= Solution
The target dianion consumes two electrons per molecule, so
$$
\eta_F=\frac{2n_{\rm product}}{n_{e^-}}
=\frac{2(7.8)}{18.2}
=0.857,
$$
giving \b[$85.7\%$ Faradaic yield]. Of the passed electron equivalents, $2.6\,\mathrm{mmol}$ 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 $50\,\mathrm{mmol}$ 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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