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Question #44540

Difficulty: 8/10 | 🟢 Free

The Question

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

$\ce{Ag(s, P_1) | Ag+(aq, 0.1 M, P_1) || Ag+(aq, 0.01 M, P_2) | Ag(s, P_2)}$

The left compartment is maintained at a standard pressure $P_1 = 1\text{ bar}$ . To completely halt the flow of electrons (i.e., $E_{\text{cell}} = 0$ ), a specific hydrostatic pressure $P_2$ is applied to the right compartment.

Given: $RT \ln 10 = 5700\text{ J mol}^{-1}$ Partial molar volume of $\ce{Ag+(aq)}$ , $\bar{V}_{\ce{Ag+}} = 67\text{ cm}^3\text{ mol}^{-1}$ Molar volume of $\ce{Ag(s)}$ , $V_{\ce{Ag}} = 10\text{ cm}^3\text{ mol}^{-1}$ $1\text{ bar} = 10^5\text{ Pa}$

Assuming the liquids and solids are incompressible, what is the required pressure $P_2$ ?

Options

A.
$1001\text{ bar}$
B.
$852\text{ bar}$
C.
$741\text{ bar}$
D.
$1000\text{ bar}$

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

Osmotic Pressure vs Diffusion|Standard Electrode Potential Sign Conventions

Why This Question Is Tricky

Common Mistakes

This problem tests the integration of electrochemical potential with mechanical pressure.

Students typically memorize the Nernst equation as purely concentration-dependent, ignoring the $PV$ work term in the Gibbs free energy ( $dG = VdP - SdT$ ).

The 'Hidden Variable' trap is the molar volume of the solid electrode ( $V_{\ce{Ag}}$ ).

Even if students account for the partial molar volume of the aqueous ion, they often forget that the solid electrode in the right compartment is also subjected to the pressure $P_2$ , meaning its chemical potential also changes by $V_{\ce{Ag}} \Delta P$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Osmotic Pressure vs Diffusion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free