$id:A\to A^{op}$ is completely positive iff $A$ is abelian Let $A$ be a $C^*$-algebra and $A^{op}$ it's opposite $C^*$-algebra. Let $id:A\to A^{op}$ be the identity map. $id$ is positive. 
The claim is: $id$ is completely positive iff $A$ is abelian.
I need this statement for further studies.
The direction $\Leftarrow$ is easier to understand (I hope that everything is correct): If $A$ is abelian, then $A^{op}$ is abelian too. One can identify $A^{op}$ with $C_0(X)$ for a locally compact Hausdorff space $X$ and consider $id$ as a map $A\to C_0(X)$. This map is positive, hence completely positive (it's a standard fact about completely positive maps). It follows that $id:A\to A^{op}$ is completely positive. 
But I'm stuck to prove $\Rightarrow$.  Can anybody give me some hints or does anybody know good references?  It will be greatly appreciated.
 A: In the unital case, it is a result of Choi that a unital 2-positive map $f:A\to B$ is a $*$-homomorphism if and only if $f(a^2) = f(a)^2$ for all self-adjoint $a\in A$. In the present case, this implies that if $\mathrm{id}:A\to A^{\mathrm{op}}$ is 2-positive, then it is a $*$-homomorphism, i.e. $\mathrm{id}(ab)=\mathrm{id}(a)\mathrm{id}(b)$, which means that $ab=ba$, so that $A$ is abelian.
By the way, the $\Leftarrow$ direction does not need Gelfand duality. $A$ being abelian means precisely that $A=A^{\mathrm{op}}$, and $\mathrm{id}:A\to A$ is trivially completely positive.

Conclusion: If $A$ is unital, then $A$ is abelian if and only if $\mathrm{id}:A\to A^{\mathrm{op}}$ is 2-positive.

I don't know whether the above argument can be generalized to cover the non-unital case as well. One way to go about this would be to show that the extension of $\mathrm{id}:A\to A^{\mathrm{op}}$ to the unitization is still 2-positive, or otherwise to go through Choi's arguments and see to what extent they actually require unitality.
A: I'm not sure where I saw the following argument: it might have been mentioned in a book, or a paper, or a lecture.
I seem to remember that every non-abelian ${\rm C}^*$-algebra contains a *-subalgebra isomorphic to ${\bf M}_2$.
EDIT: this is not the case, as pointed out in comments: however, every non-abelian von Neumann algebra does contain a *-subalgebra isomorphic to ${\bf M}_2$. Since the double-adjoint of a c.p. map $A\to A^{\rm op}$ will be a c.p. map $A^{**} \to (A^{\rm op})^{**} = (A^{**})^{\rm op}$, one merely needs to show the identity map ${\rm j}: {\bf M}_2 \to {\bf M}_2^{\rm op}$ is not completely positive.
Well, since the transpose map is a *-isomorphism from ${\bf M}_2^{\rm op}$ onto ${\bf M}_2$,  so if ${\rm j}$ were c.p. then the transpose map on ${\bf M}_2$ would also be c.p., and this is well known to be false.
There should be a more intrinsic argument, that doesn't require this "embedding of a copy of ${\bf M}_2$" result, but I can't see how to do it right now.
