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Question #44747

Difficulty: 8/10 | 🟢 Free

The Question

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

Pt(s)∣H2(g,1 bar)∣HA(aq,4×10−4 M)+NaA(aq,2×10−4 M)∣∣HA(aq,6×10−4 M)∣H2(g,1 bar)∣Pt(s)Pt(s) | H_2(g, 1 \text{ bar}) | HA(aq, 4 \times 10^{-4} \text{ M}) + NaA(aq, 2 \times 10^{-4} \text{ M}) || HA(aq, 6 \times 10^{-4} \text{ M}) | H_2(g, 1 \text{ bar}) | Pt(s)

Given:

  • The acid dissociation constant of the weak acid HAHA is Ka=10−4K_a = 10^{-4}.
  • 2.303RTF=0.06 V\frac{2.303 RT}{F} = 0.06 \text{ V} at 298 K298 \text{ K}.
  • log⁡2=0.30\log 2 = 0.30, log⁡3=0.48\log 3 = 0.48.

Calculate the electromotive force (EcellE_{cell}) of the cell in millivolts (mV).

Options

  • A.

    18.0 mV18.0 \text{ mV}

  • B.

    0.0 mV0.0 \text{ mV}

  • C.

    23.4 mV23.4 \text{ mV}

  • D.

    5.4 mV5.4 \text{ mV}

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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

Degree of Dissociation (Alpha)|Electrode Potential Dependencies|Gibbs Energy and Spontaneity

Why This Question Is Tricky

Common Mistakes

This problem tests the limits of standard ionic equilibrium approximations within an electrochemical context.

Students typically rely on the Henderson-Hasselbalch equation for buffers and the Ostwald dilution approximation ([H+]=KaC[H^+] = \sqrt{K_a C}) for weak acids.

However, when the acid concentration is comparable to its KaK_a (here, 10−410^{-4}), the degree of dissociation (α\alpha) is significant, and these approximations fail catastrophically.

The 'hidden variable' is the necessity to solve the exact quadratic equations for both half-cells to find the true [H+][H^+] before applying the Nernst equation.

Blindly applying the Henderson-Hasselbalch equation to the anode yields [H+]=2×10−4 M[H^+] = 2 \times 10^{-4} \text{ M}, leading to an erroneous Ecell=0.0 mVE_{cell} = 0.0 \text{ mV} (Option B).

Blindly applying the Ostwald approximation to the cathode yields [H+]=6×10−4 M[H^+] = \sqrt{6} \times 10^{-4} \text{ M}, leading to 23.4 mV23.4 \text{ mV} (Option C).

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterElectrochemistry - Class 12
ConceptDegree of Dissociation (Alpha)
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree