**Appendix A. Calculations of the Surfactant Concentration Depletion Due to Adsorption on Emulsion Drops**

After the formation of a new interface with area *A*, in a surfactant solution with the initial concentration *c*0, an adsorption process starts and the system achieves a new equilibrium state, characterized by an adsorption Γ and a bulk concentration *c*, satisfying the set of equations provided by the mass balance and by the equilibrium adsorption isotherm:

$$\begin{cases} A\Gamma + \mathcal{c}V = \mathfrak{c}\_0 V \\ \mathcal{c} = \mathfrak{c}(\Gamma) \end{cases} \tag{A1}$$

where *V* is the volume of the surfactant solution. Within the Langmuir model, the adsorption isotherm can be made explicit and the system becomes:

$$\begin{cases} A\Gamma + \mathcal{c}V = \mathcal{c}\_0 V \\ \mathcal{c} = \frac{\omega \Gamma}{b(1 - \omega \Gamma)} \end{cases} \tag{A2}$$

From which, by substitution in the mass balance, the following second-degree equation in Γ can be obtained:

$$A b \omega \Gamma^2 - (b A + \omega V + b \csc \psi V) \Gamma + b \csc V = 0 \tag{A3}$$

and solved with the condition, ωΓ < 1, which is satisfied by the solution with the minus sign.

For more complicated isotherms (Frumkin, 2-states), where an explicit relationship between *c* and Γ is not provided, numerical schemes must instead be adopted to solve the above set of Equation (A1).

Concerning the complete emulsification of an oil volume *V*oil in a volume *V* of water, as a first approximation we can assume the total area of the droplets given by:

$$A = \frac{3}{R} V\_{\text{oil}} \tag{A4}$$

where *R* is the average radius of the oil droplets. Introducing this area, together with the volume fraction Φ =*V*oil/*V*, the mass balance reads:

$$
\mathfrak{c} + \frac{3}{R} \Phi \Gamma = \mathfrak{c}\_0 \tag{A5}
$$

In the case the adsorption process co-exists with surfactant partitioning between the volume of water *V* and the volume of oil *V*oil, the mass balance is:

$$A\Gamma + cV + kcV\_{\text{oil}} = c\_0V \tag{A6}$$

where *k* = *c*oil/*c*water is the partition coefficient and *c*<sup>0</sup> and *c* refer to the concentrations in water.

After defining α = (1 + *kV*oil/*V*), this equation can be recast as:

$$A\Gamma + \mathfrak{a}\mathcal{L}V = \mathfrak{c}\_0 V \tag{A7}$$

Which allows the above equations for the depletion in the absence of partitioning to be easily adapted to account for partitioning.

*Coatings* **2020**, *10*, 397

The values of Γ, calculated using the described approach, allow estimating the realistic surfactant coverage, ωΓ, for the droplets in the emulsion, which for the systems investigated here are reported in Table A1.

More accurate values for *A* can be obtained by considering the drop size distribution. If the number of drops *N*<sup>i</sup> with the radius *R*i, normalized on the total number of drops *NT*, is distributed accordingly to *Ni*/*NT* = *fi(Ri)*, the total drop volume is given by:

$$V\_{\rm coll} = \sum\_{i} N\_{i} \frac{4}{3} \pi R\_{i}^{3} = N\_{T} \sum\_{i} f\_{i} \frac{4}{3} \pi R\_{i}^{3} \tag{A8}$$

Thus

$$N\_T = \frac{V\_{\text{oil}}}{\sum\_{i} f\_i \frac{4}{3} \pi R\_i^{\text{-}3}} \tag{A9}$$

The total droplet area is thus given by:

$$A = \sum\_{i} N\_{i} 4\pi R\_{i}^{2} = N\_{T} \sum\_{i} f\_{i} 4\pi R\_{i}^{2} = 3V\_{\text{coll}} \frac{\sum\_{i} f\_{i} R\_{i}^{2}}{\sum\_{i} f\_{i} R\_{i}^{3}} \tag{A10}$$

So that, the equation equivalent to the mass balance A5 reads:

$$\mathbf{c} + 3\Phi \frac{\sum\_{i} f\_i \mathbf{R}\_i^2}{\sum\_{i} f\_i \mathbf{R}\_i^3} \,\Gamma = \mathbf{c}\_0 \tag{A11}$$

For the emulsions studied here, the resulting size distributions are however always quite narrow, so that the use of Equation (A11), does not modify substantially the values of the coverage reported in Table A1.

**Table A1.** Depleted concentrations of the continuous phase (*c*) and the corresponding surface coverages of the droplets after emulsification. *c*<sup>0</sup> is the concentration of the aqueous solution before emulsification.

