Match List I with List II.

| List I (Conversion) | List II (Number of Faraday required) |
|---|---|
| A. 1 mol of \mathrm{H2O} to \mathrm{O2} | I. 3 F |
| B. 1 mol of \mathrm{MnO4^-} to \mathrm{Mn^{2

Question

Match List I with List II.

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

Choose the correct answer from the options given below:

Vishal Gupta · Accepted Answer

D. Faradays needed equal electrons transferred. \mathrm{H2O 	o 	frac12 O2}: 2 e⁻ per mole (II). \mathrm{MnO4^- 	o Mn^{2+}}: 5 e⁻ (IV). \mathrm{Ca^{2+} 	o Ca}: 2 e⁻ per mole, so 1.5 mol needs 3 F (I). \mathrm{FeO 	o Fe2O3}: Fe goes +2 → +3, 1 e⁻ per Fe, 1 F (III). Hence A-II, B-IV, C-I, D-III.

Electrochemistry

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