(Below by "$2$-category" I mean "bicategory.")

It's more fun to think about opposites, not in the $2$-category of categories, functors, and natural transformations, but in the $2$-category of categories, bimodules / distributors / profunctors, and natural transformations of bimodules. I prefer the bimodule terminology so let me use that. Recall that a $(C, D)$-bimodule, where $C, D$ are categories, is a functor $M : C^{op} \times D \to \text{Set}$. These naturally form a category. There is a tensor product operation taking a $(C, D)$-bimodule and a $(D, E)$-bimodule and returning a $(C, E)$-bimodule which gives the composition in this $2$-category. Moreover, the product of categories gives this $2$-category a symmetric monoidal structure.

The punchline is now that $C^{op}$ is the monoidal dual of $C$ in this symmetric monoidal $2$-category. The evaluation map is given by

$$\text{Hom}(-, -) : (C^{op} \times C) \times 1 \to \text{Set}$$

(regarded as a $(C^{op} \times C, 1)$-bimodule) and the coevaluation map is given by

$$\text{Hom}(-, -) : 1 \times (C \times C^{op}) \to \text{Set}$$

(regarded as a $(1, C \times C^{op})$-bimodule, and I need to switch the inputs here). In particular, every object of this $2$-category is dualizable.

The same story works with essentially no modification if we enrich everything in a reasonable symmetric monoidal category $V$. In particular, when $V = \text{Ab}$ we get a $2$-category into which the usual bimodule $2$-category of rings, bimodules, and morphisms of bimodules embeds, and we recover the observation that every ring $R$ is dualizable in this $2$-category with monoidal dual $R^{op}$. This is one reason bimodule categories are an attractive target category for TQFTs.