Match List I with List II. List I (Conversion) 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 \( Fe_2O_3 \) List II (Number of Faraday required) I. 3F II. 2F III. 1F IV. 5F Choose the correct answer from the options given below:

A. Oxidation of \( H_2O \) to \( O_2 \): \( H_2O \rightarrow \frac{1}{2}O_2 + 2H^+ + 2e^- \). Thus, 2 moles of electrons (2F) are required per mole of \( H_2O \) (A - II).\nB. Reduction of \( MnO_4^- \) to \( Mn^{2+} \): \( MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \). Thus, 5 moles of electrons (5F) are required (B - IV).\nC. Formation of Ca from \( CaCl_2 \): \( Ca^{2+} + 2e^- \rightarrow Ca \). For 1 mole of Ca, 2F is required. For 1.5 moles of Ca, \( 1.5 \times 2F = 3F \) is required (C - I).\nD. Oxidation of \( FeO \) (containing \( Fe^{2+} \)) to \( Fe_2O_3 \) (containing \( Fe^{3+} \)): \( Fe^{2+} \rightarrow Fe^{3+} + e^- \). Thus, 1 mole of electron (1F) is required (D - III).\nTherefore, the correct matching is A-II, B-IV, C-I, D-III.