Question

Match List I with List II

List-I (Conversion)		List-II  (Shape/geometry)
(A)	1 mol of H2O to O2	(I)	3F
(B)	1 mol of MnO-4 to Mn2+	(II)	2F
(C)	1.5 mol of Ca from molten CaCl2	(III)	1F
(D)	1 mol of FeO to Fe2O3	(IV)	5F

Choose the correct answer from the options given below:

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

Answer 1

The Correct Option is A

The objective is to correlate chemical transformations with the equivalent number of Faradays (F) required. Each case is analyzed sequentially:

(A) Transformation of 1 mol of $ \text{H}_2\text{O} $ to $ \frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- $:
The half-reaction for water oxidation is: $ \text{H}_2\text{O} \rightarrow \frac{1}{2}\text{O}_2 + 2\text{H}^+ + 2e^- $.
This process involves 2 moles of electrons per mole of $ \text{H}_2\text{O} $. Consequently, $ 2F $ is the required Faraday amount.
Thus, (A) corresponds to (II).

(B) Transformation of 1 mol of $ \text{MnO}_4^- $ to $ \text{Mn}^{2+} $:
The oxidation state of Mn in $ \text{MnO}_4^- $ is determined as follows: $ x + 4(-2) = -1 \implies x = +7 $.
The oxidation state of Mn in $ \text{Mn}^{2+} $ is $ +2 $.
The change in oxidation state is $ +7 - (+2) = 5 $.
The balanced half-reaction in an acidic medium is: $ \text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O} $.
For the reduction of 1 mole of $ \text{MnO}_4^- $ to $ \text{Mn}^{2+} $, 5 moles of electrons are transferred. Therefore, $ 5F $ is required.
Hence, (B) corresponds to (IV).

(C) Production of 1.5 mol of Ca from molten $ \text{CaCl}_2 $:
The reduction half-reaction for $ \text{Ca}^{2+} $ to Ca is: $ \text{Ca}^{2+} + 2e^- \rightarrow \text{Ca} $.
Producing 1 mole of Ca necessitates 2 moles of electrons ($ 2F $).
For 1.5 moles of Ca, $ 1.5 \times 2 = 3 $ moles of electrons are needed. Therefore, $ 3F $ is the required Faraday amount.
Thus, (C) corresponds to (I).

(D) Transformation of 1 mol of $ \text{FeO} $ to $ \text{Fe}_2\text{O}_3 $:
The oxidation state of Fe in $ \text{FeO} $ is $ +2 $.
The oxidation state of Fe in $ \text{Fe}_2\text{O}_3 $ is calculated as: $ 2x + 3(-2) = 0 \implies 2x = +6 \implies x = +3 $.
The change in oxidation state for Fe is $ +3 - (+2) = +1 $.
The conversion involves 2 moles of Fe in $ \text{Fe}_2\text{O}_3 $ per reaction unit: $ 2\text{FeO} + \frac{1}{2}\text{O}_2 \rightarrow \text{Fe}_2\text{O}_3 $.
For 2 moles of $ \text{FeO} $, the total change in the oxidation state of Fe is $ 2 \times (+3 - +2) = 2 $, indicating 2 moles of electrons are involved.
However, considering 1 mole of $ \text{FeO} $, the oxidation state change of Fe is $ +1 $, thus requiring 1 mole of electrons ($ 1F $).
Therefore, (D) corresponds to (III).

Final Matches:
(A) - (II)
(B) - (IV)
(C) - (I)
(D) - (III)

Final Answer:
The final answer is: $ \boxed{\text{A-II, B-IV, C-I, D-III}} $