From my class text book:

The oxidation states of Vanadium ions 5+ and 4+ are shown in the compounds $\ce{VO2^{+}}$ and $\ce{VO^{2+}}$ respectively, whereas $\ce{V^{3+}}$ and $\ce{V^{2+}}$ are shown like this in the redox reaction (reduction half equation).

Why is this? cannot $\ce{V^{5+}}$ and $\ce{V^{4+}}$ exist indepepently as ions like $\ce{V^{3+}}$ and $\ce{V^{2+}}$ (when measuring the reduction potential)?


When an atom $M$ looses one or more electrons, it becomes positively charged and becomes the ion ${M^{z+}}$. In aqueous solution this positive charge attracts the unused doublets of the Oxygen atom from $H_2O$, and it repells the proton of $H^+$.

If $Z$ = $1$, the ion $M^+$ attracts several molecules water around it, and it makes a Debye layer of oriented $H_2O$ molecules around the ion, and no chemical changes.

If $Z = 2$, the attraction is strong enough to repell one proton $H^+$ from one of the molecules adsorbed around the $Z^{2+}$ ion. The solution becomes a little bit acidic, and the $M^{2+}$ion is partially hydrolyzed into the ion [${M(OH)]^+}$. It is an equilibrium :$$M^{2+} + H_2O \ce{<=>}[M(OH)]^+ + H^+$$

If $Z = 3$, the attraction between $Z^{3+}$ and water is strong enough to at least transform all $M^{3+}$ions into $[M(OH)]^{2+}$, and part of them to [$M(OH)_2]^+$ with the two reactions : $$M^{3+} + H_2O\ce{->} [M(OH)]^{2+} + H^+$$ $$[M(OH)]^{2+} + H_2O \ce{<=>} [M(OH)_2]^+ + H^+$$ This is the case if $M^{3+}$ is $Al^{3+}$ or $Fe^{3+}$

If $Z = 4$, all ions $M^{4+}$ are at least hydrolyzed into $[M(OH)_3]^+$ and even into $M(OH)_4$, according to : $$M^{4+} + 4 H_2O \ce{->}M(OH)_4 + 4 H^+$$It should be mentioned that $M(OH)_4$ has a tendency to loose water and produce a structure that has acidic properties. So it is often presented under the formula $H_4MO_4$ or $H_2MO_3$. That is the case with $M = Si$. Silicic acid is often presented as $H_4SiO_4$ or $H_2SiO_3$

With Vanadium $V^{4+}$, it goes another way. The molecule $V(OH)_4$ does not exist, as the ion $V^{4+}$ is too small to fix $4$ groups $OH^-$. As a consequence, $V^{4+}$ gives another positive ion $VO^{2+}$ according to : $$V^{4+} + H_2O \ce{->} [VO]^{2+}+2H^+$$ Strangely enough, it looks as if the non-existent $V(OH)_4$ behaves like a basic hydroxide and gets dissociated according to : $$V(OH)_4\ce{->}[VO]^{2+}+ 2 OH^–$$ If $Z=5$, the ion $M^{5+}$ can be hydrolyzed according to $$M^{5+} + 3 H_2O \ce{->} [MO_3]^{-}+6H^+$$$$M^{5+} + 4 H_2O \ce{->} [MO_4]^{3-}+8H^+$$ But with Vanadium $V$, the ion $V^{5+}$ may be hydrolyzed according to :$$V^{5+} + 2 H_2O \ce{->} [VO_2]^{+}+4H^+$$$$V^{5+} + 3 H_2O \ce{->} [VO_3]^{-}+6H^+$$The second reaction is favored at high pH.

This theoretical development is taken from General Chemistry, by Linus Pauling, W. F. Freemann and Co., San Francisco, 1953, Chapter XXI.

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