Let $\mathcal{A}$ be a C*-algebra and $x \in \mathcal{A}$ a normal element. Can you show that $\left\{ \phi(x) : \phi \text{ is a state on } \mathcal{A} \right\}$ is the closed convex hull of the spectrum of $x$?
If you want you can use the result above for sef-adjoint elements in C*-algebras. That is, if $a \in \mathcal{A}$ is a self-adjoint element, then $[a,b]=\left\{ \phi(x) : \phi \text{ is a state on } \mathcal{A} \right\}$, where $a=\min{\sigma(a)}$ e $b=\max{\sigma(a)}$. One proof of this fact can be found in Proposition 7.8 of Conway, for example.
Since the spectrum does not depend on the (C$^*$) algebra, we can assume that $\mathcal A=C^*(x)$. Using the Gelfand transform, we can identify $C^*(x)$ with $C(\sigma(x))$, with $x$ mapped to the function $z\mapsto z$.
The point evaluation states $f\mapsto f(t)$ are precisely the pure states The pure states are the extremal points of the set of states of $\mathcal A$. So any state $\phi$ is a limit of convex combinations of points evaluations, which means that $\phi(x)$ is a limit of convex combinations of points in $\sigma(x)$.
As Jonas rightfully mentions, one needs to check that the set does not depend on the algebra. This is verified by checking that every state on $C^*(x)$ extends to a state on $\mathcal A$. So $$ \{\phi(x):\ \phi\ \text{ is a state on }\mathcal A\}=\{\phi(x):\ \phi\ \text{ is a state on }C^*(x)\}. $$