Commutativity of the spatial tensor product Let $A, B$ be two arbitrary $C^{*}$-algebras (not neccesary unital). 
Let also $A \otimes B$ be an algebraic tensor product, whilst $A \otimes_{\min} B$ stands for the minimal tensor product, which is constructed as follows: if $\pi_{A}$ and $\pi_{B}$ are two representations of $A, B$ on Hilbert spaces $\mathcal{H_{1}}, \mathcal{H_{2}}$ respectively, then one can construct the map $\pi(a \otimes b) = \pi(a) \otimes \pi(b)$.
Furthermore, if one endows $A \otimes B$ with the norm $\| \cdot \|$ defined as $$\| \sum_{i}{a_{i} \otimes b_{i}} \| = \sup \| (\pi_{A} \otimes \pi_{B})(\sum_{i}{a_{i} \otimes b_{i}}) \|$$ where the supremum is taken over the all representations $\pi_{A}$ of $A$ and $\pi_{B}$ of $B$, then the completition of the algebraic tensor product w.r.t to the given normed is called a spatial tensor product of star-algebras.
I would like to show that in fact there exists a $*$-isomorphism $A \otimes_{min} B = B \otimes_{min} A$.
In fact, the very first candidate to be such a map is $a \otimes b \mapsto b \otimes a$. 
How to rigorously check that the map above is indeed a $*$-isomorphim?
(i can show that for a given non-degenerate representation $\pi$ of $A \otimes B$ on $\mathcal{H}$ there exist $\pi_{A}$, $\pi_{B}$ such that $\pi(a \otimes b) = \pi_{A}(a) \otimes \pi_{B}(b) = \pi_{B}(b) \otimes \pi_{A}(a)$ $-$ in unital case it is clear, in a non-unital one it is clear as well via the construction using approximaing units. 
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
A \otimes B & \ra{a \otimes b \mapsto b \otimes a} & B \otimes A   \\
 \da{\pi} & & \da{???}   \\
B(\mathcal{H}) & \ra{\pi_{A} \otimes \pi_{B} \mapsto \pi_{B} \otimes \pi_{A}} & B(\mathcal{H})    \\
\end{array}
$$
Looks as if it possible to obtain $a \otimes b \mapsto b \otimes a$ as a composition of three underlying arrows but i can't figure out an easy way to do this.
 A: As you say, there is a $*$-homomorphism 
$$
 \phi: A \otimes B \to B \otimes_{\mathrm{min}} A.
$$
This morphism is indeed bounded with respect to the minimal norm on $A \otimes B$:
\begin{align*}
\left \lVert \phi \left (\sum_{i=1}^n a_i \otimes b_i \right ) \right \rVert_{\mathrm{min}} & = \left \lVert \sum_{i=1}^n b_i \otimes a_i   \right \rVert_{\mathrm{min}} = \left \lVert \sum_{i=1}^n \pi_B(b_i) \otimes \pi_A(a_i) \right \rVert_{\mathbb B(H_B \otimes H_A)} \\
& =  \left \lVert \sum_{i=1}^n \pi_A(a_i) \otimes \pi_B(b_i) \right \rVert_{\mathbb B(H_A \otimes H_B)}  \\
& = \left \lVert \sum_{i=1}^n a_i \otimes b_i \right \rVert _{\mathrm{min}}
\end{align*}
So in fact, $\phi$ is isometric. This shows that we get a $*$-homomorphism 
$$
 \phi : A \otimes_{\mathrm{min}} B \to B \otimes_{\mathrm{min}} A
$$
which is isometric and clearly has dense image. That means that $\phi$ is an isomorphism.
Of course, I cheated a bit by using that 
$$
\mathbb B(H_A \otimes H_B) \cong \mathbb B(H_B \otimes H_A),
$$
where $\cong$ means isometric isomorphism of Banach spaces.
