Match List I with List II. List I (Conversion) List II(Number of Faraday required) A. 1 mol of $H_{2

Question

Match List I with List II. List I (Conversion) List II(Number of Faraday required) A. 1 mol of $H_{2}O$ to $O_{2}$ I. 3F B. 1 mol of ${MnO}_{4}^{-}$to ${Mn}^{2 +}$ II. 2 F C. $\ 1.5	ext{ }mol$ of Ca from molten ${CaCl}_{2}$ III. 1F D. 1 mol of FeO to ${Fe}_{2}O_{3}$ IV. 5 F Choose the correct answer from the options given below:

Accepted Answer

Sol. [Key Concept: Faraday requirements from electron counts]A: 2H₂O → O₂ + 4H⁺ + 4e⁻, so oxidation of 1 mol H₂O requires 2F (A–II).B: Mn changes from +7 to +2, requiring 5 mol electrons per mol MnO₄⁻,hence 5F (B–IV).C: Ca²⁺ + 2e⁻ → Ca, so 1.5 mol Ca requires 3F (C–I).D: Fe changes from +2 in FeO to +3 in Fe₂O₃, so each mole FeO corresponds to 1F (D–III). Hence, option (3).

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More Electrochemistry Questions

List I (Conversion)	List II(Number of Faraday required)
A. 1 mol of $H_{2}O$ to $O_{2}$	I. 3F
B. 1 mol of ${MnO}_{4}^{-}$to ${Mn}^{2 +}$	II. 2 F
C. $\ 1.5\text{ }mol$ of Ca from molten ${CaCl}_{2}$	III. 1F
D. 1 mol of FeO to ${Fe}_{2}O_{3}$	IV. 5 F