Match List-I with List-II: List-I (Conversion) (A) $1$ mol of $\mathrm{H}{}_2\mathrm{O}$ to $\mathrm{O}{}_2$ (B) $1$ mol of $\mathrm{MnO}{}_4{}^{-}$ to $\mathrm{Mn}{}^{2+}$ (C) $1.5$ mol of $\mathrm{Ca}$ from molten $\mathrm{CaCl}{}_2$ (D) $1$ mol of $\mathrm{FeO}$ to $\mathrm{Fe}{}_2\mathrm{O}{}_3$ List-II (No. of Faraday required) (i) $3F$ (ii) $2F$ (iii) $1F$ (iv) $5F$ Choose the correct match:

Solution & Explanation

\textbf{Answer:} (D) A-(ii), B-(iv), C-(i), D-(iii) \textbf{Solution:} The number of Faradays equals the total number of electrons transferred per the stated amount. (A) $\mathrm{H}{}_2\mathrm{O}\to \mathrm{O}{}_2$: oxidation of water, $2\mathrm{H}{}_2\mathrm{O}\to \mathrm{O}{}_2+4\mathrm{H}{}^{+}+4\mathrm{e}{}^{-}$; per $1$ mol of $\mathrm{H}{}_2\mathrm{O}$, $2e^{-}$ are lost $\Rightarrow 2F$ $\to$ (ii). (B) $\mathrm{MnO}{}_4{}^{-}\to \mathrm{Mn}{}^{2+}$: $\mathrm{Mn}$ goes from $+7$ to $+2$, a $5e^{-}$ gain; $1$ mol $\Rightarrow 5F$ $\to$ (iv). (C) $\mathrm{Ca}{}^{2+}+2\mathrm{e}{}^{-}\to \mathrm{Ca}$: $1.5$ mol $\mathrm{Ca}$ needs $1.5\times 2=3F$ $\to$ (i). (D) $\mathrm{FeO}\to \mathrm{Fe}{}_2\mathrm{O}{}_3$: $\mathrm{Fe}$ goes from $+2$ to $+3$, a $1e^{-}$ loss; $1$ mol $\mathrm{FeO}$ $\Rightarrow 1F$ $\to$ (iii). Hence A-(ii), B-(iv), C-(i), D-(iii).