Answer:
\[2{{F}_{2}}+2{{H}_{2}}O\to
4HF+{{O}_{2}}\]
\[\underset{Oxidant}{\mathop{3{{F}_{2}}}}\,+\underset{\text{Reductant}}{\mathop{3{{H}_{2}}O}}\,\to
6HF+{{O}_{3}}\]
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