 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4/51/c/vi/solution

The target dianion consumes two electrons per molecule, so

η_{F} = \frac{2 n _{product}}{n _{e^{-}}} = \frac{2 ( 7.8 )}{18.2} = 0.857,

giving $85.7%$ Faradaic yield. Of the passed electron equivalents,

2.6 mmol

are not accounted for by the quantified bis-adduct.

They can reside in incompletely captured reduced quinones, other reduction products or parasitic solvent/proton/CO

_{2}

chemistry. The data do not identify one unique loss pathway. The starting solution contains

50 mmol

BQ, so the target product represents only

15.6%

of that inventory; low bulk conversion alone does not explain a sub-unity Faradaic yield, because unreacted BQ has consumed no charge.

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