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

H_{2}

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F_{2}

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O_{2}

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all of the above

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Solution

The correct option is A $H_{2}$
As we know,
Ion with higher potential will go to corresponding electrode and discharge.
So,
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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