Which reaction occurs in calcium vanadate when it comes in contact with water?

I am a physics student currently studying the electrical properties of $\ce{CaVO3}$ and I have noticed that the surface of my thin film samples, which were originally smooth and transparent, now have blue and orange spots all over. These spots vary in size, some being only a few $\pu{\mu m}$ or smaller and the largest being approximately $\pu{200 \mu m}$ across but consisting of many spots combined. Visually it appears like it might have been water condensation droplets that caused this after having been inside of a helium flow cryostat at temperatures between $\pu{4 K}$ and room temperature. I have also noticed a change in the electrical properties of my samples, which I might be able to understand if I know what sort of compound these spots are. I do not know for certain that these spots were caused by a reaction with water, I am purely basing this on the visual appearance of droplets.


The change in the electrical properties has happened over time. I first measured the properties when the samples were new and a year later I repeated these measurements and noticed an increase in the resistivity for the entire temperature range as well as a change in the charge carrier density, accompanied by these colorful spots, which makes me believe that they may be responsible for it. The spots are too small to be measured individually with my methods.




If you mean $\ce{Ca}\left( \ce{V}\ce{O\_{3}} \right)\_{2}$ aka $\ce{Ca} \ce{V\_{2}}\ce{O\_{6}}$ then $\ce{Ca} \ce{V\_{2}}\ce{O\_{6}} + 2 \cdot \ce{H\_{2}}\ce{O} \to \ce{Ca^{2+}} + 2 \cdot \ce{H\_{2}}\ce{V}\ce{O\_{4}}^{-}$ should apply.
You can also find the reaction [here](https://www.sciencedirect.com/topics/earth-and-planetary-sciences/calcium-vanadate) in equation $81$: $$
\ce{Ca\_{n}}\ce{V\_{2}}\ce{O\_{5 + n}} + \left( 3 - n \right) \cdot \ce{H\_{2}}\ce{O} \to n \cdot \ce{Ca^{2+}} + 2 \cdot \ce{H\_{3 - n}}\ce{V}\ce{O\_{4}}^{n-}
$$ where $n = 1$.


So your calcium vanadate ($\ce{Ca}\left( \ce{V}\ce{O\_{3}} \right)\_{2}$) reacts with water ($\ce{H\_{2}}\ce{O}$) to calcium ions ($\ce{Ca^{2+}}$) and [dihydroxy-oxido-oxovanadium](https://pubchem.ncbi.nlm.nih.gov/compound/dihydrogenvanadate) ($\ce{H\_{2}}\ce{V}\ce{O\_{4}}^{-}$ aka $\ce{V}\ce{O\_{2}}\left( \ce{O}\ce{H} \right)\_{2}^{-}$).


The whole thing could theoretically continue to react at high enough temperatures: see [here](https://www.rsc.org/suppdata/c7/dt/c7dt02702h/c7dt02702h2.pdf) (page $8$)):
$$
4 \cdot \ce{V}\ce{O\_{2}}\left( \ce{O}\ce{H} \right)\_{2}^{-} \to \ce{V\_{4}}\ce{O\_{12}}^{-4} + 4 \cdot \ce{H\_{2}}\ce{O}
$$


But predicting the color of the whole thing is not that easy. On the one hand, there can be many other reactions, especially if your substance contains other particles. On the other hand, it's really difficult to find references for this. Also, more information would be useful, like its new properties.




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> 
> My material is actually $\ce{Ca}\ce{V}\ce{O\_{3}}$
> 
> 
> 


I would assume that $\ce{Ca}\ce{V}\ce{O\_{3}}$ dissociates first in water:
$$
n \cdot \ce{Ca}\ce{V}\ce{O\_{3}} + m \cdot \ce{H\_{2}}\ce{O} \rightleftharpoons n \cdot \left( \ce{Ca^{2+}} + \ce{V}\ce{O\_{3}}^{2-} \right) + m \cdot \ce{H\_{2}}\ce{O}
$$


Because of the $\ce{V}\ce{O\_{3}}^{2-}$ ion's peculiar electron configuration, I would assume that it would "steal" the hydrogen ($\ce{H}$) from the water ($\ce{H\_{2}}\ce{O}$). The resulting oxygen in turn wants to react and could also react with the ion. This would result in something of the form:
$$
n \cdot \ce{Ca^{2+}} + m \cdot \ce{H\_{2}}\ce{O} + n \cdot \ce{V}\ce{O\_{3}}^{2-} \to n \cdot \ce{Ca^{2+}} + \left( n - w \right) \cdot \ce{V\_{x}}\ce{O\_{y}}^{z-} + \ce{R\_{1}}
$$


If we let it dry again, we should just get a new calcium vanadate again (and some remaining part $\ce{R\_{1}}$). Some vanadium-oxides are kinda yellow-orange or dark blue like [vanadium($\rm{IV}$)-oxide](https://en.wikipedia.org/wiki/Vanadium(IV)_oxide). Here $\left\{ w,\, x,\, y,\, z \right\} \in \mathbb{N\_{0}}$, $n - w$ is the number of this special kind of vanadate (the vanadate $\ce{V\_{x}}\ce{O\_{y}}$), $x$ and $y$ are forming something which is supposed to imply the type of vanadate, $z$ is the electric charge of this vanadate / vanadate-ion and $\ce{R\_{1}}$ is a remaining part resulting from the reaction, which in turn can be another vanadate or it can be a completely different compound.


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