Does ZFC pin down precisely which theorems PA can and cannot prove? We can show (in $\mathrm{ZFC})$ that $\mathrm{PA}$ has a model. Thus:
$$\mathrm{ZFC} \vdash (\mathrm{PA} \not\vdash \bot)$$
This is probably a silly question, but is it true that for all sentences $\phi$ in the language of $\mathrm{PA}$, we have the following? $$(\mathrm{ZFC} \vdash (\mathrm{PA} \vdash \phi)) \vee (\mathrm{ZFC} \vdash (\mathrm{PA} \not\vdash \phi))$$
I'm guessing "no."
 A: First, if ZFC were inconsistent then the answer would trivially be "yes", so the question is only of interest if ZFC is consistent. I will freely use the consistency of ZFC for the rest of the answer.
In that case, the answer is "no". 
It is a standard fact in computability theory that PA is an "effectively inseparable" theory. This means that if we let $$V = \{ \phi : PA \vdash \phi\}$$ and $$U = \{ \phi : PA \vdash \lnot \phi\}$$ then there is no computable "separating set" $C$ with $V \subseteq C$ and $C \cap U = \emptyset$. (Both $U$ and $V$ are r.e. sets because PA is an effective theory; inseparability implies among other things that neither of them is computable). 
Note that (*): if $PA \vdash \phi$ then $ZFC \vdash \ulcorner PA \vdash \phi\urcorner$. This is because if we have a derivation of $\phi$ in PA we can convert it to a derivation of $\ulcorner PA \vdash \phi\urcorner$ in ZFC.
Now let $$S = \{\phi : ZFC \vdash \ulcorner PA \vdash \phi\urcorner\}.$$ Then $S$ is an r.e. set also, and $V \subseteq S$ because of (*). Moreover, $U \cap S = \emptyset$. Suppose otherwise - then there would be a $\phi$ such that $ZFC\vdash \ulcorner PA \vdash \phi \urcorner$ and $PA \vdash \lnot \phi$. But then, using (*), $ZFC \vdash \ulcorner PA \vdash \lnot \phi\urcorner$. But ZFC proves 
$$
\lnot(\ulcorner PA \vdash \phi\urcorner \land \ulcorner PA \vdash \lnot \phi\urcorner)
$$
because ZFC proves PA is consistent (e.g. Gentzen's consistency proof goes through formalized in ZFC). Therefore, because ZFC is consistent, $U \cap S$ is empty.
If the question had a positive answer, $S$ would be computable. The algorithm would be: given $\phi$, enumerate derivations in ZFC until you find a derivation of $\ulcorner PA \vdash \phi\urcorner$ or a derivation of $\lnot \ulcorner PA \vdash \phi\urcorner$. An affirmative answer to the question says this search will always terminate, and since ZFC is consistent exactly one of the two options will happen. Then we can tell whether $\phi \in S$ by seeing which option occurs.  
So, in summary, if the question had a positive answer then $S$ would be a computable separating set for $V$ and $U$, which is impossible. 
Also, for those who are interested, it is not necessary to use the fact that ZFC has an $\omega$-model, just the fact that ZFC is consistent. That should be expected, because effective inseparability is a very strong form of incompleteness. 
If we use the fact that ZFC has an $\omega$-model, we get a much shorter argument, as mentioned in the comments. In that case, we have for all $\phi$ that $ZFC \vdash \ulcorner PA \vdash \phi\urcorner$ if and only if $PA \vdash \phi$. Then, if the question had an affirmative answer, we could decide for arbitrary $\phi$ whether $\phi \in V$ by searching for a ZFC derivation of $\ulcorner PA \vdash \phi\urcorner$ or a derivation of $\lnot \ulcorner PA \vdash \phi \urcorner$; the first option would mean $PA \vdash \phi$ and the second would mean $PA \not\vdash \phi$. Since $V$ is not computable, that contradiction means the question has a negative answer. This proof is easier than the one using effective inseparability, because it leverages an extra property of ZFC (having an $\omega$-model, or more specifically $\Sigma^0_1$ soundness) that is not used in the proof by inseparability. 
