Context
Let $(A_\lambda)_{\lambda \in \Lambda}$ a family of unital C*-algebras. I am trying to see that they have a coproduct. In order to construct it, I understand that we take the coproduct of $(A_\lambda)_{\lambda \in \Lambda}$ as unital *-algebras, which I know exists and corresponds to the free product. Then, define a $C^*$-seminorm as
$\left\lVert{x}\right\rVert:=\sup\{p(x) \hspace{1mm} | \hspace{1mm} p \hspace{1mm} \text{is a C*-seminorm in} \ast_{\lambda \in \Lambda} A_\lambda \hspace{1mm} \text{such that it coincides with} \hspace{1mm} \left\lVert\cdot \right\rVert_{\lambda} \hspace{1mm} \text{in} \hspace{1mm} A_\lambda \}^*$.
Then the set $I=\{x\in \ast_{\lambda \in \Lambda}A_\lambda \hspace{1mm} \text{s.t.} \hspace{1mm} \left\lVert x\right\rVert=0\}$ is a two-sided closed self-adjoint ideal which does not$^*$ cointain $A_\lambda$ for any $\lambda \in \Lambda$, and so the quotient $\ast_{\lambda \in \Lambda}A_\lambda / I$ is a C*-algebra.
$^*$This set is not empty, as the following is a C*-seminorm which satisfies the condition
$p(x):=\begin{cases} \left\lVert{a}\right\rVert_\lambda & \text{if} \hspace{1mm} \exists\lambda \in \Lambda, a\in A_\lambda \hspace{1mm} \text{s.t.} \hspace{1mm} f_\lambda(a)=x \\ 0 & \text{otherwise} \end{cases}$
where $f_\lambda : A_\lambda \hookrightarrow \ast_{\lambda \in \Lambda} A_\lambda$.
Question
Is the seminorm just defined a norm, as it is the case for the tensor product? Or in other words, is the ideal $I$ the zero ideal? How is this justified?
If it is not, how can we justify that $\ast_{\lambda \in \Lambda}A_\lambda/ I$ has the universal property of the coproduct? After all, introducing relations restricts the possibilities of construction of a free product of *-homomorphisms.