Calculating the cell potential of a half cell from two half cell equations I understand the premise of the question and how to solve it but I don't quite understand why I'm a factor of 2 out:

Construct a potential diagram (at $\mathrm{pH} = 0$) for the reduction of aqueous $\ce{HNO2}$ to $\ce{NO}$ and then to $\ce{N2O}$ given that $E^0$ for the $\ce{HNO2/NO}$ and $\ce{NO/N2O}$ couples are $\pu{+0.98 V}$ and $\pu{+1.59 V}$ respectively. Calculate $E^0$ for the following half-reaction:
$$\ce{2HNO2 + 4H+ +4e- &lt;=&gt; N2O + 3H2O}$$

So by constructing the two half-equations:
$$\ce{HNO2 + H+ +e- &lt;=&gt; NO + H2O~~~~~~~~~~~~(E^\circ= +0.98V)}$$
$$\ce{2NO + 2H+ +2e- &lt;=&gt;N2O +H2O~~~~~~~(E^\circ= +1.59V)}$$
And so by adding 2 times the first equation to the second, you obtain the desired overall equation. And so you would add the individual reduction potentials to obtain the overall potential:
$$\pu{+0.98 V + 1.59 V} = \pu{+2.57 V}$$
The answer stated however is $\pu{+1.29 V}$, ie, half of my answer.
Why do we need to divide by $2$ at the end here? The standard reduction potential doesn't depend of stoichiometry, and when doing similar questions combining two half cells into a full equation (with no electrons in the equation), simply subtraction or adding the equations is sufficient. Why the difference here?
Josh, you and others have missed the key point. It is not a simple question! They are asking you to draw a potential diagram not Ecell, basically a Latimer diagram.
So take the approach of using a Latimer diagram. The book answer is correct. Nitrous acid when goes to nitrous oxide, the potential must be +1.29 with respect to SHE. Check the web for solving Latimer diagrams.
You would work on the portion starting from nitrous acid to nitrous oxide.