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Question #45025

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The Question

A $1.0\text{ L}$ aqueous solution at $\text{pH} = 0$ and $298\text{ K}$ initially contains $0.3\text{ moles}$ of $\ce{EuCl_2(aq)}$ and $0.1\text{ moles}$ of $\ce{Ce(SO_4)_2(aq)}$ . Given the standard reduction potentials: $E^\circ(\ce{Ce^{4+}}/\ce{Ce^{3+}}) = +1.74\text{ V}$ $E^\circ(\ce{Eu^{3+}}/\ce{Eu^{2+}}) = -0.43\text{ V}$ $E^\circ(\ce{O_2}/\ce{H_2O}) = +1.23\text{ V}$ $E^\circ(\ce{H^+}/\ce{H_2}) = 0.00\text{ V}$ Faraday's constant $F = 96500\text{ C mol}^{-1}$ .

Assume the oxidation of water by $\ce{Ce^{4+}}$ is kinetically extremely slow and can be neglected. All other thermodynamically favorable redox reactions proceed to completion. Calculate the standard Gibbs free energy change ( $\Delta G^\circ$ ) for the oxidation of $1\text{ mole}$ of $\ce{Eu^{2+}}$ by $\ce{Ce^{4+}}$ , and determine the total moles of $\ce{H_2(g)}$ evolved from the solution.

Options

A.
-209.4 kJ/mol, 0.1 mol
B.
-418.8 kJ/mol, 0.2 mol
C.
-126.4 kJ/mol, 0.15 mol
D.
-209.4 kJ/mol, 0.0 mol

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

Metal Activity Series Logic|E° (M³⁺/M²⁺) Redox Potential|Lanthanoid Contraction cause

Why This Question Is Tricky

Common Mistakes

The question tests the kinetic vs. thermodynamic stability of anomalous lanthanide oxidation states.

Students often fall into the trap of assuming that $\ce{Ce^{4+}}$ ( $4f^0$ , noble gas core) and $\ce{Eu^{2+}}$ ( $4f^7$ , half-filled core) are thermodynamically stable in water due to their exchange energy and electronic configurations.

However, the +3 state is universally the most stable for lanthanides in aqueous media due to the dominant effect of hydration enthalpy.

Thus, $\ce{Ce^{4+}}$ acts as a strong oxidant and $\ce{Eu^{2+}}$ acts as a strong reductant, even reducing $\ce{H^+}$ to $\ce{H_2}$ .

The problem forces students to sequentially apply these potentials to find the limiting reagent and the subsequent reaction with the solvent, bypassing the kinetic trap of water oxidation by $\ce{Ce^{4+}}$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Redox Reactions - Class 11 Part 2
Concept	Metal Activity Series Logic
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free