Question

Based on the following reaction,
$Xe{O}_6^{4-}(aq)+2F^{-}(aq)+6H^{+}(aq)\to Xe{O}_3(g)+F_2(g)+3H_{2}O\left(\ell \right)(\Delta G^{\circ }<0)$
It can be concluded that

• A. oxidising power of $F^{-}$ is greater than that of $Xe{O}_6^{4-}$
• B. it is not a redox reaction
• C. it is a disproportionation reaction
• D. oxidising power of $X{e}_6^{4-}$ is greater than that of $F^{-}$

Answer 1

Answer: (D)

Since, $\Delta G^{\circ }<0$, hence it is spontaneous in forward direction Oxidation number of Xe decreases, hence, it is an oxidising agent; and oxidation number of F increases, hence it is a reducing agent