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Question #44199

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The Question

Calculate the standard reduction potential ( $E^\circ$ ) of the \ce{NO+(aq) / NO(g)} half-cell relative to the Standard Hydrogen Electrode (SHE) at 298 K, and identify the molecular orbital from which the electron is removed during the first ionization of \ce{NO}.

Given the following thermodynamic data at 298 K:

First ionization energy of \ce{NO(g)} = $9.26 \text{ eV}$
Hydration enthalpy of \ce{NO+(g)} = $-4.16 \text{ eV}$
Hydration enthalpy of \ce{NO(g)} $\approx 0 \text{ eV}$
Absolute standard reduction potential of SHE = $4.44 \text{ V}$
Assume entropic contributions ( $T\Delta S^\circ$ ) are negligible.

Options

A.
$+0.66 \text{ V}, \pi_{2p}^*$
B.
$-0.66 \text{ V}, \pi_{2p}^*$
C.
$+4.82 \text{ V}, \pi_{2p}^*$
D.
$+0.66 \text{ V}, \pi_{2p}$

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

Molecular Orbital Mixing|Electrode Potential Sign Conventions

Why This Question Is Tricky

Common Mistakes

This problem tests the intersection of Molecular Orbital Theory and Electrochemistry.

Students often mistakenly equate a low gas-phase ionization energy (due to the electron occupying an antibonding $\pi^*$ orbital) directly with a high oxidation potential in solution.

The 'hidden variable' is the hydration enthalpy of the resulting cation, which must be incorporated via a Born-Haber cycle to find the absolute electrode potential.

Furthermore, it tests the strict sign convention required to convert absolute potentials to relative standard reduction potentials ( $E^\circ$ ) versus the SHE.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Bonding and Molecular Structure - Class 11 Part 1
Concept	Molecular Orbital Mixing
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free