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Question #44756

Difficulty: 8/10 | 🟢 Free

The Question

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

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

To the anodic compartment, solid $\ce{KCN}$ is added such that the overall analytical concentration of $\ce{CN-}$ becomes $0.3\text{ M}$ . Assuming no change in the volume of the solution, calculate the new electromotive force ( $E_{\text{cell}}$ ) of the cell.

[Given: Formation constant ( $K_f$ ) of $\ce{[Ag(CN)2]-}$ is $10^{21}$ ; $\frac{2.303RT}{F} = 0.059\text{ V}$ ]

Options

A.
1.121 V
B.
1.177 V
C.
1.180 V
D.
0.000 V

Check Your Answer

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

Concentration Cells|Electrode Potential Dependencies|Crystal Field Splitting (Octahedral)

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Electrochemistry and Coordination Chemistry.

The primary trap is the 'Hidden Variable' of equilibrium ligand concentration.

Students frequently use the initial analytical concentration of the ligand ( $0.3\text{ M}$ ) in the $K_f$ expression instead of subtracting the stoichiometrically consumed amount ( $0.2\text{ M}$ ), leading to the distractor 1.177 V.

Another trap is ignoring the stoichiometric coefficient of 2 for cyanide, leading to a failure to square the $[\ce{CN-}]$ term (yielding 1.180 V).

Finally, conceptually weak students might assume $E_{\text{cell}}$ remains 0 V because the total analytical concentration of silver in both half-cells is technically unchanged, failing to realize that electrodes only respond to the thermodynamic activity of free ions.

Quick Facts About This Question

Question Information

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