Question

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

• A. oxidising power of $F^{-}$ is greater than that of $XeO^{4-}_{6}$
• B. it is not a redox reaction
• C. it is a disproportionation reaction
• D. oxidising power of $Xe^{4-}_{6}$ 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