I can't for the life of me figure out the oxidations state for oxygen is Al(OH)4^-1 2Al + 8H2O → 2^+1 + 3H2 + 2Al(OH)4^-1 One oxygen atom in Al(OH)4^-1

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I can't for the life of me figure out the oxidations state for oxygen is Al(OH)4^-1 2Al + 8H2O → 2^+1 + 3H2 + 2Al(OH)4^-1 One oxygen atom in Al(OH)4^-1

Chemistry
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For starters it's an ion so whenever you have an ion the oxidation states sum up to whatever number that ion is. So in that case they sum up to -1
Al(OH)4- Step one count up the atoms for each element Al = 1 O = 4 H = 4 Charge = -1 so They must add up to that. Then multiply #aroms by oxidation state: 4 atoms of H x 1 =+4 4 atoms of O x 2 = -8 Add them to find the remaining charge +4-8 = -4 At this point we know that -1 must be the sum since it's an ion so we need what number added to -4 gives us -1? X -4 = -1 x = +3 +3 = one of aluminum oxidation state.

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