Vanadium species in oxidation states IV, III and II are formed by the…

The idea

Zinc in acid is a strong enough reducing agent to walk vanadium down the whole ladder, and each step has its own colour. Yellow VO₂⁺, vanadium(V), becomes blue VO²⁺, vanadium(IV), then green V³⁺ and finally violet V²⁺. The half-equations need the acid: VO₂⁺ + 2H⁺ + e⁻ → VO²⁺ + H₂O, then VO²⁺ + 2H⁺ + e⁻ → V³⁺ + H₂O, then the simpler V³⁺ + e⁻ → V²⁺. Zinc supplies the electrons through Zn → Zn²⁺ + 2e⁻, so one mole of zinc reduces two moles of vanadium at each step. A green colour part way through can be genuine V³⁺, or a mixture of blue vanadium(IV) with yellow vanadium(V), so watch the sequence rather than one snapshot.

What earns the marks

• The correct sequence of colours, yellow then blue then green then violet.
• Each correct half-equation, balanced with H⁺ and H₂O as well as electrons.
• The oxidation half-equation Zn → Zn²⁺ + 2e⁻.
• Combining the half-equations so that the electrons cancel in the overall equation.

Common mistakes and how to put them right

• Writing V⁵⁺ and V⁴⁺ as simple ions instead of the oxoions VO₂⁺ and VO²⁺.

• Omitting H⁺ and H₂O from the half-equations, so leaving oxygen unbalanced.

• Giving the wrong charge on VO²⁺, often writing VO⁺ by forgetting that oxygen contributes −2.

• Assuming a green solution is always V³⁺ rather than possibly a mixture of the yellow and blue species.

Exam tips

• Balance oxygen with water first, then hydrogen with H⁺, then charge with electrons.
• Scale the half-equations so the electrons cancel exactly before adding them together.
• Learn the four colours as a sequence, because questions often ask you to predict the next change.