Start Learning Free

Before You Begin

Question #44212

Difficulty: 8/10 | 🟢 Free

The Question

An electrochemical cell is constructed at 298 K:

\ce{Pt(s) | I^- (0.1 M) | I_2(s) || Fe^{2+} (0.1 M), Fe(OH)_3(s) | Pt(s)}

The right half-cell is buffered at pH=3.0\text{pH} = 3.0.

Given: E\ceFe3+/Fe2+∘=0.77 VE^\circ_{\ce{Fe^{3+}/Fe^{2+}}} = 0.77\text{ V} E\ceI2/I−∘=0.53 VE^\circ_{\ce{I_2/I^-}} = 0.53\text{ V} Ksp(\ceFe(OH)3)=10−38K_{sp}(\ce{Fe(OH)_3}) = 10^{-38} 2.303RTF=0.06 V\frac{2.303 RT}{F} = 0.06\text{ V} 1 F=96500 C/mol1\text{ F} = 96500\text{ C/mol}

Calculate the cell potential (EcellE_{cell}) and the Gibbs free energy change (ΔG\Delta G) for the cell reaction as written: \ce{2 Fe^{3+}(aq) + 2 I^-(aq) -> 2 Fe^{2+}(aq) + I_2(s)}

Options

  • A.

    Ecell=−0.06 V,ΔG=+11.58 kJ/molE_{cell} = -0.06\text{ V}, \Delta G = +11.58\text{ kJ/mol}

  • B.

    Ecell=+0.06 V,ΔG=−11.58 kJ/molE_{cell} = +0.06\text{ V}, \Delta G = -11.58\text{ kJ/mol}

  • C.

    Ecell=+0.54 V,ΔG=−104.22 kJ/molE_{cell} = +0.54\text{ V}, \Delta G = -104.22\text{ kJ/mol}

  • D.

    Ecell=−0.06 V,ΔG=+5.79 kJ/molE_{cell} = -0.06\text{ V}, \Delta G = +5.79\text{ kJ/mol}

Check Your Answer

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

IUPAC Sign Convention|Reaction Quotient vs. Equilibrium Constant|Nernstian Redox Equilibrium

Why This Question Is Tricky

Common Mistakes

This problem tests the integration of solubility equilibria with Nernstian redox systems and the IUPAC sign convention.

The primary trap is the cognitive dissonance of a negative cell potential: students often assume galvanic cells must have Ecell>0E_{cell} > 0 and artificially invert the sign (Option B).

Furthermore, the reaction quotient QQ (101010^{10}) heavily exceeds the equilibrium constant KK (10810^8), driving the reaction non-spontaneously in the forward direction.

A secondary trap involves the stoichiometry of the free energy calculation: failing to recognize that the balanced equation provided requires n=2n=2 electrons leads to a halved ΔG\Delta G value (Option D).

Finally, incorrectly formulating the reaction quotient leads to Option C.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterThermodynamics - Class 11 Part 1
ConceptIUPAC Sign Convention
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree