Answer:
\[2{{F}_{2}}(g)+2{{H}_{2}}O(l)\to
{{O}_{2}}(g)+4HF(aq)\]
\[3{{F}_{2}}(g)+3{{H}_{2}}O(l)\to
{{O}_{3}}(g)+6HF(aq)\]
In these reactions, water acts as a reducing agent as it
gets oxidised to either \[{{O}_{2}}\] or \[{{O}_{3}}\] while fluorine acts as an
oxidising agent as it is reduced to \[{{F}^{-}}\] ion.
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