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Question #44812

Difficulty: 8/10 | 🟢 Free

The Question

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

\ceAg(s)∣Ag+(aq,1.0×10−3M)∣∣Ag+(aq,0.10M)∣Ag(s)\ce{Ag(s) | Ag+(aq, 1.0 \times 10^{-3} M) || Ag+(aq, 0.10 M) | Ag(s)}

The \ceAg+\ce{Ag+} ions in the cathodic compartment are continuously consumed by a pseudo-first-order chemical reaction with a rate constant k=2.303×10−4 s−1k = 2.303 \times 10^{-4} \text{ s}^{-1}. The concentration of \ceAg+\ce{Ag+} in the anodic compartment is maintained constant.

Assuming standard Nernstian behavior and 2.303RTF=0.059 V\frac{2.303 RT}{F} = 0.059 \text{ V} at 298 K298 \text{ K}, calculate the time required for the cell potential (EcellE_{\text{cell}}) to drop to exactly half of its initial value.

Options

  • A.

    1.00×104 s1.00 \times 10^4 \text{ s}

  • B.

    3.01×103 s3.01 \times 10^3 \text{ s}

  • C.

    2.00×104 s2.00 \times 10^4 \text{ s}

  • D.

    4.34×103 s4.34 \times 10^3 \text{ s}

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

First Order Half-Life Independence|Concentration Cells

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Chemical Kinetics and Electrochemistry.

The core trap exploits the student's intuition regarding exponential decay.

Because the concentration of the reactant decays exponentially in a first-order reaction, students frequently assume the cell potential (EMF) also decays exponentially, leading them to erroneously apply the half-life formula directly to the EMF (t=ln⁡(2)/k≈3010 st = \ln(2)/k \approx 3010 \text{ s}, Option B).

However, the Nernst equation applies a logarithmic transformation to the concentration.

Consequently, an exponential decay in concentration results in a strictly linear decay in cell potential over time.

The time for the EMF to halve is not the half-life of the reaction, but rather the time required for the concentration's logarithmic exponent to halve.

Quick Facts About This Question

Question Information

SubjectChemistry
ChapterChemical Kinetics - Class 12
ConceptFirst Order Half-Life Independence
Difficulty8
Cognitive Load7
Target Solve Time160 sec
AccessFree