Disproportionation of silver(I) in aqueous solution
It is convenient to solve problems like that with a Latimer diagram, which is a great tool for predicting conditions for the reactions of disproportionation and synproportionation.
A generic Latimer diagram
$$\ce{A -&gt;[$E_1$] B -&gt;[$E_2$] C}$$
posesses the following properties:

If $E_2 &gt; E_1,$ then $\ce{B}$ is thermodynamically unstable and disproportionates to $\ce{A}$ and $\ce{C}.$
If $E_2 &lt; E_1,$ then the mixture of $\ce{A}$ and $\ce{C}$ is thermodynamically unstable and synproportionates to $\ce{B}.$

Now we can visualize your problem using a Latimer diagram
$$\ce{Ag^2+ -&gt;[\pu{+2.00 V}] Ag+ -&gt;[\pu{+0.80 V}] Ag}$$
and the condition of disproportionation resulting from the application of Nernst equation:

if the potential to the right of the species is higher than the potential on the left, it will disproportionate.

Since $E^\circ(\ce{Ag+(aq)/Ag}) &lt; E^\circ(\ce{Ag^2+(aq)/Ag+(aq)}),$ disproportionation is thermodynamically unfavorable, and silver(I) can be considered stable in solution.
You can get the same result from a linear combination of both equations written for the disproportionation:
$$
\begin{align}
\ce{Ag+(aq) + e- &amp;-&gt; Ag} &amp;\quad E^\circ_1 = \pu{+0.80 V} &amp; \tag{1}\\
\ce{Ag^2+(aq) + e- &amp;-&gt; Ag+(aq)} &amp;\quad E^\circ_2 = \pu{+2.00 V} &amp;\quad|\cdot (-1)\tag{2}\\
\hline
\ce{2 Ag+(aq) &amp;-&gt; Ag + Ag^2+(aq)} &amp;\quad E^\circ = \pu{-1.20 V}
\end{align}
$$
Since resulting $E^\circ = \pu{-1.20 V} &lt; 0,$ free Gibbs energy $Δ_\mathrm{r}G^\circ = -nFE^\circ &gt; 0,$ and the disproportionation of silver(I) in solution can be considered a thermodynamically unfavorable process.