Standard model of ZFC and existence of model of ZFC 
The assumption that there exists a standard model of ZFC in a given universe is stronger than the assumption that there exists a model.

So, existence of a model of ZFC and existence of a standard model of ZFC are only assumption, not theorem?
 A: By model theory, the existence of a model of ZFC is equivalent to ZFC being consistent. By the incompleteness theorem, ZFC cannot prove its own consistency. Therefore, the existence of a model of ZFC is not a theorem of ZFC---if that is what you are asking about. 
I am not sure whether "standard model" is a rigorous term. For example, the term standard model of the Peano's arithmetic refers to $\omega$, the usual natural numbers. I guess vaguely the term "standard" would mean the structure from which some theory was inspired by. However, by the above, it is not known whether ZFC has a model, so I can't imagine what a standard model would mean. However, if ZFC does have a model, then there are models with special properties like $V = L$. 
A: We usually assume that there is a grand universe of sets which is not a set, and we often assume that this universe is a model of ZFC. It does not prove ZFC is consistent, though, because it is too large for us to capture.
It may be the case, however, that inside our universe of sets there is a set $M$ and a binary relation $E$ over $M$ such that $\langle M,E\rangle$ is a model of set theory. This is to say that ZFC is consistent, or that it has a model.
From the point of view of $M$ the relation $E$ is $\in$. However, we as all knowing beings, know that $E$ may be something else.
We say that $M$ is well-founded if $E$ is well-founded as a relation in the universe. It is important to remark that $M$ always thinks that $E$ is well-founded, but it might be the case that $M$ does not know about any decreasing sequence.
Well-founded models are good. They are good because we have a theorem known as the Mostowski collapse lemma which tells us that if $\langle A,R\rangle$ is well-founded then there is some $B$ such that $\langle A,R\rangle\cong\langle B,\in\rangle$ as ordered sets, and $B$ is transitive (i.e. $x\in B$ and $y\in x$ imply that $y\in B$).
Now suppose that $\langle M,E\rangle$ is a well-founded model of ZFC, namely it is a model of ZFC and $E$ is really well-founded, in this case we can collapse it and have some $\langle N,\in\rangle$ which is a model of ZFC and $\in$ is the restriction of $\in$ to $N$. Such models are standard models of ZFC. 
Now comes the punch: the existence of a well-founded model is stronger than simply the existence of a model. (Since the true $\in$ is well-founded, standard models are always well-founded.)
