The color of the solution changes from yellow to blue, green and purple
name: Reduction of V(V) to oxidation states II, III and IV
VO3- + 2H+ → VO2+ + H2O
2VO2+ + SO32- + 2H+ → 2VO2+ + SO42- + H2O
2VO2+ + 2Sn + 8H+ → 2V3+ + 2Sn2+ + 4H2O
2VO2+ + 3Zn + 8H+ → 2V2+ + 3Zn2+ + 4H2O
can be represented as three reactions
2VO2+ + Zn + 4H+ → 2VO2+ + Zn2+ + 2H2O
2VO2+ + Zn + 4H+ → 2V3+ + Zn2+ + 2H2O
2V3+ + Zn → 2V2+ + Zn2+
Vanadium is the only element that may be in four different oxidation states, all of which are stable in aqueous solutions.
Solution becomes yellow due to the presence of ions VO2+.
Sulfite anion is a weak reducing agent that can restore V (V) to a V (IV), giving a solution of a blue color.
A stronger reducing agent is tin. It can restore to V (III), giving a solution of a green color.
Finally, most of the proposed strong reductants, zinc, step by step restores V (V) to V (II), giving violet solution.
Ce (IV) - a strong enough oxidizing agent to convert V (II) to V (V).