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Question #44526

Difficulty: 8/10 | 🟢 Free

The Question

Two aqueous solutions are prepared at $298 \text{ K}$ using $1.0 \text{ kg}$ of water each:

Solution A: $0.10 \text{ mol}$ \ce{HCl} Solution B: $0.10 \text{ mol}$ \ce{HCl} and $1.80 \text{ mol}$ Sucrose (\ce{C12H22O11})

The solutions are placed in separate open beakers inside a sealed desiccator and allowed to reach thermodynamic equilibrium via isothermal solvent vapor transfer. Assume ideal behavior for all solutes ( $\gamma = 1$ ) and complete dissociation of \ce{HCl}.

After equilibrium is established, the two solutions are used to construct the following concentration cell with a liquid junction:

$Pt(s) | \ce{H2}(g, 1 \text{ bar}) | \text{Solution B}_{eq} \vdots \text{Solution A}_{eq} | \ce{H2}(g, 1 \text{ bar}) | Pt(s)$

Given the transport number of \ce{Cl^-} is $t_- = 0.20$ (assume constant across the junction) and $\frac{2.303 RT}{F} = 0.060 \text{ V}$ at $298 \text{ K}$ . What is the measured electromotive force ( $E_{cell}$ ) of this cell?

Options

A.
$+0.024 \text{ V}$
B.
$+0.060 \text{ V}$
C.
$+0.096 \text{ V}$
D.
$0.000 \text{ V}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Vapor Pressure Lowering (Mole Fraction)|Electrode Potential Dependencies|Liquid Junction Potential

Why This Question Is Tricky

Common Mistakes

This problem tests two hidden variables often overlooked by students.

First, Colligative Coupling: Students fail to realize that a non-electrolyte (sucrose) lowers the solvent's chemical potential, driving isothermal distillation of water until osmolalities equalize, which drastically alters the electroactive \ce{HCl} concentrations.

Second, Liquid Junction Potential (LJP): Even if concentrations are found, students typically apply the standard Nernst equation ( $E = 0.06 \log(C_1/C_2)$ ), ignoring that in a cell with transference (no salt bridge), the unequal mobilities of \ce{H+} and \ce{Cl-} generate an opposing LJP.

The true potential depends strictly on the transport number of the anion ( $E = 2t_- \frac{RT}{F} \ln \frac{a_1}{a_2}$ ).

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Vapor Pressure Lowering (Mole Fraction)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free