Question

Which of the following reactions is an example of a redox reaction?

• A. ${\mathrm{XeF}}_6\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{XeOF}}_4\left(\mathrm{aq}\right)+2\mathrm{HF}\left(\mathrm{l}\right)$
• B. ${\mathrm{XeF}}_6\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{XeO}}_2{\mathrm{F}}_2\left(\mathrm{aq}\right)+4\mathrm{HF}\left(\mathrm{l}\right)$
• C. ${\mathrm{XeF}}_4\left(\mathrm{aq}\right)+{\mathrm{O}}_2{\mathrm{F}}_2\left(\mathrm{aq}\right)\to {\mathrm{XeF}}_6\left(\mathrm{aq}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$
• D. ${\mathrm{XeF}}_2\left(\mathrm{aq}\right)+{\mathrm{PF}}_5\left(\mathrm{aq}\right)\to {\left[\mathrm{XeF}\right]}^{+}{\mathrm{PF}}_6^{-}\left(\mathrm{aq}\right)$

Answer 1

The correct option is C

${\mathrm{XeF}}_4\left(\mathrm{aq}\right)+{\mathrm{O}}_2{\mathrm{F}}_2\left(\mathrm{aq}\right)\to {\mathrm{XeF}}_6\left(\mathrm{aq}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

Redox reaction

Electrons travel between reactants that are already in touch in a redox reaction. Changes in the oxidation states of the organisms can then be used to measure how so many atoms are transferred.

The explanation for the correct option:

Option (C):

${\mathrm{XeF}}_4\left(\mathrm{aq}\right)+{\mathrm{O}}_2{\mathrm{F}}_2\left(\mathrm{aq}\right)\to {\mathrm{XeF}}_6\left(\mathrm{aq}\right)+{\mathrm{O}}_2\left(\mathrm{g}\right)$

Where Xenon (Xe) is oxidized and Oxygen (O₂) is reduced. The oxidation number of Xe grows from +4 to +6 in the chemical equation above, while the oxidation number of oxygen reduces from +1 to 0. Therefore, it is an example of a redox reaction.

The explanation for the incorrect options:

Option (A): ${\mathrm{XeF}}_6\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{XeOF}}_4\left(\mathrm{aq}\right)+2\mathrm{HF}\left(\mathrm{l}\right)$. In this reaction, there is no change in the oxidation state of Xe, hence this doesn't count for redox reaction. $\mathrm{Xe}$ has +6 oxidation state in ${\mathrm{XeF}}_6$ on the reactant side and ${\mathrm{XeOF}}_4$ on the product side.

Option (B): ${\mathrm{XeF}}_6\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\to {\mathrm{XeO}}_2{\mathrm{F}}_2\left(\mathrm{aq}\right)+4\mathrm{HF}\left(\mathrm{l}\right)$. In this reaction, there is no change in the oxidation state of Xe, hence this doesn't count for redox reaction. $\mathrm{Xe}$ has +6 oxidation state in ${\mathrm{XeF}}_6$ on the reactant side and ${\mathrm{XeO}}_2{\mathrm{F}}_2$ on the product side.

Option (D): ${\mathrm{XeF}}_2\left(\mathrm{aq}\right)+{\mathrm{PF}}_5\left(\mathrm{aq}\right)\to {\left[\mathrm{XeF}\right]}^{+}{\mathrm{PF}}_6^{-}\left(\mathrm{aq}\right)$. In this reaction, there is no change in the oxidation state of Xe, hence this doesn't count for redox reaction. $\mathrm{Xe}$ has +2 oxidation state in ${\mathrm{XeF}}_6$ on the reactant side and ${\left[\mathrm{XeF}\right]}^{+}{{\mathrm{PF}}_6}^{-}$ on the product side.

Hence, these options are incorrect.

Therefore, option (C) is the correct answer.