How do we derive half equations with OH-?

[crayolé]

These types are giving me a bit of trouble at the moment *sadfase*

For example, how would you go about turning this into two half equations?

Are there any specific methods you guys use?

4Al (s) + 3O₂ (g) + 6 H₂O (l) ---------> 4 Al(OH)₃ (s)

and

H(absorbed on M) + NiO(OH)(s) → Ni(OH)₂(s)

[akira88]

The oxidation number of Al has changed from 0 to +3, therefore it has undergone oxidation
So go Al ---> Al3+
Balance the charges:
Al ---> Al3+ + 3e
Next, we have oxygen and water going to hydroxide ions
O2 + H2O ---> OH-
Balance the equation and its charges
O2 + 2H2O + 4e ---> 4OH-

Now you have two half equations. Multiply the first by 4 and the second by 3 so the electrons are equal so you can cancel them out. Note that in my half equations I didn't take notice of the coefficients of the substances of the overall equation.

You can test it out, and if your half equations equal the overall equation, you're in business

[crayolé]

Oh alright thanks ;]

Solutions say its
O2 (g) + 2 H2O (l) + 4 e- --------> 4 OH- (aq)                           
Al (s) + 3 OH- (aq) -------------> Al(OH)3 (s) + 3 e.

Are both ways correct?