Is "ZC + Reflection" equivalent to ZFC? By reflection I mean the schema:
if $\phi$ is a sentence, and if $\phi^{V_\alpha}$ is the formula obtained by merely bounding every quantifier in $\phi$ by $V_\alpha$, then: $$\phi \implies\exists \alpha \exists V_\alpha \ ( \phi^{V_\alpha})$$
Where as usual $V_\alpha$ is defined as: $$x=V_\alpha \iff \exists f: \\function(f) \land \\ dom(f)=\alpha \land \\ \forall \beta \in \alpha ( f(\beta ) = \bigcup \{P(f(\lambda)): \lambda < \beta\}) \land \\ x= \bigcup rng(f) $$
By parametric reflection it's meant:
if $\phi$ is a formula whose free variables are among $y_1,…,y_n,z_1,…,z_m,w$, then:
$for \ \ m,n=1,2,3,...\\ \forall y_1,...,y_n \exists \alpha \exists V_\alpha ( y_1,...,y_n \in V_\alpha \land \\ \forall z_1,..,z_m \in V_\alpha [\exists w (\phi) \to \exists w \in V_\alpha (\phi^{V_\alpha})])$

Is "ZC + reflection" equivalent to ZFC?
Is "ZC + parametric reflection" equivalent to ZFC?

 A: (I am assuming that the last line in the definition of $V_\alpha$,
$x=\bigcup \{(f(\lambda))\mid\lambda<\alpha\}$.)
ZC+"ℵ1 exists" proves the formula Con("ZC + reflection"). We will call an ordinal  sentence-determined if there is a   such that " is a sentence" holds,and  is the least ordinal such that " holds in " holds. Since there are uncountably many limit ordinals in ℵ1 and only countably many ordinals which are sentence-determined, there must be a limit ordinal  which is not sentence-determined. Then the formula ""ZC + reflection" holds in " must hold.  
A: No your version of "reflection" does not imply full ZFC (not even consistency wise).
Suppose we live in a universe of ZFC+CH. There is a proper class of ordinals $\alpha$ (namely all limit ordinals $>\omega$) such that $V_\alpha$ satisfies ZC. On the other hand there are only (thanks to CH) $\omega_1$ many $\in$-theories. We thus can find $\alpha<\beta<\omega_2$ so that $V_\alpha$ and $V_\beta$ are elementarily equivalent and models of ZC. Clearly $V_\beta$ is a model of your version of "reflection", every instance is witnessed by $V_\alpha$ (note that $V_\alpha^{V_\beta}=V_\alpha$). However, from the point of view of $V_\beta$, $\omega_2$ does not exist (because $\beta<\omega_2$ and $V_\beta$ contains the true $\mathcal P(\mathcal P(\omega))$ ) and thus it is not a model of ZFC. 
I am sure the use of CH can be avoided.
Edit: To avoid CH, simply work in the minimal $V_\gamma$ that is a model of ZFC (if it exists) and find any elementarily equivalent $V_\alpha$ and $V_\beta$.
