms of atoms and charge. 
Steve
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