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If an aqueous solution of $\mathrm{NaF}$ is electrolysed between inert electrodes, the product obtained at anode is
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$\mathrm{O}_{2}$
Aqueous NaF, when electrolysed between inert electrodes we always get $\mathrm{O}_{2}$ at anode, as: At cathode $2 \mathrm{H}^{+}+2 e^{-} \longrightarrow \mathrm{H}_{2} \uparrow$
At anode $4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 e^{-}$
Also, if any fluorine is liberated, that too reacts with $\mathrm{H}_{2} \mathrm{O}$.
$2 \mathrm{~F}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{HF}+\mathrm{O}_{2}$
So, ultimate product is oxygen.
At anode $4 \mathrm{OH}^{-} \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}+\mathrm{O}_{2}+4 e^{-}$
Also, if any fluorine is liberated, that too reacts with $\mathrm{H}_{2} \mathrm{O}$.
$2 \mathrm{~F}_{2}+2 \mathrm{H}_{2} \mathrm{O} \longrightarrow 4 \mathrm{HF}+\mathrm{O}_{2}$
So, ultimate product is oxygen.
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