$\text{Ordinal Replacement:}$ if $\phi(x,y)$ is a formula in two free variables $x,y$, then: 


$\forall x \ [ordinal(x) \to \exists! y \ (ordinal (y) \wedge \phi(x,y)) ]
\to 
\forall A \ (\forall x \in A (ordinal(x)) \to \exists B \ \forall y \ (y \in B \leftrightarrow \exists x \in A \ \phi(x,y))) $


is an axiom. 


In English for every set $A$ of von Neumann ordinals and a function $F$ from von Neumann ordinals to von Neumann ordinals, then we can get a set $B$ whose elements are all the $F$-images of elements of $A$. 


i.e. we can replace elements of a set of ordinal by ordinals. 


Now is full Replacement provable in $\text{Z + Ordinal Replacement}$? 



Is full Replacement provable in Z + Ordinal Replacement?
No. Consider the model $\langle V\_{\omega\_1},\in\rangle$. This is a model of Zermelo's theory, but all ordinals in it are countable. Thus, it satisfies the ordinal-replacement axiom, since the image of a countable ordinal under any function will be countable, and $V\_{\omega\_1}$ contains all its countable subsets, since $\omega\_1$ is regular. 


In fact, in this case, we needn't restrict $B$ to consist of ordinals. Rather, we get the general version of replacement for functions whose domain is a set contained in the ordinals. 


Since this model does not satisfy full replacement, it shows that full replacement is not provable from ordinal replacement over Z.


