[johnnyjohn993]

O₂(g) + H₂O (l) + Pb  Pb(OH)₂ (s)

my solution : 
 I don't know if half reaction is a good idea

2(OH)^- + Pb  Pb(OH)₂ + 2e^-

O₂(g) + H₂O (l) + Pb  Pb(OH)₂ (s) + 2e^-

I'm pretty much stuck with this  please help me!

[Borek]

Your lead half reaction looks OK to me.

Your other half reaction should start with O₂ without Pb. In water oxygen gets reduced to OH^-...

[Hunter2]

First you have to seperate the Redoxpair
Which are it?

Oxidation :.?

Reduction I give you is oxygen /hydroxide.

O₂ + 2 H₂O + 4 e^- => 4 OH^-

If you know the oxidation equation balance the amount of electrons and do addition of both equations.

[Hunter2]

Your lead half reaction looks OK to me.

Your other half reaction should start with O₂ without Pb. In water oxygen gets reduced to OH^-...

It works , but I would make it more simple Pb => Pb²⁺ + 2 e^-

This combined with the reduction equation guides to the result as well.