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Question #44882

Difficulty: 8/10 | 🟢 Free

The Question

Consider the following electrochemical cell at 298 K298\text{ K}:

\cePt(s)∣Br2(l)∣Br−(0.1M)∣∣MnO4−(0.1M),Mn2+(0.001M),H+(pH=x)∣Pt(s)\ce{Pt(s) | Br_2(l) | Br^{-}(0.1 M) || MnO_4^{-}(0.1 M), Mn^{2+}(0.001 M), H^{+}(pH = x) | Pt(s)}

Given the standard reduction potentials: E\ceMnO4−/Mn2+∘=1.51 VE^\circ_{\ce{MnO_4^{-}/Mn^{2+}}} = 1.51\text{ V} E\ceBr2/Br−∘=1.09 VE^\circ_{\ce{Br_2/Br^{-}}} = 1.09\text{ V}

Assume 2.303RTF=0.06 V\frac{2.303 RT}{F} = 0.06\text{ V}. If the cell potential is exactly zero, what is the value of xx?

Options

  • A.

    4.00

  • B.

    5.25

  • C.

    4.375

  • D.

    3.75

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

KMnO₄ Redox pH Dependency|Nernst Equation Concentration Dependence|Standard Electrode Potential Sign Conventions

Why This Question Is Tricky

Common Mistakes

This problem tests the Nernst equation's concentration dependence, specifically the high sensitivity of the permanganate reduction potential to pH due to the 8th-power dependence on [\ceH+][\ce{H^{+}}].

A common trap is incorrectly formulating the reaction quotient QQ for the anode (e.g., using [\ceBr−]2[\ce{Br^{-}}]^2 in the numerator for oxidation, leading to a pH of 5.25) or ignoring the non-standard concentrations of the manganese species (leading to a pH of 3.75).

It forces students to rigorously apply logarithmic properties and stoichiometric coefficients.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterD and F Block Elements - Class 12
ConceptKMnO₄ Redox pH Dependency
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree