This is probably a very simple question, and but I'm stuck on a loophole I can't get wrap my head around. A (co)chain complex of $R$ modules (here $R$ is $k$ algebra, where $k$ is a field) $C^{\bullet}$ is called contractible if there exists a map $h : C^{\bullet} \rightarrow C^{\bullet}[-1]$ such that $d_{C}h + hd_{C} = id_{C}$. Contractible complexes are acyclic, even though acyclicity doesn't imply contractibility. Moreover, contractibility is preserved by additive functors.
Here's my doubt. Say you have a complex $C^{\bullet}$ which is contractible. Then, if we tensor $C^{\bullet}$ with a module $M$, as tensor product is additive, from what we said above we get that $C^{\bullet} \otimes M$ is contractible, hence acyclic. In particular, we could apply this reasoning to the bar resolution of $R$ (here by bar resolution I mean the resolution plus the augmentation $R \otimes_k R \rightarrow R$), and a $R$-$R$ bimodule $M$. As the bar resolution is contractible, this would seem to imply that the Hochschild homology of $M$ vanishes in all non zero degrees. What I am doing wrong?