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Topic: Electrolytic and galvanic cells, electrode potentials· Redox Reactions and Electrochemistry

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_3List II (Number of Faraday required): I. 3F II. 2F III. 1F IV. 5FChoose the correct answ…

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:

1. A-III, B-IV, C-II, D-I

2. A-II, B-IV, C-I, D-III

3. A-III, B-IV, C-I, D-II

4. A-II, B-III, C-I, D-IV

1 answer

Best answer

Answer: 2. A-II, B-IV, C-I, D-III

Key idea: Faradays = moles of electrons transferred.

  • A: \(\mathrm{2H_2O\to O_2+4H^++4e^-}\), so 1 mol \(\mathrm{H_2O}\) needs 2 F (II).
  • B: \(\mathrm{MnO_4^-\to Mn^{2+}}\): +7 to +2, 5 F (IV).
  • C: \(\mathrm{Ca^{2+}+2e^-\to Ca}\); 1.5 mol needs 3 F (I).
  • D: \(\mathrm{FeO\to Fe_2O_3}\): Fe +2 to +3, 1 F per FeO (III).

A-II, B-IV, C-I, D-III: option 2.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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