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Question #45009

Difficulty: 8/10 | 🟢 Free

The Question

A galvanic cell is constructed at 298 K298 \text{ K} with the following cell representation:

\ceSn(s)∣Sn2+(aq,1.0M)∣∣Sn4+(aq,1.0M),Sn2+(aq,1.0M)∣Pt(s)\ce{Sn(s) | Sn^{2+}(aq, 1.0 M) || Sn^{4+}(aq, 1.0 M), Sn^{2+}(aq, 1.0 M) | Pt(s)}

The standard reduction potentials are E\ceSn4+/Sn2+∘=0.15 VE^\circ_{\ce{Sn^{4+}/Sn^{2+}}} = 0.15 \text{ V} and E\ceSn2+/Sn∘=−0.14 VE^\circ_{\ce{Sn^{2+}/Sn}} = -0.14 \text{ V}. Initially, the cell operates spontaneously. A ligand \ceL−\ce{L^-} is then added to the cathode compartment, which selectively forms the highly stable complex \ce[SnL6]2−\ce{[SnL_6]^{2-}} with \ceSn4+\ce{Sn^{4+}} (Formation constant β6=1022\beta_6 = 10^{22}). The concentration of free \ceL−\ce{L^-} is maintained at 0.1 M0.1 \text{ M}, and the volume change is negligible.

What is the magnitude of the new cell potential, and what is the spontaneous cell reaction after the addition of the ligand? (Assume 2.303RTF=0.06 V\frac{2.303RT}{F} = 0.06 \text{ V})

Options

  • A.

    0.19 V0.19 \text{ V}, Disproportionation of \ce{Sn^{2+}}

  • B.

    0.19 V0.19 \text{ V}, Comproportionation of \ce{Sn} and \ce{Sn^{4+}}

  • C.

    0.34 V0.34 \text{ V}, Disproportionation of \ce{Sn^{2+}}

  • D.

    0.47 V0.47 \text{ V}, Comproportionation of \ce{Sn} and \ce{Sn^{4+}}

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Comproportionation Reversibility|Concentration Cells|Standard State Conditions

Why This Question Is Tricky

Common Mistakes

This problem tests the Nernstian shift in a concentration/complexation cell.

Students must calculate the drastically reduced free Sn(IV) concentration due to complexation, which shifts the cathode potential from +0.15 V to -0.33 V.

The hidden trap lies in interpreting the resulting negative cell potential (E_cell = -0.19 V).

A negative potential indicates that the original spontaneous reaction (comproportionation of Sn and Sn(IV)) has reversed.

Students who only calculate the magnitude will incorrectly choose Option B, failing to recognize that the new spontaneous process is the disproportionation of Sn(II).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterRedox Reactions - Class 11 Part 2
ConceptComproportionation Reversibility
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree