Question

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

Answer 1

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

Answer 2

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.