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?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.