How to balance more complex redox reactions?
Once you know the method, balancing any redox equation that you will get in AP chem will be quite easy. All you have to do is follow this simple method:


Balance all the elements in the equation except for $\ce{O}$ and $\ce{H}$. For polyatomic ions that are spectator ions, just think as them as one big element.
Check if $\ce{O}$ are balanced. If they aren't, add water molecules to the side, witch doesn't have enough $\ce{O}$.
Check if $\ce{H}$ are balanced. If they aren't, add $\ce{H+}$ ions to the side, witch doesn't have enough $\ce{H}$.
Now that all the elements are balanced, make sure that the charges are balanced. If not balanced, add electrons to the side which is more positive.
Check your equation now to make sure that everything is balanced correctly.


I will do the first equation as an example for you. You should be able to balance rest of the equations by following this procedure.


Balanced elements except for O and H: Here the sulfate ion is a spectator ion, so you can imagine it as one big element: $$\ce{Pb + PbO2 + 2SO4^2- -&gt; 2PbSO4 }$$
Balance oxygen: $$\ce{Pb + PbO2 + 2SO4^2- -&gt; 2PbSO4 + 2H2O }$$
Balance hydrogen: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -&gt; 2PbSO4 + 2H2O }$$
Balance charge: charge is already balanced. Don't need to add electrons.
Check equation to make sure everything is balanced correctly: all the elements and charges are balanced correctly.


So your balanced redox equation for the first example is: $$\ce{Pb + PbO2 + 2SO4^2- + 4H+ -&gt; 2PbSO4 + 2H2O }$$

Note: If the question asks to provide the redox equation in basic conditions there are two more steps.


Add the same number of $\ce{OH-}$ ions as $\ce{H+}$ to the products and reactants. The $\ce{OH-}$ ions and $\ce{H+}$ will react to form water.
Cancel out the water molecules from both sides (a bit like algebra).


That should give the redox equation in basic conditions.