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Question #44788

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following concentration cell with transference at $298\text{ K}$ :

$\ce{Ag(s) | AgNO3(aq, 0.01 M) \vdots AgNO3(aq, 0.1 M) | Ag(s)}$

The vertical dotted line ( $\vdots$ ) represents a liquid junction. The transport number of $\ce{NO3^-}$ in the given concentration range is $0.55$ . Assuming activity coefficients are unity and $\frac{2.303 RT}{F} = 0.060\text{ V}$ , calculate the total Gibbs free energy change ( $\Delta G$ ) for the cell process per mole of electrons transferred. (Given: $F = 96500\text{ C/mol}$ )

Options

A.
$-6.37\text{ kJ/mol}$
B.
$-5.79\text{ kJ/mol}$
C.
$-5.21\text{ kJ/mol}$
D.
$-0.58\text{ kJ/mol}$

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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

Liquid Junction Potential|Gibbs Energy and Spontaneity|Standard Electrode Potential Sign Conventions

Why This Question Is Tricky

Common Mistakes

This problem tests the integration of Liquid Junction Potential (LJP) into the total Gibbs free energy of a concentration cell.

A common trap is ignoring the LJP and calculating only the Nernstian potential ( $E_{cell} = 0.060\text{ V}$ ), leading to $-5.79\text{ kJ/mol}$ .

Another trap is using the transport number of the cation ( $t_+ = 1 - 0.55 = 0.45$ ) instead of the anion for a cell reversible to cations, yielding $-5.21\text{ kJ/mol}$ .

The student must recognize that spontaneity and total work in a non-ideal junction cell are governed by the net EMF ( $E_{wt}$ ), which explicitly depends on the migration of the counter-ion.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Electrochemistry - Class 12
Concept	Liquid Junction Potential
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free