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Question #44213

Difficulty: 8/10 | 🟢 Free

The Question

Consider the oxidation of the triiodide ion (\ce{I_3^-}) by permanganate (\ce{MnO_4^-}) in a basic medium at 298 K, producing iodate (\ce{IO_3^-}) and solid manganese dioxide (\ce{MnO_2(s)}).

The standard reduction potentials at 298 K are: E\ceMnO4−/\ceMnO2,\ceOH−∘=+0.600 VE^\circ_{\ce{MnO_4^-}/\ce{MnO_2}, \ce{OH^-}} = +0.600 \text{ V} E\ceIO3−/\ceI3−,\ceOH−∘=+0.615 VE^\circ_{\ce{IO_3^-}/\ce{I_3^-}, \ce{OH^-}} = +0.615 \text{ V}

An electrochemical cell is constructed using these half-reactions. All soluble species except \ce{OH^-} are maintained at a concentration of 1.0 M.

Calculate the Gibbs Free Energy change (ΔG\Delta G) for the cell reaction, scaled to exactly one mole of \ce{I_3^-}, when the pH of the solution is maintained at 10.0.

(Given: 2.303RTF=0.060 V\frac{2.303 RT}{F} = 0.060 \text{ V} at 298 K, F=96500 C mol−1F = 96500 \text{ C mol}^{-1})

Options

  • A.

    −54.04 kJ mol−1-54.04 \text{ kJ mol}^{-1}

  • B.

    +23.16 kJ mol−1+23.16 \text{ kJ mol}^{-1}

  • C.

    −162.12 kJ mol−1-162.12 \text{ kJ mol}^{-1}

  • D.

    +216.16 kJ mol−1+216.16 \text{ kJ mol}^{-1}

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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

IUPAC Sign Convention|Fractional Oxidation State Misconception|KMnO₄ Redox pH Dependency

Why This Question Is Tricky

Common Mistakes

The problem tests the 'Fractional Oxidation State Misconception' and 'pH Dependency of Redox Potentials'.

Students often miscalculate the n-factor for polyatomic ions with fractional oxidation states (like \ce{I_3^-}), leading to incorrect electron transfer values.

Furthermore, they frequently apply standard potentials without correcting for the Nernstian shift caused by the pH in basic media.

Here, the reaction is non-spontaneous under standard basic conditions (E∘=−0.015E^\circ = -0.015 V) but becomes spontaneous at pH 10 due to the decreased concentration of the product \ce{OH^-} relative to standard conditions, flipping the sign of ΔG\Delta G.

Failing to scale the reaction to exactly one mole of \ce{I_3^-} or incorrectly balancing with \ce{H^+} leads to the distractors.

Quick Facts About This Question

Question Information

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