Standard electrode potential of copper disproportionation
Alas, the confusions related to signs in electrochemistry will never vanish. You mention that
$\ce{2Cu+ (aq) -&gt; Cu (s) + Cu^2+ (aq)}$ is $\pu{0.36 V}$.
Let me start with a single equation, x-y = 10; There can be indefinite solutions if you can simultaneously change the value of x and y. However, the moment you fix the value of x, the value of y is fixed.
You stated that the overall cell potential is $\pu{+0.36 V}$. Electrochemically, this means that this reaction is spontaneous.
Now you also know that,
$$E_\text{cell} = E_\text{reduction} - E_\text{anode} \tag{1}$$
You are not supposed to change any sign of the half-cell from the electrode potential tables. People should stop teaching this nonsense to relatively innocent students. Suppose, I write
\begin{align}
&amp;\ce{H2O (liquid) -&gt; H2O (gas)} &amp;T &amp;= \pu{100 ^\circ C} \\
&amp;\ce{H2O (gas) -&gt; H2O (liquid)},    &amp;T &amp;=\pu{ -100 ^\circ C} ??
\end{align}
The equation (1), itself takes care of all sign flipping and all.
Your half-cell corresponding to the reduction is
$\ce{Cu+ + e- -&gt; Cu}$ (potential $=  \pu{0.52 V}$)
And your half-cell potential for the oxidation is
$\ce{Cu^2+ + e- -&gt; Cu+}$ (potential $=  \pu{0.16 V}$)
Using equation (1), what you do get (remember no sign flipping) = $\pu{+0.36 V}$