QNEET-2024-Ch-17: NEET 2024 Chemistry - Class 12 NEET-2024-Ch-17 Solution: Match List I with List II. List I (Conversion) List II (Number…

To answer this question, we need to calculate the number of Faraday's required for each of the reactions in List I. Reaction A: Conversion of H 2 O to O 2 The reaction for electrolysis of water to produce oxygen can be written as: 2H 2 O (l) _ O 2 (g) + 4 H + + 4 e _ Each mole of O 2 requires the transfer of 4 moles of electrons. Therefore, to produce 1 mole of O 2 , you need 4 Faraday's of charge. However, since the reaction shown in the table suggests the production of only 1 mole, it would require 2 Faraday's. This is because we are dealing with a half reaction in which only 2 moles of electrons are required for every mole of O 2 produced from 1 mole of H 2 O. Reaction B: Conversion of MnO 4 - to Mn 2+ The reduction reaction for permanganate to manganese ion is: MnO 4 _ + 8 H + + 5 e _ _ Mn 2 + + 4 H 2 O This reaction indicates that 5 moles of electrons are involved in the reduction of 1 mole of MnO 4 - to Mn 2+ . Therefore, 5 Faraday's are required. Reaction C: Production of 1.5 moles of Ca from molten CaCl 2 The electrolytic production of calcium from its chloride can be illustrated as follows: Ca 2 + + 2 e _ _ Ca Each mole of Ca produced requires 2 Faraday's of electrons considering the transfer of 2 moles of electrons. For 1.5 moles of Ca, the total Faraday's required would be 3 Faraday's, because, 1.5 * 2 = 3F. Reaction D: Oxidation of FeO to Fe 2 O 3 The reaction can be considered in terms of iron oxidation states: 4FeO _ 2Fe 2 O 3 + O 2 + 4 e _ Each mole of O 2 requires 2 Faraday's of charge. If the products include 2 moles of Fe 2 O 3 and 1 mole of O 2 , thus, only 2 Faraday's will be required for 1 mole of FeO to react (considering FeO as just a part of the full reaction). Correct Matching: A-II, B-IV, C-I, D-III The mappings are A-II (2F for H 2 O to O 2 ), B-IV (5F for MnO 4 - to Mn 2+ ), C-I (3F for 1.5 mol of Ca from CaCl 2 ), and D-III (2F for FeO to Fe 2 O 3 ). Therefore, the correct answer is Option A : A-II, B-IV, C-I, D-III.