What additional property does the antipode give on the category of all modules over an Hopf algebra? It is well known that many constructions involving bialgebras extends to monoidal categories, and often becomes more natural in that framework.
If one cares about the category of finite dimensional modules, then the existence of an antipode on a given bialgebra is equivalent to the rigidity of its category of finite dimensional modules. 
I wonder what happens if one cares instead about the category of all modules on some Hopf algebra $H$. At first I thought existence of an antipode was equivalent to existence of internal Hom's in $H-mod$, which clearly generalizes being rigid: if $V,W$ are $H$-modules, then the internal Hom from $V$ to $W$ is the space of linear maps between those, equipped with the adjoint action (which does of course involve the antipode). But it turns out that modules over a bialgebra also have internal Hom's for abstract non-sense reasons: the functors "tensoring with $V$" on either side are cocontinuous so they have right adjoints. WHat is true is that a bialgebra is an Hopf algebra iff the fiber functor to Vect is compatible with internal Hom's, and in fact the same holds for Hopf monads.
But this characterization does involve the fiber functor, which I don't want. 
Note that if one looks at categories of comodules instead, then it is generated under filtered colimits by the finite dimensional ones, and those are dualizable iff we start with something Hopf. This is the "correct" generalization of being rigid in the large setting. 

What is special about the category of all modules over an Hopf algebra, which doesn't involve the fiber functor ?

 A: Here's a possible answer, or rather a non-answer, for which I claim no originality but isn't written anywhere as far as I know.
It's indeed well known that when dealing with an infinite dimensional Hopf/bi algebra (or some generalization thereof) $H$  there often are issues in matching properties of $H$ with properties of its category of modules. One possible explanation is that one should really take the duality all the way and think of $H-mod$ as a comonoidal category. Indeed bialgebras are coalgebras in algebras in $Vect$ say, so applying the symmetric monoidal functor $-mod$ one gets naturally a coalgebra in linear (locally presentable) categories. In that way I believe one would have all the dual statements of those for comodules, e.g.


*

*$B-mod$ for a bialgebra is co-rigid (which should be a statement about the existence of a nice adjoint to the cotensor product) iff $B$ is Hopf, 

*$H-mod$ is co-braided iff there exists an automorphism (as opposed to an element) $R$ of $H \otimes H$ satisfying appropriate axioms. This would apply e.g. to the example of quantum groups where the $R$-matrix do not belongs to $U_q(\mathfrak g)^{\otimes 2}$, hence doesn't acts on arbitrary modules, while its action by conjugation on $U_q(\mathfrak g)^{\otimes 2}$ is well-defined. Interestingly it should also encompass a Poisson version of the $R$-matrix as considered here which does not gives rise to a quasi-triangular structure according to the usual definition. 

*Similarly a quasi-Hopf algebra should really be defined using an automorphism of $H^{\otimes 3}$ rather than an element

*etc....

