*4.2. Performance Calculation*

#### 4.2.1. Hydrogen Production Rate and Coulombic Efficiency

The system's efficiency was measured in terms of the rate of hydrogen production, the rate of hydrogen recovery, the Coulombic efficiency, the volumetric density, and the energy recovered [61].

On the basis of COD elimination, the total potential number of moles generated, nth, is

$$\mathbf{n}\_{\rm th} = \frac{\mathbf{b}\_{\rm H\_{2/s}} \mathbf{v}\_{\rm L} \Delta \mathbf{C}\_{\rm S}}{\mathbf{M}\_{\rm s}},\tag{5}$$

where bH2/s = 4 mol/mol denotes the maximum amount of hydrogen that can be stoichiometrically generated from the substrate, vL denotes the volume of liquid in the reactor, ΔS (g COD·L−1) denotes the change in concentration of the substrate during one batch cycle, and Ms denotes the molecular weight of the substrate. The COD concentration (g COD·L−1) was converted into moles of acetate using a conversion factor of 0.78 g COD·g<sup>−</sup><sup>1</sup> sodium acetate. According to the measured current, the moles of hydrogen recovered by nCE can be calculated as follows:

$$\mathbf{n\_{CE}} = \frac{\int\_{t=0}^{t} \mathbf{Id} \mathbf{t}}{2\mathbf{F}} \,, \tag{6}$$

where I = V/Rex is the current computed from the voltage across the resistor, and 2 is the conversion factor used to transform moles of electrons to hydrogen. F = 96,485 C/mol e<sup>−</sup> is Faraday's constant, and dt (s) is the interval over which data were collected. The Coulombic hydrogen recovery is expressed as

$$
\mathbf{r\_{CE}} = \frac{\mathbf{n\_{CE}}}{\mathbf{n\_{th}}} = \mathbf{C\_{E}}.\tag{7}
$$

where CE is the Coulombic efficiency. The hydrogen recovery (in moles) at the cathode rCat is calculated as

$$\text{trCat} = \frac{\mathbf{n}\_{\text{H}\_2}}{\mathbf{n}\_{\text{CE}}},\tag{8}$$

where nH2 is the number of moles of hydrogen recovered over a batch cycle. The maximum volumetric hydrogen production rate (Q) measured in m<sup>3</sup> H2·m<sup>−</sup><sup>3</sup> of reactor per day (m<sup>3</sup> H2·m<sup>−</sup>3·day−1) is calculated as

$$\mathrm{Q\_{H\_2}}\left(\mathrm{m}^3 \,\mathrm{m}^{-3} \cdot \mathrm{day}\right) = \frac{\mathrm{I\_V}\left(\mathrm{A}/\mathrm{m}^3\right)\mathrm{r\_{\mathrm{Cat}}}\left[1\mathrm{C}/\mathrm{s}\right]/\mathrm{A}}{\left(\mathrm{F} = 9.65 \times 10^4 \frac{\mathrm{C}}{\mathrm{mol}} \cdot \mathrm{e}^-\right)\mathrm{C\_6}\left(\mathrm{mol} \cdot \frac{\mathrm{H\_2}}{\mathrm{L}}\right)\left(10^3 \frac{\mathrm{L}}{\mathrm{m}^3}\right)} = \frac{43.2 \,\mathrm{I\_V}\mathrm{r\_{\mathrm{Cat}}}}{\mathrm{Fc\_g}(\mathrm{T})}.\tag{9}$$
