Na₄XeO_6(aq) + x${\mathrm{F}}_{\left(\mathrm{aq}\right)}^{-}$ + y${\mathrm{H}}_{\left(\mathrm{aq}\right)}^{+}$ $\to$A_(g) +z F_2(g) + bH₂O_(l). The incorrect statement regarding balanced equation is

Balanced equation is  $2{\mathrm{F}}^{-}+{\mathrm{XeO}}_6^{-4}+6{\mathrm{H}}^{+}\to {\mathrm{F}}_2+{\mathrm{XeO}}_3+3{\mathrm{H}}_2\mathrm{O}$

In ${\mathrm{XeO}}_6^{-4}$, Xe is in higest oxidation state = +8, it is stronger oxidizing agent than F₂ ;

SRP value of Na₄XeF₆/XeO₃ is more positive than F₂/½F^- ;

Direction of any redox reaction :

 strong oxidizing agent + strong reducing agent$\to$ weak oxidizing agent + weak reducing agent