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ChemistryNEET 2024Electrolysis and Faraday's laws; batteries and corrosion

Electrochemistry

Match List I with List II.

List I (Conversion) List II (Number of Faraday required)
A. 1 mol of H2O\mathrm{H_2O} to O2\mathrm{O_2} I. 3 F
B. 1 mol of MnO4−\mathrm{MnO_4^-} to Mn2+\mathrm{Mn^{2+}} II. 2 F
C. 1.5 mol of Ca from molten CaCl2\mathrm{CaCl_2} III. 1 F
D. 1 mol of FeO to Fe2O3\mathrm{Fe_2O_3} IV. 5 F

Choose the correct answer from the options given below:

  1. A.A-III, B-IV, C-I, D-II
  2. B.A-II, B-III, C-I, D-IV
  3. C.A-III, B-IV, C-II, D-I
  4. D.A-II, B-IV, C-I, D-III✓

Solution

Faradays needed equal electrons transferred. H2O→12O2\mathrm{H_2O \to \tfrac12 O_2}: 2 e⁻ per mole (II). MnO4−→Mn2+\mathrm{MnO_4^- \to Mn^{2+}}: 5 e⁻ (IV). Ca2+→Ca\mathrm{Ca^{2+} \to Ca}: 2 e⁻ per mole, so 1.5 mol needs 3 F (I). FeO→Fe2O3\mathrm{FeO \to Fe_2O_3}: Fe goes +2 → +3, 1 e⁻ per Fe, 1 F (III). Hence A-II, B-IV, C-I, D-III.

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