A dilute aqueous solution of sodium fluoride is electrolyzed, the products at the anode and cathode are:

O_{2}, H_{2}

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F_{2}, N a

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O_{2}, N a

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F_{2}, H_{2}

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Solution

The correct option is A $O_{2}, H_{2}$
Ion with higher potential will go to corresponding electrode and discharge.
So,
$N a F$ :
Cathode : $2 H^{\oplus} + 2 e^{-} \to H_{2}$
Anode : $4 ⊖ O H \to O_{2} + 2 H_{2} O + 4 e^{-}$

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