Disproportionation of plutonium species - Latimer's diagram

Why (according to provided answers) $\ce{Pu(IV)}$ disproportionates into adjacent species whilst $\ce{Pu(V)}$ does not in acid aqueous solution.

$$\ce{\underset{(+6)}{Pu}O2^2+ ->[\pu{+1.02 V}] \underset{(+5)}{Pu}O2+ ->[\pu{+1.04 V}]\underset{(+4)}{Pu}^4+ ->[\pu{+1.01V}] \underset{(+3)}{Pu}^3+}$$

I thought it is the other way - the more positive potential of $\pu{+1.04V}$ on the right of $\ce{PuO2+}$ species will drive the disproportionation while in $\ce{Pu^4+}$ the higher value of $\pu{+1.04V}$ to the left renders the species by itself thermodynamically stable.

In acidic solutions ($\mathrm{pH} = 0$), the values of the relevant reduction potentials of $\ce{Pu^3+/Pu^4+}$ ($\pu{+1.01 V}$), $\ce{Pu^3+/PuO2+}$ ($\pu{+1.03 V}$), $\ce{Pu^3+/PuO2^2+}$ ($\pu{+1.02 V}$), $\ce{Pu^4+/PuO2+}$ ($\pu{+1.04 V}$), $\ce{Pu^4+/PuO2^2+}$ ($\pu{+1.03 V}$), and $\ce{PuO2+/PuO2^2+}$ ($\pu{+1.02 V}$) are all very similar. Therefore, plutonium has the remarkable feature that all four important oxidation states can be present simultaneously in aqueous solutions.

In aqueous solution, $\ce{Pu^4+}$ disproportionates into a mixture of oxidation states: $$\ce{2Pu^4+ + 2H2O <=> Pu^3+ + PuO2+ + 4H+}$$ $$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$ Thus, the net reaction of the disproportionation of $\ce{Pu^4+}$ is approximately:
$$\ce{3Pu^4+ + 2H2O <=> 2Pu^3+ + PuO2^2+ + 4H+}$$ Nevertheless, $\ce{Pu^4+}$ may still be the predominant oxidation state in the mixture. Besides, the equations imply that the disproportionation of $\ce{Pu^4+}$ dcreseases as the acidity of the solution further increases.

Concerning pentavalent plutonium ($\ce{PuO2+}$), your expectations are correct. In typical acidic solutions, $\ce{PuO2+}$ indeed quickly disproportionates into a mixture of oxidation states: $$\ce{2PuO2+ + 4H+ <=> Pu^4+ + PuO2^2+ + 2H2O}$$ $$\ce{Pu^4+ + PuO2+ <=> Pu^3+ + PuO2^2+}$$

However, the given values for the reduction potentials apply to pH = 0, and the various reduction potentials of plutonium depend on pH. Furthermore, the stability of individual oxidation states can be strongly influenced by hydrolysis, or the formation of complexes or compounds with low solubility products. Therefore, the predominant oxidation state can be changed by means of multiple factors. Furthermore, the valence of plutonium solutions may change as a consequence of radiolysis.

Tetravalent plutonium as the ion $\ce{Pu^4+}$ does not disproportionate according to the equation:

$$\ce{3Pu^4+ + 2H2O -> 2Pu^3+ + PuO2^2+ + 4H+}\label{a}\tag{1}$$

The products on the right-hand side of Eq. $\ref{a}$ inevitably react according to Eq. $\ref{b}$.

$$\ce{Pu^3+ + PuO2^2+ -> Pu^4+ + PuO2^+}\label b\tag2$$

The true disproportionation equation contains all the disproportionation products in their stoichiometric proportions. Those products are $\ce{Pu^3+}$, $\ce{PuO2^+}$ and $\ce{PuO2^2+}$, and $\ce{H+}$.

Equation (1) is the limiting form of the disproportionation equation as the acid concentration increases. The limiting form of the disproportionation equation as the acidity decreases is Eq. $\ref{c}$.

$$\ce{2Pu^4+ + 2H2O -> Pu^3+ + PuO2+ + 4H+}\label{c}\tag3$$

The disproportionation equation at intermediate values between the limiting cases of "high" and "low" acidities is more complicated. It contains $\ce{Pu^4+}$ on the left-hand side and the three species $\ce{Pu^3+}$, $\ce{PuO2^+}$, and $\ce{PuO2^2+}$ on the right-hand side. The coefficients in the disproportionation equation are integers followed by decimal fractions.