Which tetrahedra are scissor congruent to a cube?

Question

Question: Which Euclidean tetrahedra are scissor congruent to cubes?

Consider a Euclidean tetrahedron $T$ in $\mathbb{R}^3$ with edge lengths $l_1,\ldots, l_6$ and dihedral angles $\alpha_1,\ldots, \alpha_6.$ According to Sydler's theorem this is equivalent to vanishing of the Dehn invariant $$ D(T)=\sum l_i \otimes \alpha_i \in \mathbb{R}\otimes \mathbb{R}/2\pi \mathbb{Z}. $$ Denote by $r(T)$ the dimension of the $\mathbb{Q}-$span $\langle \alpha_1,\ldots,\alpha_6, \pi \rangle.$ It is easy to see that $1 \leq r(T)\leq 6.$

A list of partial results:

1. If $r(T)=1$ then all angles of $T$ are rational multiples of $\pi,$ such tetrahedra were classified by Coxeter (this is explained here).

2. Some examples of $1-$dimensional families of tetrahedra with $r(T)=2,3$ are presented here.

3. A similar question about right-angled pyramid appeared on MOF recently and was beautifully solved. It is easy to see that such pyramid is
   scissor congruent to a centrally symmetric Schläfli orthoscheme.

It might be simpler to restrict to a certain family of examples, like Schläfli orthoschemes.

Simpler questions:

1. What examples of Euclidean tetrahedra scissor congruent to cubes are known?

2. What is the biggest possible dimension of an algebraic family of tetrahedra scissor congruent to cubes?

3. Which tetrahedra are known not to be scissor congruent to cubes?

4. Can a partial answer be obtained for certain values of $r(T)?$