When ammonium vanadate is heated with oxalic acid solution, a substance Z containing two oxalate groups per vanadium is formed. A sample of Z was treated with KMn O 4 solution in hot acidic solution

Question

When ammonium vanadate is heated with oxalic acid solution, a substance Z containing two oxalate groups per vanadium is formed. A sample of Z was treated with $KMn O_{4}$ solution in hot acidic solution. The resulting liquid was reduced with $S O_{2}$ , the excess $S O_{2}$ boiled off and the liquid again titrated with the same $KMn O_{4}$ . The ratio of the volumes of $KMn O_{4}$ used in the two titrations was 4:1. What is the oxidation state of vanadium in substance Z? ( $KMn O_{4}$ oxidizes all oxidation states of vanadium to V(+5) and $S O_{2}$ reduces V(+5) to V(+4).)

Answer:

5.00

Solution:

Per V atom: first titration takes 2 × 2 = 4 $e^{-}$ for the two oxalates plus (5-x) $e^{-}$ for V; the second takes 1 $e^{-}$ (V +4 $\to$ +5). Ratio: 4 + (5-x) = 4 $\Rightarrow$ $x$ = 5. Check: V already +5, so $KMn O_{4}$ oxidizes only the oxalates first (4e) and V(+4) second (1e): 4:1. Ans: 5.

TestHub · Accepted Answer

Per V atom: first titration takes 2 × 2 = 4 e − for the two oxalates plus (5-x) e − for V; the second takes 1 e − (V +4 → +5). Ratio: 4 + (5-x) = 4 ⇒ x = 5. Check: V already +5, so KMn O 4 ​ oxidizes only the oxalates first (4e) and V(+4) second (1e): 4:1. Ans: 5.

Chemistry - Redox Reactions Question with Solution | TestHub

Doubts & Discussion