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Application of Electrolysis

The electroly...

Question

The electrolysis of molten NaOH liberates O $_{2}$ at anode.
If true enter 1, else enter 0.

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Solution

At anode: $2 O H^{-} \to \frac{1}{2} O_{2} + H_{2} O + 2 e^{-}$
At cathode: $N a^{+} + e^{-} \to N a$
Hence the statement is correct.

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