A morphism $S' \to S$ is a morphism of descent for $(Sch \to Sch)$ iff it is a universal effective epimorphism for $(Sch)$? In SGA 1, VIII there is the following theorem (roughly translated by me)

Theorem 5.2 Let $\mathcal F$ be the fibred category of arrows over the category of schemes (i.e. objects are morphisms $X \to S$ and morphisms are commutative squares). If $g: S' \to S$ is a fpqc morphism, then $g$ is a morphism of descent for $\mathcal F$.
This means that for any $X, Y \in \operatorname{Sch}/S$, the diagram
$$\DeclareMathOperator{\Hom}{Hom} \Hom_S(X,Y) \to \Hom_{S'}(X', Y') \rightrightarrows \Hom_{S''}(X'', Y'')$$
is an equalizer diagram (where $X' = X \times_S S'$, $S'' = S' \times_S S'$, $X'' = X \times_S S''$ and similarly for $Y$).

Then Grothendieck writes that according to Giraud [D], Theorem 5.2 is equivalent to

Corollary 5.3 Any fpqc morphism is a universal effective epimorphism.

The proof of the later statement can be found in the stacks project, 023P (assuming the definitions are the same).
However I don't know about the equivalence of Theorem 5.2 and Corollary 5.3, and didn't find anything from skimming Giraud. Is that easy, or covered somewhere in the stacks project?
[D] J. Giraud, Méthode de la descente, 1964
 A: $\DeclareMathOperator{\Mor}{Mor}$The following shows the implication $5.3 \implies 5.2$.
Suppose $S' \to S$ is a universal effective epimorphism, and $X, Y \in \text{Sch}/S$ are given. Then by definition, $X' \to X$ is an effective epimorphism, i.e.
$$\Mor(X, Y) \to \Mor(X', Y) \rightrightarrows \Mor(X'', Y)$$
is an equalizer diagram. (Note that $X'' = X \times_S S'' = X' \times_X X'$). Now clearly we have an inclusion $\Mor_S(X,Y) \subset \Mor(X,Y)$. We also have canonical maps $\Mor_{S'}(X', Y') \to \Mor(X', Y)$ and $\Mor_{S''}(X'', Y'') \to \Mor(X'', Y)$. One easily verifies using the universal property of products that those maps are injective with image $\Mor_S(X', Y)$ and $\Mor_S(X'', Y)$. So we get a "commutative" diagram
\begin{align}
\Mor(X,Y) & \quad \rightarrow &\Mor(X', Y) & \quad \rightrightarrows & \Mor(X'', Y) \\
&& \cup \quad&& \cup \quad \\
\cup \quad && \Mor_S(X', Y) && \Mor_S(X'', Y)\\
&& \uparrow \text{bijective}  && \uparrow \text{bijective}  \\
\Mor_S(X,Y) & \quad \rightarrow & \Mor_{S'}(X', Y') & \quad \rightrightarrows &\Mor_{S''}(X'', Y'')
\end{align}
By "commutative" I mean that if we choose the upper arrows for each $\rightrightarrows$ it becomes commutative and the same for the lower arrows.
Now the upper row is an equalizer diagram by assumption. We have to show that the lower row is also an equalizer diagram.
Clearly, the map $\Mor_S(X,Y) \to \Mor_{S'}(X', Y')$ is injective, because the composite $$\Mor_S(X,Y) \to \Mor(X,Y) \to \Mor(X', Y)$$ is injective and the diagram commutes. It is also clear that $\Mor_S(X,Y)$ is contained in the equalizer of $\Mor_{S'}(X', Y') \rightrightarrows \Mor_{S''}(X'', Y'')$.
Now suppose $f' \in \Mor_{S'}(X', Y')$ is contained in that equalizer. Then the composite morphism $X' \xrightarrow{f'} Y' \to Y$ is induced by a morphism $g: X \to Y$, because the upper row is an equalizer diagram. This means we have a commutative diagram
$\require{AMScd}$
\begin{CD}
X' @>f'>> Y' \\
@VgVV @VVhV \\
X @>f>> Y,
\end{CD}
and we have to show that $f$ is a morphism over $S$. By assumption, $f'$ is a morphism over $S'$, which means that $h \circ f'$ is a morphism over $S$. Let $p: X \to S$, $q: Y \to S$ be the given structure maps. Then
$$p \circ f = q \circ h \circ f' = q \circ f \circ g.$$
Now $g$ is an (effective) epimorphism, and hence $p = q \circ g$.
I don't know how to prove the other direction.
