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(D) $1$. In an aqueous solution of $NaCl$, the ions present are $Na^+$, $Cl^-$, $H^+$, and $OH^-$.$2$. At the cathode, the reduction reaction competes

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$H_{2(g)}$

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(D) $1$. In an aqueous solution of $NaCl$, the ions present are $Na^+$, $Cl^-$, $H^+$, and $OH^-$.$2$. At the cathode, the reduction reaction competes between $Na^+ + e^- ightarrow Na(s)$ $(E^o = -2.71 	ext{ V})$ and $2H_2O(l) + 2e^- ightarrow H_{2(g)} + 2OH^-(aq)$ $(E^o = -0.83 	ext{ V})$.$3$. Since the reduction potential of water is higher than that of $Na^+$, water is reduced preferentially.$4$. Therefore, $H_{2(g)}$ is produced at the cathode.

During the electrolysis of aqueous $NaCl$ solution, the product formed at the cathode is?

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