The products of electrolysis of dilute aqueous solution of sodium hydride are (a) \(\mathrm{Na}\) at cathode and \(\mathrm{H}_{2}\) at anode (b) \(\mathrm{H}_{2}\) at cathode and \(\mathrm{O}_{2}\) at cathode (c) \(\mathrm{Na}\) at cathode and \(\mathrm{O}_{2}\) at anode (d) \(\mathrm{H}_{2}\) at both cathode and anode

Sodium hydride (NaH) is an inorganic compound composed of sodium (Na) cations and hydride (H-) anions. When NaH is dissolved in water, it dissociates into its respective ions: \(\mathrm{NaH} \rightarrow \mathrm{Na}^+ + \mathrm{H}^-\) The water molecule (H2O) also dissociates slightly into hydronium (H3O+) and hydroxide (OH-) ions: \(\mathrm{H}_2\mathrm{O} \rightleftharpoons \mathrm{H}_3\mathrm{O}^+ + \mathrm{OH}^-\) So, in the electrolyte solution, we have the following species: \(\mathrm{Na}^+\), \(\mathrm{H}^-\), \(\mathrm{H}_3\mathrm{O}^+\), and \(\mathrm{OH}^-\).

At the cathode, a reduction process takes place (gain of electrons). The most common reduction reactions involve cations being reduced to their elemental form, or protons (H+) being reduced to form hydrogen gas (H2). In this case, the possible reactions at the cathode are: 1. Sodium ions (Na+) being reduced to sodium metal (Na): \(\mathrm{Na}^+ + \mathrm{e}^- \rightarrow \mathrm{Na}\) 2. Hydronium ions (H3O+) being reduced to form hydrogen gas (H2): \(\mathrm{H}_3\mathrm{O}^+ + \mathrm{e}^- \rightarrow \dfrac{1}{2}\mathrm{H}_2 + \mathrm{H}_2\mathrm{O}\) Since sodium is an alkali metal with a low reduction potential, it does not get reduced in aqueous solution. Instead, hydronium ions are more easily reduced to form hydrogen gas. Therefore, the correct reaction at the cathode is the reduction of hydronium ions to produce hydrogen gas (H2).

At the anode, an oxidation process takes place (loss of electrons). The most common oxidation reactions in aqueous solutions involve anions being oxidized to their elemental form, or water being oxidized to produce oxygen gas (O2). In this case, the possible reactions at the anode are: 1. Hydride ions (H-) being oxidized to form hydrogen gas (H2): \(\mathrm{H}^- - \mathrm{e}^- \rightarrow \dfrac{1}{2}\mathrm{H}_2\) 2. Hydroxide ions (OH-) being oxidized to form oxygen gas (O2): \(\mathrm{2OH}^- - \mathrm{4e}^- \rightarrow \mathrm{O}_2 + \mathrm{2H}_2\mathrm{O}\) Since hydride ions have a high tendency to reduce protons (H+), they do not get oxidized in aqueous solution. Instead, hydroxide ions are more easily oxidized to form oxygen gas. Therefore, the correct reaction at the anode is the oxidation of hydroxide ions to produce oxygen gas (O2).

Based on our analysis, the correct products of electrolysis in a dilute aqueous solution of sodium hydride are: hydrogen gas (H2) at the cathode, and oxygen gas (O2) at the anode. Therefore, the correct option is: (b) \(\mathrm{H}_{2}\) at cathode and \(\mathrm{O}_{2}\) at anode