In electrolysis of very dilute $N a O H$ solution using platinum electrodes:

H_{2}

is evolved at cathode

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H_{2}

is produced at anode

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N a

is obtained at cathode

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O_{2}

is produced at anode

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Solution

The correct options are
A $H_{2}$ is evolved at cathode
D $O_{2}$ is produced at anode
In aqueous NaOH solution, on electrolysis, we have the reduction potential of $H^{+}$ is much more than $N a^{+}$ , so the reaction takes place as;

At anode: $4 O H^{-} \to 2 H_{2} O + O_{2} + 4 e^{-}$

At cathode: $4 H^{+} + 4 e^{-} \to 2 H_{2}$

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In electrolysis of very dilute NaOH solution using platinum electrodes:

The correct options are A H2 is evolved at cathode D O2 is produced at anodeIn aqueous NaOH solution, on electrolysis, we have the reduction potential of H+ is much more than Na+, so the reaction takes place as; At anode: 4OH−→2H2O+O2+4e−At cathode: 4H++4e−→2H2

In electrolysis of very dilute $N a O H$ solution using platinum electrodes: