Since the questioner starts off asking about $C^*$ algebras, I am going to assume that he only cares about Hopf algebras over $\mathbb{C}$. Every finitely generated, commutative $\mathbb{C}$-Hopf-algebra is the polynomial functions on an algebraic group $G$. As Ben says, we can just take $G$ to be Spec of the Hopf algebra, and then the comultiplication gives a group structure on this Spec.
There are several ways that working over $\mathbb{C}$ makes things nicer than working over arbitrary fields. Other answers have pointed them out but, IMO, have done so in a way that makes things sound more confusing. Let me instead point out how nice algebraic groups over $\mathbb{C}$ are:
- You might worry that $G$ would not be reduced. This happens over fields of characteristic $p$, but not characteristic zero. See my earlier question. Also, you might worry that $G$ has singularities, but it doesn't. In short, $G$ is a complex Lie group.
- Over non-algebraically closed fields $k$, the behavior of the $k$-points of $G$ may not determine the behavior of $G$. For example, let $X = \mathrm{Spec} \mathbb{R}[x]/(x^n-1)$. Then $X$ can be equipped with the structure of an algebraic group of order $n$; the coproduct is $x \mapsto x \otimes x$. Intuitively, you should think $x_1 \times x_2 \mapsto x_1 x_2$, so this group is the $n$-th roots of $1$ under multiplication. By the Nullstellansatz, this can't happen over $\mathbb{C}$. The $\mathbb{C}$-points will be dense (in both the ordinary and Zariski topologies) and any map will be determined by what it does to the $\mathbb{C}$-points.