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. 2F C. 1.5 mol of Ca from molten CaCl 2 III. 1F D. 1 mol of FeO to Fe 2 O 3 IV. 5F Choose the correct answer from the options given below:

Chemistry · Electrochemistry

NEET (UG) 2024 — Question 74

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. 2F
C. 1.5 mol of Ca from molten CaCl $_2$	III. 1F
D. 1 mol of FeO to Fe $_2$ O $_3$	IV. 5F

Choose the correct answer from the options given below:

Option A:
A-II, B-IV, C-I, D-III
Correct
Option B:
A-III, B-IV, C-I, D-II
Option C:
A-II, B-III, C-I, D-IV
Option D:
A-III, B-IV, C-II, D-I

Answer: A

Step-by-step solution

For A: Oxidation of water: $\ce{2H2O -> O2 + 4H+ + 4e-}$ . So 4 electrons for 2 mol H2O, thus 2 F for 1 mol H2O. → A-II For B: Reduction of MnO4−: $\ce{MnO4- + 8H+ + 5e- -> Mn^{2+} + 4H2O}$ . So 5 F per mol. → B-IV For C: Reduction of Ca2+: $\ce{Ca^{2+} + 2e- -> Ca}$ . So 2 F per mol; for 1.5 mol, 3 F. → C-I For D: Oxidation of FeO: $\ce{Fe^{2+} -> Fe^{3+} + e-}$ . So 1 F per mol FeO. → D-III Thus the correct matching is A-II, B-IV, C-I, D-III, which corresponds to option A.

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Paper: 2024 paper
Subject: Chemistry
Chapter: Electrochemistry
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