Match the redox conversions in List-I with the corresponding number of Faradays required in List-II.

**List-I

Question

Match the redox conversions in List-I with the corresponding number of Faradays required in List-II.

**List-I (Redox Conversion)**
A. $1	ext{ mol}$ of $	ext{H}_2	ext{O}$ to $	ext{O}_2$
B. $1	ext{ mol}$ of $	ext{MnO}_4^-$ to $	ext{Mn}^{2+}$
C. $1.5	ext{ mol}$ of $	ext{Ca}$ from molten $	ext{CaCl}_2$
D. $1	ext{ mol}$ of $	ext{FeO}$ to $	ext{Fe}_2	ext{O}_3$

**List-II (Number of Faraday required)**
I. $3	ext{F}$
II. $2	ext{F}$
III. $1	ext{F}$
IV. $5	ext{F}$

Accepted Answer

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

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

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

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

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

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