Why exactly are standard potentials additive?
Following on Derek's great answer, it is very important to remind that the conventional way we use to add half-cell potentials is a consequence of the conservation of energy. Therefore, we should look at this from the perspective of Hess's Law.

How so?

Well, if we add two reactions, no matter which ones, there's one certainty we keep in our souls: the new free energy will equal the sum of the individual ones.

This is a critical issue when trying to figure out the potential of a new reaction that is not the sum of an oxidation and a reduction.

For instance, take a look at these reactions:
$$ \ce{Fe^3+ + e- -&gt; Fe^2+} \qquad {E^\circ}_1 = 0.771\, \text{V}_{SHE}  $$
$$ \ce{Fe^2+ + 2e- -&gt; Fe_{(s)}} \qquad {E^\circ}_2 = -0.447 \, \text{V}_{SHE}  $$
$$ \ce{Fe^3+ + 3e- -&gt; Fe_{(s)}} \qquad {E^\circ}_3 =\, ? \, \text{V}_{SHE}  $$

It is tempting to simply add these potentials and get the third value, but we would be incurring on a major error. The proper way to proceed would be to calculate each reaction free energy and then figure out the standard potential.

$$ \ce{Fe^3+ + e- -&gt; Fe^2+} \qquad {\Delta G^\circ}_1 = -(1)\mathrm{F}{E^\circ}_1  $$
$$ \ce{Fe^2+ + 2e- -&gt; Fe_{(s)}} \qquad {\Delta G^\circ}_2 = -(2)\mathrm{F}{E^\circ}_2  $$
$$ \ce{Fe^3+ + 3e- -&gt; Fe_{(s)}} \qquad {\Delta G^\circ}_3 = -(3)\mathrm{F}{E^\circ}_3  $$

Since we know that ${\Delta G^\circ}_3 = {\Delta G^\circ}_1 + {\Delta G^\circ}_2$, we have:
$${E^\circ}_3=\frac{{E^\circ}_1 +2{E^\circ}_2}{3} = -0.041\, \text{V}_{SHE}  $$

Why don't we go through all this way when working with cell potentials? Because you always work your equations so that the number of electrons transferred on each of them has the same value. That way, the sum of the potentials is an indirect way of summing the free energies, which are now normalized by a common value.

This trick is evidently very useful, but sometimes it may be very misleading.