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Match List-I with List-II regarding the number of Faraday required for the given conversions.
Column I
- A. 1 mol of $H_2O$ to $O_2$
- B. 1 mol of $MnO_4^-$ to $Mn^{2+}$
- C. 1.5 mol of $Ca$ from molten $CaCl_2$
- D. 1 mol of $FeO$ to $Fe_2O_3$
Column II
- I. 3F
- II. 2F
- III. 1F
- IV. 5F
Correct answer: A → II, B → IV, C → I, D → III
Detailed Solution
(A) $H_2O \rightarrow 2H^+ + \frac{1}{2}O_2 + 2e^-$; Q = nF = 2F
(B) $MnO_4^- + 5e^- \rightarrow Mn^{2+}$; Q = 5F
(C) $Ca^{2+} + 2e^- \rightarrow Ca$; 1 mol Ca needs $2e^-$, so 1.5 mol needs $3e^-$; Q = 3F
(D) Oxidation number of Fe in FeO is +2 and in $Fe_2O_3$ is +3; 1 electron per Fe is required, so Q = 1F
Correct match: A-II, B-IV, C-I, D-III.