Question: 61. Match List-I with List-II.
| List-I (Conversion) | List-II (Number of Faraday required) |
|---|---|
| (A) 1 mol of \mathrm{H}_{2} \mathrm{O} to \mathrm{O}_{2} | (I) 3 F |
| (B) 1 mol of \mathrm{MnO}_{4}^{-} to \mathrm{Mn}^{2+} | (II) 2 F |
| (C) 1.5 mole of Ca from molten \mathrm{CaCl}_{2} | (III) 1 F |
| (D) 1 mol of FeO to \mathrm{Fe}_{2} \mathrm{O}_{3} | (IV) 5 F |
(1) A-II, B-IV, C-I, D-III
(2) A-III, B-IV, C-I, D-II
(3) A-II, B-III, C-I, D-IV
(4) A-III, B-IV, C-II, D-I
Answer: Option (1)
Explanation:
Faraday is the amount of electric charge required to carry out one mole of electrons in an electrochemical reaction.
For (A), conversion of water to oxygen occurs as:
2 \mathrm{H_2O} \rightarrow \mathrm{O_2} + 4\mathrm{H}^+ + 4e^-Thus, 2 moles of water require 4 moles of electrons, so 1 mole of water requires 2F.
Hence, A-II.
For (B), reduction of permanganate ion in acidic medium occurs as:
\mathrm{MnO_4^-} + 8\mathrm{H}^+ + 5e^- \rightarrow \mathrm{Mn^{2+}} + 4\mathrm{H_2O}One mole of \mathrm{MnO_4^-} requires 5 moles of electrons, that is 5F. Hence, B-IV.
For (C), deposition of calcium from molten \mathrm{CaCl_2} occurs as:
\mathrm{Ca^{2+}} + 2e^- \rightarrow \mathrm{Ca}One mole of calcium requires 2F. Therefore, 1.5 moles of calcium require 3F. Hence, C-I.
For (D), oxidation of FeO to \mathrm{Fe_2O_3} occurs as:
2\mathrm{FeO} \rightarrow \mathrm{Fe_2O_3} + 2e^-Thus, 2 moles of FeO involve loss of 2 moles of electrons,
so 1 mole of FeO involves 1 mole of electrons, that is 1F.
Hence, D-III.
Therefore, the correct matching is A-II, B-IV, C-I, D-III and the correct answer is Option (1).