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To aquire the yellow colour of vanadium ion, it must be present in a 5+ oxdiation state.

This can be acheived (according to my textbook) by adding dilute sulfuric acid to $\ce{NH4VO3}$, then followed by zinc.

$$\ce{VO3- (aq) + 2H+ (aq) -> VO2^{+} (aq) + H2O (l)}$$ where $\ce{VO2^{+}}$ gives the yellow colour from the $\ce{V^{5+}}$ present in the compound

However, is not $\ce{VO3-}$(aq) yellow in colour?

If it is, then why is $\ce{NH4VO3}$ not yellow in colour but white?

Side question: what is the purpose of zinc in this reaction explained above - not including reduction to lower oxidation states?

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