Question

Question 8.16 Why does the following reaction occur?

XeO6(4-) (aq) + 2F– (aq) + 6H+(aq)→XeO3(g) + F2(g) + 3H2O(l)

What conclusion about the compound Na4XeO6 (of which XeO6(4-) is a part) can be drawn from the reaction.

Class XI Redox Reactions Page 273

Answer 1

Hey!

Given, Reaction:

XeO6(4-) (aq) + 2F– (aq) + 6H+(aq)→XeO3(g) + F2(g) + 3H2O(l)

In this reaction, we can conclude that the ON of Xe decreases from + 8 of XeO6(4-) to form + 6 of XeO3.

The ON of F also increases from -1 of F- to form 0 in F2.

So, we can say that Na4XeO6 is stronger oxidizing agent as compared to F.  

Answer 2

concept :- for finding which substance is oxidising or reducing . find the oxidation state of each element , the substance, in which oxidation number of the element is increasing , acts as reducing agent and if the substance , in which oxidation number of the element is decreasing , acts as oxidising agent .

$XeO_6^{4-}+2F^-+6H^+---->XeO_3+F_2+3H_2O$


in above reaction, oxidation number of Xe in XeO6^4- decreases from +8 to +6 in XeO3 and oxidation number of F increases -1 in F^- ion to zero ( in F2 ) . Hence , XeO6^4- is reduced and F^- is oxidised . This reaction occurs due to XeO6^4- is a stronger oxidizing agent than Fluorine ion.