A. For the oxidation of $1\text{ mol}$ of $\text{H}_2\text{O}$ to $\text{O}_2$: $\text{H}_2\text{O} \rightarrow \frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2\text{e}^-$ 2 moles of electrons are involved, so $2\text{ Faradays}$ ($2\text{F}$) are required.

B. For the reduction of $1\text{ mol}$ of $\text{MnO}_4^-$ to $\text{Mn}^{2+}$: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$ 5 moles of electrons are involved, so $5\text{ Faradays}$ ($5\text{F}$) are required.

C. For the reduction of $1.5\text{ mol}$ of $\text{Ca}$ from $\text{CaCl}_2$: $\text{Ca}^{2+} + 2\text{e}^- \rightarrow \text{Ca}$ To produce $1\text{ mol}$ of $\text{Ca}$, $2\text{F}$ are required. To produce $1.5\text{ mol}$ of $\text{Ca}$, $1.5 \times 2\text{ F} = 3\text{ F}$ are required.

D. For the oxidation of $1\text{ mol}$ of $\text{FeO}$ to $\text{Fe}_2\text{O}_3$: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + \text{e}^-$ 1 mole of electrons is involved, so $1\text{ Faraday}$ ($1\text{F}$) is required.

Therefore, the correct matching is A - II, B - IV, C - I, D - III.