If electrolysis of water is carried out for a time duration of 2 hours, how much electric current in amperes would be required to liberate $100 \mathrm{~ml}$ of $\mathrm{O}_2$ gas measured under standard conditions of temperature and pressure?

The electrolysis of water produces oxygen gas at the anode according to the reaction: $2H_2O(l) \rightarrow O_2(g) + 4H^{+}(aq) + 4e^{-}$.

From the stoichiometry of the reaction, $4$ moles of electrons are required to produce $1$ mole of $O_2$ gas. At standard temperature and pressure (STP), $1$ mole of any gas occupies $22.4$ liters or $22400$ ml.

The volume of $O_2$ given is $100$ ml. The number of moles of $O_2$ produced is $n_{O_2} = \dfrac{100}{22400} = \dfrac{1}{224}$ moles.

The total charge $Q$ required is $n_{e^{-}} \times F$, where $n_{e^{-}} = 4 \times n_{O_2} = 4 \times \dfrac{1}{224} = \dfrac{1}{56}$ moles of electrons and $F = 96500$ C/mol.

$Q = \dfrac{96500}{56} \approx 1723.21$ Coulombs.

The time duration is $2$ hours, which is $2 \times 3600 = 7200$ seconds.

The current $I$ is given by $I = \dfrac{Q}{t} = \dfrac{1723.21}{7200} \approx 0.2393$ A.

Answer: 0.2393 A