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Question #44212
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 .
Given:
Calculate the cell potential () and the Gibbs free energy change () for the cell reaction as written: \ce{2 Fe^{3+}(aq) + 2 I^-(aq) -> 2 Fe^{2+}(aq) + I_2(s)}
Options
- A.
- B.
- C.
- D.
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 and artificially invert the sign (Option B).
Furthermore, the reaction quotient () heavily exceeds the equilibrium constant (), 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 electrons leads to a halved value (Option D).
Finally, incorrectly formulating the reaction quotient leads to Option C.
Quick Facts About This Question
Question Information
| Subject | Chemistry |
| Chapter | Thermodynamics - Class 11 Part 1 |
| Concept | IUPAC Sign Convention |
| Difficulty | 8 |
| Cognitive Load | 7 |
| Target Solve Time | 160 sec |
| Access | Free |